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Original questions in the style of each exam boardFull worked mark schemes with method and accuracy marksCreated by Evan Kotronis with AI assistanceUpdated 16 October 2026 · new questions added continuouslyFree access until 31 December 2026How every question is checked →
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1IBMI-032-03-r2-01Vectors and kinematicsPaper 1Easy6 marks
A fishing boat leaves a mooring and moves with constant velocity. Relative to a harbour at the origin O, its position vector \(t\) hours after 08:00 is \(\mathbf{r} = \begin{pmatrix}2\\-1\end{pmatrix} + t\begin{pmatrix}3\\4\end{pmatrix}\) where distances are in kilometres, the first component is east of O and the second is north of O.
(a)
Find the speed of the boat.
(2)
(b)
Find the position vector of the boat at 10:30.
(2)
(c)
Find the distance of the boat from the harbour at 10:30.
Common error: Using t = 10.5 or t = 2.3 for 10:30. The parameter counts hours after 08:00, so 10:30 is 2.5 hours later.
2IBMI-032-03-r2-02Vectors and kinematicsPaper 1Easy6 marks
A survey drone takes off from the roof of a building. Relative to a point O at ground level directly below the take-off point, the drone's position \(t\) seconds after take-off is \(\mathbf{r} = \begin{pmatrix}0\\0\\20\end{pmatrix} + t\begin{pmatrix}4\\-3\\1.5\end{pmatrix}\) where distances are in metres and the third component is the height above the ground.
(a)
Find the speed of the drone.
(2)
(b)
Find the time at which the drone is 50 m above the ground.
(2)
(c)
Find the horizontal distance of the drone from O at this time.
(2)
Show mark scheme
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Part (a)
\(\sqrt{4^2+(-3)^2+1.5^2} = \sqrt{27.25}\) M1
✓ 1
= 5.22 m s−1 A1
✓ 1
Part (b)
\(20 + 1.5t = 50\) M1
✓ 1
\(t = 20\) s A1
✓ 1
Part (c)
Horizontal components at \(t=20\): 80 and −60, so \(\sqrt{80^2+60^2}\) M1
✓ 1
= 100 m A1
✓ 1
Exam tip: Horizontal distance uses only the first two components. Include the vertical component only when the full straight-line distance is asked for.
3IBMI-032-04-r2-01Graph theory and networksPaper 1Easy6 marks
A road inspector is responsible for the roads joining four villages, A, B, C and D. There are direct roads between A and B, A and C, B and C, B and D, and C and D. There are no other roads. The road network is modelled by a graph G whose vertices are the villages and whose edges are the roads.
(a)
Write down the adjacency matrix M for G, with the vertices in the order A, B, C, D.
(2)
(b)
Find the number of different walks of length 4 that start at A and finish at D.
(2)
(c)
The inspector wants to drive along every road exactly once. Determine whether this is possible. If it is, state the villages where the route must start and finish.
A1 for a symmetric 4×4 matrix with zeros on the leading diagonal
✓ 1
A1 for all entries correct
✓ 1
Part (b)
M1 for using the (A, D) entry of \(M^4\)
✓ 1
A1 for 10 walks
✓ 1
Part (c)
R1 for degrees A = 2, B = 3, C = 3, D = 2, so exactly two vertices (B and C) have odd degree
✓ 1
A1 for: it is possible (an Eulerian trail exists), but the route must start at B and finish at C, or start at C and finish at B
✓ 1
Exam tip: Use your calculator's matrix mode to find \(M^4\). Take the entry in row A and column D; do not add up a whole row.
4IBMI-032-04-r2-02Graph theory and networksPaper 1Easy6 marks
A college is laying fibre-optic cable to connect six buildings, A, B, C, D, E and F. The cable can only be laid along the following routes. Each length is given in metres. AB 120, AC 90, AD 150, BC 70, BE 110, CD 100, CE 140, DE 80, DF 130, EF 60. Every building must be connected to the network, and the total length of cable must be as small as possible.
(a)
Use Kruskal's algorithm to find a minimum spanning tree for the network. List the edges in the order in which they are added.
(3)
(b)
Write down the total length of cable required.
(1)
(c)
The cable costs $18.50 per metre to lay. Find the total cost of laying the cable.
(2)
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Part (a)
M1 for considering edges in increasing order of length
✓ 1
A1 for EF, BC, DE added first, in that order
✓ 1
A1 for AC then CD, with the algorithm stopping once 5 edges are included
✓ 1
Part (b)
A1 for 400 m
✓ 1
Part (c)
M1 for 400 × 18.50
✓ 1
A1 for $7400.00
✓ 1
Common error: A spanning tree on six vertices has exactly five edges. Once five edges have been added without forming a cycle, stop: adding a sixth edge always creates a cycle.
5IBMI-032-08-r1-01Vectors and kinematicsPaper 1Easy5 marks
A rescue boat leaves a harbour buoy. Its position vector, in kilometres relative to a lighthouse at the origin O, is given by \(\mathbf{r} = \begin{pmatrix} 2 \\ -1 \end{pmatrix} + t\begin{pmatrix} 3 \\ 4 \end{pmatrix}\) where t is the time in hours after leaving the buoy. The unit vectors point east and north respectively.
(a)
Find the speed of the boat.
(2)
(b)
Find the position vector of the boat when t = 2.5.
(1)
(c)
Find the distance of the boat from the lighthouse when t = 2.5.
(2)
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Part (a)
M1: speed = \(\sqrt{3^2+4^2}\)
✓ 1
A1: 5 km h−1
✓ 1
Part (b)
A1: \(\begin{pmatrix} 9.5 \\ 9 \end{pmatrix}\)
✓ 1
Part (c)
M1: \(\sqrt{9.5^2+9^2}\)
✓ 1
A1: 13.1 km (13.086…)
✓ 1
Exam tip: The speed comes from the direction (velocity) vector only; the distance from O comes from the position vector. Do not mix the two.
6IBMI-032-08-r1-02Vectors and kinematicsPaper 1Easy6 marks
A survey drone takes off from a point with position vector \(\begin{pmatrix} 10 \\ 5 \\ 0 \end{pmatrix}\) m, measured from a ground station. It flies with constant velocity \(\begin{pmatrix} 4 \\ -2 \\ 1.5 \end{pmatrix}\) m s−1, where the third component is vertical.
(a)
Calculate the speed of the drone.
(2)
(b)
Find the time taken for the drone to reach a height of 30 m.
(2)
(c)
Find the distance the drone has travelled at this time.
Common error: Finding the distance from the ground station instead of the distance travelled from the take-off point. The drone did not start at the origin.
7IBMI-032-08-r2-01Vectors and kinematicsPaper 1Easy5 marks
A fishing boat leaves a harbour. Relative to a lighthouse at the origin O, the position vector of the boat t hours after leaving the harbour is \(\boldsymbol{r} = \begin{pmatrix}3\\-2\end{pmatrix} + t\begin{pmatrix}4\\3\end{pmatrix}\) where distances are in kilometres, \(\boldsymbol{i}\) points east and \(\boldsymbol{j}\) points north.
(a)
Find the speed of the boat.
(2)
(b)
Find the position vector of the boat 2.5 hours after it leaves the harbour.
(1)
(c)
Hence find the distance of the boat from the lighthouse at this time.
Exam tip: Speed is the magnitude of the direction (velocity) vector only. The starting position vector plays no part in finding the speed.
8IBMI-032-08-r2-06Vectors and kinematicsPaper 1Easy7 marks
A plane is pointed due north and flies at 300 km h−1 relative to the air. A wind blows with velocity \(\begin{pmatrix}40\\-30\end{pmatrix}\) km h−1, where \(\boldsymbol{i}\) points east and \(\boldsymbol{j}\) points north.
(a)
Write down the velocity of the plane relative to the ground.
(1)
(b)
Find the ground speed of the plane.
(2)
(c)
Find the bearing of the plane's actual path.
(2)
(d)
The plane keeps this ground velocity for 500 km. Find the time taken, in hours.
(2)
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Part (a)
\(\begin{pmatrix}0\\300\end{pmatrix} + \begin{pmatrix}40\\-30\end{pmatrix} = \begin{pmatrix}40\\270\end{pmatrix}\) km h−1 A1
✓ 1
Part (b)
\(\sqrt{40^2+270^2} = \sqrt{74500}\) M1
✓ 1
= 273 km h−1 A1
✓ 1
Part (c)
\(\tan\theta = \frac{40}{270}\) M1
✓ 1
Bearing 008.4° A1
✓ 1
Part (d)
Time = 500 ÷ 272.95 M1
✓ 1
= 1.83 h A1
✓ 1
Common error: Do not subtract the wind or simply add the speeds 300 + 50. Velocities combine as vectors: add the components, then find the magnitude.
9IBMI-032-03-r1-01Vectors and kinematicsPaper 1Easy5 marks
A delivery drone moves in a straight line after take-off. Relative to an origin at the base of its charging station, its position vector t seconds after take-off is r = (3, −2, 1) + t(4, 5, 2), where the components are in metres, x is east, y is north and z is vertically upwards. Ground level is z = 0.
(a)
Find the speed of the drone.
(2)
(b)
Find the position vector of the drone when t = 6.
(1)
(c)
The drone must reach a height of 15 m above the ground before it may cross a road. Find the time taken for the drone to reach this height.
(2)
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Part (a)
M1: finds the magnitude of the velocity vector, \(\sqrt{4^2+5^2+2^2}=\sqrt{45}\)
✓ 1
A1: 6.71 m s−1
✓ 1
Part (b)
A1: (27, 28, 13) m
✓ 1
Part (c)
M1: sets the vertical component equal to 15, 1 + 2t = 15
✓ 1
A1: t = 7 s
✓ 1
Exam tip: Speed is the magnitude of the direction (velocity) vector, not of the position vector. Height comes from the z-component only, including its starting value.
10IBMI-032-04-r1-01Graph theory and networksPaper 1Easy6 marks
Four villages, A, B, C and D, are linked by roads. There are direct roads between A and B, A and C, B and C, B and D, and C and D. There are no other direct roads. The road network is modelled by a graph G whose vertices are the villages and whose edges are the roads.
(a)
Write down the adjacency matrix M for G, with the vertices in the order A, B, C, D.
(2)
(b)
Find the entry in row B, column B of M2, and interpret this value in context.
(2)
(c)
A delivery van travels along exactly 4 roads, starting at A and finishing at D. Roads and villages may be revisited. Find the number of different possible journeys.
R1 there are 3 different journeys along exactly two roads that start and finish at B (equal to the number of roads meeting at B)
✓ 1
Part (c)
M1 uses the (A, D) entry of M4
✓ 1
A1 10 journeys
✓ 1
Exam tip: The number of walks of length n between two vertices is read straight from Mⁿ. Enter the matrix into your calculator and raise it to the power. Do not try to list the walks by hand.
11IBMI-032-04-r1-02Graph theory and networksPaper 1Easy7 marks
A university plans to connect six buildings, P, Q, R, S, T and U, with fibre-optic cable. Cable can only be laid along the routes listed below. The lengths are in metres. PQ 120, PR 90, QR 70, QS 150, RS 110, RT 130, ST 80, SU 100, TU 95, QU 200 Every building must be connected to the network, and the total length of cable must be as small as possible.
(a)
Use Kruskal's algorithm to find a minimum spanning tree. List the edges in the order in which they are considered, and state clearly any edge that is rejected.
(4)
(b)
Write down the total length of cable required.
(1)
(c)
The cable costs $18.50 per metre to install. Find the total installation cost.
(2)
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Part (a)
M1 edges considered in increasing order of length
✓ 1
A1 QR (70), ST (80), PR (90), TU (95) accepted
✓ 1
R1 SU (100) rejected because it would form a cycle S–T–U
✓ 1
A1 RS (110) accepted, and the tree is complete with 5 edges
✓ 1
Part (b)
A1 445 m
✓ 1
Part (c)
M1 445 × 18.50
✓ 1
A1 $8232.50
✓ 1
Common error: Stop as soon as you have accepted n − 1 edges for n vertices. When you reject an edge, name the cycle it would create, because a stated reason is what earns the R mark.
12IBMI-032-03-r1-02Vectors and kinematicsPaper 1Easy6 marks
A ferry leaves a harbour, taken as the origin, and travels at a constant speed of 12 km h−1 on a bearing of 060°. Position vectors are given as (east component, north component) in kilometres, and t is the time in hours after the ferry leaves the harbour. A lighthouse is located at (20, 4).
(a)
Find the velocity vector of the ferry, giving its components to 3 significant figures.
(2)
(b)
Find the position vector of the ferry after 1.5 hours.
(2)
(c)
Find the distance between the ferry and the lighthouse at this time.
(2)
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Part (a)
M1: uses east = 12 sin 60°, north = 12 cos 60°
✓ 1
A1: (10.4, 6.00) km h−1 (accept 6)
✓ 1
Part (b)
M1: multiplies the velocity vector by 1.5
✓ 1
A1: (15.6, 9.00) km (accept 9)
✓ 1
Part (c)
M1: finds the magnitude of the displacement from ferry to lighthouse, e.g. \(\sqrt{(20-15.59)^2+(4-9)^2}\)
✓ 1
A1: 6.67 km (unrounded values); accept 6.66 km from the rounded position (15.6, 9)
✓ 1
Common error: Bearings are measured clockwise from north, so the east component uses sine and the north component uses cosine. Swapping them gives (6, 10.4), a bearing of 030°. Keep unrounded values in your calculator for later parts.
13IBMI3H-3-06-01Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy4 marks
A survey drone flies in a straight line from A(2, −1, 5) to B(8, 3, 17), where coordinates are in metres relative to a base station.
(a)
Find the vector \(\overrightarrow{AB}\).
(1)
(b)
Find the distance AB.
(2)
(c)
The drone flies at a constant 3.5 m s−1. Find the time taken to fly from A to B.
(1)
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Part (a)
A1 \(\begin{pmatrix}6\\4\\12\end{pmatrix}\)
✓ 1
Part (b)
M1 \(\sqrt{6^2+4^2+12^2}\)
✓ 1
A1 14 m
✓ 1
Part (c)
A1 \(14\div3.5=4.0\) s
✓ 1
Exam tip: The distance between two points is the magnitude of the vector between them: subtract position vectors as end minus start.
14IBMI3H-3-06-02Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy5 marks
Two tugboats pull a barge. The forces they exert are \(\mathbf{F}_1=\begin{pmatrix}8\\6\end{pmatrix}\) kN and \(\mathbf{F}_2=\begin{pmatrix}5\\-12\end{pmatrix}\) kN.
(a)
Calculate the scalar product \(\mathbf{F}_1\cdot\mathbf{F}_2\).
M1 evaluates \(\mathbf{M}^3\) (entry in row A, column D)
✓ 1
A1 2
✓ 1
Part (c)
A1 A–B–C–D or A–C–B–D
✓ 1
Exam tip: The entry in row i, column j of \(\mathbf{M}^n\) counts walks of length n from vertex i to vertex j.
16IBMI3H-3-06-12Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy5 marks
Three villages are at A(0, 0), B(6, 0) and C(0, 8) on a map (1 unit = 1 km). A wind turbine is to be built at the point that is the same distance from all three villages.
(a)
Find the equations of the perpendicular bisectors of AB and of AC.
(2)
(b)
Write down the coordinates of the Voronoi vertex where the three bisectors meet.
(1)
(c)
Find the distance from this point to each village.
(2)
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Part (a)
A1 \(x=3\)
✓ 1
A1 \(y=4\)
✓ 1
Part (b)
A1 (3, 4)
✓ 1
Part (c)
M1 \(\sqrt{3^2+4^2}\)
✓ 1
A1 5 km
✓ 1
Exam tip: The vertex of a Voronoi diagram is equidistant from the three sites, so one distance calculation gives all three.
17IBMI3H-3-06b-01Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy5 marks
In a retail warehouse, the directions of two conveyor belts are given by the vectors a = 3i − 2j + 4k and b = i + 5j − 2k.
Exam tip: A negative scalar product means the angle between the vectors is obtuse.
18IBMI3H-3-06b-02Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy5 marks
A retailer's sign shows a triangle with vertices A(2, 1), B(6, 1) and C(6, 4). The triangle is transformed by the matrix \(\begin{pmatrix}0&-1\\1&0\end{pmatrix}\).
(a)
Find the coordinates of the image of each vertex.
(3)
(b)
Describe fully the single transformation represented by this matrix.
(2)
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Part (a)
A1 A′(−1, 2)
✓ 1
A1 B′(−1, 6)
✓ 1
A1 C′(−4, 6)
✓ 1
Part (b)
M1 identifies a rotation
✓ 1
A1 90° anticlockwise about the origin
✓ 1
Common error: Multiplying the matrix by the row of coordinates in the wrong order; write each point as a column vector and put the matrix on the left.
19IBMI3H-3-06b-03Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy5 marks
The contacts between four patients A, B, C and D on a ward are modelled by a graph. The edges are AB, AC, BC and CD.
(a)
Write down the adjacency matrix for the graph, with rows and columns in the order A, B, C, D.
(2)
(b)
Find the number of different walks of length 4 from A to D.
M1 attempt to find the fourth power of the adjacency matrix
✓ 1
A1 entry in row A, column D of M4
✓ 1
A1 4
✓ 1
Exam tip: The (i, j) entry of the n-th power of an adjacency matrix counts walks of length n from vertex i to vertex j.
20IBMI3H-3-06b-04Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy5 marks
Five solar arrays P, Q, R, S and T are to be connected by cable. The possible cable lengths in km are: PQ = 4, PR = 7, PS = 9, QR = 3, QS = 8, QT = 10, RS = 5, RT = 6, ST = 2.
(a)
Use Kruskal's algorithm to find a minimum spanning tree. List the edges in the order chosen and state the total length of cable.
(4)
(b)
Explain why edge QS is not in the tree.
(1)
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Part (a)
M1 edges considered in increasing order of length
✓ 1
A1 ST (2), QR (3), PQ (4)
✓ 1
A1 RS (5) completes the tree; edges that would form a cycle are rejected
✓ 1
A1 total length = 14 km
✓ 1
Part (b)
R1 QS would join Q and S which are already connected through R, so it forms a cycle
✓ 1
Common error: Adding an edge that creates a cycle just because it is the next shortest.
21IBMI3H-3-03r6-01Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy5 marks
Calculator
A light aircraft leaves airfield A and flies 240 km on a bearing of 065° to airfield B. It then flies 180 km on a bearing of 150° to airfield C.
(a)
Find the size of angle ABC.
(2)
(b)
Find the direct distance AC. Use the unrounded value of angle ABC and give your answer to the nearest kilometre.
(3)
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Part (a)
M1 uses the back-bearing from B to A, 245°, or co-interior angles at B
✓ 1
A1 angle ABC = 245° − 150° = 95°
✓ 1
Part (b)
M1 attempts to use the cosine rule in triangle ABC
Common error: A bearing is measured clockwise from north at the point you leave, so the angle inside the triangle comes from the back-bearing, not from the two bearings directly.
22IBMI3H-3-03r6-04Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy4 marks
Calculator
A triangular field PQR has PQ = 420 m, PR = 350 m and angle QPR = 72°.
(a)
Find the area of the field in m2. Give your answer to the nearest m2.
(2)
(b)
Seed costs €0.035 per m2. Using the unrounded area, find the cost of seeding the whole field, giving your answer in euros to 2 decimal places.
(2)
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Part (a)
M1 uses area = ½ ab sin C
✓ 1
A1 area = 69903 m2
✓ 1
Part (b)
M1 multiplies the area by 0.035
✓ 1
A1 cost = €2446.59
✓ 1
Exam tip: Keep the unrounded value in your calculator when a later part says to use it, and round only the final answer.
23IBMI3H-3-03r6-06Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy5 marks
Calculator
Two aircraft leave the same airport at the same time. Relative to the airport, with i pointing east, j north and k vertically upwards, their velocities are v1 = (200, 150, 20) km h−1 and v2 = (180, −60, 15) km h−1.
(a)
Find v1 · v2.
(2)
(b)
Find the angle between the two flight paths. Give your answer to the nearest 0.1°.
(3)
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Part (a)
M1 multiplies corresponding components and adds
✓ 1
A1 200 × 180 + 150 × (−60) + 20 × 15 = 27300
✓ 1
Part (b)
M1 uses cos θ = (v1 · v2) ÷ (|v1||v2|)
✓ 1
A1 |v1| = 250.8… and |v2| = 190.3… (both correct to 1 d.p.)
✓ 1
A1 θ = 55.1°
✓ 1
Common error: The scalar product of two vectors is a number, not a vector, and it is the angle between the direction vectors (not the position vectors) that gives the angle between paths.
24IBMI3H-3-03r6-10Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy4 marks
Calculator
An airline operates direct routes between five airports A, B, C, D and E. The adjacency matrix below shows the number of direct routes between each pair of airports, with rows and columns in the order A, B, C, D, E. 0 1 2 0 0 1 0 1 1 0 2 1 0 1 1 0 1 1 0 2 0 0 1 2 0
(a)
Write down the number of direct routes from airport C.
(1)
(b)
Find the number of different journeys from A to E that use exactly three flights. Each distinct route counts separately.
(3)
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Part (a)
A1 2 + 1 + 1 + 1 = 5 (sum of row C)
✓ 1
Part (b)
M1 attempts to find the matrix cube M3
✓ 1
A1 identifies the row A, column E entry
✓ 1
A1 number of journeys = 7
✓ 1
Exam tip: The (i, j) entry of the n-th power of an adjacency matrix counts walks of length n from vertex i to vertex j, so read the right row and column.
25IBMI3H-3-04r9-01Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy6 marks
Positions on a mountain are given relative to base camp, in metres, using i for east, j for north and k for vertically upwards. Camp 1 is at (120, 350, 640) and Camp 2 is at (−80, 510, 930).
(a)
Find the vector from Camp 1 to Camp 2.
(2)
(b)
Find the straight-line distance between the two camps, to the nearest metre.
(2)
(c)
Find the average gradient between the camps, defined as vertical rise divided by horizontal distance, to 3 significant figures.
(2)
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Part (a)
M1 Attempt to subtract position vectors
✓ 1
A1 (−200, 160, 290), that is −200i + 160j + 290k
✓ 1
Part (b)
M1 Attempt to find the magnitude √(2002 + 1602 + 2902)
✓ 1
A1 387 m
✓ 1
Part (c)
M1 Horizontal distance = √(2002 + 1602)
✓ 1
A1 Gradient = 290 ÷ √65600 = 1.13
✓ 1
Exam tip: Horizontal distance uses only the i and j components; the k component is the vertical rise.
26IBMI3H-3-04r9-11Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy6 marks
A bakery designs a decoration stencil in the shape of triangle PQR, where P = (1, 1), Q = (4, 1) and R = (1, 3), measured in centimetres. A printing process transforms every point (x, y) using the matrix \(M=\begin{pmatrix}3&1\\-1&2\end{pmatrix}\).
(a)
Find the coordinates of the images P′, Q′ and R′.
(3)
(b)
The area of triangle PQR is 3 cm2. Find the determinant of M and hence the area of triangle P′Q′R′.
(3)
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Part (a)
A1 P′ = (4, 1)
✓ 1
A1 Q′ = (13, −2)
✓ 1
A1 R′ = (6, 5)
✓ 1
Part (b)
A1 det M = 3 × 2 − 1 × (−1) = 7
✓ 1
M1 Area scale factor is |det M|, so area = 7 × 3
✓ 1
A1 = 21 cm2
✓ 1
Exam tip: The area of the image is the original area multiplied by the absolute value of the determinant, regardless of the shape.
27IBMI3H-3-04r9-13Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy8 marks
Mountaineers leave a hut H and walk 5.2 km on a bearing of 072° to a point P. They then walk 3.8 km on a bearing of 150° to a point Q. Give lengths to 3 significant figures and bearings to the nearest degree.
(a)
Show that angle HPQ is 102°.
(2)
(b)
Find the distance HQ.
(3)
(c)
Find the bearing of H from Q.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 Bearing of H from P is 072° + 180° = 252°
✓ 1
A1 Angle HPQ = 252° − 150° = 102°
✓ 1
Part (b)
M1 Use of the cosine rule with 5.2, 3.8 and 102°
✓ 1
A1 HQ2 = 5.22 + 3.82 − 2 × 5.2 × 3.8 × cos 102°
✓ 1
A1 HQ = 7.05 km
✓ 1
Part (c)
M1 Use of the sine rule to find angle HQP, sin(HQP)/5.2 = sin 102°/HQ, giving 46.2°
✓ 1
A1 Bearing of P from Q is 330°
✓ 1
A1 Bearing of H from Q = 330° − 46.2° = 284°
✓ 1
Exam tip: Draw a north line at each point before reading off angles; back-bearings differ by exactly 180°.
28IBMI3H-3-04r12-03Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy6 marks
Calculator
Two lighting beams from a rig at the origin O on a theatre ceiling point in the directions \(\mathbf{u}=\begin{pmatrix}3\\-2\\6\end{pmatrix}\) and \(\mathbf{v}=\begin{pmatrix}2\\4\\-1\end{pmatrix}\).
(a)
Calculate \(\mathbf{u}\cdot\mathbf{v}\).
(1)
(b)
Find the angle between the two beams, giving your answer in degrees to 3 significant figures.
(3)
(c)
A third beam has direction \(\mathbf{w}=\begin{pmatrix}k\\1\\-2\end{pmatrix}\) and is perpendicular to \(\mathbf{u}\). Find the value of \(k\), giving your answer to 3 significant figures.
A1 \(|\mathbf{u}|=7\) and \(|\mathbf{v}|=\sqrt{21}\) (ECF from (a))
✓ 1
A1 \(\theta=104^\circ\)
✓ 1
Part (c)
M1 perpendicular so \(\mathbf{u}\cdot\mathbf{w}=0\): \(3k-2-12=0\)
✓ 1
A1 \(k=\frac{14}{3}=4.67\)
✓ 1
Common error: A negative scalar product means the angle is obtuse; do not take the absolute value unless the question asks for the acute angle between the lines.
29IBMI3H-3-04r12-07Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy6 marks
A shipping company operates direct routes between five ports A, B, C, D and E. The direct routes are AB, AC, BC, BD, CD and DE; each can be sailed in either direction.
(a)
Write down the adjacency matrix for the graph, with the ports in alphabetical order.
(2)
(b)
Find the number of different walks of length 3 from A to E.
(2)
(c)
Find the number of different walks of length 4 from A to D.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 symmetric \(5\times5\) matrix with entries 0 or 1
M1 uses the (A, E) entry of the cube of the adjacency matrix (or lists walks)
✓ 1
A1 2 (walks ABDE and ACDE)
✓ 1
Part (c)
M1 uses the (A, D) entry of the fourth power of the adjacency matrix
✓ 1
A1 12
✓ 1
Exam tip: The \((i,j)\) entry of the nth power of an adjacency matrix counts walks of length n from vertex i to vertex j; a walk may revisit vertices and edges.
30IBMI3H-3-04r12-08Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy6 marks
A port authority plans to connect six container terminals A to F by rail links so that every terminal can be reached from every other. The possible links and their construction costs, in millions of euros, are: AB 12, AC 9, BC 7, BD 15, CD 11, CE 14, DE 8, DF 10, EF 6.
(a)
Use Kruskal's algorithm to find a minimum spanning tree. List the links in the order in which they are added and state the minimum total cost.
(4)
(b)
Planning permission for the link CD is refused. Find the minimum total cost, in millions of euros, of a network that avoids CD.
(2)
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Mark
Notes
Part (a)
M1 considers links in increasing order of cost and rejects any that would form a cycle
✓ 1
A1 links added in order: EF (6), BC (7), DE (8), AC (9) (with DF rejected as it forms a cycle)
✓ 1
A1 then CD (11)
✓ 1
A1 total cost €41 million
✓ 1
Part (b)
M1 the two groups \(\{A,B,C\}\) and \(\{D,E,F\}\) must be joined by another link: cheapest available is CE (14), (BD costs 15)
✓ 1
A1 \(41-11+14=44\), so €44 million
✓ 1
Exam tip: State which edges you reject as well as which you accept: this shows the cycle check that Kruskal's algorithm requires.
31IBMI3H-3-04r12-14Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy6 marks
Calculator
A volcano is modelled as a cone with a circular base of radius 1.8 km and a vertical height of 0.9 km.
(a)
Calculate the slant height of the cone, giving your answer in km to 3 significant figures.
(2)
(b)
Calculate the angle that the slope of the volcano makes with the horizontal, giving your answer in degrees to 3 significant figures.
(2)
(c)
Calculate the area of the curved surface of the cone, giving your answer in km2 to 3 significant figures. Use the unrounded slant height.
(2)
Show mark scheme
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Mark
Notes
Part (a)
M1 \(\sqrt{1.8^2+0.9^2}\) (Pythagoras)
✓ 1
A1 2.01 km
✓ 1
Part (b)
M1 \(\tan\theta=\frac{0.9}{1.8}\) (or equivalent using slant height)
✓ 1
A1 26.6\(^\circ\)
✓ 1
Part (c)
M1 curved surface area \(=\pi rl\) with \(r=1.8\)
✓ 1
A1 \(\pi\times1.8\times2.0125=11.4\) km2
✓ 1
Common error: Use the slant height, not the vertical height, in the formula \(\pi r l\) for the curved surface area of a cone.
32IBMI3H-3-03r15-01Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy5 marks
A sprinter in a training session is tracked on a flat field. Her position vector relative to a marker at the origin is \(\mathbf{r} = \begin{pmatrix}20\\5\end{pmatrix} + t\begin{pmatrix}3.5\\-1.2\end{pmatrix}\), where distances are in metres and t is the time in seconds, \(t \geq 0\).
(a)
Write down her position when \(t = 10\).
(1)
(b)
Find the time at which her x-coordinate is 90 m.
(1)
(c)
Calculate her speed. Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
\(\begin{pmatrix}55\\-7\end{pmatrix}\) or (55, −7) A1
✓ 1
Part (b)
\(20 + 3.5t = 90 \Rightarrow t = 20\) s A1
✓ 1
Part (c)
M1 speed = magnitude of the direction vector
✓ 1
M1 \(\sqrt{3.5^2+(-1.2)^2}\) = \(\sqrt{13.69}\)
✓ 1
A1 3.70 m s−1
✓ 1
Exam tip: Speed is the magnitude of the velocity vector, so ignore the starting position when you calculate it.
33IBMI3H-3-03r15-02Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy6 marks
Two thrusters on a space probe produce forces (in newtons) represented by the vectors \(\mathbf{F}_1 = \begin{pmatrix}2\\3\\-1\end{pmatrix}\) and \(\mathbf{F}_2 = \begin{pmatrix}5\\-1\\4\end{pmatrix}\).
(a)
Calculate \(\mathbf{F}_1 \cdot \mathbf{F}_2\).
(2)
(b)
Find the angle between the two forces. Give your answer in degrees to 1 decimal place.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(2\times5 + 3\times(-1) + (-1)\times4\)
✓ 1
A1 3
✓ 1
Part (b)
M1 use of \(\cos\theta = \dfrac{\mathbf{F}_1\cdot\mathbf{F}_2}{|\mathbf{F}_1||\mathbf{F}_2|}\)
✓ 1
A1 \(|\mathbf{F}_1| = \sqrt{14}\) and \(|\mathbf{F}_2| = \sqrt{42}\)
✓ 1
M1 \(\theta = \cos^{-1}\!\left(\dfrac{3}{\sqrt{14}\sqrt{42}}\right)\) (ECF from (a))
✓ 1
A1 82.9°
✓ 1
Common error: Do not forget to divide by both magnitudes; the scalar product on its own is not a cosine.
34IBMI3H-3-03r15-10Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy4 marks
On a coaching board, a player's position (x, y) is transformed by a rotation of 90° anticlockwise about the origin, followed by an enlargement with scale factor 2 centred at the origin.
(a)
Write down the matrix for the rotation of 90° anticlockwise about the origin.
(1)
(b)
Find the single matrix T that represents the combined transformation.
M1 \(\begin{pmatrix}0&-2\\2&0\end{pmatrix}\begin{pmatrix}3\\5\end{pmatrix}\) (ECF from (b))
✓ 1
A1 (−10, 6)
✓ 1
Common error: Enlargement and rotation commute here, but in general the order of matrices matters: the first transformation is the matrix on the right.
35IBMI3H-3-03r15-14Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy5 marks
A sports hall is a cuboid 40 m long, 25 m wide and 12 m high. A cable is stretched in a straight line from floor corner A to the opposite roof corner G (diagonally opposite through the hall).
(a)
Calculate the length of the cable, to 3 significant figures.
(2)
(b)
Find the angle the cable makes with the floor, to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\sqrt{40^2+25^2+12^2}\)
✓ 1
A1 48.7 m
✓ 1
Part (b)
M1 right-angled triangle with the floor diagonal \(\sqrt{40^2+25^2} = 47.2\) m
Common error: The angle with the floor uses the floor diagonal, not a side of the base, as the adjacent side.
36IBMI3H-3-03r16-03Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy7 marks
A hospital has five departments: Admissions (A), Biochemistry (B), Cardiology (C), Dispensary (D) and Exit (E). Corridors join the following pairs: AB, AC, BC, BD, CD and DE. Each corridor is a single edge of an undirected graph and the adjacency matrix M uses the order A, B, C, D, E.
(a)
Write down the adjacency matrix M.
(2)
(b)
Find the number of different walks of length 3 from A to E, and list them.
(3)
(c)
Find the number of walks of length 4 from B to C.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 symmetric 5×5 matrix with zeros on the leading diagonal
Exam tip: The (i, j) entry of the nth power of an adjacency matrix counts walks of length n from vertex i to vertex j; a calculator can raise the matrix to a power directly.
37IBMI3H-3-03r16-07Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy7 marks
A hospital logo is drawn as triangle PQR with P(2, 1), Q(5, 1) and R(5, 4). The matrix \(\mathbf{M}=\begin{pmatrix}0 & -1\\ 1 & 0\end{pmatrix}\) transforms the logo.
(a)
Describe fully the transformation represented by M.
(2)
(b)
Find the coordinates of P′, Q′ and R′, the images of P, Q and R.
(2)
(c)
The image is then reflected in the y-axis, whose matrix is \(\mathbf{N}=\begin{pmatrix}-1 & 0\\ 0 & 1\end{pmatrix}\). Find the single matrix that represents the combined transformation, and describe it geometrically.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 rotation
✓ 1
A1 90° anticlockwise about the origin
✓ 1
Part (b)
M1 multiply M by each position vector
✓ 1
A1 P′(−1, 2), Q′(−1, 5), R′(−4, 5)
✓ 1
Part (c)
M1 order of multiplication NM
✓ 1
A1 \(\begin{pmatrix}0&1\\1&0\end{pmatrix}\)
✓ 1
A1 reflection in the line y = x
✓ 1
Common error: For a transformation A followed by B the combined matrix is BA, not AB.
38IBMI3H-3-03r16-11Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy7 marks
An air ambulance is based at H. A hospital A is 18 km from H on a bearing of 060°, and an accident site B is 25 km from H on a bearing of 140°.
(a)
Write down the size of angle AHB.
(1)
(b)
Find the distance AB, in kilometres to 3 significant figures.
(3)
(c)
Find the bearing of B from A, to the nearest degree. Use the unrounded value of AB.
M1 sine rule \(\dfrac{\sin\angle HAB}{25}=\dfrac{\sin80^\circ}{AB}\)
✓ 1
A1 angle HAB = 61.0°
✓ 1
A1 bearing = 240° − 61.0° = 179°
✓ 1
Exam tip: Draw a north line at each point and mark the bearing angles before using the sine or cosine rule; the back bearing of a line is 180° more or less than the forward bearing.
39IBMI3H-3-03r16-13Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy7 marks
A shelter over a row of hives is a pyramid with a horizontal square base ABCD of side 6 m. The apex V is 4 m vertically above the centre O of the base.
(a)
Find the length of the sloping edge VA, in metres to 3 significant figures.
(3)
(b)
Find the angle between VA and the base.
(2)
(c)
Find the angle between the triangular face VAB and the base.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(OA=\sqrt{3^2+3^2}\) (\(=4.243\ldots\))
✓ 1
M1 \(VA=\sqrt{OA^2+4^2}\)
✓ 1
A1 5.83 m
✓ 1
Part (b)
M1 \(\tan\theta=\dfrac{4}{OA}\)
✓ 1
A1 43.3°
✓ 1
Part (c)
M1 midpoint M of AB has \(OM=3\), so \(\tan\phi=\dfrac43\)
✓ 1
A1 53.1°
✓ 1
Common error: The angle between a sloping face and the base is measured along a line perpendicular to the shared edge, so use the half-side, not the half-diagonal.
40IBMI3H-3-04r16-01Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy8 marks
Two Formula 1 cars are on different straights. At a particular instant, car P has velocity \(\begin{pmatrix}k\\60\end{pmatrix}\) m s−1 and car Q has velocity \(\begin{pmatrix}90\\-120\end{pmatrix}\) m s−1, where the components are east and north. The two velocities are perpendicular.
(a)
Find the value of k.
(3)
(b)
Find the speed of car P.
(2)
(c)
A third car R has velocity \(\begin{pmatrix}50\\120\end{pmatrix}\) m s−1. Find the angle between the directions of travel of P and R, in degrees to 1 decimal place.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 scalar product of the two velocities equals zero
✓ 1
A1 \(90k-7200=0\)
✓ 1
A1 k = 80
✓ 1
Part (b)
M1 \(\sqrt{80^2+60^2}\)
✓ 1
A1 100 m s−1
✓ 1
Part (c)
M1 \(80(50)+60(120)=11200\)
✓ 1
M1 \(\cos\theta=\dfrac{11200}{100\times130}\)
✓ 1
A1 30.5°
✓ 1
Common error: Use the magnitudes of both vectors in the denominator of the angle formula; 130 is the speed of car R, not 120.
41IBMI3H-3-04r16-03Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy7 marks
Five wards P, Q, R, S and T of a hospital are joined by corridors. The adjacency matrix M, with rows and columns in the order P, Q, R, S, T, is (0 1 0 1 0) (1 0 1 0 1) (0 1 0 1 1) (1 0 1 0 1) (0 1 1 1 0). Each corridor joins two wards and is shown by 1 in the matrix.
(a)
Find the number of corridors.
(2)
(b)
Find the number of walks of length 4 from P to R.
(2)
(c)
Find the number of closed walks of length 3 that start and end at Q, and explain your answer with reference to the graph.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sum of all entries is 14
✓ 1
A1 14 ÷ 2 = 7 corridors
✓ 1
Part (b)
M1 (P, R) entry of \(\mathbf{M}^4\)
✓ 1
A1 14
✓ 1
Part (c)
M1 (Q, Q) entry of \(\mathbf{M}^3\)
✓ 1
A1 2
✓ 1
R1 the only triangle through Q is Q–R–T, which can be walked in two directions
✓ 1
Common error: Summing all entries of an adjacency matrix counts each corridor twice, so divide by 2 to get the number of edges.
42IBMI3H-3-04r16-11Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy8 marks
An ambulance leaves base A and drives 12 km on a bearing of 075° to a hospital B. It then drives 9 km on a bearing of 160° to a clinic C.
(a)
Show that angle ABC = 95°.
(2)
(b)
Find the straight-line distance AC, in kilometres to 3 significant figures.
(3)
(c)
Find the bearing of C from A, to the nearest degree. Use the unrounded value of AC.
M1 sine rule \(\dfrac{\sin\angle BAC}{9}=\dfrac{\sin95^\circ}{AC}\)
✓ 1
A1 angle BAC = 35.0°
✓ 1
A1 bearing = 075° + 35.0° = 110°
✓ 1
Exam tip: State the angle at B using parallel north lines (co-interior angles) rather than just reading it from a sketch, and keep bearings as three figures.
43IBMI3H-3-04r16-13Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy7 marks
A hospital ward is a cuboid of length 12 m, width 8 m and height 3.5 m. A sensor is fixed at corner A on the floor, and a second sensor is fixed at corner B on the ceiling, diagonally opposite A.
(a)
Find the length of the diagonal of the floor, in metres to 3 significant figures.
(2)
(b)
Find the distance AB, in metres to 3 significant figures.
(2)
(c)
Find the angle between AB and the floor, in degrees to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\sqrt{12^2+8^2}\)
✓ 1
A1 14.4 m
✓ 1
Part (b)
M1 \(\sqrt{12^2+8^2+3.5^2}\) (or Pythagoras with the floor diagonal)
✓ 1
A1 14.8 m
✓ 1
Part (c)
M1 right-angled triangle with height 3.5 and base the floor diagonal
✓ 1
M1 \(\tan\theta=\dfrac{3.5}{\sqrt{208}}\)
✓ 1
A1 13.6°
✓ 1
Exam tip: To find the angle between a line and a plane, drop a perpendicular to the plane and use the right-angled triangle formed by the line, its projection and the perpendicular.
44IBMI3H-3-06r20-01Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy4 marks
Calculator
Two straight railway lines leave a junction J. Station A is 12.4 km from J on a bearing of 072°. Station B is 9.8 km from J on a bearing of 145°.
(a)
Show that angle AJB is 73°.
(1)
(b)
Calculate the direct distance AB, to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(145^\circ-72^\circ=73^\circ\)
✓ 1
Part (b)
M1 use of the cosine rule \(AB^2=12.4^2+9.8^2-2(12.4)(9.8)\cos73^\circ\)
✓ 1
A1 \(AB^2=178.7\) (accept 178.74)
✓ 1
A1 \(AB=13.4\) km
✓ 1
Common error: Bearings are measured clockwise from north, so the angle between two bearings from the same point is found by subtracting them. Make sure your calculator is in degree mode.
45IBMI3H-3-06r20-02Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy5 marks
Calculator
A university observatory building is a vertical cylinder of radius 6 m and height 4 m, topped by a hemispherical dome of radius 6 m that sits exactly on the cylinder's top face.
(a)
Calculate the total volume of the building, in m3, to 3 significant figures.
(3)
(b)
A student stands at ground level at a point P that is 25 m horizontally from the vertical axis of the building. Find the angle of elevation of the highest point of the dome from P, to the nearest 0.1°.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 volume of cylinder \(\pi(6)^2(4)=144\pi\)
✓ 1
M1 volume of hemisphere \(\frac{2}{3}\pi(6)^3=144\pi\)
✓ 1
A1 total \(=288\pi=905\) m3
✓ 1
Part (b)
M1 height of highest point \(=4+6=10\) m and \(\tan\theta=\frac{10}{25}\)
✓ 1
A1 \(\theta=21.8^\circ\)
✓ 1
Exam tip: Add the height of the cylinder to the radius of the hemisphere to find the height of the top of the dome, and measure horizontal distances to the central axis, not to the wall.
46IBMI3H-3-06r20-04Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy6 marks
Calculator
In a sports science study, the position of a marker on a pitch is recorded relative to the centre spot, in metres. A marker is at \(P(4,2)\). The matrix \(\mathbf{R}=\begin{pmatrix}0&-1\\1&0\end{pmatrix}\) represents an anticlockwise rotation of 90° about the centre spot.
(a)
Find the coordinates of the image of \(P\) under the rotation.
(2)
(b)
Write down the matrix \(\mathbf{E}\) for an enlargement about the centre spot with scale factor 1.5.
(1)
(c)
Find the single matrix that represents the rotation followed by the enlargement.
(2)
(d)
Hence write down the coordinates of the final image of \(P\).
M1 correct order of multiplication, \(\mathbf{E}\mathbf{R}\)
✓ 1
A1 \(\begin{pmatrix}0&-1.5\\1.5&0\end{pmatrix}\)
✓ 1
Part (d)
A1 \((-3,6)\)
✓ 1
Common error: For a sequence of transformations, the matrix of the first transformation is written on the right, next to the vector. The rotation then enlargement is \(\mathbf{E}\mathbf{R}\), not \(\mathbf{R}\mathbf{E}\).
47IBMI3H-3-06r20-05Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy5 marks
Calculator
A cyclist in a training session has position vector \(\mathbf{r}=\begin{pmatrix}30\\10\end{pmatrix}+t\begin{pmatrix}4.5\\6.0\end{pmatrix}\) metres relative to a coach, where \(t\) is the time in seconds after the start. The unit vectors point east and north respectively.
(a)
Find the speed of the cyclist.
(2)
(b)
Find the position vector of the cyclist when \(t=20\).
(1)
(c)
Find the bearing on which the cyclist is travelling, to the nearest 0.1°.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\sqrt{4.5^2+6.0^2}\)
✓ 1
A1 \(7.5\) m s−1
✓ 1
Part (b)
A1 \(\begin{pmatrix}120\\130\end{pmatrix}\) m
✓ 1
Part (c)
M1 \(\tan\theta=\frac{4.5}{6.0}\) with \(\theta\) measured from north
✓ 1
A1 \(\theta=036.9^\circ\)
✓ 1
Common error: A bearing is measured clockwise from north, so with east as the first component the angle from north satisfies \(\tan\theta=\frac{\text{east}}{\text{north}}\).
48IBMI3H-3-04r23-01Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy6 marks
A swimmer crosses a straight river that is 60 m wide. In still water she swims at 1.2 m s−1. She aims directly north, perpendicular to the banks. The river flows east at 0.5 m s−1. Use unit vectors i (east) and j (north).
(a)
Write down her resultant velocity as a column vector and find her speed over the ground.
(2)
(b)
Find the bearing of her actual direction of travel, to the nearest 0.1°.
(2)
(c)
Find how far downstream she is carried by the time she reaches the far bank.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 resultant velocity = (0.5, 1.2) m s−1
✓ 1
A1 speed = √(0.52 + 1.22) = 1.3 m s−1
✓ 1
Part (b)
M1 uses tan−1(0.5/1.2) (angle measured from north)
✓ 1
A1 022.6°
✓ 1
Part (c)
M1 time to cross = 60 / 1.2 = 50 s
✓ 1
A1 downstream distance = 0.5 × 50 = 25 m
✓ 1
Common error: The bearing is measured clockwise from north, so for a velocity (east, north) use tan⁻¹(east/north), not north/east.
49IBMI3H-3-04r23-06Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy7 marks
Five libraries A, B, C, D and E are connected by direct roads. The adjacency matrix M for the network is shown, with rows and columns in the order A, B, C, D, E. A: (0 1 1 0 0) B: (1 0 1 1 0) C: (1 1 0 1 0) D: (0 1 1 0 1) E: (0 0 0 1 0)
(a)
Find the number of roads in the network.
(2)
(b)
The matrix M3 has entry 2 in the row for A and column for E. Explain what this entry represents and list the corresponding routes.
(3)
(c)
Use the matrix M2 to find the total number of walks of length 2 that start at library D.
(2)
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Notes
Part (a)
M1 sum of all entries of M is 12 (each road counted twice)
✓ 1
A1 6 roads
✓ 1
Part (b)
R1 the number of walks of length 3 (three roads) from A to E
✓ 1
A1 A–B–D–E
✓ 1
A1 A–C–D–E
✓ 1
Part (c)
M1 finds the row for D in M2: (2, 1, 1, 3, 0)
✓ 1
A1 2 + 1 + 1 + 3 + 0 = 7 walks
✓ 1
Exam tip: Entry (i, j) of Mⁿ counts walks of exactly n edges from vertex i to vertex j; a walk may repeat edges and vertices.
50IBMI3H-3-04r23-12Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy7 marks
A wind farm has three turbines A, B and C on flat ground. Turbine B is 6.4 km from A on a bearing of 070°. Turbine C is 4.8 km from A on a bearing of 150°.
(a)
Write down the size of angle BAC.
(1)
(b)
Find the distance BC, to 3 significant figures.
(3)
(c)
Find the bearing of C from B, to the nearest 0.1°.
(3)
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Part (a)
A1 150° − 70° = 80°
✓ 1
Part (b)
M1 uses the cosine rule: BC2 = 6.42 + 4.82 − 2(6.4)(4.8) cos 80°
✓ 1
A1 BC2 = 53.33…
✓ 1
A1 BC = 7.30 km
✓ 1
Part (c)
M1 uses the sine rule: sin(ABC) / 4.8 = sin 80° / 7.3028…, so angle ABC = 40.3°
✓ 1
M1 bearing of A from B is 250°; subtracts angle ABC
Exam tip: Draw a north line at each point and mark the bearing angles before using a rule; the bearing back from B to A is the forward bearing plus 180°.
51IBMI3H-3-04r23-13Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy6 marks
A swimming pool is 25 m long and 12 m wide. The floor slopes uniformly from a depth of 1.2 m at the shallow end to 3.8 m at the deep end.
(a)
Find the angle the floor makes with the horizontal, to 3 significant figures.
(2)
(b)
Find the volume of water in the full pool.
(2)
(c)
Find the length of the sloping floor from the shallow end to the deep end, to 3 significant figures.
Common error: The depth difference (2.6 m), not the greater depth (3.8 m), is the vertical side of the right-angled triangle that gives the slope.
52IBMI3H-3-05r23-01Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy7 marks
Calculator
Two thrusters on a space probe produce forces (in kN) F1 = 2i − j + 3k and F2 = 4i + 2j − k.
(a)
Find F1 · F2.
(2)
(b)
Find the angle between the two forces, correct to 1 decimal place.
(3)
(c)
A third thruster produces the force F3 = 4i + 2j + pk kN. Given that F3 is perpendicular to F1, find the value of p.
(2)
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Part (a)
M1 attempt to multiply corresponding components and add
✓ 1
A1 2×4 + (−1)×2 + 3×(−1) = 3
✓ 1
Part (b)
M1 use of cos θ = (F1·F2)/(|F1||F2|)
✓ 1
A1 |F1| = √14 and |F2| = √21
✓ 1
A1 θ = 79.9° (ECF from (a))
✓ 1
Part (c)
M1 set F1·F3 = 0, i.e. 8 − 2 + 3p = 0
✓ 1
A1 p = −2
✓ 1
Common error: Perpendicular vectors have a scalar product of zero; setting the vector product to zero instead would test for parallel vectors.
53IBMI3H-3-05r23-02Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy6 marks
Calculator
A rover explores a lunar plain. Its position relative to the landing module is r(t) = (t3 − 2t)i + (5t − t2)j, where distances are in metres and t is the time in seconds, t ≥ 0.
(a)
Find an expression for the velocity of the rover at time t.
(2)
(b)
Find the speed of the rover when t = 2, giving your answer to 3 significant figures.
(2)
(c)
Find the magnitude of the acceleration of the rover when t = 2, giving your answer to 3 significant figures.
(2)
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Part (a)
M1 differentiate each component with respect to t
✓ 1
A1 v = (3t2 − 2)i + (5 − 2t)j
✓ 1
Part (b)
M1 v(2) = 10i + j and attempt at magnitude √(102 + 12)
✓ 1
A1 10.0 m s−1
✓ 1
Part (c)
M1 a = 6ti − 2j so a(2) = 12i − 2j
✓ 1
A1 √(122 + 22) = 12.2 m s−2
✓ 1
Exam tip: Speed is the magnitude of the velocity vector, so a final answer for speed is never a vector and is never negative.
54IBMI3H-3-05r23-09Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy7 marks
Calculator
A satellite design is drawn on a coordinate grid. A rotation of 90° anticlockwise about the origin is represented by the matrix R = \(\begin{pmatrix}0 & -1\\1 & 0\end{pmatrix}\), and a reflection in the x-axis is represented by S = \(\begin{pmatrix}1 & 0\\0 & -1\end{pmatrix}\).
(a)
Find the image of the point B(4, 1) under the rotation R.
(2)
(b)
Find the single matrix that represents the rotation R followed by the reflection S.
(2)
(c)
The point P is transformed by R followed by S to give the image P′(−2, −5). Find the coordinates of P.
(3)
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Part (a)
M1 multiply R by the column vector (4, 1)
✓ 1
A1 (−1, 4)
✓ 1
Part (b)
M1 the product SR, in this order
✓ 1
A1 correct matrix with rows (0, −1) and (−1, 0)
✓ 1
Part (c)
M1 use of the inverse matrix, or solve −y = −2 and −x = −5
✓ 1
A1 y = 2
✓ 1
A1 x = 5, so P is (5, 2)
✓ 1
Common error: For "R followed by S" the combined matrix is SR, not RS; the matrix applied first is written on the right.
55IBMI3H-3-05r23-11Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy8 marks
Calculator
A drone inspects a suspension bridge. It leaves tower T and flies 340 m on a bearing of 065° to a point P. It then flies 280 m on a bearing of 150° to a point Q.
(a)
Show that angle TPQ is 95°.
(2)
(b)
Find the distance TQ, correct to the nearest metre.
(3)
(c)
Find the bearing of Q from T, correct to the nearest degree.
(3)
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Part (a)
M1 the bearing of T from P is 065° + 180° = 245°
✓ 1
A1 245° − 150° = 95°
✓ 1
Part (b)
M1 use of the cosine rule: TQ2 = 3402 + 2802 − 2×340×280×cos 95°
✓ 1
A1 correct substitution
✓ 1
A1 TQ = 459 m
✓ 1
Part (c)
M1 use of the sine rule to find angle PTQ, using the unrounded value of TQ
✓ 1
A1 angle PTQ = 37.4°
✓ 1
A1 bearing 065° + 37.4° = 102°
✓ 1
Common error: A bearing is measured clockwise from north; to find the angle inside the triangle, draw a north line at each vertex and use the fact that co-interior angles add to 180°.
56IBMI3H-3-03r24-01Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy5 marks
An underwater inspection robot at a fish farm moves from point P to point Q. Relative to a fixed buoy, the position vectors are \(\overrightarrow{OP}=\begin{pmatrix}2\\-1\\-4\end{pmatrix}\) m and \(\overrightarrow{OQ}=\begin{pmatrix}8\\3\\-16\end{pmatrix}\) m, where the third component is the height relative to the surface (negative below the surface).
(a)
Find the vector \(\overrightarrow{PQ}\).
(1)
(b)
Find the distance between P and Q.
(2)
(c)
The robot travels from P to Q in a straight line at a constant speed of 0.70 m s−1. Calculate the time taken.
M1 time = distance ÷ speed, \(14\div0.70\) (ECF from (b))
✓ 1
A1 20 s
✓ 1
Exam tip: For a distance between two points, subtract the position vectors first, then take the magnitude of the difference.
57IBMI3H-3-03r24-02Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy6 marks
Two segments of a robotic arm are modelled by the vectors \(\mathbf{a}=\begin{pmatrix}2\\k\\-3\end{pmatrix}\) and \(\mathbf{b}=\begin{pmatrix}4\\-1\\2\end{pmatrix}\), where \(k\) is a constant.
(a)
The segments are perpendicular. Find the value of \(k\).
(2)
(b)
The arm is adjusted so that \(k=5\). Find the angle between the segments, giving your answer in degrees to 3 significant figures.
(4)
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Part (a)
M1 set \(\mathbf{a}\cdot\mathbf{b}=0\), so \(8-k-6=0\)
✓ 1
A1 \(k=2\)
✓ 1
Part (b)
A1 \(\mathbf{a}\cdot\mathbf{b}=-3\)
✓ 1
A1 \(|\mathbf{a}|=\sqrt{38}\) and \(|\mathbf{b}|=\sqrt{21}\)
✓ 1
M1 \(\cos\theta=\dfrac{-3}{\sqrt{38}\sqrt{21}}\)
✓ 1
A1 \(\theta=96.1^\circ\)
✓ 1
Common error: A negative scalar product means the angle is obtuse; do not take the modulus of the scalar product before finding the angle.
58IBMI3H-3-03r24-11Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy4 marks
A triangular fabric print has vertices A(1, 1), B(4, 1) and C(1, 3), measured in cm. A design software applies the transformation represented by the matrix \(\mathbf{T}=\begin{pmatrix}2&1\\0&3\end{pmatrix}\).
(a)
Find the coordinates of the images A′, B′ and C′.
(2)
(b)
Find the area of the image triangle A′B′C′.
(2)
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Part (a)
M1 multiply \(\mathbf{T}\) by the position vectors
✓ 1
A1 A′(3, 3), B′(9, 3), C′(5, 9)
✓ 1
Part (b)
M1 area of ABC is 3 cm2 and \(\det\mathbf{T}=6\), so area scales by 6
✓ 1
A1 18 cm2
✓ 1
Exam tip: The area scale factor of a matrix transformation is \(|\det\mathbf{T}|\), so you can find the image area without the image coordinates.
59IBMI3H-3-03r26-01Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy6 marks
Calculator
A robotic arm moves its gripper in a straight line from point A(2, −1, 3) to point B(8, 3, 5). Coordinates are in metres relative to the base of the arm.
(a)
Find the vector \(\overrightarrow{AB}\).
(2)
(b)
Calculate the distance AB, giving your answer to 3 significant figures.
(2)
(c)
The move takes 2.5 seconds at constant speed. Find this speed to 3 significant figures, using the unrounded distance from (b).
(2)
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Part (a)
M1 subtracts position vectors, \(\mathbf{b}-\mathbf{a}\)
Common error: Always use the unrounded value from an earlier part when a later part asks for it; rounding early can change the third significant figure.
60IBMI3H-3-03r26-05Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy6 marks
Calculator
A warehouse robot has constant acceleration \(\mathbf{a}=\begin{pmatrix}2\\-1\end{pmatrix}\) m s−2. At time \(t=0\) it is at position \(\begin{pmatrix}5\\-2\end{pmatrix}\) m with velocity \(\begin{pmatrix}1\\3\end{pmatrix}\) m s−1.
(a)
Find an expression for the velocity of the robot at time \(t\) seconds.
(2)
(b)
Find the position of the robot when \(t=4\).
(2)
(c)
Find the speed of the robot when \(t=4\), to 3 significant figures.
(2)
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Notes
Part (a)
M1 integrates acceleration and uses the initial velocity
M1 integrates \(\mathbf{v}\) and uses \(\mathbf{r}(0)\), giving \(\mathbf{r}=\begin{pmatrix}5+t+t^2\\-2+3t-\tfrac12t^2\end{pmatrix}\)
✓ 1
A1 \(\begin{pmatrix}25\\2\end{pmatrix}\) m
✓ 1
Part (c)
M1 \(\mathbf{v}(4)=\begin{pmatrix}9\\-1\end{pmatrix}\) and \(\sqrt{9^2+(-1)^2}\)
✓ 1
A1 \(9.06\) m s−1
✓ 1
Exam tip: Speed is the magnitude of the velocity vector; it is a scalar and has no direction.
61IBMI3H-3-03r26-07Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy6 marks
Calculator
Five charging stations A, B, C, D and E in a warehouse are linked by tracks used by robots. The tracks are AB, AC, BC, BD, CE and DE. Each track can be used in either direction.
(a)
Write down the adjacency matrix for the graph, with rows and columns in the order A, B, C, D, E.
(2)
(b)
Use the matrix to find the degree of vertex C and the total number of tracks.
(2)
(c)
Find the number of different walks of length 3 from A to E.
(2)
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Part (a)
A1 rows for A, B and C correct: \((0,1,1,0,0)\), \((1,0,1,1,0)\), \((1,1,0,0,1)\)
✓ 1
A1 rows for D and E correct: \((0,1,0,0,1)\), \((0,0,1,1,0)\)
✓ 1
Part (b)
A1 degree of C is the sum of its row, \(3\)
✓ 1
A1 sum of all entries is \(12\), so \(6\) tracks (half of the total)
✓ 1
Part (c)
M1 attempt to find \(\mathbf{M}^3\) (or row A of \(\mathbf{M}^2\) times column E of \(\mathbf{M}\)): \(1\times1+1\times1\)
✓ 1
A1 \(2\) walks
✓ 1
Exam tip: The entry in row \(i\), column \(j\) of \(\mathbf{M}^n\) counts the walks of length \(n\) from vertex \(i\) to vertex \(j\).
62IBMI3H-3-03r26-14Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy6 marks
Calculator
Three boutiques are located at A(2, 1), B(8, 1) and C(5, 7), with coordinates in kilometres. The Voronoi diagram for these sites has one vertex. The perpendicular bisector of AB is the line \(x=5\).
(a)
Find the equation of the perpendicular bisector of AC.
(3)
(b)
Find the coordinates of the Voronoi vertex, and the distance from the vertex to boutique A.
(3)
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Notes
Part (a)
M1 midpoint of AC \(=(3.5,\,4)\) and gradient of AC \(=2\)
✓ 1
M1 perpendicular gradient \(-\tfrac12\)
✓ 1
A1 \(y=-\tfrac12x+5.75\)
✓ 1
Part (b)
M1 substitutes \(x=5\) into the bisector of AC to get \(y=3.25\), so the vertex is \((5,\,3.25)\)
✓ 1
M1 distance \(\sqrt{(5-2)^2+(3.25-1)^2}\)
✓ 1
A1 \(3.75\) km
✓ 1
Exam tip: A Voronoi vertex is equidistant from three sites, so you can check your answer by calculating the distance to each of the three.
63IBMI3H-3-06r27-01Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy6 marks
A research ship monitors a volcanic island. Relative to a harbour at O, the position vector of the ship t hours after 09:00 is r = (4, 10) + t(6, −8), where components are in kilometres, the first being east and the second north.
(a)
Find the speed of the ship.
(2)
(b)
Find the position vector of the ship at 11:30.
(2)
(c)
Find the time at which the ship is due east of the harbour.
(2)
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Notes
Part (a)
M1 √(62 + (−8)2)
✓ 1
A1 10 km h−1
✓ 1
Part (b)
M1 substitutes t = 2.5
✓ 1
A1 (19, −10)
✓ 1
Part (c)
M1 10 − 8t = 0 (north component is zero)
✓ 1
A1 t = 1.25, so the time is 10:15 (t = 1.25 hours)
✓ 1
Common error: The speed is the magnitude of the velocity vector (6, −8), not the sum of its components.
64IBMI3H-3-06r27-02Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy6 marks
Two spotlights in a theatre shine along beams with direction vectors a = 3i − 2j + 6k and b = 2i + 3j − k.
(a)
Find a · b.
(1)
(b)
Find the angle between the two beams, giving your answer to the nearest degree.
(3)
(c)
A third beam has direction vector ki + 3j − k. It is perpendicular to the first beam. Find the value of k.
(2)
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Part (a)
A1 6 − 6 − 6 = −6
✓ 1
Part (b)
M1 cos θ = (a · b)/(|a||b|)
✓ 1
M1 |a| = 7 and |b| = √14
✓ 1
A1 θ = 103° (ECF from (a))
✓ 1
Part (c)
M1 3k − 6 − 6 = 0 (scalar product is zero)
✓ 1
A1 k = 4
✓ 1
Exam tip: The scalar product of perpendicular vectors is zero; the angle formula gives the angle between the directions, which lies between 0° and 180°.
65IBMI3H-3-06r27-12Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy6 marks
A cargo ship leaves port P and sails 12 km on a bearing of 065° to a buoy B. It then sails 9 km on a bearing of 140° to a point C.
(a)
Find the distance PC, correct to 3 significant figures.
(3)
(b)
Find the bearing of P from C, to the nearest degree.
A1 the bearing of B from C is 320°, so the bearing of P from C is 320° − 43.8° = 276°
✓ 1
Exam tip: Draw a north line at each turning point and use alternate or co-interior angles to find the angle inside the triangle.
66IBMI3H-3-06r27-13Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy6 marks
From a point A on level ground, the angle of elevation of the summit S of a volcano is 12°. An observer walks 600 m directly towards the volcano, along a straight horizontal line, to a point B. From B the angle of elevation of S is 17°. A, B and the point directly below S are in the same vertical plane.
(a)
Show that angle ASB = 5°.
(2)
(b)
Use the sine rule to find the distance BS, correct to 3 significant figures.
(2)
(c)
Find the height of the summit above the level of A and B. Use the unrounded value of BS and give your answer to 3 significant figures.
(2)
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Part (a)
M1 angle SBA = 180° − 17° = 163° (or uses the exterior angle of triangle ABS)
✓ 1
A1 angle ASB = 180° − 12° − 163° = 5°
✓ 1
Part (b)
M1 BS/sin 12° = 600/sin 5°
✓ 1
A1 BS = 1430 m
✓ 1
Part (c)
M1 height = BS × sin 17°
✓ 1
A1 418 m
✓ 1
Common error: The angle at B inside the triangle is 180° − 17°, not 17°; the angle of elevation is measured at the second point from the horizontal pointing away from the first.
67IBMI3H-3-06r29-01Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy6 marks
A robotic arm has two segments. Relative to the shoulder joint at the origin O (units in decimetres), the first segment is represented by the vector \(\mathbf{u}=\begin{pmatrix}2\\1\\2\end{pmatrix}\) and the second segment, starting at the end of the first, by \(\mathbf{v}=\begin{pmatrix}-1\\2\\2\end{pmatrix}\).
(a)
Find \(|\mathbf{u}|\) and \(|\mathbf{v}|\).
(2)
(b)
Find \(\mathbf{u}\cdot\mathbf{v}\).
(2)
(c)
Find the angle between the two vectors, in degrees, to 3 significant figures.
Common error: The angle given by the scalar product is between the two vectors placed tail to tail, not the elbow angle between the segments placed head to tail.
68IBMI3H-3-06r29-05Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy6 marks
Four fishing ports P, Q, R and S are linked by direct sea routes: P–Q, P–R, Q–R, Q–S and R–S. There is no direct route between P and S. The routes are two-way.
(a)
Write down the adjacency matrix for the network, with the ports in the order P, Q, R, S.
(2)
(b)
Find the number of different walks of length 3 from P to S.
(2)
(c)
List these walks.
(2)
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Part (a)
A1 first two rows \((0,1,1,0)\) and \((1,0,1,1)\)
✓ 1
A1 last two rows \((1,1,0,1)\) and \((0,1,1,0)\)
✓ 1
Part (b)
M1 use of the adjacency matrix cubed (entry in row P, column S)
✓ 1
A1 2
✓ 1
Part (c)
A1 P–Q–R–S
✓ 1
A1 P–R–Q–S
✓ 1
Common error: The entry in \(\mathbf{M}^n\) counts walks of length exactly \(n\), and a walk may repeat edges.
69IBMI3H-3-06r29-10Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy6 marks
A fashion print repeats a motif of area 12 cm2. A computer transforms the motif using the matrix \(\mathbf{M}=\begin{pmatrix}2&1\\-1&3\end{pmatrix}\).
(a)
Find \(\det\mathbf{M}\).
(2)
(b)
Find the area of the transformed motif.
(2)
(c)
A second matrix \(\mathbf{N}=\begin{pmatrix}2&4\\1&2\end{pmatrix}\) is tried. Explain why N cannot be used to transform the motif.
(2)
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Part (a)
M1 \(2(3)-1(-1)\)
✓ 1
A1 7
✓ 1
Part (b)
M1 area scale factor is \(|\det\mathbf{M}|\)
✓ 1
A1 84 cm2
✓ 1
Part (c)
R1 \(\det\mathbf{N}=4-4=0\)
✓ 1
R1 every point is mapped onto a line, so the motif would have zero area
✓ 1
Exam tip: The area scale factor of a transformation is the absolute value of the determinant.
70IBMI3H-3-06r29-13Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy6 marks
A cuboid fish crate has a rectangular base ABCD with AB = 60 cm and BC = 40 cm, and height 30 cm. A thermometer probe lies along the line from base corner A to the opposite top corner G, directly above C.
(a)
Find the length of the base diagonal AC, to 3 significant figures.
(2)
(b)
Find the length of the probe AG, to 3 significant figures.
(2)
(c)
Find the angle between the probe and the base of the crate.
Common error: The angle between a line and a plane is measured to the line's projection on that plane, here AC, not to an edge of the base.
71IBMI3H-3-05r30-01Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy4 marks
A crop-spraying aircraft takes off from a point O on a farm. Relative to O, its position vector in kilometres at time t hours after take-off is r = \(\begin{pmatrix}12\\-8\end{pmatrix}\) + t\(\begin{pmatrix}150\\90\end{pmatrix}\).
(a)
Find the position vector of the aircraft 40 minutes after take-off.
(2)
(b)
Find the speed of the aircraft. Give your answer in km h−1 correct to 3 significant figures.
(2)
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Notes
Part (a)
M1 substitutes t = 2/3 (or 0.6667) into the equation
✓ 1
A1 position vector (112, 52) km
✓ 1
Part (b)
M1 speed = √(1502 + 902) = √30600
✓ 1
A1 175 km h−1
✓ 1
Common error: The speed is the magnitude of the direction vector, not the magnitude of the position vector.
72IBMI3H-3-05r30-02Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy5 marks
A farmer has five sheds, A, B, C, D and E, joined by tracks. There is exactly one track between each of these pairs of sheds: AB, AC, BC, BD, CD, CE and DE. There are no other tracks.
(a)
Write down the adjacency matrix for this graph, using the sheds in alphabetical order.
(2)
(b)
Use your matrix to find the number of different walks of length 3 from B to D.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 symmetric 5×5 matrix with zeros on the leading diagonal
M1 attempts to calculate the cube of the adjacency matrix
✓ 1
A1 correct entry in row B, column D used
✓ 1
A1 7 walks
✓ 1
Exam tip: A walk of length n from vertex i to vertex j is counted by the (i, j) entry of the nth power of the adjacency matrix.
73IBMI3H-3-05r30-03Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy6 marks
A music app animates a note head on a screen using transformations of the plane. Transformation A is a rotation of 90° anticlockwise about the origin. Transformation B is an enlargement with scale factor 3 and centre the origin. The transformation A is applied first, followed by B.
(a)
Write down the single matrix that represents A followed by B.
(2)
(b)
The note head starts at P(4, −1). Find the coordinates of its final position.
(2)
(c)
A triangle on the screen has area 2.5 square units before the transformations. Find the area of its image.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 matrix for A is \(\begin{pmatrix}0&-1\\1&0\end{pmatrix}\) and matrix for B is \(\begin{pmatrix}3&0\\0&3\end{pmatrix}\)
✓ 1
A1 combined matrix BA = \(\begin{pmatrix}0&-3\\3&0\end{pmatrix}\)
✓ 1
Part (b)
M1 multiplies the combined matrix by the column vector (4, −1)
✓ 1
A1 (3, 12)
✓ 1
ECF from (a)
—
Part (c)
M1 determinant of the combined matrix = 0×0 − (−3)(3) = 9
✓ 1
A1 9 × 2.5 = 22.5 square units
✓ 1
Common error: To apply A first and then B, the matrix product is written BA, with the first transformation on the right.
74IBMI3H-3-05r30-08Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy6 marks
A farmer inspects five fields, A to E, by walking along the tracks between them. The lengths of the tracks in kilometres are: AB 4, AC 6, BC 5, BD 9, CD 3, CE 8 and DE 5. The farmer wants to walk along every track at least once, starting and finishing at the same field, with the shortest total distance.
(a)
Show that exactly two fields have an odd number of tracks meeting at them, and name them.
(2)
(b)
Find the shortest distance that the farmer must walk, and state which tracks are walked twice.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 degrees found: A 2, B 3, C 4, D 3, E 2
✓ 1
A1 B and D are the odd vertices
✓ 1
Part (b)
M1 sum of all track lengths = 40 km
✓ 1
M1 identifies shortest path between B and D
✓ 1
A1 B–C–D with length 5 + 3 = 8 km, shorter than the direct track BD of 9 km, so BC and CD are walked twice
✓ 1
A1 total 40 + 8 = 48 km
✓ 1
Common error: The shortest path between the two odd vertices is not always the direct edge between them; check the routes through other vertices.
75IBMI3H-3-05r31-01Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy6 marks
A cargo ship is tracked from a harbour at the origin O. Distances are in kilometres, with x measured east and y north. At time t = 0 hours the ship is at (−20, 30) and it moves with constant velocity (9, −12) km h−1.
(a)
Calculate the speed of the ship.
(2)
(b)
Write down the position vector of the ship at time t hours, and find the time at which the ship crosses the east–west line through the harbour (y = 0).
(2)
(c)
Find the distance from the harbour to the ship at the moment it crosses this line.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to find the magnitude of the velocity vector, e.g. √(92 + 122)
✓ 1
A1 15 km h−1
✓ 1
Part (b)
A1 r = (−20 + 9t, 30 − 12t)
✓ 1
A1 30 − 12t = 0 so t = 2.5 hours
✓ 1
Part (c)
M1 substitute t = 2.5 into the x-component: −20 + 9 × 2.5
✓ 1
A1 2.5 km
✓ 1
Common error: Speed is the magnitude of the velocity vector; a distance from the origin is the magnitude of a position vector, so check the sign of your component before quoting it.
76IBMI3H-3-05r31-02Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy7 marks
In a theatre, two spotlights hang from cables attached to a rigging point R. Taking R as the origin, the spotlights are at points A and B, where RA = (4, −3, −12) and RB = (−2, 6, −9). All coordinates are in metres.
(a)
Show that the cable RA has length 13 m.
(2)
(b)
Find the angle between the cables RA and RB. Give your answer in degrees to the nearest 0.1°.
(3)
(c)
Find the distance AB between the two spotlights. Give your answer to 3 significant figures.
Exam tip: For the angle between two vectors, find the scalar product and both magnitudes first, then use cos θ = (a·b) ÷ (|a||b|). Check the result is sensible: a positive scalar product means an acute angle.
77IBMI3H-3-05r31-12Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy8 marks
A volcano is modelled as a right circular cone with base radius 1.7 km and vertical height 0.85 km.
(a)
Calculate the slant height of the cone. Give your answer to 3 significant figures.
(2)
(b)
Calculate the angle the slope makes with the horizontal. Give your answer to the nearest 0.1°.
(2)
(c)
Calculate the volume of the cone in km3 to 3 significant figures.
(2)
(d)
The rock has density 2500 kg m−3. Use the unrounded volume to find the mass of the volcano in kilograms, in the form a × 10k where 1 ≤ a < 10.
(2)
Show mark scheme
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Mark
Notes
Part (a)
M1 √(1.72 + 0.852)
✓ 1
A1 1.90 km
✓ 1
Part (b)
M1 tan θ = 0.85 ÷ 1.7
✓ 1
A1 26.6°
✓ 1
Part (c)
M1 V = ⅓π × 1.72 × 0.85
✓ 1
A1 2.57 km3
✓ 1
Part (d)
M1 convert: 1 km3 = 109 m3, so mass = 2.5724 × 109 × 2500
✓ 1
A1 6.43 × 1012 kg
✓ 1
Common error: Convert units before multiplying by a density: 1 km³ is 10⁹ m³, not 10³ m³. Keep unrounded values in a calculation chain and round only the final answer.
78IBMI3H-3-05r33-01Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy5 marks
An aircraft approaches a small airport. Relative to the airport, which is at the origin O, the position vector of the aircraft, in km, is given by \(\mathbf{r}=\begin{pmatrix}-60\\-80\end{pmatrix}+t\begin{pmatrix}90\\120\end{pmatrix}\), where t is the time in hours after 14:00.
(a)
Find the speed of the aircraft.
(2)
(b)
Show that the aircraft flies directly over the airport, and find the time, in minutes after 14:00, at which this happens.
(3)
Show mark scheme
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Mark
Notes
Part (a)
M1 attempt to find the magnitude of the velocity vector, e.g. \(\sqrt{90^2+120^2}\)
✓ 1
A1 150 km h−1
✓ 1
Part (b)
M1 sets one component of r equal to zero, e.g. −60 + 90t = 0 giving t = 2/3
✓ 1
R1 shows the other component is also zero at the same time: −80 + 120 × 2/3 = 0
✓ 1
A1 40 minutes after 14:00
✓ 1
Exam tip: To show that a line passes through a point, solve for the parameter using one component and then check that the other component gives the same value.
79IBMI3H-3-05r33-05Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy5 marks
Five airports A, B, C, D and E are linked by direct two-way flights on these routes only: AB, AC, AE, BC, BD, CD, CE and DE.
(a)
Write down the adjacency matrix of this network, using the order A, B, C, D, E for the rows and columns.
(2)
(b)
Use the cube of the adjacency matrix to find the number of different routes from A to D that use exactly three flights.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 a 5 × 5 symmetric matrix with zeros on the diagonal
✓ 1
A1 rows: A (0 1 1 0 1), B (1 0 1 1 0), C (1 1 0 1 1), D (0 1 1 0 1), E (1 0 1 1 0)
✓ 1
Part (b)
M1 attempt to find the third power of the adjacency matrix
✓ 1
A1 identifies the entry in row A, column D
✓ 1
A1 4 routes
✓ 1
Exam tip: Entry (i, j) of the nth power of an adjacency matrix counts the walks of exactly n edges from vertex i to vertex j; walks may revisit vertices.
80IBMI3H-3-05r33-09Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy5 marks
A designer scales and turns a triangular stage backdrop. The transformation is an enlargement, scale factor 1.5 and centre the origin, followed by an anticlockwise rotation of 90° about the origin.
(a)
Find the single matrix that represents the combined transformation.
(3)
(b)
A corner of the backdrop is at (4, 6). Find the coordinates of its image.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 enlargement matrix \(\begin{pmatrix}1.5&0\\0&1.5\end{pmatrix}\) and rotation matrix \(\begin{pmatrix}0&-1\\1&0\end{pmatrix}\)
✓ 1
M1 multiplies in the correct order, rotation matrix × enlargement matrix
✓ 1
A1 \(\begin{pmatrix}0&-1.5\\1.5&0\end{pmatrix}\)
✓ 1
Part (b)
M1 multiplies the combined matrix by the column vector (4, 6)
✓ 1
A1 (−9, 6)
✓ 1
Common error: For a sequence of transformations the matrix for the first one is written on the right: the product is rotation × enlargement, in that order.
81IBMI3H-3-05r36-01Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy6 marks
A tagged dolphin is at position (2, 5) km relative to a research buoy at 08:00. It then swims with constant velocity (3, −4) km h−1. Let t be the time in hours after 08:00.
(a)
Find the speed of the dolphin.
(2)
(b)
Find the position vector of the dolphin at 10:30.
(2)
(c)
Hence find the distance of the dolphin from the buoy at 10:30, in km, correct to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to find the magnitude of the velocity, \(\sqrt{3^2+(-4)^2}\)
✓ 1
A1 5 km h−1
✓ 1
Part (b)
M1 use of r = (2, 5) + 2.5(3, −4)
✓ 1
A1 (9.5, −5) km
✓ 1
Part (c)
M1 \(\sqrt{9.5^2+(-5)^2}\) (ECF from (b))
✓ 1
A1 10.7 km
✓ 1
Common error: Speed is the magnitude of the velocity vector, not the sum of its components.
82IBMI3H-3-05r36-05Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy4 marks
Four fish-processing depots A, B, C and D are linked by direct roads: AB, AC, BC, BD and CD. Each road can be travelled in both directions. Let M be the adjacency matrix for the depots, with rows and columns in the order A, B, C, D.
(a)
Write down M.
(2)
(b)
Use M3 to find the number of different walks of length 3 from A to D.
M1 attempt to find M3 (by GDC) and read the entry in row A, column D
✓ 1
A1 2 walks (ABCD and ACBD)
✓ 1
Exam tip: The (i, j) entry of Mⁿ counts walks of length n from vertex i to vertex j.
83IBMI3H-3-05r36-09Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy5 marks
A reef map has sampling points P(2, 1), Q(5, 1) and R(2, 4). The map is rotated 90° anticlockwise about the origin, and then enlarged by scale factor 1.5 with centre the origin.
(a)
Write down the matrix for the rotation.
(1)
(b)
Find the coordinates of the image of Q after both transformations.
(2)
(c)
Find the single matrix that represents the rotation followed by the enlargement.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(\begin{pmatrix}0&-1\\1&0\end{pmatrix}\)
✓ 1
Part (b)
M1 rotation image (−1, 5), or multiplication by the combined matrix
✓ 1
A1 (−1.5, 7.5)
✓ 1
Part (c)
M1 multiply the matrices in the correct order, or scale each entry of the rotation matrix by 1.5
✓ 1
A1 \(\begin{pmatrix}0&-1.5\\1.5&0\end{pmatrix}\)
✓ 1
Common error: For two transformations, the matrix of the transformation applied first is written on the right.
84IBMI3H-3-06r36-01Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy5 marks
Two support cables on a wind turbine tower have direction vectors u = (k, 3, −2) and v = (2, k, 5), where k is a constant.
(a)
Find the value of k for which the cables are perpendicular.
(2)
(b)
When k = 1, find the angle between the directions of u and v, in degrees, correct to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 set the scalar product u·v = 2k + 3k − 10 = 0
✓ 1
A1 k = 2
✓ 1
Part (b)
M1 scalar product u·v = 2 + 3 − 10 = −5
✓ 1
M1 use of cos θ = (u·v)/(|u||v|) with |u| = √14 and |v| = √30
✓ 1
A1 104°
✓ 1
Exam tip: A scalar product of zero means the vectors are perpendicular; a negative scalar product means the angle between them is obtuse.
85IBMI3H-3-06r36-05Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy6 marks
Five retail stores A to E are linked by delivery roads. The adjacency matrix M, with rows and columns in the order A, B, C, D, E, is \(\begin{pmatrix}0&1&1&0&1\\1&0&0&1&0\\1&0&0&1&1\\0&1&1&0&0\\1&0&1&0&0\end{pmatrix}\)
(a)
Find the number of roads in the network.
(2)
(b)
Using M2, find the number of walks of length 2 from B to C, and list them.
(2)
(c)
A driver wants to use every road exactly once. Determine whether this is possible if the driver must return to the starting store.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sum of the entries is 12, and each road is counted twice (or sum of degrees ÷ 2)
✓ 1
A1 6 roads
✓ 1
Part (b)
M1 attempt to find M2 and read the entry in row B, column C
✓ 1
A1 2 walks: BAC and BDC
✓ 1
Part (c)
A1 stores A and C have odd degree (3), and the others have degree 2
✓ 1
R1 a closed route using every road once needs every vertex to have even degree, so it is not possible
✓ 1
Exam tip: The degree of a vertex is the sum of the entries in its row of the adjacency matrix.
86IBMI3H-3-06r36-09Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy5 marks
A solar array on a field plan is triangle PQR with P(1, 1), Q(4, 1) and R(1, 3). A horizontal shear is applied using the matrix S = \(\begin{pmatrix}1&2\\0&1\end{pmatrix}\).
(a)
Find the coordinates of the images P′, Q′ and R′.
(3)
(b)
Show that the area of triangle P′Q′R′ equals the area of triangle PQR.
(2)
Show mark scheme
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Mark
Notes
Part (a)
M1 multiply S by a column vector of a vertex
✓ 1
A1 P′(3, 1) and Q′(6, 1)
✓ 1
A1 R′(7, 3)
✓ 1
Part (b)
M1 det S = 1 × 1 − 2 × 0 = 1
✓ 1
R1 the area scale factor is |det S| = 1, so the area is unchanged
✓ 1
Common error: A shear changes shape but not area, so the determinant of a shear matrix is 1.
87IBMI3H-3-06r36-13Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy6 marks
A rectangular storage warehouse is 30 m long, 18 m wide and 9 m high. A cable runs from one floor corner F to the opposite roof corner G.
(a)
Find the length of the diagonal of the floor, in m, correct to 3 significant figures.
(2)
(b)
Find the length of the cable FG, in m, correct to 3 significant figures.
(2)
(c)
Find the angle between the cable and the floor, correct to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 Pythagoras: √(302 + 182)
✓ 1
A1 35.0 m
✓ 1
Part (b)
M1 √(302 + 182 + 92)
✓ 1
A1 36.1 m
✓ 1
Part (c)
M1 tan θ = 9/√1224 (unrounded floor diagonal)
✓ 1
A1 14.4°
✓ 1
Exam tip: Find the floor diagonal first, then use it as one side of a right-angled triangle with the height.
88IBMI3H-3-07r37-01Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy4 marks
Calculator
Two camps on a mountain are described using position vectors relative to a base point O, where the x-axis points east, the y-axis points north and the z-axis points vertically up. All distances are in km. Camp C1 is at (1.2, 0.8, 3.4) and camp C2 is at (2.9, 2.1, 4.6).
(a)
Find the vector \(\overrightarrow{C_1C_2}\).
(2)
(b)
Hence find the straight-line distance between the two camps. Give your answer correct to 3 significant figures.
Exam tip: Subtract in the order end point minus start point; reversing it gives a vector pointing the wrong way but the same distance.
89IBMI3H-3-07r37-06Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy6 marks
Calculator
A telecoms company connects five relay towers A, B, C, D and E with direct cable links. The links are AB, AC, BC, BD, CE and DE. Each link can carry signals in both directions.
(a)
Write down the adjacency matrix for this network, taking the towers in the order A, B, C, D, E.
(2)
(b)
Determine whether the network has an Eulerian circuit. Justify your answer.
(2)
(c)
Find the number of different walks of length 3 from tower A to tower E.
R1 two vertices (B and C) have odd degree, so there is no Eulerian circuit
✓ 1
Part (c)
M1 attempts the (A, E) entry of the cube of the adjacency matrix
✓ 1
A1 2
✓ 1
Exam tip: The entry in row i, column j of the nth power of an adjacency matrix counts walks of length n from vertex i to vertex j, so check the row and column order.
90IBMI3H-3-07r37-08Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy7 marks
Calculator
A park ranger must inspect every one of seven mountain trails. The trails join five huts A, B, C, D and E. The trail lengths in km are: AB 5, AC 7, BC 4, BD 8, CD 6, CE 9 and DE 3.
(a)
Show that B and D are the only huts at which an odd number of trails meet.
(2)
(b)
The ranger starts and ends at the same hut and walks every trail at least once. Use the Chinese postman algorithm to find the minimum distance she must walk.
(3)
(c)
On a different day the ranger may start at B and finish at D. She walks at an average speed of 2.5 km h−1. Find the minimum time needed to walk every trail.
(2)
Show mark scheme
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Mark
Notes
Part (a)
M1 finds the degrees A 2, B 3, C 4, D 3, E 2
✓ 1
R1 only B and D are odd
✓ 1
Part (b)
M1 total of all trails 5 + 7 + 4 + 8 + 6 + 9 + 3 = 42
✓ 1
M1 shortest path between B and D is the direct trail BD of 8 km (other routes are at least 10 km)
✓ 1
A1 42 + 8 = 50 km
✓ 1
Part (c)
M1 recognises no repeats are needed so the distance is 42 km, and divides by 2.5
✓ 1
A1 16.8 hours
✓ 1
Exam tip: If a graph has exactly two odd vertices, an open route between them covers every edge once, so no edge needs repeating.
91IBMI3H-3-07r37-12Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy7 marks
Calculator
From a base camp B, a summit S lies on a bearing of 040° at a distance of 5.2 km. A mountain hut H lies on a bearing of 115° from B at a distance of 3.8 km.
(a)
Find the distance between the summit and the hut. Give your answer to 3 significant figures.
(4)
(b)
Find the size of angle BSH. Use the unrounded value of SH from part (a) and give your answer to 3 significant figures.
Exam tip: A bearing difference at the base gives the angle in the triangle; draw the north line at B first so the 75° angle is not misread.
92IBMI3H-3-05r38-01Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy6 marks
A mountaineering team sets up base camp C and plans to climb to the summit S. Relative to a fixed origin, with i pointing east, j north and k vertically upwards (all distances in km), the position vectors are rC = 1.2i + 0.8j + 1.5k and rS = 3.6i + 2.0j + 3.9k.
(a)
Find the displacement vector from C to S.
(2)
(b)
Calculate the straight-line distance from C to S.
(2)
(c)
Find the angle of elevation of the line CS above the horizontal. Give your answer to the nearest 0.1°.
(2)
Show mark scheme
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Notes
Part (a)
M1 attempt to subtract position vectors, rS − rC
✓ 1
A1 2.4i + 1.2j + 2.4k (km)
✓ 1
Part (b)
M1 attempt to find the magnitude, √(2.4² + 1.2² + 2.4²) (ECF from (a))
✓ 1
A1 3.6 km
✓ 1
Part (c)
M1 use of sin θ = 2.4 / 3.6 (or tan θ = 2.4 / √(2.4² + 1.2²)) (ECF from (a) and (b))
✓ 1
A1 41.8°
✓ 1
Exam tip: The k component of a displacement is its vertical rise, so the angle of elevation comes straight from the ratio of that component to the magnitude.
93IBMI3H-3-05r38-06Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy5 marks
Four telecom towers A, B, C and D are linked by cables. The number of direct cables between each pair of towers is given by the adjacency matrix below (rows and columns in the order A, B, C, D). A: 0 1 1 0 B: 1 0 1 1 C: 1 1 0 1 D: 0 1 1 0
(a)
Write down the degree of tower C and the total number of cables.
(2)
(b)
Use the matrix M3 to find the number of different walks of length 3 from A to D.
(3)
Show mark scheme
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Mark
Notes
Part (a)
A1 degree of C = 3
✓ 1
A1 5 cables (sum of degrees is 10, so 5 edges)
✓ 1
Part (b)
M1 attempt to find M3 (by technology)
✓ 1
A1 row A, column D entry of M3 identified
✓ 1
A1 2
✓ 1
Exam tip: The (i, j) entry of Mⁿ counts walks of length n from vertex i to vertex j, so you only need one entry, not the whole matrix.
94IBMI3H-3-05r38-10Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy5 marks
On a plan of a telecom site, each mast is a point (x, y). The plan is first rotated 90° anticlockwise about the origin and then enlarged by scale factor 2 with centre the origin.
(a)
Write down the matrix R for the rotation.
(1)
(b)
Find the single matrix M that represents the rotation followed by the enlargement.
(2)
(c)
A mast is at P(3, 5). Find the coordinates of its image.
(2)
Show mark scheme
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Mark
Notes
Part (a)
A1 R = ((0, −1), (1, 0)) (rows listed)
✓ 1
Part (b)
M1 enlargement matrix multiplied on the left of R, i.e. ((2, 0), (0, 2)) R
✓ 1
A1 M = ((0, −2), (2, 0))
✓ 1
Part (c)
M1 multiply M by the column vector (3, 5)
✓ 1
A1 (−10, 6)
✓ 1
Exam tip: In a product, the transformation applied first is written closest to the point or vector, on the right.
95IBMI3H-3-05r39-01Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy6 marks
In a beekeeping apiary, three hives A, B and C stand on a hillside. Relative to the entrance gate, their position vectors, in metres, are \(\mathbf{a}=\begin{pmatrix}2\\3\\1\end{pmatrix}\), \(\mathbf{b}=\begin{pmatrix}8\\-1\\4\end{pmatrix}\) and \(\mathbf{c}=\begin{pmatrix}-1\\5\\2\end{pmatrix}\).
(a)
Find the vector \(\overrightarrow{AB}\) and hence the distance between hive A and hive B, giving the distance to 3 significant figures.
(3)
(b)
Find the size of angle BAC, giving your answer to the nearest tenth of a degree.
A1 \(\overrightarrow{AC}=\begin{pmatrix}-3\\2\\1\end{pmatrix}\) and \(\overrightarrow{AB}\cdot\overrightarrow{AC}=-23\)
✓ 1
M1 uses \(\cos\theta=\dfrac{\overrightarrow{AB}\cdot\overrightarrow{AC}}{|\overrightarrow{AB}||\overrightarrow{AC}|}\) with \(|\overrightarrow{AB}|=\sqrt{61}\), \(|\overrightarrow{AC}|=\sqrt{14}\)
✓ 1
A1 angle BAC \(=141.9^\circ\)
✓ 1
Common error: Using the position vectors a and c instead of the displacement vectors from A gives the angle at the origin, not the angle at hive A.
96IBMI3H-3-05r39-02Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy5 marks
A Formula 1 car is driven along a straight practice stretch. Relative to a marker post, its position vector at time \(t\) seconds is \(\mathbf{r}=\begin{pmatrix}40t+3t^2\\25t-t^2\end{pmatrix}\) metres. Give answers to 3 significant figures where they are not exact.
(a)
Find the velocity of the car when \(t=2\) and hence its speed at that time.
(3)
(b)
Find the magnitude of the acceleration of the car.
(2)
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Part (a)
M1 differentiates each component with respect to \(t\)
✓ 1
A1 \(\mathbf{v}=\begin{pmatrix}40+6t\\25-2t\end{pmatrix}\), so at \(t=2\), \(\mathbf{v}=\begin{pmatrix}52\\21\end{pmatrix}\) m s−1
✓ 1
A1 speed \(=\sqrt{52^2+21^2}=56.1\) m s−1
✓ 1
Part (b)
M1 differentiates the velocity to obtain \(\mathbf{a}=\begin{pmatrix}6\\-2\end{pmatrix}\) (constant)
✓ 1
A1 \(|\mathbf{a}|=\sqrt{40}=6.32\) m s−2
✓ 1
Exam tip: Speed is the magnitude of the velocity vector, so square and add both components before taking the square root.
97IBMI3H-3-05r39-03Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy4 marks
A hospital has five wards A, B, C, D and E joined by corridors. The adjacency matrix \(\mathbf{M}\) below has rows and columns in the order A, B, C, D, E, and each corridor can be walked in either direction. \(\mathbf{M}=\begin{pmatrix}0&1&1&0&0\\1&0&1&1&0\\1&1&0&1&0\\0&1&1&0&1\\0&0&0&1&0\end{pmatrix}\)
(a)
Write down the number of corridors that meet at ward D, and find the total number of corridors in the hospital.
(2)
(b)
Given that \(\mathbf{M}^3=\begin{pmatrix}2&5&5&2&2\\5&4&5&6&1\\5&5&4&6&1\\2&6&6&2&3\\2&1&1&3&0\end{pmatrix}\), find the number of different walks of length 3 from ward A to ward E, and list them.
(2)
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Part (a)
A1 3 corridors at D (sum of row D is \(0+1+1+0+1=3\))
✓ 1
A1 sum of all entries is 12, so there are 6 corridors
✓ 1
Part (b)
A1 the entry in row A, column E of \(\mathbf{M}^3\) is 2, so there are 2 walks
✓ 1
A1 A–B–D–E and A–C–D–E
✓ 1
Exam tip: The (i, j) entry of the k-th power of an adjacency matrix counts walks of length k from vertex i to vertex j.
98IBMI3H-3-06r41-01Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy6 marks
A light aircraft leaves a holding point and flies at constant velocity. Its position vector relative to a small airfield at the origin O is r = (−45, 30) + t(120, −90), where components are in kilometres (east, north) and t is the time in hours after leaving the holding point.
(a)
Find the speed of the aircraft in km h−1.
(2)
(b)
Find the time, in hours, at which the aircraft is due south of the airfield.
(2)
(c)
Find the distance of the aircraft from the airfield at this time.
(2)
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Part (a)
M1 √(1202 + 902)
✓ 1
A1 150 km h−1
✓ 1
Part (b)
M1 −45 + 120t = 0 (east component zero)
✓ 1
A1 t = 0.375 hours (22.5 minutes)
✓ 1
Part (c)
M1 north component 30 − 90 × 0.375
✓ 1
A1 3.75 km (south of the airfield)
✓ 1
Exam tip: 'Due south of' means the east component is zero and the north component is negative, so set one component of the position vector equal to zero.
99IBMI3H-3-06r41-03Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy5 marks
A sound engineer sits at the mixing desk, which is taken as the origin O on a plan of a concert hall. The positions of the left and right stage speakers, in metres relative to O (east, north), are L(−3.0, 5.0) and R(4.0, 6.0).
(a)
Find the scalar product of the vectors OL and OR.
(2)
(b)
Find the size of angle LOR, the angle that the two speakers make at the mixing desk. Give your answer in degrees to 3 significant figures.
(3)
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Part (a)
M1 (−3)(4) + (5)(6)
✓ 1
A1 18
✓ 1
Part (b)
M1 |OL| = √34 and |OR| = √52
✓ 1
M1 cos θ = 18 / (√34 √52)
✓ 1
A1 θ = 64.7°
✓ 1
Common error: Both vectors must start at the same point (here the mixing desk); then cos θ = (a·b) / (|a||b|) gives the angle between them.
100IBMI3H-3-06r41-11Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy5 marks
Two crop-spraying drones fly at constant velocity. Relative to the farm gate, with components in kilometres (east, north) and t in hours after 08:00, their positions are drone 1: r1 = (10, 5) + t(3, 2) drone 2: r2 = (30, −5) + t(−4, 9).
(a)
Find the coordinates of the point where the two flight paths cross.
(3)
(b)
Determine whether the drones collide at this point. Give a reason.
(2)
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Part (a)
M1 equating the paths with different parameters: (10 + 3λ, 5 + 2λ) = (30 − 4μ, −5 + 9μ)
✓ 1
M1 solving simultaneously: λ = 4 and μ = 2
✓ 1
A1 (22, 13)
✓ 1
Part (b)
M1 drone 1 reaches (22, 13) when t = 4 and drone 2 when t = 2
✓ 1
R1 No, they reach the crossing point at different times
✓ 1
Common error: Two paths crossing does not mean the objects collide: you must use a different parameter for each path when finding the crossing point, then compare the times.
101IBMI3H-3-07r41-01Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy6 marks
A space probe is moving through a region of space in a straight line at constant velocity. Relative to a mission origin O, its position vector at time \(t=0\) is \(\mathbf{r}_0 = \begin{pmatrix}120\\-45\\300\end{pmatrix}\) km and its velocity is \(\mathbf{v} = \begin{pmatrix}2.4\\-1.8\\3.0\end{pmatrix}\) km s−1.
(a)
Calculate the speed of the probe, giving your answer to 3 significant figures.
(2)
(b)
Find the position vector of the probe when \(t = 50\) s.
(2)
(c)
Hence calculate the distance of the probe from O when \(t=50\) s. Give your answer to 3 significant figures.
(2)
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Part (a)
M1 attempt to find the magnitude, \(\sqrt{2.4^2+(-1.8)^2+3.0^2}\)
✓ 1
A1 \(4.24\) km s−1
✓ 1
Part (b)
M1 use of \(\mathbf{r}=\mathbf{r}_0+t\mathbf{v}\) with \(t=50\)
✓ 1
A1 \(\begin{pmatrix}240\\-135\\450\end{pmatrix}\) km
✓ 1
Part (c)
M1 magnitude of the position vector from (b), \(\sqrt{240^2+135^2+450^2}\)
✓ 1
A1 \(528\) km (ECF from (b))
✓ 1
Common error: The speed is the magnitude of the velocity vector, not the sum of its components, so square each component before adding.
102IBMI3H-3-07r41-11Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy5 marks
Two thrusters on a probe produce forces in the directions of the vectors \(\mathbf{u}=\begin{pmatrix}p\\2\\-3\end{pmatrix}\) and \(\mathbf{w}=\begin{pmatrix}4\\p\\1\end{pmatrix}\), where \(p\) is a constant.
(a)
Find the value of \(p\) for which the two forces act at right angles.
(2)
(b)
For \(p=2\), find the angle between the two forces. Give your answer to 3 significant figures.
(3)
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Part (a)
M1 set \(\mathbf{u}\cdot\mathbf{w}=0\), \(4p+2p-3=0\)
✓ 1
A1 \(p=0.5\)
✓ 1
Part (b)
M1 use of \(\cos\theta=\dfrac{\mathbf{u}\cdot\mathbf{w}}{|\mathbf{u}||\mathbf{w}|}\)
Common error: Perpendicular vectors have a scalar product of zero; setting the vector product to zero would instead make them parallel.
103IBMI3H-3-07r41-12Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy5 marks
A cargo ship carrying green coffee beans leaves port P and sails 48 km on a bearing of 065° to a buoy Q. It then sails 35 km on a bearing of 140° to a port R.
(a)
Show that angle PQR is 105°.
(2)
(b)
Calculate the distance PR, giving your answer to 3 significant figures.
(3)
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Part (a)
M1 bearing of P from Q is \(065^\circ+180^\circ=245^\circ\)
✓ 1
A1 \(245^\circ-140^\circ=105^\circ\)
✓ 1
Part (b)
M1 use of the cosine rule
✓ 1
A1 \(PR^2=48^2+35^2-2(48)(35)\cos105^\circ\)
✓ 1
A1 \(PR=66.3\) km
✓ 1
Exam tip: Draw a north line at each point; angles on a straight line and alternate angles then give the angle inside the triangle.
104IBMI3H-3-07r41-14Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy5 marks
A conveyor cable runs across a cuboid-shaped roasting hall ABCDEFGH. The base ABCD is horizontal with AB = 12 m and BC = 5 m, and the vertical edge CG = 4 m. The cable runs in a straight line from A to G.
(a)
Calculate the length of the diagonal AC of the base.
(2)
(b)
Find the angle that the cable AG makes with the horizontal base. Give your answer to 3 significant figures.
(3)
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Part (a)
M1 use of Pythagoras in triangle ABC, \(\sqrt{12^2+5^2}\)
✓ 1
A1 13 m
✓ 1
Part (b)
M1 identify the angle GAC in right-angled triangle ACG
✓ 1
A1 \(\tan\theta=\dfrac{4}{13}\)
✓ 1
A1 \(\theta=17.1^\circ\)
✓ 1
Common error: The angle between a line and a plane is measured to the line’s projection on the plane, here AC, not to a base edge.
105IBMI3H-3-05r44-01Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy6 marks
A rooftop solar panel has normal vector n = (1, 2, 2), where the axes point east, north and vertically upwards. At a certain moment sunlight travels in the direction of the vector d = (−3, −4, −12).
(a)
Find the angle between −d and n, giving your answer to the nearest 0.1°. This is the angle of incidence of the sunlight on the panel.
(4)
(b)
The power output of the panel is proportional to cos θ, where θ is the angle of incidence. Calculate the power output as a percentage of the maximum possible output.
(2)
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Part (a)
M1 attempt to find the scalar product of −d = (3, 4, 12) and n
✓ 1
A1 scalar product = 35
✓ 1
M1 use of cos θ = (a·b)/(|a||b|) with |n| = 3 and |d| = 13
✓ 1
A1 θ = 26.2°
✓ 1
Part (b)
M1 cos θ × 100% (ECF from (a))
✓ 1
A1 89.7%
✓ 1
Exam tip: Use the unrounded value of cos θ if you can; the percentage follows directly from the ratio 35/39.
106IBMI3H-3-05r44-05Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy5 marks
During a disease outbreak, close contacts between five people P, Q, R, S and T are recorded as an undirected graph. The pairs who have been in close contact are PQ, PR, QR, QS, RS and ST.
(a)
Write down the adjacency matrix M for the graph, with the vertices in the order P, Q, R, S, T.
(2)
(b)
Find the number of different walks of length 3 from P to T.
(2)
(c)
State what this number represents in terms of the spread of the infection.
(1)
Show mark scheme
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Notes
Part (a)
M1 a symmetric 5 × 5 matrix with zeros on the leading diagonal
M1 attempt to find M3 (GDC) and read the entry in row P, column T (ECF from (a))
✓ 1
A1 2
✓ 1
Part (c)
R1 the number of different routes by which infection could pass from P to T through exactly three successive contacts
✓ 1
Exam tip: The (i, j) entry of Mⁿ counts walks of length n, and a walk may revisit a vertex or reuse an edge.
107IBMI3H-3-05r44-09Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy6 marks
A shop's logo is a triangle with vertices A(1, 2), B(4, 2) and C(4, 6) on a design screen, where one unit is 1 cm. The designer applies the transformation represented by the matrix T = \(\begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}\).
(a)
Find the coordinates of the images A′, B′ and C′ of the three vertices.
(3)
(b)
Find the area of the image triangle A′B′C′ by using the determinant of T.
(3)
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Part (a)
A1 A′ = (4, 6)
✓ 1
A1 B′ = (10, 6)
✓ 1
A1 C′ = (14, 18)
✓ 1
Part (b)
A1 area of ABC = ½ × 3 × 4 = 6 cm2
✓ 1
M1 area scale factor = |det T| = 6
✓ 1
A1 area of A′B′C′ = 36 cm2
✓ 1
Exam tip: The area of the image equals the area of the object multiplied by the absolute value of the determinant of the transformation matrix.
108IBMI3H-3-05r44-13Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy7 marks
A telecoms equipment shelter is a right pyramid. Its base ABCD is a horizontal rectangle with AB = 12 m and BC = 9 m. The apex V is 7 m vertically above the centre O of the base.
(a)
Calculate the length of the diagonal AC.
(2)
(b)
Find the angle between the edge VA and the base, giving your answer to the nearest 0.1°.
(2)
(c)
Find the angle between the sloping face VAB and the base, giving your answer to the nearest 0.1°.
(3)
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Part (a)
M1 Pythagoras: AC2 = 122 + 92
✓ 1
A1 AC = 15 m
✓ 1
Part (b)
M1 tan θ = VO ÷ AO with AO = 7.5 (ECF from (a))
✓ 1
A1 θ = 43.0°
✓ 1
Part (c)
M1 let M be the midpoint of AB; then OM = 4.5 m and VM is perpendicular to AB
✓ 1
M1 tan φ = 7 ÷ 4.5
✓ 1
A1 φ = 57.3°
✓ 1
Common error: The angle between a face and the base is measured along the line of greatest slope, which is perpendicular to the shared edge, not along an edge of the pyramid.
109IBMI3H-3-08r44-02Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy4 marks
Two straight fissures start at the same vent on a volcano. Using unit vectors east, north and up, their directions are given by u = (2, 3, −1) and v = (4, −1, 2).
(a)
Calculate the scalar product u·v.
(2)
(b)
Find the angle between the two fissures, giving your answer in degrees to 3 significant figures.
(2)
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Part (a)
M1 attempt to multiply corresponding components and add
✓ 1
A1 3
✓ 1
Part (b)
M1 attempt to use cos θ = u·v / (|u||v|) with |u| = √14 and |v| = √21
✓ 1
A1 79.9°
✓ 1
Common error: The scalar product of two vectors is a number, not a vector, and the angle formula needs the magnitudes of both vectors in the denominator.
110IBMI3H-3-08r44-06Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy6 marks
Five ports, A, B, C, D and E, are linked by direct shipping routes. The routes are A–B, A–C, B–C, B–D, C–E and D–E. A route can be travelled in either direction.
(a)
Write down the adjacency matrix for this graph, with the ports in alphabetical order.
(2)
(b)
Use the adjacency matrix to find the number of different journeys of exactly three routes that start at B and finish at D.
(2)
(c)
Find the number of different journeys of exactly three routes that start at A and finish at A.
(2)
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Part (a)
M1 attempt at a 5 × 5 symmetric matrix of 0s and 1s with zeros on the diagonal
M1 attempt to find the cube of the adjacency matrix and read the entry for B and D
✓ 1
A1 5
✓ 1
Part (c)
M1 attempt to read the entry for A and A in the cube of the matrix
✓ 1
A1 2
✓ 1
Exam tip: Entry (i, j) of Mⁿ counts the walks of length n from vertex i to vertex j; a walk may repeat edges and vertices.
111IBMI3H-3-08r44-11Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy5 marks
A turntable on a stage rotates scenery anticlockwise through 40° about the origin of a coordinate grid, with distances in metres. After the rotation, a corner of a prop is at the point (1.5, 4.2).
(a)
Write down the matrix for an anticlockwise rotation of 40° about the origin, with entries to 3 decimal places.
(2)
(b)
Find the original coordinates of the corner before the rotation, each to 3 significant figures.
(3)
Show mark scheme
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Notes
Part (a)
M1 matrix of the form rows (cos θ, −sin θ), (sin θ, cos θ)
✓ 1
A1 rows (0.766, −0.643), (0.643, 0.766)
✓ 1
Part (b)
M1 attempt to apply a rotation of −40° (the inverse matrix) to (1.5, 4.2), using unrounded matrix entries
✓ 1
A1 x = 3.85
✓ 1
A1 y = 2.25
✓ 1
Common error: To undo a rotation, rotate through the same angle in the opposite direction (use the inverse matrix); do not apply the same matrix again.
112IBMI3H-3-08r44-12Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy8 marks
A ship leaves port P and sails 12 km on a bearing of 072° to point Q. It then sails 9 km on a bearing of 150° to point R.
(a)
Show that angle PQR is 102°.
(2)
(b)
Find the distance PR, in km to 3 significant figures.
(3)
(c)
Find the bearing of R from P, to the nearest degree.
(3)
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Part (a)
M1 use of the back bearing 072° + 180° = 252° at Q (or co-interior angles with the north lines)
✓ 1
A1 252° − 150° = 102°
✓ 1
Part (b)
M1 attempt to use the cosine rule: PR² = 12² + 9² − 2 × 12 × 9 × cos 102°
✓ 1
A1 PR² = 269.9…
✓ 1
A1 16.4 km
✓ 1
Part (c)
M1 attempt to use the sine rule to find angle QPR: sin(QPR)/9 = sin 102°/16.4…
Exam tip: Draw a north line at each point and mark the bearings on it; angles in a triangle then come from alternate or co-interior angles, or from back bearings.
113IBMI3H-3-07r46-01Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy5 marks
Two tagged tiger sharks are tracked in a marine reserve. Relative to a fixed origin, with i east, j north and k vertically upwards, their velocities are constant. Shark P has velocity u = 3i − 2j + 2k and shark Q has velocity w = i + 4j − k, both in km h−1.
(a)
Calculate the scalar product u·w.
(2)
(b)
Find the angle between the directions of travel of the two sharks. Give your answer in degrees to 3 significant figures.
(3)
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Part (a)
M1 attempt to multiply corresponding components and add, e.g. 3×1 + (−2)×4 + 2×(−1)
✓ 1
A1 u·w = −7
✓ 1
Part (b)
M1 attempt to find a magnitude, e.g. |u| = √17 or |w| = √18
✓ 1
M1 substitution into cos θ = (u·w)/(|u||w|) = −7/(√17 × √18) (ECF from (a))
✓ 1
A1 θ = 114° (113.6…°, accept 114 or 113.6)
✓ 1
Common error: If the scalar product is negative the angle is obtuse; do not take the modulus before using the inverse cosine.
114IBMI3H-3-07r46-11Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy5 marks
A research boat leaves harbour H and sails 14 km on a bearing of 062° to a buoy B. It then sails 9 km on a bearing of 150° to a tagging site C.
(a)
Show that angle HBC is 92°.
(2)
(b)
Find the distance HC. Give your answer in kilometres to 3 significant figures.
(3)
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Part (a)
A1 the bearing of H from B is 062° + 180° = 242°
✓ 1
A1 angle HBC = 242° − 150° = 92°
✓ 1
Part (b)
M1 use of the cosine rule, HC² = 14² + 9² − 2×14×9×cos 92°
✓ 1
A1 HC² = 285.8… (accept 286)
✓ 1
A1 HC = 16.9 km
✓ 1
Exam tip: Draw a north line at each turning point; to find the return bearing, add or subtract 180°.
115IBMI3H-3-07r46-12Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy5 marks
In triangle ABC, a steel roof truss has AB = 12.5 m, angle BAC = 38° and angle ABC = 64°.
(a)
Find the length of BC. Give your answer in metres to 3 significant figures.
(3)
(b)
Find the area of triangle ABC. Give your answer in square metres to 3 significant figures.
(2)
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Part (a)
A1 angle ACB = 180° − 38° − 64° = 78°
✓ 1
M1 sine rule, BC/sin 38° = 12.5/sin 78°
✓ 1
A1 BC = 7.87 m
✓ 1
Part (b)
M1 area = ½ × AB × BC × sin 64° (using BC from (a), accept rounded or unrounded)
✓ 1
A1 area = 44.2 m2
✓ 1
Common error: Check that your calculator is in degree mode before using the sine rule.
116IBMI3H-3-06r47-01Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy6 marks
In a computer game the origin O is the spawn point and one unit is 1 metre. Character P stands at (2, 1, 3) and character Q stands at (−1, 4, 2).
(a)
Find the vector \(\overrightarrow{PQ}\).
(1)
(b)
Find the distance between the two characters, to 3 significant figures.
(2)
(c)
Find the angle \(\angle POQ\), giving your answer to 1 decimal place.
M1 use \(\cos\theta=\frac{\mathbf{p}\cdot\mathbf{q}}{|\mathbf{p}||\mathbf{q}|}\) with \(|\mathbf{p}|=\sqrt{14}\) and \(|\mathbf{q}|=\sqrt{21}\)
✓ 1
A1 \(62.2^\circ\)
✓ 1
Common error: Subtract the position vectors in the correct order: \(\overrightarrow{PQ}=\mathbf{q}-\mathbf{p}\), not \(\mathbf{p}-\mathbf{q}\). The angle between the vectors from O needs the position vectors themselves.
117IBMI3H-3-06r47-02Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy7 marks
A drone in a game moves so that its position, in metres relative to its base, at time t seconds is given by \(\mathbf{r}(t)=\begin{pmatrix}2t+1\\t^2-4t\end{pmatrix}\) for \(t\ge0\).
(a)
Find the velocity vector \(\mathbf{v}(t)\).
(2)
(b)
Find the time at which the drone is moving parallel to the x-axis, and its position at that time.
(3)
(c)
Find the speed of the drone when \(t=3\), to 3 significant figures.
(2)
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Part (a)
A1 x-component \(=2\)
✓ 1
A1 y-component \(=2t-4\)
✓ 1
Part (b)
M1 set \(2t-4=0\)
✓ 1
A1 \(t=2\) s
✓ 1
A1 position \((5,-4)\)
✓ 1
Part (c)
M1 \(\sqrt{2^2+2^2}\) (ECF from (a))
✓ 1
A1 \(2.83\) m s−1
✓ 1
Exam tip: Velocity is the derivative of position and speed is the magnitude of velocity. Speed is never negative and has no direction.
118IBMI3H-3-06r47-03Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy6 marks
A water supply network has five pumping stations A, B, C, D and E. Pipes connect the pairs AB, AC, BC, BD, CD, CE, DE. Each pipe can carry water in either direction. The adjacency matrix \(\mathbf{M}\) of the graph has rows and columns in the order A, B, C, D, E.
(a)
Write down \(\mathbf{M}\).
(2)
(b)
Given that \(\mathbf{M}^3\) has entry 7 in row B, column D, interpret this value. Give a reason why this entry is the number of walks of length 3 from B to D.
(2)
(c)
The diagonal entry of \(\mathbf{M}^3\) for station C is 6. Use the graph to explain this value.
(2)
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Part (a)
A1 row A: \((0\;1\;1\;0\;0)\) and row B: \((1\;0\;1\;1\;0)\)
A1 there are 7 different walks of length 3 (using 3 pipes) from B to D
✓ 1
R1 the \((i,j)\) entry of \(\mathbf{M}^n\) counts walks of length n from vertex i to vertex j
✓ 1
Part (c)
A1 closed walks of length 3 from C are those around a triangle: ABC, BCD and CDE are the triangles containing C
✓ 1
R1 each triangle can be travelled in 2 directions, giving \(3\times2=6\)
✓ 1
Exam tip: To count walks of length n, raise the adjacency matrix to the power n and read off the entry for the two vertices.
119IBMI3H-3-06r47-04Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy6 marks
A water engineer measures from a pumping station P. A tower T is 8.4 km from P on a bearing of 065°. A reservoir R is 5.2 km from P on a bearing of 140°.
(a)
Show that angle \(\angle TPR=75^\circ\).
(2)
(b)
Find the distance TR, to 3 significant figures.
(2)
(c)
Find angle \(\angle PTR\), to 3 significant figures. Use the unrounded value of TR.
(2)
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Part (a)
M1 \(140^\circ-65^\circ\)
✓ 1
A1 \(75^\circ\) (the bearings are measured from the same north line at P)
M1 sine rule \(\frac{\sin\angle PTR}{5.2}=\frac{\sin75^\circ}{TR}\)
✓ 1
A1 \(35.5^\circ\)
✓ 1
Common error: Always draw the north line at the starting point of each bearing before finding the angle in the triangle. Check that your calculator is in degree mode.
120IBMI3H-3-06r48-01Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy6 marks
Calculator
A light aircraft's velocity relative to the ground is \(\mathbf{v} = \begin{pmatrix} 120 \\ 90 \end{pmatrix}\) km h−1, where the first component is the eastward speed and the second is the northward speed. The velocity is constant.
(a)
Find the speed of the aircraft.
(2)
(b)
Find the bearing on which the aircraft is flying, correct to the nearest 0.1°.
(2)
(c)
Find the distance flown in 40 minutes.
(2)
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Part (a)
M1 attempt to use \(\sqrt{120^2 + 90^2}\)
✓ 1
A1 \(150\) km h−1
✓ 1
Part (b)
M1 attempt to use \(\tan^{-1}\left(\frac{120}{90}\right)\) for the angle measured from north
✓ 1
A1 \(053.1°\)
✓ 1
Part (c)
M1 attempt to use \(150 \times \frac{40}{60}\)
✓ 1
A1 \(100\) km
✓ 1
Common error: A bearing is measured clockwise from north, so use the northward component as the adjacent side, not the eastward component.
121IBMI3H-3-06r48-02Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy6 marks
Calculator
A spraying drone flies along two straight rows of an orchard, with direction vectors \(\mathbf{a} = \begin{pmatrix} 3 \\ 4 \\ 0 \end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}\), where the components are in metres.
(a)
Calculate \(\mathbf{a} \cdot \mathbf{b}\).
(2)
(b)
Find \(|\mathbf{a}|\) and \(|\mathbf{b}|\).
(2)
(c)
Find the angle between the two rows, correct to the nearest 0.1°.
(2)
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Part (a)
M1 attempt to use \(3\times1 + 4\times2 + 0\times2\)
✓ 1
A1 \(11\)
✓ 1
Part (b)
A1 \(|\mathbf{a}| = 5\)
✓ 1
A1 \(|\mathbf{b}| = 3\)
✓ 1
Part (c)
M1 attempt to use \(\cos\theta = \frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|} = \frac{11}{15}\)
✓ 1
A1 \(42.8°\)
✓ 1
Exam tip: Find the scalar product and both magnitudes separately before substituting into \(\cos\theta\); this keeps the working easy to follow and check.
122IBMI3H-3-06r48-05Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy6 marks
Calculator
Five practice rooms A, B, C, D and E in a music school are joined by corridors AB, AC, BC, BD, CD and DE. The school is modelled as a graph with the rooms as vertices and the corridors as edges.
(a)
Write down the adjacency matrix of the graph, with the rows and columns in the order A, B, C, D, E.
(2)
(b)
Use the adjacency matrix to find the number of different walks of length 3 from A to E.
(2)
(c)
Explain why the (E, E) entry of the cube of the adjacency matrix is 0.
(2)
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Part (a)
M1 a symmetric matrix of 0s and 1s with zeros on the leading diagonal
M1 attempt to find the (A, E) entry of the cube of the adjacency matrix
✓ 1
A1 \(2\) (the walks A–B–D–E and A–C–D–E)
✓ 1
Part (c)
R1 a closed walk of length 3 from E would need a triangle containing E
✓ 1
R1 E is joined only to D, so no such triangle exists
✓ 1
Common error: The entries of the \(n\)th power of an adjacency matrix count walks, which may repeat edges and vertices; they do not count paths.
123IBMI3H-3-06r48-09Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Easy5 marks
Calculator
A vineyard plan uses coordinates in metres. A transformation maps points on the plan by first enlarging with scale factor 2 about the origin, and then reflecting in the \(y\)-axis.
(a)
Write down the matrix \(E\) for the enlargement and the matrix \(R\) for the reflection.
(2)
(b)
Find the single matrix \(M\) that represents the enlargement followed by the reflection.
(2)
(c)
Find the image of the point \(P(3, 5)\) under this transformation.
Glass for the panel costs USD 85 per m2. Find the cost of the glass, to the nearest cent.
(2)
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Part (a)
A1 for \(\overrightarrow{AB}=(8,1,0)\)
✓ 1
A1 for \(\overrightarrow{AC}=(3,6,3)\)
✓ 1
Part (b)
M1 for a correct method for the vector product, e.g. at least two correct components
✓ 1
A1 for \((3,-24,45)\)
✓ 1
Part (c)
M1 for \(\frac{1}{2}\sqrt{3^2+(-24)^2+45^2}=\frac{1}{2}\sqrt{2610}\)
✓ 1
A1 for 25.5 m2 (25.5440…)
✓ 1
Part (d)
M1 for multiplying the unrounded area by 85, e.g. \(\frac{1}{2}\sqrt{2610}\times85\) or \(25.54408\ldots\times85\)
✓ 1
A1 for USD 2171.25
✓ 1
Common error: Rounding the area to 25.544 before multiplying gives USD 2171.24, not USD 2171.25. Keep the full calculator value of the area when the answer is needed to the nearest cent. Also remember that the magnitude of the vector product gives the area of a parallelogram, so halve it for a triangle.
125IBMI-032-03-r2-03Vectors and kinematicsPaper 1Medium7 marks
Two cargo ships, A and B, move with constant velocities. Relative to a fixed buoy, their position vectors \(t\) hours after 13:00 are \(\mathbf{r}_A = t\begin{pmatrix}12\\5\end{pmatrix}\) and \(\mathbf{r}_B = \begin{pmatrix}30\\-10\end{pmatrix} + t\begin{pmatrix}4\\9\end{pmatrix}\) where distances are in kilometres.
(a)
Show that the displacement vector from A to B at time \(t\) is \(\begin{pmatrix}30-8t\\-10+4t\end{pmatrix}\).
(2)
(b)
Find the time of day at which the ships are closest together.
(3)
(c)
Find the minimum distance between the ships.
(2)
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Part (a)
Uses \(\mathbf{r}_B - \mathbf{r}_A\) M1
✓ 1
\(\begin{pmatrix}30+4t-12t\\-10+9t-5t\end{pmatrix}\) leading to the given vector A1
Exam tip: Minimising d² gives the same t as minimising d and avoids differentiating a square root. Give the answer as a time of day when the question asks for one.
126IBMI-032-03-r2-05Vectors and kinematicsPaper 1Medium7 marks
An air traffic controller monitors two light aircraft. Relative to a radar station, their positions \(t\) minutes after 14:00 are \(\mathbf{r}_A = \begin{pmatrix}0\\0\\6\end{pmatrix} + t\begin{pmatrix}3\\4\\0\end{pmatrix}\) and \(\mathbf{r}_B = \begin{pmatrix}24\\10\\4.8\end{pmatrix} + t\begin{pmatrix}-2\\1\\0.2\end{pmatrix}\) where distances are in kilometres and the third component is altitude.
(a)
Show that the flight paths of the two aircraft intersect, and find the coordinates of the point of intersection.
(3)
(b)
Determine whether the two aircraft collide. Justify your answer.
(2)
(c)
Find the distance between the aircraft at 14:04.
(2)
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Part (a)
Uses different parameters, e.g. \(3\lambda = 24 - 2\mu\), \(4\lambda = 10 + \mu\), \(6 = 4.8 + 0.2\mu\) M1
✓ 1
\(\mu = 6\), \(\lambda = 4\), and all three equations are satisfied A1
✓ 1
Point (12, 16, 6) A1
✓ 1
Part (b)
A is at the point when t = 4, but B is there when t = 6 R1
✓ 1
They are not at the point at the same time, so they do not collide A1
✓ 1
Part (c)
\(\mathbf{r}_B(4) = \begin{pmatrix}16\\14\\5.6\end{pmatrix}\), so \(\mathbf{r}_B - \mathbf{r}_A = \begin{pmatrix}4\\-2\\-0.4\end{pmatrix}\) M1
✓ 1
\(\sqrt{20.16}\) = 4.49 km A1
✓ 1
Common error: Using the same parameter t for both lines when testing whether the paths intersect. Paths can cross at different times, so use two parameters. Only a collision needs the same t.
127IBMI-032-08-r1-03Vectors and kinematicsPaper 1Medium7 marks
Two ships are tracked by a port radar. Their positions, in kilometres relative to the radar, t hours after 09:00 are Ship A: \(\mathbf{r}_A = t\begin{pmatrix} 12 \\ 5 \end{pmatrix}\) Ship B: \(\mathbf{r}_B = \begin{pmatrix} 30 \\ -10 \end{pmatrix} + t\begin{pmatrix} -6 \\ 9 \end{pmatrix}\) A collision-risk alert is triggered if the ships come within 3 km of each other.
(a)
Show that the position of ship B relative to ship A is \(\begin{pmatrix} 30-18t \\ -10+4t \end{pmatrix}\).
(2)
(b)
Find the time at which the ships are closest together. Give your answer to the nearest minute.
(2)
(c)
Find the minimum distance between the ships and determine whether the alert is triggered.
(3)
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Part (a)
M1: \(\mathbf{r}_B - \mathbf{r}_A\) attempted
✓ 1
A1: \(\begin{pmatrix} 30-6t-12t \\ -10+9t-5t \end{pmatrix}\) leading to given answer
✓ 1
Part (b)
M1: minimising \(d^2=(30-18t)^2+(-10+4t)^2\) (by GDC or differentiation: 340t = 580)
✓ 1
A1: t = 1.706 h, i.e. 10:42
✓ 1
Part (c)
M1: substituting t = 1.706 into \(\sqrt{(30-18t)^2+(-10+4t)^2}\)
✓ 1
A1: 3.25 km
✓ 1
R1: 3.25 > 3 so the alert is not triggered
✓ 1
Exam tip: Keep the unrounded value of t in your calculator when finding the minimum distance; rounding t to 1.7 first changes the third significant figure.
128IBMI-032-08-r1-04Vectors and kinematicsPaper 1Medium6 marks
A light aircraft flies with an airspeed of 220 km h−1 on a heading of 060°. A steady wind of 40 km h−1 blows from the north. Take the unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) to point east and north respectively.
(a)
Write down the velocity of the aircraft relative to the ground as a column vector.
(2)
(b)
Find the ground speed of the aircraft.
(2)
(c)
Find the bearing of the aircraft's actual path over the ground.
(2)
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Part (a)
M1: air velocity \(\begin{pmatrix} 220\sin 60° \\ 220\cos 60° \end{pmatrix}\) added to wind \(\begin{pmatrix} 0 \\ -40 \end{pmatrix}\)
M1: \(\tan^{-1}\left(\frac{190.53}{70}\right)\) measured clockwise from north
✓ 1
A1: 069.8°
✓ 1
Common error: A wind 'from the north' blows towards the south, so its vector is (0, −40), not (0, 40).
129IBMI-032-08-r2-03Vectors and kinematicsPaper 1Medium7 marks
Two ships, A and B, are observed from a port at the origin. At time t hours after 09:00 their position vectors, in kilometres, are \(\boldsymbol{r}_A = t\begin{pmatrix}10\\5\end{pmatrix}\) and \(\boldsymbol{r}_B = \begin{pmatrix}20\\-5\end{pmatrix} + t\begin{pmatrix}2\\9\end{pmatrix}\).
(a)
Find the position vector of B relative to A, \(\boldsymbol{r}_B - \boldsymbol{r}_A\), in terms of t.
(2)
(b)
Show that the square of the distance between the ships is \(d^2 = 80t^2 - 360t + 425\).
(2)
(c)
Find the minimum distance between the ships and the time of day at which it occurs.
Exam tip: Minimising \(d^2\) gives the same time as minimising \(d\), and the algebra is far easier. Remember to take the square root at the end and to convert t into a clock time.
130IBMI-032-03-r1-03Vectors and kinematicsPaper 1Medium7 marks
Two fishing boats, A and B, move with constant velocities. At time t hours after 08:00, their position vectors relative to a harbour are rA = (2, 1) + t(3, 4) and rB = (14, −3) + t(−1, 6), where distances are in kilometres and the components are east and north respectively.
(a)
Find the distance between the boats at 08:00.
(1)
(b)
Show that the position of B relative to A at time t is (12 − 4t, −4 + 2t).
(2)
(c)
Find the time of day at which the boats are closest together, and the minimum distance between them.
M1: valid method to minimise distance, e.g. minimises D2 = (12 − 4t)2 + (−4 + 2t)2 by differentiating or using a GDC, or sets (12 − 4t, −4 + 2t)·(−4, 2) = 0
✓ 1
A1: t = 2.8
✓ 1
A1: time of day 10:48
✓ 1
A1: minimum distance 1.79 km
✓ 1
Exam tip: Convert the decimal time into hours and minutes: 0.8 h is 48 minutes, not 80 minutes. A final answer given in context, as a time of day, earns the interpretation mark.
131IBMI-032-03-r1-06Vectors and kinematicsPaper 1Medium7 marks
A hot-air balloon drifts with constant velocity. Relative to its launch site, with x east, y north and z vertically upwards and all distances in metres, it is observed at position (200, 150, 300) at t = 0 and at position (260, 70, 340) at t = 20 seconds. A monument stands on level ground at (329, −22, 0).
(a)
Find the velocity vector of the balloon.
(2)
(b)
Find the angle at which the balloon is rising, measured from the horizontal.
(2)
(c)
Show that the balloon passes directly above the monument, and find its height at that moment.
(3)
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Part (a)
M1: displacement (60, −80, 40) divided by 20
✓ 1
A1: (3, −4, 2) m s−1
✓ 1
Part (b)
M1: horizontal speed \(\sqrt{3^2+4^2}\) = 5 and tan θ = \(\frac{2}{5}\)
✓ 1
A1: θ = 21.8°
✓ 1
Part (c)
M1: solves 200 + 3t = 329 and 150 − 4t = −22
✓ 1
R1: both give t = 43, so the balloon is directly above the monument at that time
✓ 1
A1: height = 300 + 2(43) = 386 m
✓ 1
Common error: Finding the time from the x-component alone is not enough. Both horizontal components must give the same t for the balloon to be directly above the point.
132IBMI-032-04-r1-05Graph theory and networksPaper 1Medium6 marks
A town park has 8 junctions, J1 to J8, joined by 12 walkways. The network is modelled by a connected simple graph. The degrees of J1 to J7 are 3, 3, 2, 4, 2, 3 and 4 respectively. The degree of J8 is k.
(a)
Find the value of k.
(2)
(b)
A jogger wants to run along every walkway exactly once, without needing to return to the start. Explain why this is not possible.
(2)
(c)
To reduce maintenance, the council will close some walkways while making sure every junction can still be reached from every other junction. Find the greatest number of walkways that can be closed.
(2)
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Part (a)
M1 uses the handshaking lemma: the sum of degrees = 2 × 12 = 24
✓ 1
A1 21 + k = 24, so k = 3
✓ 1
Part (b)
A1 J1, J2, J6 and J8 have odd degree, so there are four odd vertices
✓ 1
R1 an Eulerian trail exists only if there are 0 or 2 odd vertices
✓ 1
Part (c)
M1 the remaining network must be a spanning tree with 8 − 1 = 7 edges
✓ 1
A1 12 − 7 = 5 walkways
✓ 1
Exam tip: The handshaking lemma (sum of degrees = 2 × number of edges) and the fact that a tree on n vertices has n − 1 edges are quick results that each earn a method mark when you state them.
133IBMI-032-04-r2-04Graph theory and networksPaper 1Medium7 marks
A courier must visit five depots, A, B, C, D and E. The courier starts and finishes at A and visits every other depot exactly once. Every pair of depots is joined by a direct road. The road lengths in km are: AB 14, AC 20, AD 17, AE 11, BC 13, BD 22, BE 18, CD 16, CE 19, DE 15.
(a)
Use the nearest neighbour algorithm, starting at A, to find an upper bound for the length of the courier's route.
(2)
(b)
Delete vertex A. Use the deleted vertex algorithm to find a lower bound for the length of the route.
(4)
(c)
Hence comment on the route found using the nearest neighbour algorithm.
(1)
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Part (a)
M1 for applying nearest neighbour to give the cycle A–E–D–C–B–A
✓ 1
A1 for upper bound 11 + 15 + 16 + 13 + 14 = 69 km
✓ 1
Part (b)
M1 for finding a minimum spanning tree of B, C, D, E
✓ 1
A1 for edges BC, DE, CD with total weight 13 + 15 + 16 = 44
✓ 1
M1 for adding the two shortest edges from A, AE = 11 and AB = 14
✓ 1
A1 for lower bound 44 + 25 = 69 km
✓ 1
Part (c)
R1 for: the lower bound equals the upper bound, so 69 km is the minimum length and route A–E–D–C–B–A is optimal
✓ 1
Common error: In the deleted vertex algorithm, add the two shortest edges that join the deleted vertex to the tree. Do not add just one edge, and do not use any edge between the remaining vertices.
134IBMI-032-08-r2-02Vectors and kinematicsPaper 1Medium6 marks
A light aircraft is climbing after take-off. Its velocity is constant and equal to \(\boldsymbol{v} = \begin{pmatrix}200\\150\\10\end{pmatrix}\) km h−1 where \(\boldsymbol{i}\) points east, \(\boldsymbol{j}\) points north and \(\boldsymbol{k}\) points vertically upwards.
(a)
Find the speed of the aircraft.
(2)
(b)
Find the angle at which the aircraft is climbing, measured from the horizontal.
(2)
(c)
Find the bearing on which the aircraft is travelling.
Horizontal speed = \(\sqrt{200^2+150^2}\) = 250, so \(\tan\theta = \frac{10}{250}\) (or equivalent, e.g. \(\sin\theta = \frac{10}{\sqrt{62600}}\)) M1
✓ 1
θ = 2.29° A1
✓ 1
Part (c)
\(\tan\alpha = \frac{200}{150}\), where α is the angle measured clockwise from north M1
✓ 1
Bearing = 053.1° A1
✓ 1
Common error: The angle of climb compares the vertical component with the horizontal speed \(\sqrt{200^2+150^2}\). Using the full 3D speed in a tangent ratio gives the wrong angle. The full speed belongs in a sine ratio instead.
135IBMI3H-3-06-04Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium6 marks
A delivery robot moves in a warehouse. Its position vector relative to the charging point is \(\mathbf{r}=\begin{pmatrix}3t\\2t^2-8t\end{pmatrix}\) metres, where \(t\) is the time in seconds, \(t\ge0\).
(a)
Find the velocity vector of the robot at time \(t\) and its speed when \(t=3\).
(2)
(b)
Find the time at which the robot is moving parallel to the x-axis.
(2)
(c)
Find the distance of the robot from the charging point at this time.
Common error: Speed is the magnitude of the velocity vector, so square and add both components before taking the root.
136IBMI3H-3-06-07Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
A maintenance vehicle must travel along every road in a rural network joining junctions A, B, C, D and E, starting and finishing at the same junction. Road lengths in km: AB 5, BC 4, CD 6, DE 3, EA 7, BD 8.
(a)
Identify the vertices of odd degree and explain why the vehicle cannot complete the route without repeating any road.
(2)
(b)
Use the Chinese postman algorithm to find the length of the shortest route that uses every road and returns to the start.
(3)
(c)
The vehicle may instead start at one junction and finish at a different junction. State the best start and finish junctions and the length of the shortest route.
(2)
Show mark scheme
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Notes
Part (a)
A1 B and D have odd degree (3)
✓ 1
R1 a closed route using every edge once needs all vertices even
✓ 1
Part (b)
M1 shortest path from B to D is BD = 8 (via C would be 10)
✓ 1
M1 total road length \(5+4+6+3+7+8=33\) is added to 8
✓ 1
A1 41 km
✓ 1
Part (c)
M1 two odd vertices, so start at B and end at D (or reverse)
✓ 1
A1 33 km
✓ 1
Common error: Add the repeated path only between the odd vertices; adding the whole network length twice is not needed.
137IBMI3H-3-06-13Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
A storage container is a cuboid 8.0 m long, 5.0 m wide and 4.0 m high. A cable runs from corner A on the floor to the opposite corner G at the roof, in a straight line through the inside of the container.
(a)
Calculate the length of the diagonal of the floor.
Common error: The angle with the floor is found from the floor diagonal, not from an edge of the floor.
138IBMI3H-3-06b-05Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium6 marks
A drone flies so that its position, in metres relative to a launch point, at time t seconds is \(\boldsymbol{r} = \begin{pmatrix}3t+1\\2t^2-4t\end{pmatrix}\) for \(t \geq 0\).
(a)
Find an expression for the velocity vector of the drone.
(2)
(b)
Find the time at which the drone is moving parallel to the x-axis.
M1 set the vertical component equal to zero: 4t − 4 = 0
✓ 1
A1 t = 1 s
✓ 1
Part (c)
M1 attempt to find the magnitude of v(3) = (3, 8)
✓ 1
A1 √73 = 8.54 m s−1
✓ 1
Common error: Differentiating the magnitude of r instead of the components; speed is the magnitude of the velocity, not of the position.
139IBMI3H-3-06b-06Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
Two survey drones fly along straight paths. Relative to a base station, the paths are modelled by the lines \(l_1: \boldsymbol{r} = \begin{pmatrix}1\\2\\0\end{pmatrix} + s\begin{pmatrix}2\\1\\1\end{pmatrix}\) and \(l_2: \boldsymbol{r} = \begin{pmatrix}5\\0\\3\end{pmatrix} + t\begin{pmatrix}-1\\2\\-1\end{pmatrix}\).
(a)
Show that the two paths do not intersect.
(4)
(b)
Find the acute angle between the two paths.
(3)
Show mark scheme
Marking point
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Notes
Part (a)
M1 equate components: 1 + 2s = 5 − t, 2 + s = 2t, s = 3 − t
✓ 1
A1 solving two equations, e.g. t = 5/3 and s = 4/3
✓ 1
M1 substitute into the remaining equation
✓ 1
R1 1 + 8/3 ≠ 5 − 5/3, so the lines do not intersect
✓ 1
Part (b)
M1 attempt to use the scalar product of the direction vectors
✓ 1
A1 d₁ · d₂ = −1 and |d₁| = |d₂| = √6
✓ 1
A1 acute angle = 80.4°
✓ 1
Exam tip: Lines in three dimensions can be skew: parallel is checked with directions, intersection with the position equations.
140IBMI3H-3-06b-07Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
A triangular solar panel on a roof has vertices A(1, 0, 2), B(4, 2, 1) and C(2, 5, 3), with coordinates in metres.
Find the Cartesian equation of the plane containing the panel.
(2)
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Notes
Part (a)
A1 AB = (3, 2, −1) and AC = (1, 5, 1)
✓ 1
M1 attempt at the vector product
✓ 1
A1 (7, −4, 13)
✓ 1
Part (b)
M1 area = ½|AB × AC|
✓ 1
A1 ½√234 = 7.65 m2
✓ 1
Part (c)
M1 use (7, −4, 13) as the normal vector and point A
✓ 1
A1 7x − 4y + 13z = 33
✓ 1
Common error: Forgetting the factor of ½ when converting the magnitude of the vector product to the area of a triangle.
141IBMI3H-3-06b-08Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium6 marks
A courier must travel along every street shown at least once, starting and finishing at the same junction. The streets and their lengths in km are: AB = 5, BC = 7, CD = 4, DE = 6, EA = 8, BD = 9, CE = 10.
(a)
Identify the vertices of odd degree.
(1)
(b)
Find the length of the shortest route that the courier can take.
(5)
Show mark scheme
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Mark
Notes
Part (a)
A1 B, C, D and E each have degree 3 (A has degree 2)
✓ 1
Part (b)
M1 consider the three pairings of odd vertices
✓ 1
A1 BC + DE = 7 + 6 = 13
✓ 1
A1 BD + CE = 19 and BE + CD = 13 + 4 = 17
✓ 1
A1 sum of all street lengths = 49
✓ 1
A1 minimum route = 49 + 13 = 62 km
✓ 1
Exam tip: For each pairing of odd vertices use the shortest path between them, which may pass through other vertices.
142IBMI3H-3-06b-09Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium6 marks
A mobile vaccination team must visit five villages A, B, C, D and E, starting and finishing at A and visiting each village exactly once. The travel times in minutes between villages are: AB = 12, AC = 9, AD = 15, AE = 11, BC = 8, BD = 10, BE = 14, CD = 7, CE = 13, DE = 6.
(a)
Use the nearest neighbour algorithm starting at A to find an upper bound for the shortest time.
(3)
(b)
Delete vertex A to find a lower bound for the shortest time.
(3)
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Notes
Part (a)
M1 attempt to apply the nearest neighbour algorithm
✓ 1
A1 A → C → D → E → B → A
✓ 1
A1 9 + 7 + 6 + 14 + 12 = 48 minutes
✓ 1
Part (b)
M1 minimum spanning tree of B, C, D, E: DE (6), CD (7), BC (8) giving 21
✓ 1
A1 add the two shortest edges at A: 9 + 11 = 20
✓ 1
A1 lower bound = 41 minutes
✓ 1
Common error: Quoting the nearest neighbour total as the optimal tour; it is only an upper bound.
143IBMI3H-3-06b-10Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
A logo on a screen is transformed in two stages. First, a stretch with matrix \(\begin{pmatrix}3&0\\0&2\end{pmatrix}\) is applied, and then a rotation of 90° anticlockwise about the origin with matrix \(\begin{pmatrix}0&-1\\1&0\end{pmatrix}\) is applied.
(a)
Find the single matrix T that represents the combined transformation.
(2)
(b)
Find the image of the point P(4, 2).
(1)
(c)
Find the coordinates of the point whose image is (−6, 9).
(2)
(d)
A triangle in the original logo has area 5 cm2. Find the area of its image.
(2)
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Part (a)
M1 multiply in the correct order, rotation matrix × stretch matrix
✓ 1
A1 T = \(\begin{pmatrix}0&-2\\3&0\end{pmatrix}\)
✓ 1
Part (b)
A1 (−4, 12)
✓ 1
Part (c)
M1 attempt to use the inverse of T, with det T = 6
✓ 1
A1 (3, 3)
✓ 1
Part (d)
M1 area scale factor = |det T| = 6
✓ 1
A1 30 cm2
✓ 1
Common error: Multiplying the matrices in the order the transformations are applied; the first transformation must be the right-hand matrix.
144IBMI3H-3-03r6-02Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
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A hay barn has a horizontal rectangular base ABCD with AB = 14 m and BC = 10 m. The roof is a pyramid with apex V directly above the centre of the base, at a height of 4 m above the base.
(a)
Find the length of the edge VA. Give your answer to 3 significant figures.
(3)
(b)
Find the angle between the edge VA and the base. Give your answer to the nearest 0.1°.
(2)
(c)
Find the angle between the roof face VAB and the base. Give your answer to the nearest 0.1°.
(2)
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Notes
Part (a)
M1 attempts to find the distance from the centre of the base to A (half of the diagonal)
✓ 1
A1 half-diagonal = √(72 + 52) = √74 m
✓ 1
A1 VA = √(74 + 42) = √90 = 9.49 m
✓ 1
Part (b)
M1 uses tan θ = 4/√74 (or a correct trigonometric ratio with VA)
✓ 1
A1 θ = 24.9°
✓ 1
Part (c)
M1 identifies the perpendicular distance from the centre to AB as 5 m and uses tan φ = 4/5
✓ 1
A1 φ = 38.7°
✓ 1
Exam tip: In 3D problems, sketch the right-angled triangle that contains the angle and label it with lengths you have already found.
145IBMI3H-3-03r6-03Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
Calculator
Three wells are located on a farm. Using a map with 1 unit = 1 km, the wells are at A(2, 1), B(8, 4) and C(3, 9). A Voronoi diagram is drawn for the three wells.
(a)
Find the equation of the perpendicular bisector of AB, giving your answer in the form y = mx + c.
(3)
(b)
The perpendicular bisector of BC is y = x + 1. Find the coordinates of the vertex V where the two bisectors meet. Give each coordinate to 3 significant figures.
(2)
(c)
A sprinkler is to be placed at V. Find its distance from well A, using unrounded coordinates of V and giving your answer to 3 significant figures.
(2)
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Notes
Part (a)
M1 midpoint of AB is (5, 2.5)
✓ 1
M1 gradient of AB is 0.5 so the perpendicular gradient is −2
✓ 1
A1 y = −2x + 12.5
✓ 1
Part (b)
M1 equates −2x + 12.5 = x + 1
✓ 1
A1 V = (3.83, 4.83)
✓ 1
Part (c)
M1 uses the distance formula between V and A
✓ 1
A1 distance = 4.25 km
✓ 1
Common error: A Voronoi vertex is equidistant from three sites, so you can check your answer by finding the distance to a different site and comparing.
146IBMI3H-3-03r6-05Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium6 marks
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An air traffic control tower stands on level ground. From a point A on the ground the angle of elevation of the top T of the tower is 20°. From a point B, 200 m closer to the tower and in the same straight line, the angle of elevation of T is 30°.
(a)
Find the sizes of angle ABT and angle ATB.
(2)
(b)
Use the sine rule to find the length BT. Give your answer to the nearest metre.
(2)
(c)
Find the height of the tower, using the unrounded value of BT and giving your answer to the nearest metre.
(2)
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Notes
Part (a)
A1 angle ABT = 180° − 30° = 150°
✓ 1
A1 angle ATB = 30° − 20° = 10°
✓ 1
Part (b)
M1 attempts the sine rule in triangle ABT
✓ 1
A1 BT = 200 × sin 20° ÷ sin 10° = 394 m
✓ 1
Part (c)
M1 height = BT × sin 30°
✓ 1
A1 height = 197 m
✓ 1
Exam tip: When two angles of elevation are taken from the same line, the angle at the top of the tower is the difference of the two elevations.
147IBMI3H-3-03r6-14Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium8 marks
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A lighting designer projects a triangular logo onto a stage floor. The corners of the logo are A(1, 1), B(3, 1) and C(2, 4), with coordinates in metres. The projector effect applies the transformation represented by the matrix M = [[0, −2], [2, 0]] (rows listed in order) to every point.
(a)
Find the coordinates of the images A′, B′ and C′.
(2)
(b)
Describe fully the single transformation represented by M.
(2)
(c)
Find M−1.
(2)
(d)
Find the area of the image triangle A′B′C′.
(2)
Show mark scheme
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Notes
Part (a)
M1 multiplies M by the position vectors of the points
✓ 1
A1 A′ = (−2, 2), B′ = (−2, 6), C′ = (−8, 4)
✓ 1
Part (b)
A1 a rotation of 90° anticlockwise about the origin
✓ 1
A1 combined with an enlargement of scale factor 2, centre the origin
M1 area of ABC = ½ × 2 × 3 = 3 m2, and area scale factor |det M| = 4
✓ 1
A1 area of image = 12 m2
✓ 1
Exam tip: Areas are multiplied by the absolute value of the determinant of the transformation matrix, so you do not need the image coordinates to find the new area.
148IBMI3H-3-04r9-02Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
A climbing rope is anchored at point P. Relative to P, two climbers are at positions given in metres by vectors a = (6, 2, −3) and b = (2, 5, −4), where the third component is height. Give angles to 3 significant figures.
(a)
Find a·b.
(2)
(b)
Find the angle between the two ropes at P.
(3)
(c)
A third rope from P has direction vector (k, −2, 1). Given that this rope is perpendicular to the rope to the first climber, find k to 3 significant figures.
(2)
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Notes
Part (a)
M1 6 × 2 + 2 × 5 + (−3)(−4)
✓ 1
A1 = 34
✓ 1
Part (b)
A1 |a| = 7 and |b| = √45
✓ 1
M1 Use of cosθ = a·b/(|a||b|)
✓ 1
A1 θ = 43.6°
✓ 1
Part (c)
M1 Perpendicular so scalar product with a equals zero, 6k − 4 − 3 = 0
✓ 1
A1 k = 7/6 = 1.17
✓ 1
Exam tip: Perpendicular vectors have a scalar product of zero; do not set the vectors themselves equal to anything.
149IBMI3H-3-04r9-03Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium8 marks
A drone inspects a telecoms mast. Its position, in metres relative to the mast base, at time t seconds (0 ≤ t ≤ 10) is r(t) = (2t2 − 8t)i + 6tj + (60 + 4t − 0.5t2)k, where k is vertically upwards.
(a)
Find the velocity v(t).
(2)
(b)
Find the time at which the drone is moving horizontally.
(2)
(c)
Find the speed of the drone at this time.
(2)
(d)
Find the magnitude of the drone's acceleration, to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 Attempt to differentiate each component
✓ 1
A1 v = (4t − 8)i + 6j + (4 − t)k
✓ 1
Part (b)
M1 Vertical component of velocity equals zero, 4 − t = 0
✓ 1
A1 t = 4 s
✓ 1
Part (c)
A1 v(4) = (8, 6, 0)
✓ 1
A1 Speed = √(64 + 36) = 10 m/s
✓ 1
Part (d)
A1 a = (4, 0, −1), constant
✓ 1
A1 |a| = √17 = 4.12 m/s2
✓ 1
Common error: Speed is the magnitude of the velocity vector; the velocity itself is a vector and cannot be compared with a single number.
150IBMI3H-3-04r9-05Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium8 marks
Two drones fly in straight lines above a telecoms site. Relative to a fixed origin, in metres, their position vectors at time t seconds are r1 = (20, 20, 45) + t(3, 2, −1) and r2 = (80, 70, 20) + t(−2, −2, 1).
(a)
Show that the two flight paths intersect, and find the coordinates of the point of intersection.
(4)
(b)
Hence explain why the drones do not collide.
(1)
(c)
Find the distance between the drones at t = 10.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 Equate paths using separate parameters, e.g. (20, 20, 45) + s(3, 2, −1) = (80, 70, 20) + u(−2, −2, 1)
✓ 1
A1 Solve two equations, 3s + 2u = 60 and 2s + 2u = 50, giving s = 10 and u = 15
✓ 1
R1 Check the third component: 45 − 10 = 35 and 20 + 15 = 35
✓ 1
A1 Intersection at (50, 40, 35)
✓ 1
Part (b)
R1 Drone 1 reaches the point at t = 10 s but drone 2 at t = 15 s, so they are never there at the same time
✓ 1
Part (c)
A1 At t = 10, r2 = (60, 50, 30)
✓ 1
M1 Separation vector from (50, 40, 35) is (10, 10, −5)
✓ 1
A1 Distance = √225 = 15 m
✓ 1
Common error: Two lines intersecting does not mean the objects collide: the parameter values at the common point must be equal for the same time.
151IBMI3H-3-04r9-06Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium8 marks
A triangular snow face on a mountain has corners A, B and C. Relative to a survey point, in metres, A = (0, 0, 100), B = (60, 20, 130) and C = (30, 80, 150), where the third coordinate is height.
(a)
Find the vectors AB and AC.
(2)
(b)
Find AB × AC.
(2)
(c)
Hence find the area of the snow face.
(2)
(d)
Find the angle between the snow face and the horizontal, to 3 significant figures.
(2)
Show mark scheme
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Notes
Part (a)
A1 AB = (60, 20, 30)
✓ 1
A1 AC = (30, 80, 50)
✓ 1
Part (b)
M1 Attempt at the vector product
✓ 1
A1 (−1400, −2100, 4200)
✓ 1
Part (c)
M1 Area = ½|AB × AC|
✓ 1
A1 ½ × √(14002 + 21002 + 42002) = ½ × 4900 = 2450 m2
✓ 1
Part (d)
R1 Angle between the normal and the vertical k is the required angle
✓ 1
A1 cosθ = 4200/4900 and θ = 31.0°
✓ 1
Exam tip: The vector product gives a vector perpendicular to the face; its magnitude is the parallelogram area, so halve it for the triangle.
152IBMI3H-3-04r9-07Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium6 marks
Five relay stations A, B, C, D and E in a telecoms network are joined by direct links. The adjacency matrix M, with rows and columns in the order A, B, C, D, E, is \(M=\begin{pmatrix}0&1&1&0&0\\1&0&1&1&0\\1&1&0&1&1\\0&1&1&0&1\\0&0&1&1&0\end{pmatrix}\) and \(M^{2}=\begin{pmatrix}2&1&1&2&1\\1&3&2&1&2\\1&2&4&2&1\\2&1&2&3&1\\1&2&1&1&2\end{pmatrix}\).
(a)
Write down the degree of vertex C, and find the number of links in the network.
(2)
(b)
Use M2 to state the number of walks of length 2 from A to D, and list them.
(2)
(c)
Find the number of walks of length 3 from A to E.
(2)
Show mark scheme
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Notes
Part (a)
A1 Degree of C = 4
✓ 1
A1 Sum of degrees = 14, so 7 links
✓ 1
Part (b)
A1 2 walks
✓ 1
A1 A–B–D and A–C–D
✓ 1
Part (c)
M1 Use of M3 = M2M
✓ 1
A1 Entry (A, E) = 1 × 1 + 2 × 1 = 3 walks
✓ 1
Exam tip: The (i, j) entry of Mⁿ counts walks of length exactly n; walks may revisit vertices and edges.
153IBMI3H-3-04r9-09Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
Mountain huts A, B, C, D and E are joined by trails with these lengths in kilometres: AB 4, AD 9, AE 7, BC 5, BE 3, CD 6, CE 8, DE 4. A warden must walk along every trail at least once, starting and finishing at hut A, and wants the shortest total distance (the Chinese postman problem).
(a)
Find the degree of each vertex and state which vertices have odd degree.
(2)
(b)
By considering all pairings of the odd vertices, find the minimum extra distance that must be walked, stating which trails are repeated.
(3)
(c)
Find the length of the shortest route.
(2)
Show mark scheme
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Notes
Part (a)
A1 Degrees: A 3, B 3, C 3, D 3, E 4
✓ 1
A1 Odd vertices are A, B, C and D
✓ 1
Part (b)
M1 Three pairings considered: AB+CD, AC+BD, AD+BC
✓ 1
A1 Shortest paths used: AB = 4, CD = 6, AC = 9, BD = 7, AD = 9, BC = 5, giving totals 10, 16 and 14
✓ 1
A1 Minimum extra distance 10 km, repeating AB and CD
✓ 1
Part (c)
A1 Total of all trails = 46 km
✓ 1
A1 46 + 10 = 56 km
✓ 1
Exam tip: Check the shortest path between each pair of odd vertices, which may pass through other vertices, before choosing a pairing.
154IBMI3H-3-04r9-14Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium8 marks
A telecoms mast QP is vertical, with Q at ground level and P at the top, height 42 m. Point R is on level ground 28 m due east of Q, and point S is on level ground 45 m due north of Q. Give lengths and angles to 3 significant figures.
(a)
Find the length of the cable PR.
(2)
(b)
Find the angle that PR makes with the horizontal ground.
(2)
(c)
Find angle PRS.
(4)
Show mark scheme
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Notes
Part (a)
M1 PR2 = 422 + 282
✓ 1
A1 PR = 50.5 m
✓ 1
Part (b)
M1 tanθ = 42/28
✓ 1
A1 θ = 56.3°
✓ 1
Part (c)
A1 RS = √(282 + 452) = 53 m and PS = √(422 + 452) = 61.55 m
✓ 1
M1 Use of the cosine rule in triangle PRS
✓ 1
A1 cos(PRS) = (PR2 + RS2 − PS2)/(2 × PR × RS)
✓ 1
A1 Angle PRS = 73.0°
✓ 1
Exam tip: Sketch the vertical triangle and the horizontal triangle separately; every right angle in 3D problems sits at the base of the vertical mast.
155IBMI3H-3-04r12-01Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium8 marks
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During an eruption, a volcanic bomb is thrown from a vent at the origin O. Axes are chosen so that the x-axis points east, the y-axis points north and the z-axis points vertically upwards, with distances in metres. At time t seconds after launch, the position vector of the bomb is \(\mathbf{r}=\begin{pmatrix}18t\\24t\\40t-4.9t^2\end{pmatrix}\), for \(t\geq0\) until it lands on level ground at \(z=0\). Air resistance is ignored.
(a)
Find the velocity vector of the bomb when \(t=2\) and hence its speed at that instant. Give the speed to 3 significant figures.
(3)
(b)
Find the maximum height reached by the bomb. Give your answer to 3 significant figures.
(2)
(c)
Find the horizontal distance from O to the point where the bomb lands. Give your answer to 3 significant figures.
M1 vertical velocity zero: \(40-9.8t=0\), so \(t=4.0816\) s
✓ 1
A1 \(z=40t-4.9t^2=81.6\) m
✓ 1
Part (c)
M1 lands when \(40t-4.9t^2=0\), so \(t=\frac{40}{4.9}=8.1633\) s (reject \(t=0\))
✓ 1
M1 horizontal displacement \(\begin{pmatrix}18t\\24t\end{pmatrix}\) has magnitude \(\sqrt{18^2+24^2}\,t=30t\)
✓ 1
A1 245 m
✓ 1
Exam tip: The velocity is the derivative of position, and the speed is the magnitude of the velocity vector; the speed is a single number, not a vector.
156IBMI3H-3-04r12-04Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium8 marks
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Three light fixtures on a theatre's lighting truss are at the points \(A(1,0,2)\), \(B(4,2,3)\) and \(C(2,5,1)\), where coordinates are in metres.
(a)
Find the vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\).
M1 attempt at the vector product (cross product) formula
✓ 1
A1 \(\begin{pmatrix}-7\\4\\13\end{pmatrix}\)
✓ 1
Part (c)
M1 uses the vector product as a normal vector: \(-7x+4y+13z=d\) (ECF from (b))
✓ 1
A1 substitutes a point: \(d=-7+0+26=19\), so \(-7x+4y+13z=19\) (or equivalent)
✓ 1
Part (d)
M1 area \(=\frac12\left|\overrightarrow{AB}\times\overrightarrow{AC}\right|\) (ECF from (b))
✓ 1
A1 \(\frac12\sqrt{49+16+169}=7.65\) m2
✓ 1
Exam tip: The magnitude of the vector product of two sides gives the area of the parallelogram; halve it for the triangle.
157IBMI3H-3-04r12-06Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium8 marks
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A monitoring drone follows a cloud of volcanic ash. The acceleration of a piece of ash is modelled by \(\mathbf{a}=\begin{pmatrix}6t\\-4\end{pmatrix}\) m s−2 at time t seconds. At \(t=0\) the ash is at the origin with velocity \(\begin{pmatrix}0\\20\end{pmatrix}\) m s−1.
(a)
Show that the velocity of the ash at time t is \(\begin{pmatrix}3t^2\\20-4t\end{pmatrix}\) and find its speed when \(t=3\), giving your answer to 3 significant figures.
(2)
(b)
Find the position vector of the ash at time t and the coordinates of the ash at the moment it is moving horizontally.
(3)
(c)
Find the distance between the positions of the ash at \(t=1\) and \(t=4\). Give your answer to 3 significant figures.
(3)
Show mark scheme
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Notes
Part (a)
M1 integrates the acceleration: \(\begin{pmatrix}3t^2+c_1\\-4t+c_2\end{pmatrix}\) and uses initial velocity to find \(c_1=0,\ c_2=20\) (AG)
✓ 1
A1 \(\mathbf{v}(3)=\begin{pmatrix}27\\8\end{pmatrix}\), speed \(=28.2\) m s−1
M1 moving horizontally when vertical component of velocity is zero: \(20-4t=0\), so \(t=5\)
✓ 1
A1 position \((125,\,50)\)
✓ 1
Part (c)
M1 positions at \(t=4\) and \(t=1\) are \(\begin{pmatrix}64\\48\end{pmatrix}\) and \(\begin{pmatrix}1\\18\end{pmatrix}\), so displacement \(=\begin{pmatrix}63\\30\end{pmatrix}\)
✓ 1
M1 magnitude \(\sqrt{63^2+30^2}\)
✓ 1
A1 69.8 m
✓ 1
Common error: Distance between two positions is the magnitude of the displacement vector between them; it is not the difference of the two speeds or of the two position magnitudes.
158IBMI3H-3-04r12-09Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium8 marks
A volcano observatory has a network of footpaths linking five junctions A, B, C, D and E. The lengths of the paths, in kilometres, are: AB 4, AC 3, BC 5, BD 6, CD 2, CE 7, DE 4. A ranger must walk along every path at least once, and wants the shortest possible total distance.
(a)
Identify the vertices of odd degree and explain why a route that uses every path exactly once and returns to its start does not exist.
(2)
(b)
Find the shortest path between the two odd vertices, giving its length.
(2)
(c)
The ranger must start and finish at the same junction. Find the length of the shortest such route.
(2)
(d)
The ranger may instead start at B and finish at D. Find the length of the shortest route that uses every path at least once, and explain why no path needs repeating.
(2)
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Notes
Part (a)
A1 B and D have odd degree (3)
✓ 1
R1 a closed route using every edge exactly once (Eulerian circuit) requires all vertices to have even degree
✓ 1
Part (b)
M1 compares routes between B and D: BD = 6, BCD = 7, BACD = 9
✓ 1
A1 shortest is the direct path BD, length 6 km
✓ 1
Part (c)
M1 total length of all paths is \(4+3+5+6+2+7+4=31\) km, plus repeated path of 6 km
✓ 1
A1 37 km
✓ 1
Part (d)
R1 only B and D have odd degree, so a route from B to D can use each path exactly once
✓ 1
A1 31 km
✓ 1
Exam tip: In a Chinese postman problem, pair up odd vertices with the shortest connecting paths; for a route that need not return to its start, begin and end at the two odd vertices.
159IBMI3H-3-04r12-11Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
Calculator
A theatre's revolving set piece is represented by triangle \(PQR\) with \(P(2,1)\), \(Q(6,1)\) and \(R(6,4)\) on a plan, where units are metres. A transformation is applied in two stages: first a rotation of \(90^\circ\) anticlockwise about the origin, with matrix \(\begin{pmatrix}0&-1\\1&0\end{pmatrix}\), then a stretch by factor 2 in the x-direction, with matrix \(\begin{pmatrix}2&0\\0&1\end{pmatrix}\).
(a)
Find the single matrix M that represents the combined transformation.
(2)
(b)
A point X is mapped by the combined transformation to \((8,3)\). Find the coordinates of X.
(2)
(c)
Find the area, in m2, of the image of triangle PQR.
(3)
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Part (a)
M1 multiplies in the correct order: \(\begin{pmatrix}2&0\\0&1\end{pmatrix}\begin{pmatrix}0&-1\\1&0\end{pmatrix}\)
M1 \(\mathbf{M}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}8\\3\end{pmatrix}\), that is \(-2y=8\) and \(x=3\) (or uses \(\mathbf{M}^{-1}\))
✓ 1
A1 \(X=(3,\,-4)\)
✓ 1
Part (c)
M1 area of \(PQR\) is \(\frac12\times4\times3=6\)
✓ 1
M1 area scale factor is \(|\det\mathbf{M}|=|0\times0-(-2)\times1|=2\)
✓ 1
A1 12 m2
✓ 1
Common error: For a combined transformation, the first transformation is applied first, so its matrix is written on the right in the product.
160IBMI3H-3-04r12-13Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium8 marks
Calculator
Two seismic monitoring stations, A and B, are placed near a volcano V. From A, the volcano is 8.4 km away on a bearing of \(052^\circ\) and station B is 6.1 km away on a bearing of \(140^\circ\). Assume the ground is flat.
(a)
Find the size of angle VAB.
(1)
(b)
Calculate the distance BV, giving your answer to 3 significant figures.
(3)
(c)
Find the bearing of V from B, giving your answer in degrees to 3 significant figures.
M1 sine rule: \(\frac{\sin\angle ABV}{8.4}=\frac{\sin88^\circ}{BV}\) (using the unrounded \(BV\))
✓ 1
A1 \(\angle ABV=55.3^\circ\) (acute, because it is opposite the shorter side AV, so it is smaller than \(88^\circ\))
✓ 1
M1 the bearing of A from B is \(140^\circ+180^\circ=320^\circ\), and V lies clockwise of A as seen from B, so bearing \(=320^\circ+\angle ABV-360^\circ\)
✓ 1
A1 15.3\(^\circ\) (that is, 015.3\(^\circ\))
✓ 1
Common error: Back-bearings differ by 180°: to continue from the bearing of A from B, add or subtract 180° first, then add the interior angle and reduce to the range 0° to 360°.
161IBMI3H-3-03r15-03Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
Three stars A, B and C are used to define a plane in a 3D star map. Relative to the Sun, their positions (in light years) are \(A(1, 2, 0)\), \(B(4, 2, 3)\) and \(C(2, 5, 1)\).
(a)
Find the vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\).
Exam tip: The magnitude of the vector product gives the area of the parallelogram, so halve it for the triangle.
162IBMI3H-3-03r15-04Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
A javelin thrown in a sports science experiment has position vector \(\mathbf{r} = \begin{pmatrix}18t\\1.8 + 14t - 4.9t^2\end{pmatrix}\) metres at time t seconds after release. The x-axis is horizontal and the y-axis is vertically upwards.
(a)
Find an expression for the velocity vector \(\mathbf{v}\) at time t.
(2)
(b)
Calculate the speed of the javelin at \(t = 0.5\). Give your answer to 3 significant figures.
(2)
(c)
Find the maximum height of the javelin above the ground.
A1 \(14 - 9.8t = 0 \Rightarrow t = \tfrac{14}{9.8} \approx 1.43\) s
✓ 1
A1 \(y = 11.8\) m
✓ 1
Common error: The speed uses both components of the velocity; using only the vertical component gives the wrong answer.
163IBMI3H-3-03r15-06Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium6 marks
A rail company inspects the track on a network of six junctions A to F. The track sections and their lengths in kilometres are: AB 7, AC 5, BC 4, BD 6, CD 8, CE 9, DE 3, DF 5, EF 7. An inspection train must travel along every section at least once, starting and ending at junction A.
(a)
Identify the two junctions with odd degree and explain why the inspection route cannot use every section exactly once.
(2)
(b)
Find the shortest route between the two odd junctions, and state its length.
(2)
(c)
Hence find the minimum total distance the inspection train must travel.
(2)
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Part (a)
A1 junctions B and E (degree 3)
✓ 1
R1 a closed route using each edge once needs every vertex to have even degree
✓ 1
Part (b)
M1 comparing the routes B–D–E and B–C–E
✓ 1
A1 BDE, length 9 km
✓ 1
Part (c)
M1 total of all sections (54) plus the repeated shortest path
✓ 1
A1 63 km
✓ 1
Exam tip: List every vertex degree first; the sections to repeat are those on the shortest path between the odd vertices.
164IBMI3H-3-03r15-07Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium5 marks
Four stations P, Q, R and S on a light-rail line are joined by direct tracks. The adjacency matrix for the network, with the stations in the order P, Q, R, S, is \(\mathbf{M} = \begin{pmatrix}0&1&1&0\\1&0&1&1\\1&1&0&1\\0&1&1&0\end{pmatrix}\).
(a)
State the degree of station Q.
(1)
(b)
Explain why the matrix M is symmetric.
(1)
(c)
Find the number of different walks of length 3 from P to Q.
(3)
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Part (a)
A1 3
✓ 1
Part (b)
R1 each track can be travelled in both directions (the graph is undirected), so entry (i, j) = entry (j, i)
✓ 1
Part (c)
M1 attempt to find \(\mathbf{M}^3\)
✓ 1
A1 first row of \(\mathbf{M}^3\) is (2, 5, 5, 2)
✓ 1
A1 5 walks
✓ 1
Exam tip: The (i, j) entry of \(\mathbf{M}^n\) counts the walks of length n from vertex i to vertex j.
165IBMI3H-3-03r15-08Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium6 marks
A sports science laboratory has six monitoring stations A to F that must be linked by cable. The possible cable runs and their lengths in metres are: DE 6, BC 7, CD 8, AC 9, EF 10, BD 11, AB 12, CE 13, DF 14, AF 15, BE 16. Every station must be connected using the least total length of cable.
(a)
Use Kruskal's algorithm to find the cable runs that should be used. State the order in which you accept them, and the runs you reject.
(4)
(b)
The cable costs £14.50 per metre. Calculate the minimum cost of the cable.
(2)
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Part (a)
M1 edges considered in increasing order of length
✓ 1
A1 accept DE, BC, CD in that order
✓ 1
A1 accept AC and EF
✓ 1
R1 BD rejected because it would form a cycle with BC and CD (and all longer runs are then not needed)
✓ 1
Part (b)
A1 total length 40 m
✓ 1
A1 £580
✓ 1
Common error: Stop when you have chosen one fewer edge than there are vertices; a tree on 6 vertices has exactly 5 edges.
166IBMI3H-3-03r15-11Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
An astronomer rotates a star map by 35° anticlockwise about the origin. The rotation is represented by the matrix \(\mathbf{R} = \begin{pmatrix}\cos 35^\circ & -\sin 35^\circ\\ \sin 35^\circ & \cos 35^\circ\end{pmatrix}\). Coordinates are in centimetres.
(a)
A star is at (6, 2) on the original map. Find its position after the rotation, to 3 significant figures.
(2)
(b)
After the rotation, a second star is at (−1.5, 4.2). Find its position on the original map, to 3 significant figures.
(3)
(c)
Find the value of \(\det \mathbf{R}\) and interpret it in this context.
(2)
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Part (a)
M1 \(\mathbf{R}\begin{pmatrix}6\\2\end{pmatrix}\)
✓ 1
A1 (3.77, 5.08)
✓ 1
Part (b)
M1 use of \(\mathbf{R}^{-1}\) (rotation of 35° clockwise) or solving \(\mathbf{R}\mathbf{x} = \begin{pmatrix}-1.5\\4.2\end{pmatrix}\)
Exam tip: Set your calculator to degrees before evaluating \(\cos 35^\circ\) and \(\sin 35^\circ\).
167IBMI3H-3-03r16-01Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium8 marks
A beekeeper’s farmhouse F is the origin of a coordinate plane, with units in metres, x pointing east and y pointing north. Two hives are at A(60, 20) and B(−30, 80).
(a)
Find the vector \(\overrightarrow{AB}\) and hence the distance between the two hives, giving your answer to 3 significant figures.
(3)
(b)
Use a scalar product to find angle AFB, in degrees to 1 decimal place.
(3)
(c)
A third hive is at C(p, −30), where FC is perpendicular to FA. Find the value of p.
Common error: Two vectors are perpendicular when their scalar product is zero; do not set the product of the gradients equal to zero or forget that the product of two vectors is a number, not a vector.
168IBMI3H-3-03r16-02Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium8 marks
During a practice session a Formula 1 car moves along a straight section of track. Relative to a marker post at the origin O, its position at time t seconds is \(\mathbf{r}=\begin{pmatrix}-180+60t\\150-45t\end{pmatrix}\), where the components are in metres.
(a)
Find the speed of the car.
(2)
(b)
Find the y-coordinate of the car when it crosses the line x = 0.
(2)
(c)
Find the smallest distance between the car and the marker post, giving your answer to 3 significant figures. Use the exact value of t in your working.
(4)
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Part (a)
M1 \(\sqrt{60^2+(-45)^2}\)
✓ 1
A1 75 m s−1
✓ 1
Part (b)
M1 \(-180+60t=0\) so t = 3
✓ 1
A1 y = 15 m
✓ 1
Part (c)
M1 closest approach when \(\mathbf{r}\cdot\mathbf{v}=0\), i.e. \(60(-180+60t)-45(150-45t)=0\)
✓ 1
A1 t = 3.12 s
✓ 1
M1 substitute into r to obtain (7.2, 9.6)
✓ 1
A1 12.0 m
✓ 1
Exam tip: At the closest point to a fixed point, the position vector from that point is perpendicular to the velocity, so r·v = 0 avoids differentiating a square root.
169IBMI3H-3-03r16-04Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium8 marks
A beekeeper wants to link six apiary sites A to F with an electric fence along existing paths. The path lengths in metres are: AB 12, AC 9, BC 7, BD 15, CD 11, CE 14, DE 6, DF 10, EF 8. The fence must connect all six sites using as little fence as possible, and costs €4.50 per metre.
(a)
Use Kruskal’s algorithm to find the paths to be fenced, stating the order in which edges are accepted or rejected, and the total length of fence.
(4)
(b)
Find the total cost of the fence.
(2)
(c)
A new path BE of whole-number length x metres is proposed. Find the greatest value of x for which every minimum spanning tree contains BE.
(2)
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Part (a)
M1 edges considered in increasing order of length
✓ 1
A1 DE (6), BC (7), EF (8), AC (9) accepted
✓ 1
A1 DF (10) rejected as it forms a cycle D–E–F; CD (11) accepted
✓ 1
A1 total length 41 m
✓ 1
Part (b)
M1 41 × 4.50
✓ 1
A1 €184.50
✓ 1
Part (c)
M1 BE would form a cycle B–C–D–E whose longest edge is CD (11), so BE replaces CD only if x < 11
✓ 1
A1 x = 10
✓ 1
Common error: When a new edge is added to a spanning tree, compare it only with the edges on the cycle it creates, not with every edge in the graph.
170IBMI3H-3-03r16-09Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
A beekeeper models a sloping flower meadow as the triangle ABC with A(10, 20, 0), B(50, 50, 20) and C(30, 80, 40), where the coordinates (x, y, z) are in metres and z is the height above the valley floor.
(a)
Find the vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\).
Hence find the area of the meadow, to 3 significant figures.
(3)
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Part (a)
A1 \(\overrightarrow{AB}=(40,30,20)\)
✓ 1
A1 \(\overrightarrow{AC}=(20,60,40)\)
✓ 1
Part (b)
M1 attempt at the vector product
✓ 1
A1 \((0,-1200,1800)\)
✓ 1
Part (c)
M1 area = \(\frac12|\overrightarrow{AB}\times\overrightarrow{AC}|\)
✓ 1
M1 \(\frac12\sqrt{0^2+1200^2+1800^2}\)
✓ 1
A1 1080 m2
✓ 1
Exam tip: The magnitude of the vector product of two sides equals the area of the parallelogram they span, so halve it for a triangle.
171IBMI3H-3-03r16-14Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
Part of a Formula 1 circuit is a corner in the shape of a circular arc of radius 120 m, which subtends an angle of 70° at the centre O of the circle. A straight chord joins the two ends of the arc, and the region between the arc and the chord is a gravel run-off area.
(a)
Find the length of the arc, in metres to 1 decimal place.
(2)
(b)
Find the length of the chord, in metres to 1 decimal place.
(2)
(c)
Find the area of the run-off region, to the nearest square metre.
Exam tip: Keep the angle in radians (or use \(\frac12r^2\theta\) with θ in radians) when finding a segment area, and check the calculator mode before using sine.
172IBMI3H-3-04r16-02Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium8 marks
A hospital drone carries a sample between two buildings. Relative to a launch pad O, its position at time t seconds is \(\mathbf{r}=\begin{pmatrix}2t^2+t\\5t-t^2\end{pmatrix}\) metres, for \(t\ge0\).
(a)
Find expressions for the velocity v and the acceleration a of the drone.
(2)
(b)
Find the speed of the drone when t = 2, in m s−1 to 3 significant figures.
(2)
(c)
At one instant the drone is moving parallel to the x-axis. Find its distance from O at that instant.
M1 vertical component of velocity is zero: \(5-2t=0\)
✓ 1
A1 t = 2.5 s
✓ 1
M1 position is \((15,\ 6.25)\)
✓ 1
A1 \(\sqrt{15^2+6.25^2}=16.25\) m
✓ 1
Exam tip: Velocity is the derivative of the position vector component by component; “parallel to the x-axis” means the y-component of velocity is zero.
173IBMI3H-3-04r16-09Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium8 marks
An overhead camera wire at a Formula 1 circuit follows the line l with equation \(\mathbf{r}=\begin{pmatrix}0\\4\\1\end{pmatrix}+t\begin{pmatrix}2\\-1\\2\end{pmatrix}\), where coordinates are in metres. A lighting mast has its top at P(5, 0, 3).
(a)
Show that P does not lie on l.
(2)
(b)
Find the coordinates of the point Q on l that is closest to P.
(4)
(c)
Hence find the shortest distance from the top of the mast to the wire.
(2)
Show mark scheme
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Part (a)
M1 equate components, e.g. \(2t=5\) and \(4-t=0\)
✓ 1
R1 these give t = 2.5 and t = 4, which are inconsistent
✓ 1
Part (b)
M1 \(\overrightarrow{PQ}=\begin{pmatrix}2t-5\\4-t\\2t-2\end{pmatrix}\) for a general point Q on l
Exam tip: The shortest distance from a point to a line is along the perpendicular, so set the scalar product of the direction vector and the vector to a general point on the line equal to zero.
174IBMI3H-3-04r16-10Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
Two beehive-inspection drones fly in a field. Relative to a fixed origin, the position (in metres) of drone 1 at time t seconds is \(\mathbf{r}_1=\begin{pmatrix}10+5t\\20+t\end{pmatrix}\) and of drone 2 is \(\mathbf{r}_2=\begin{pmatrix}90-3t\\90-5t\end{pmatrix}\).
(a)
Show that the displacement of drone 2 from drone 1 is \(\begin{pmatrix}80-8t\\70-6t\end{pmatrix}\).
(2)
(b)
Find the time at which the drones are closest together.
(3)
(c)
Find the minimum distance between the drones.
(2)
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Part (a)
M1 \(\mathbf{r}_2-\mathbf{r}_1\)
✓ 1
A1 \((90-3t-10-5t,\ 90-5t-20-t)\) giving the given result (AG)
✓ 1
Part (b)
M1 \(D^2=(80-8t)^2+(70-6t)^2=100t^2-2120t+11300\)
✓ 1
M1 minimise, e.g. \(200t-2120=0\) or use a graphing calculator
✓ 1
A1 t = 10.6 s
✓ 1
Part (c)
M1 substitute t = 10.6, giving displacement (−4.8, 6.4)
✓ 1
A1 8 m
✓ 1
Common error: Minimise the squared distance rather than the distance itself; both have the same minimum position and the squared form is a simple quadratic.
175IBMI3H-3-04r16-12Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium8 marks
Three apiaries are modelled as sites A(0, 0), B(8, 0) and C(4, 6) on a map in which one unit is 1 km. Bees from a flower field forage at the nearest apiary. The Voronoi diagram of the sites is drawn.
(a)
Find the equation of the perpendicular bisector of AC, in the form y = mx + c.
(3)
(b)
The perpendicular bisector of AB is x = 4. Find the coordinates of the Voronoi vertex where the three cells meet.
(2)
(c)
A flower field is at P(5, 3). Find which apiary it is closest to, and the distance to that apiary, in kilometres to 3 significant figures.
(3)
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Part (a)
M1 midpoint of AC is (2, 3)
✓ 1
M1 gradient of AC is 1.5 so the perpendicular gradient is \(-\frac23\)
✓ 1
A1 \(y=-\frac23x+\frac{13}{3}\)
✓ 1
Part (b)
M1 substitute x = 4 into the answer to (a)
✓ 1
A1 \(\left(4,\ \frac53\right)\)
✓ 1
Part (c)
M1 distances PA = \(\sqrt{34}\), PB = \(\sqrt{18}\), PC = \(\sqrt{10}\)
✓ 1
A1 closest to C
✓ 1
A1 3.16 km
✓ 1
Exam tip: Every point in a Voronoi cell is closer to that cell’s site than to any other, so to find the nearest site for a point you can compare squared distances without taking square roots.
176IBMI3H-3-04r16-14Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
Marshal posts are placed at A, B and C beside a Formula 1 circuit. AB = 250 m, BC = 200 m and angle BAC = 40°. Post C can be in either of two positions along the same line from A, so two different triangles ABC are possible.
(a)
Use the sine rule to find the acute value of angle ACB, in degrees to 1 decimal place.
(3)
(b)
Find the obtuse value of angle ACB, in degrees to 1 decimal place.
(1)
(c)
For the triangle with the obtuse angle at C, find the length AC, in metres to 3 significant figures. Use unrounded values from earlier parts.
Common error: When the sine rule is used to find an angle, check whether the obtuse angle (180° minus the calculator value) is also possible, especially when the given side opposite the known angle is shorter than the other given side.
177IBMI3H-3-06r20-03Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium6 marks
Calculator
A rail operator runs five stations A, B, C, D and E connected by the following direct tracks: AB, AC, BC, BD, CD, CE and DE. Each track can be travelled in both directions.
(a)
Write down the adjacency matrix for this network, with the stations in alphabetical order.
(2)
(b)
Use the adjacency matrix to find the number of different walks of length 3 from A to E.
(2)
(c)
An engineer wants to inspect every track exactly once. Determine whether this is possible, giving a reason, and state where the inspection could start and finish.
(2)
Show mark scheme
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Notes
Part (a)
M1 a symmetric \(5\times5\) matrix of 0s and 1s with zeros on the leading diagonal
R1 degrees are A 2, B 3, C 4, D 3, E 2, so exactly two vertices (B and D) have odd degree
✓ 1
A1 possible: an Eulerian trail exists, starting at B and finishing at D (or the reverse)
✓ 1
Exam tip: Entry \((i,j)\) of the \(n\)th power of an adjacency matrix counts walks of length \(n\) from vertex \(i\) to vertex \(j\). A walk may revisit vertices and edges.
178IBMI3H-3-06r20-06Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium6 marks
Calculator
Three relay satellites A, B and C have positions, in thousands of kilometres relative to a ground station, of \(A(12,8,20)\), \(B(15,12,23)\) and \(C(13,9,25)\).
(a)
Find the vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\).
M1 attempt at the vector product, e.g. \((4\cdot5-3\cdot1,\ 3\cdot1-3\cdot5,\ 3\cdot1-4\cdot1)\)
✓ 1
A1 \(\begin{pmatrix}17\\-12\\-1\end{pmatrix}\)
✓ 1
Part (c)
M1 area \(=\frac{1}{2}\left|\overrightarrow{AB}\times\overrightarrow{AC}\right|=\frac{1}{2}\sqrt{17^2+12^2+1^2}\)
✓ 1
A1 \(10.4\) (units: (103 km)2)
✓ 1
Common error: The magnitude of the vector product equals the area of the parallelogram, so the triangle area needs the factor \(\frac12\).
179IBMI3H-3-06r20-07Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium5 marks
Calculator
A rail company plans to connect six towns A to F so that every town can be reached from every other. The possible direct links and their lengths in kilometres are: AB 4.2, AC 6.5, BC 3.8, BD 7.1, CD 5.4, CE 8.3, DE 2.9, DF 6.2, EF 4.6. Building costs €2.4 million per kilometre.
(a)
Use Kruskal's algorithm to find the minimum spanning tree. List the links in the order in which they are accepted, and state one link that is rejected, with a reason.
(3)
(b)
Find the minimum total cost of the network, in € million.
(2)
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Part (a)
M1 links considered in increasing order of length, skipping any that make a cycle
✓ 1
A1 accepted in order: DE, BC, AB, EF, CD
✓ 1
R1 e.g. DF rejected because D–E–F is already connected (forms a cycle); accept AC, BD or CE with a valid cycle
✓ 1
Part (b)
A1 total length \(=2.9+3.8+4.2+4.6+5.4=20.9\) km
✓ 1
A1 cost \(=20.9\times2.4=\) €50.16 million
✓ 1
Exam tip: In Kruskal's algorithm, a link is rejected only if its two ends are already joined. Check that the final tree has exactly one fewer links than the number of towns.
180IBMI3H-3-06r20-08Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
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A track inspector must walk along every section of a rail network with five junctions A to E and return to the starting junction. The sections and their lengths in kilometres are: AB 5, AC 7, BC 4, BD 9, CD 3, CE 8, DE 5.
(a)
Identify the junctions of odd degree and explain why some sections must be walked twice.
(2)
(b)
Find the shortest path between the two odd junctions.
(2)
(c)
Find the length of the shortest closed inspection route.
(2)
(d)
The inspector instead starts at B and is allowed to finish at D. Find the length of the shortest route that covers every section.
(1)
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Notes
Part (a)
A1 B and D have odd degree (3)
✓ 1
R1 a closed route covering every section once needs all degrees even, so the odd vertices must be joined by repeated sections
✓ 1
Part (b)
M1 compare BD \(=9\) with B–C–D \(=4+3\)
✓ 1
A1 B–C–D, length 7 km
✓ 1
Part (c)
M1 sum of all sections \(=41\) plus the repeated path 7
✓ 1
A1 48 km
✓ 1
Part (d)
A1 41 km (no repeated sections needed)
✓ 1
Exam tip: In the Chinese postman problem, pair up odd vertices using the shortest connecting paths, and add this extra length to the total of all edges. If the route need not be closed and there are exactly two odd vertices, no repeats are needed.
181IBMI3H-3-06r20-09Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium8 marks
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Three sports halls are located at \(A(2,1)\), \(B(8,1)\) and \(C(5,7)\), where the units are kilometres. The Voronoi diagram for these sites is drawn.
(a)
Find the equation of the perpendicular bisector of AB.
(2)
(b)
Find the equation of the perpendicular bisector of AC, giving it in the form \(y=mx+c\).
(3)
(c)
Hence find the coordinates of the Voronoi vertex where the three cells meet.
(2)
(d)
A new café is placed at this vertex. Find its distance from each sports hall.
(1)
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Part (a)
M1 midpoint \((5,1)\) and AB is horizontal
✓ 1
A1 \(x=5\)
✓ 1
Part (b)
M1 midpoint of AC \(=(3.5,4)\)
✓ 1
M1 gradient of AC \(=2\), so perpendicular gradient \(=-\frac12\)
✓ 1
A1 \(y=-0.5x+5.75\)
✓ 1
Part (c)
M1 substitute \(x=5\) into the equation from part (b)
✓ 1
A1 \((5,\ 3.25)\)
✓ 1
Part (d)
A1 \(\sqrt{3^2+2.25^2}=3.75\) km
✓ 1
Exam tip: A Voronoi vertex is equidistant from the three nearest sites, so it is found where two perpendicular bisectors cross. Check by calculating the distance to the third site.
182IBMI3H-3-06r20-10Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
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The path of a comet is modelled by the line \(\mathbf{r}_1=\begin{pmatrix}1\\2\\3\end{pmatrix}+s\begin{pmatrix}2\\-1\\1\end{pmatrix}\) and the path of a space probe by \(\mathbf{r}_2=\begin{pmatrix}0\\-2\\4\end{pmatrix}+t\begin{pmatrix}1\\1\\0\end{pmatrix}\), where positions are in millions of kilometres relative to an observatory.
(a)
Show that the two paths cross, and find the coordinates of the crossing point.
(4)
(b)
Find the acute angle between the two paths, to the nearest 0.1°.
R1 check that the second equation holds: \(2-1=1=-2+3\)
✓ 1
A1 crossing point \((3,1,4)\)
✓ 1
Part (b)
M1 scalar product of direction vectors \(=2\times1-1\times1+0\times0=1\)
✓ 1
M1 \(\cos\theta=\frac{1}{\sqrt6\sqrt2}\)
✓ 1
A1 \(\theta=73.2^\circ\)
✓ 1
Common error: Two lines in 3D cross only if one pair of parameter values satisfies all three component equations. Solve using two equations and always check in the third.
183IBMI3H-3-06r20-11Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium8 marks
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A tracking camera maps pitch coordinates \((x,y)\), in metres, to image coordinates using the matrix \(\mathbf{M}=\begin{pmatrix}2.4&1.8\\-1.8&2.4\end{pmatrix}\). The mapping is a rotation combined with an enlargement.
(a)
Find the scale factor of the enlargement, and the angle and direction of the rotation, giving the angle to the nearest 0.1°.
(3)
(b)
A training zone on the pitch has area 17 m2. Find the area of its image.
(2)
(c)
A player appears in the image at \((6,3)\). Find the player's coordinates on the pitch.
(3)
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Part (a)
A1 scale factor \(=\sqrt{2.4^2+1.8^2}=3\)
✓ 1
M1 \(\tan\theta=\frac{1.8}{2.4}\)
✓ 1
A1 \(36.9^\circ\) clockwise (the image of \((1,0)\) is \((2.4,-1.8)\))
✓ 1
Part (b)
M1 area scale factor \(=\det\mathbf{M}=2.4^2+1.8^2=9\)
Common error: For a rotation with enlargement, the scale factor is the square root of the determinant, and the area scale factor is the determinant itself.
184IBMI3H-3-04r23-02Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
In a library atrium two straight steel rails are fixed in the roof. Their directions are given by the vectors a = (2, −1, 2) and b = (1, 2, −2), with units in metres.
(a)
Find a · b.
(2)
(b)
Find the acute angle between the two rails, to the nearest 0.1°.
(3)
(c)
A third rail has direction c = (k, 3, −2) and is perpendicular to the first rail. Find the value of k.
(2)
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Notes
Part (a)
M1 2 × 1 + (−1) × 2 + 2 × (−2)
✓ 1
A1 −4
✓ 1
Part (b)
A1 |a| = 3 and |b| = 3
✓ 1
M1 cos θ = |a · b| / (|a||b|) = 4/9 (or obtuse angle 116.4° found and subtracted from 180°)
✓ 1
A1 63.6°
✓ 1
Part (c)
M1 a · c = 0: 2k − 3 − 4 = 0
✓ 1
A1 k = 3.5
✓ 1
Exam tip: Two lines are perpendicular exactly when the scalar product of their direction vectors is zero; set it equal to zero rather than using angles.
185IBMI3H-3-04r23-05Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium8 marks
A library roof contains a flat triangular glass panel with vertices A(1, 0, 2), B(4, 2, 2) and C(3, 1, 5), where coordinates are in metres.
(a)
Find the vectors AB and AC.
(2)
(b)
Find AB × AC.
(2)
(c)
Hence find the area of the panel, in m2, to 3 significant figures.
(2)
(d)
Glass costs €145 per m2. Use the area to 3 significant figures from part (c) to find the cost of the panel.
(2)
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Notes
Part (a)
A1 AB = (3, 2, 0)
✓ 1
A1 AC = (2, 1, 3)
✓ 1
Part (b)
M1 uses the cross product formula with correct components
✓ 1
A1 (6, −9, −1)
✓ 1
Part (c)
M1 area = ½|AB × AC| = ½ √(36 + 81 + 1)
✓ 1
A1 5.43 m2
✓ 1
Part (d)
M1 5.43 × 145
✓ 1
A1 €787.35
✓ 1
Common error: The cross product gives the area of the parallelogram; remember to halve the magnitude to obtain the triangle's area.
186IBMI3H-3-04r23-08Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
A mobile library van delivers books along every road in a rural network with junctions A, B, C, D and E. The road lengths in km are: AB 5, AC 8, BC 6, BD 7, CD 4, CE 9, DE 5.
(a)
Find the degree of each junction and explain why the van cannot travel along every road exactly once and return to its starting point.
(2)
(b)
The van must travel along every road at least once and return to its starting point. Find the length of the shortest such route.
(3)
(c)
The van instead starts at B and finishes at D. Find the length of the shortest route that uses every road at least once.
(2)
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Part (a)
A1 degrees: A 2, B 3, C 4, D 3, E 2
✓ 1
R1 an Eulerian circuit needs every vertex to have even degree, but B and D are odd
✓ 1
Part (b)
M1 pairs the odd vertices B and D and finds their shortest connection: BD = 7 (BCD = 10 is longer)
✓ 1
A1 total length of all roads = 44 km
✓ 1
A1 shortest route = 44 + 7 = 51 km
✓ 1
Part (c)
M1 recognises that B and D are the only odd vertices, so every road can be used exactly once with no repeats
✓ 1
A1 44 km
✓ 1
Common error: When a route must start and end at different vertices, those two vertices should be the odd ones; do not add a repeated path between them.
187IBMI3H-3-04r23-10Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium8 marks
A designer transforms a triangular tile design for a swimming pool floor using the matrix T = (2 1; 1 3), meaning first row 2, 1 and second row 1, 3. The original triangle has vertices P(0, 0), Q(3, 0) and R(0, 4), in metres, and area 6 m2.
(a)
Find the determinant of T.
(1)
(b)
Find the coordinates of Q′ and R′, the images of Q and R.
(2)
(c)
A point X′ = (7, 11) is the image of a point X. Find the coordinates of X.
(3)
(d)
Find the area of the image triangle P′Q′R′.
(2)
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Part (a)
A1 2 × 3 − 1 × 1 = 5
✓ 1
Part (b)
A1 Q′ = (6, 3)
✓ 1
A1 R′ = (4, 12)
✓ 1
Part (c)
M1 finds the inverse T−1 = (1/5)(3 −1; −1 2)
✓ 1
M1 multiplies T−1 by the column vector (7, 11)
✓ 1
A1 X = (2, 3)
✓ 1
Part (d)
M1 area of image = |det T| × area of the original
✓ 1
A1 5 × 6 = 30 m2
✓ 1
Exam tip: The area scale factor of a transformation matrix is the absolute value of its determinant; check by comparing with the area of the image triangle from its vertices.
188IBMI3H-3-05r23-04Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium8 marks
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A bridge designer models a triangular steel gusset plate with vertices A(1, 0, 2), B(4, 2, 3) and C(2, 5, 1). Coordinates are in metres.
(a)
Write down the vectors AB and AC.
(2)
(b)
Find AB × AC.
(2)
(c)
Hence find the area of the plate, correct to 3 significant figures.
(2)
(d)
The plate is 12 mm thick and the steel has density 7850 kg m−3. Using the unrounded area, find the mass of the plate, correct to 3 significant figures.
(2)
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Part (a)
A1 AB = (3, 2, 1)
✓ 1
A1 AC = (1, 5, −1)
✓ 1
Part (b)
M1 attempt at the vector product
✓ 1
A1 (−7, 4, 13)
✓ 1
Part (c)
M1 area = ½|AB × AC| = ½√(49 + 16 + 169)
✓ 1
A1 7.65 m2
✓ 1
Part (d)
M1 volume = area × 0.012
✓ 1
A1 mass = volume × 7850 = 720 kg
✓ 1
Exam tip: The vector product gives a vector that is perpendicular to the plate, and its magnitude is twice the area of the triangle; remember to halve it.
189IBMI3H-3-05r23-05Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
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Five islands A, B, C, D and E are joined by footbridges. There is one bridge between each of the pairs AB, AC, BC, BD, CD and DE, and there are no other bridges.
(a)
Write down the adjacency matrix M for the islands, in the order A, B, C, D, E.
(2)
(b)
The matrix M3 is found. Use it to find the number of different walks of length 3 from A to E, and list them.
(3)
(c)
The entry in row B, column B of M3 is 4. Explain why this means that there are exactly 2 triangles (cycles of length 3) that include island B.
(2)
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Notes
Part (a)
M1 5 × 5 symmetric matrix with zeros on the leading diagonal
R1 the entry counts closed walks of length 3 that start and end at B
✓ 1
R1 each triangle through B gives two closed walks (one in each direction), so 4 ÷ 2 = 2 triangles, namely ABC and BCD
✓ 1
Common error: Entry (i, j) of Mⁿ counts walks of length exactly n, which may revisit vertices; it is not a count of paths.
190IBMI3H-3-03r24-03Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium8 marks
The position of a fishing boat relative to a harbour at time \(t\) hours is \(\mathbf{r}=\begin{pmatrix}t^2-4t+3\\2t^2-12t\end{pmatrix}\) km, for \(t\ge0\).
(a)
Find the velocity of the boat when \(t=0\) and hence its speed at that instant, in km h−1, to 3 significant figures.
(2)
(b)
Find the acceleration of the boat, and state what it shows about the motion.
(2)
(c)
Find the time at which the speed of the boat is a minimum, and the minimum speed to 3 significant figures.
A1 \(\mathbf{a}=\begin{pmatrix}2\\4\end{pmatrix}\) km h−2
✓ 1
R1 the acceleration is constant (it does not depend on \(t\))
✓ 1
Part (c)
M1 \(|\mathbf{v}|^2=(2t-4)^2+(4t-12)^2=20t^2-112t+160\) (or use GDC on the speed)
✓ 1
M1 differentiate and set to zero, or find minimum on GDC
✓ 1
A1 \(t=2.8\) h
✓ 1
A1 minimum speed \(=\sqrt{3.2}=1.79\) km h−1
✓ 1
Exam tip: To minimise a speed, minimise the speed squared; it has the same turning point and avoids differentiating a square root.
191IBMI3H-3-03r24-04Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium6 marks
A triangular panel of a garment is modelled in three dimensions with vertices A(1, 0, 2), B(4, 2, 1) and C(2, 5, 3), where all coordinates are in decimetres.
(a)
Find the vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\).
M1 area \(=\tfrac12|\overrightarrow{AB}\times\overrightarrow{AC}|=\tfrac12\sqrt{7^2+(-4)^2+13^2}\)
✓ 1
A1 \(7.65\) dm2
✓ 1
Common error: The vector product is not commutative: swapping the order reverses the direction, but the magnitude (and so the area) is unchanged.
192IBMI3H-3-03r24-05Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium8 marks
Two patrol vessels in a fishery move with constant velocities. Relative to a lighthouse, their position vectors at time \(t\) hours are \(\mathbf{r}_A=\begin{pmatrix}1\\4\end{pmatrix}+t\begin{pmatrix}2\\-1\end{pmatrix}\) and \(\mathbf{r}_B=\begin{pmatrix}14\\-7\end{pmatrix}+t\begin{pmatrix}-3\\3\end{pmatrix}\), with distances in km.
(a)
Show that the paths of the two vessels cross at the point (5, 2).
(3)
(b)
Hence explain why the vessels do not collide.
(2)
(c)
Find the distance between the vessels at the instant when vessel A reaches (5, 2), giving your answer to 3 significant figures.
(3)
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Part (a)
M1 equate the two lines using different parameters, e.g. \(1+2s=14-3t\) and \(4-s=-7+3t\)
✓ 1
A1 solving gives \(s=2\) and \(t=3\)
✓ 1
A1 substituting gives the point (5, 2)
✓ 1
Part (b)
M1 vessel A is at (5, 2) when \(t=2\) and vessel B is there when \(t=3\)
✓ 1
R1 they are at the crossing point at different times, so they are never at the same place at the same time
✓ 1
Part (c)
A1 at \(t=2\), \(\mathbf{r}_B=(8,-1)\)
✓ 1
M1 distance \(=\sqrt{(8-5)^2+(-1-2)^2}\)
✓ 1
A1 4.24 km
✓ 1
Common error: Crossing paths do not mean a collision: the two parameters must take the same value, because they both represent the same time.
193IBMI3H-3-03r24-07Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium6 marks
Five fish-landing ports A, B, C, D and E are linked by direct ferry routes AB, AC, BC, BD, CD, CE and DE. Each route can be used in either direction.
(a)
Write down the adjacency matrix \(\mathbf{M}\) for this network, with the vertices in alphabetical order.
(2)
(b)
Use \(\mathbf{M}^3\) to find the number of different walks of length 3 from A to E.
(2)
(c)
State, with a reason, whether it is possible to travel along every route exactly once and finish where you started.
(2)
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Notes
Part (a)
A1 first two rows correct: A is (0, 1, 1, 0, 0) and B is (1, 0, 1, 1, 0)
✓ 1
A1 all five rows correct: \(\begin{pmatrix}0&1&1&0&0\\1&0&1&1&0\\1&1&0&1&1\\0&1&1&0&1\\0&0&1&1&0\end{pmatrix}\)
✓ 1
Part (b)
M1 attempt to find \(\mathbf{M}^3\) (or the product of \(\mathbf{M}^2\) row A with column E)
✓ 1
A1 3 walks
✓ 1
Part (c)
M1 degrees are A 2, B 3, C 4, D 3, E 2 (row sums of \(\mathbf{M}\))
✓ 1
R1 two vertices (B and D) have odd degree, so there is no closed trail using every route once
✓ 1
Exam tip: The entry in row \(i\), column \(j\) of \(\mathbf{M}^n\) counts the walks of length \(n\) from vertex \(i\) to vertex \(j\); read off the single entry you need.
194IBMI3H-3-03r24-09Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
A cleaning robot must travel along every corridor of an exhibition hall at least once. The corridors join junctions A, B, C, D and E, and their lengths in metres are: AB 5, BC 4, CD 6, DA 7, AC 9, DE 3, EC 8.
(a)
Find the degree of each vertex and hence identify the vertices of odd degree.
(2)
(b)
Find the shortest distance between the two odd vertices.
(2)
(c)
The robot starts and finishes at A. Find the length of the shortest route that covers every corridor.
(2)
(d)
The robot is now allowed to finish at a different junction from where it starts. State the length of the shortest route covering every corridor, and a pair of start and finish junctions that achieves it.
(1)
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Part (a)
A1 degrees A 3, B 2, C 4, D 3, E 2
✓ 1
A1 odd vertices are A and D
✓ 1
Part (b)
M1 compare routes, e.g. AD = 7, ACD = 15, ABCD = 15
✓ 1
A1 7 m
✓ 1
Part (c)
M1 total of all corridors \(=42\) plus the repeated shortest path (ECF from (b))
✓ 1
A1 49 m
✓ 1
Part (d)
A1 42 m, starting at A and finishing at D (or the reverse)
✓ 1
Exam tip: In a route-inspection problem, pair up the odd-degree vertices and repeat the shortest connecting path; if there are only two odd vertices, an open route between them needs no repetition.
195IBMI3H-3-03r24-12Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
A robot gripper moves in a plane. First it is rotated \(90^\circ\) clockwise about the origin, then the result is reflected in the x-axis.
(a)
Write down the matrix \(\mathbf{R}\) for the rotation and the matrix \(\mathbf{F}\) for the reflection.
(2)
(b)
Find the single matrix \(\mathbf{T}\) that represents the combined transformation.
(2)
(c)
Describe the single transformation represented by \(\mathbf{T}\).
(1)
(d)
After both transformations the gripper is at (9, 4). Find its original position.
M1 sine rule \(\dfrac{\sin\angle BHC}{9}=\dfrac{\sin102^\circ}{HC}\)
✓ 1
A1 \(\angle BHC=29.0^\circ\)
✓ 1
A1 bearing \(=072+29.0=101^\circ\)
✓ 1
Exam tip: Draw a north line at each point on your sketch; the back bearing is the forward bearing plus or minus 180°.
197IBMI3H-3-03r26-02Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
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A fishing boat leaves a position 2 km east and 5 km north of a harbour H, which is at the origin. Relative to H, the position vector of the boat at time \(t\) hours after leaving is \(\mathbf{r}=\begin{pmatrix}2+3t\\5-4t\end{pmatrix}\) km, where the unit vectors point east and north.
(a)
Find the speed of the boat.
(2)
(b)
Find the time at which the boat is closest to H, and the distance between the boat and H at that time.
(4)
(c)
The harbour radar detects boats within 4 km of H. State, with a reason, whether the boat is ever detected.
(1)
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Part (a)
M1 velocity \(\begin{pmatrix}3\\-4\end{pmatrix}\) and magnitude \(\sqrt{3^2+(-4)^2}\)
✓ 1
A1 \(5\) km h−1
✓ 1
Part (b)
M1 sets \(\mathbf{r}\cdot\mathbf{v}=0\) or minimises \(|\mathbf{r}|^2=25t^2-28t+29\)
✓ 1
A1 \(t=0.56\) hours
✓ 1
M1 substitutes into \(|\mathbf{r}|\), e.g. \(\sqrt{3.68^2+2.76^2}\)
✓ 1
A1 \(4.60\) km (accept \(4.6\))
✓ 1
Part (c)
R1 No, since the minimum distance \(4.60\) km is greater than \(4\) km
✓ 1
Exam tip: For closest approach, the position vector is perpendicular to the velocity at that instant, so the scalar product is zero.
198IBMI3H-3-03r26-03Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium8 marks
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A fashion designer models a triangular fabric panel on a mannequin using three points P(1, 0, 2), Q(4, −1, 4) and R(2, 3, 1), with coordinates in centimetres.
M1 attempt at \(\overrightarrow{PQ}\times\overrightarrow{PR}\)
✓ 1
A1 \(\begin{pmatrix}-5\\5\\10\end{pmatrix}\)
✓ 1
A1 area \(=\tfrac12\sqrt{150}=6.12\) cm2
✓ 1
Common error: The area of a triangle with sides given by vectors \(\mathbf{u}\) and \(\mathbf{v}\) is half the magnitude of \(\mathbf{u}\times\mathbf{v}\); do not forget the factor of one half.
199IBMI3H-3-03r26-08Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
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A fisheries agency wants to connect six landing ports, A to F, with a communications network. The possible links and their installation costs, in thousands of euros (€k), are: AB 12, AC 9, BC 7, BD 14, CD 11, CE 16, DE 8, DF 13, EF 10, BF 18.
(a)
Use Kruskal's algorithm to find a minimum spanning tree. List the edges in the order you add them, and state the total cost.
(4)
(b)
The link BC becomes unavailable. Find the new minimum total cost, and the percentage increase on the original cost, to 3 significant figures.
(3)
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Part (a)
M1 considers edges in increasing order of cost
✓ 1
A1 adds BC (7), DE (8), AC (9), EF (10) in this order
✓ 1
A1 adds CD (11), joining the two components, and stops with 5 edges
✓ 1
A1 total cost \(7+8+9+10+11=45\) (€45k)
✓ 1
Part (b)
M1 applies the algorithm again without BC
✓ 1
A1 edges DE (8), AC (9), EF (10), CD (11), AB (12) giving \(50\) (€50k)
✓ 1
A1 \(\dfrac{50-45}{45}\times100=11.1\%\)
✓ 1
Exam tip: In Kruskal's algorithm, reject any edge that would form a cycle; a spanning tree on \(n\) vertices always has exactly \(n-1\) edges.
200IBMI3H-3-03r26-09Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium8 marks
Calculator
A cleaning robot must travel along every corridor of a fashion exhibition hall at least once. The corridors link five points A, B, C, D and E, and their lengths in metres are: AB 6, AC 9, BC 5, BD 7, CD 8, CE 4, DE 10.
(a)
Find the degree of each vertex and state which vertices have odd degree.
(2)
(b)
The robot must start and finish at the same point. Find the shortest distance it must repeat, and show how you know.
(2)
(c)
Find the length of the shortest closed route that covers every corridor.
(2)
(d)
The robot may instead start at B and finish at D. State why this is possible and find the shortest route length.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 degrees A 2, B 3, C 4, D 3, E 2
✓ 1
A1 B and D are the only odd vertices
✓ 1
Part (b)
M1 only one pairing of odd vertices, BD, so finds shortest path from B to D
✓ 1
A1 BD direct is \(7\) m (less than BCD \(=13\) m)
✓ 1
Part (c)
M1 total of all corridors \(6+9+5+7+8+4+10=49\) plus repeated path
✓ 1
A1 \(56\) m
✓ 1
Part (d)
R1 exactly two odd vertices, B and D, so a trail covering each corridor once exists
✓ 1
A1 \(49\) m
✓ 1
Exam tip: For a route that need not be closed, the shortest route length equals the total of all edges if there are exactly two odd vertices and the route starts at one and ends at the other.
201IBMI3H-3-03r26-11Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium8 marks
Calculator
A robot arm end point is at P(4, 2), measured in metres. The arm rotates through 30° anticlockwise about the origin O and then extends so that distances from O are multiplied by 1.5. This is represented by the matrix \(\mathbf{M}\).
(a)
Write down \(\mathbf{M}\) in exact form and in numerical form to 3 significant figures.
(2)
(b)
Find the coordinates of the image of P, to 3 significant figures, using the unrounded matrix.
(3)
(c)
Triangle OPQ, where Q is (−1, 3), is transformed by \(\mathbf{M}\). Find the area of the image triangle.
Common error: Remember to halve the magnitude of the vector product: |AB × AC| is the area of the parallelogram, not the triangle.
204IBMI3H-3-06r27-05Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
A shipping company runs direct routes between five ports A, B, C, D and E. Each route can be sailed in either direction. The adjacency matrix M shows the number of direct routes between each pair of ports.
A
B
C
D
E
A
0
2
1
0
0
B
2
0
1
1
0
C
1
1
0
1
1
D
0
1
1
0
1
E
0
0
1
1
0
(a)
Find the total number of direct routes in the network.
(2)
(b)
Find the number of different journeys from port A to port E that use exactly three direct routes. Use the matrix M3.
(3)
(c)
Find the number of different round trips from A back to A that use exactly two direct routes.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sum of all entries of M is 16 (each route counted twice), or sums the entries above the diagonal
✓ 1
A1 8 routes
✓ 1
Part (b)
M1 attempts to find M3 (the entry in row A, column E)
✓ 1
A1 correct value from M3 (row A: 4, 15, 10, 4, 5)
✓ 1
A1 5 journeys
✓ 1
Part (c)
M1 uses the entry (M2)AA = 22 + 12 + 0 + 0 + 0
✓ 1
A1 5
✓ 1
Exam tip: The entry in row i, column j of Mⁿ counts the walks of length n from vertex i to vertex j; where two routes join the same ports they count as different walks.
205IBMI3H-3-06r27-06Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium8 marks
A shipping company plans feeder links between six ports A, B, C, D, E and F. The cost of building each possible link, in thousands of pounds, is shown in the table (a dash means no link is possible).
A
B
C
D
E
F
A
–
14
9
–
18
–
B
14
–
11
8
–
–
C
9
11
–
13
16
–
D
–
8
13
–
7
12
E
18
–
16
7
–
10
F
–
–
–
12
10
–
(a)
Use Kruskal's algorithm to find a minimum spanning tree for the network. List the links in the order in which they are accepted, and state any link that is considered but rejected.
(4)
(b)
Use Prim's algorithm starting at A to find a minimum spanning tree. List the links in the order in which they are added.
(3)
(c)
Write down the minimum total cost of linking all six ports.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 considers the links in increasing order of cost
✓ 1
A1 DE (7), BD (8), AC (9) accepted
✓ 1
A1 EF (10), BC (11) accepted
✓ 1
A1 DF (12) rejected because it would form a cycle D–E–F
✓ 1
Part (b)
M1 starts at A and adds the cheapest link from A: AC (9)
✓ 1
A1 BC (11), then BD (8)
✓ 1
A1 DE (7), then EF (10)
✓ 1
Part (c)
A1 £45 thousand (7 + 8 + 9 + 10 + 11)
✓ 1
Exam tip: Both algorithms must give a tree with one fewer link than the number of vertices; count your links to check (6 ports need 5 links).
206IBMI3H-3-06r27-07Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium6 marks
The paths on a volcano's crater rim form a network with junctions A, B, C, D, E and F. The lengths of the paths, in km, are: AB 8, AC 6, AF 11, BC 5, BD 7, CD 9, CE 10, DE 4, DF 12, EF 6. A survey team must walk along every path at least once, starting and finishing at the same junction.
(a)
State which junctions have odd degree.
(1)
(b)
Find the shortest total distance that must be repeated. Show all the ways of pairing the odd vertices and the distances involved.
(3)
(c)
Find the length of the shortest route that covers every path and returns to its starting point.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 A, B, E and F (each has degree 3)
✓ 1
Part (b)
M1 considers the three pairings of A, B, E, F
✓ 1
A1 AB + EF = 8 + 6 = 14; AE + BF = 16 + 17 = 33
✓ 1
A1 AF + BE = 11 + 11 = 22, so least repeated distance is 14 (paths AB and EF)
✓ 1
Part (c)
M1 total of all path lengths is 78, plus the repeated distance
✓ 1
A1 78 + 14 = 92 km (ECF from (b))
✓ 1
Common error: After pairing odd vertices, add the length of the shortest path between each pair, which may use several edges.
207IBMI3H-3-06r29-02Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium8 marks
A fashion designer models a triangular fabric panel on a mannequin with vertices A(1, 0, 2), B(4, 2, 3) and C(2, 5, 1), where coordinates are in decimetres.
(a)
Find the vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\).
M1 area \(=\frac{1}{2}|\overrightarrow{AB}\times\overrightarrow{AC}|\)
✓ 1
A1 \(\frac{1}{2}\sqrt{234}\)
✓ 1
A1 7.65
✓ 1
Exam tip: The magnitude of a vector product gives the area of the parallelogram; halve it for a triangle.
208IBMI3H-3-06r29-06Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
A fashion company must connect six workshops A, B, C, D, E and F with a network of delivery lanes, at minimum total cost. The costs, in hundreds of euros, of building possible lanes are: AB 7, AC 4, AD 9, BC 3, BD 6, BE 8, CD 5, CE 10, DE 2, DF 11, EF 6.
(a)
Use Kruskal's algorithm to find the minimum spanning tree, listing the edges in the order they are added.
(3)
(b)
State the minimum total cost, in euros.
(1)
(c)
The lane CD becomes unavailable. Find the new minimum total cost, in euros.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 edges considered in increasing order of cost
✓ 1
A1 DE, BC, AC in that order
✓ 1
A1 CD then EF (BD rejected as it forms a cycle)
✓ 1
Part (b)
A1 €2000
✓ 1
Part (c)
M1 re-apply the algorithm without CD
✓ 1
A1 edges DE, BC, AC, BD, EF
✓ 1
A1 €2100
✓ 1
Exam tip: State which edges are rejected because they would form a cycle; this shows the algorithm was applied correctly.
209IBMI3H-3-06r29-12Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
A fisheries patrol boat leaves port P and sails 14 km on a bearing of 062° to a buoy X. It then sails 9 km on a bearing of 150° to a net marker Q.
(a)
Show that angle PXQ is 92°.
(2)
(b)
Find the distance PQ, to 3 significant figures.
(2)
(c)
Find the bearing of Q from P, to the nearest 0.1°. Use the unrounded value of PQ.
M1 sine rule \(\frac{\sin\angle XPQ}{9}=\frac{\sin92^\circ}{PQ}\)
✓ 1
A1 \(\angle XPQ=32.1^\circ\)
✓ 1
A1 bearing \(=62+32.1=094.1^\circ\)
✓ 1
Exam tip: Sketch the path with north lines at each turning point; the interior angle at the turn comes from the back-bearing.
210IBMI3H-3-05r30-04Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium6 marks
Two aircraft fly through a region of airspace. Relative to a fixed origin, their velocities in km h−1 are v1 = \(\begin{pmatrix}240\\180\\6\end{pmatrix}\) and v2 = \(\begin{pmatrix}-90\\310\\-4\end{pmatrix}\), where the third component is the climb rate.
(a)
Calculate v1 · v2.
(2)
(b)
Find the magnitude of each velocity vector, to 4 significant figures.
(2)
(c)
Find the angle between the two flight directions. Give your answer in degrees to 3 significant figures, using unrounded values from earlier parts.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempts to multiply corresponding components and add
✓ 1
A1 240(−90) + 180(310) + 6(−4) = 34 176
✓ 1
Part (b)
A1 |v1| = √(2402+1802+62) = 300.1
✓ 1
A1 |v2| = √(902+3102+42) = 322.8
✓ 1
Part (c)
M1 cos θ = 34176 / (|v1||v2|)
✓ 1
A1 θ = 69.3°
✓ 1
Exam tip: Use the scalar product formula cos θ = (a · b)/(|a||b|); the angle found is between the direction vectors, not between positions.
211IBMI3H-3-05r30-05Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
A triangular sloping field has corners at A(0, 0, 2), B(120, 0, 8) and C(30, 80, 5), where coordinates are in metres and the third coordinate is height above a reference level.
(a)
Find the vectors AB and AC.
(2)
(b)
Calculate AB × AC.
(2)
(c)
Hence find the area of the field, in m2 to the nearest whole number.
(2)
(d)
The farmer sows seed at a cost of €0.12 per square metre of field surface. Using the area from (c) to the nearest m2, find the cost of seeding the whole field, to the nearest cent.
Exam tip: The magnitude of a vector product is the area of the parallelogram on the two vectors, so the triangle area is half of it.
212IBMI3H-3-05r30-06Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium8 marks
A survey drone flies over a field. Relative to a launch point O, its velocity at time t seconds is v = \(\begin{pmatrix}8 − 2t\\0.6t2\end{pmatrix}\) m s−1, for 0 ≤ t ≤ 6. The drone is at O when t = 0.
(a)
Find the acceleration of the drone when t = 3.
(2)
(b)
Find the value of t at which the drone is moving in a direction parallel to the y-axis.
(1)
(c)
Find the position vector of the drone when t = 4.
(3)
(d)
Find the distance of the drone from O when t = 4. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 differentiates each component: a = \(\begin{pmatrix}−2\\1.2t\end{pmatrix}\)
✓ 1
A1 \(\begin{pmatrix}−2\\3.6\end{pmatrix}\) m s−2
✓ 1
Part (b)
A1 8 − 2t = 0, so t = 4 s
✓ 1
Part (c)
M1 integrates the velocity vector
✓ 1
A1 r = \(\begin{pmatrix}8t − t2\\0.2t3\end{pmatrix}\) using r(0) = 0
✓ 1
A1 \(\begin{pmatrix}16\\12.8\end{pmatrix}\) m
✓ 1
Part (d)
M1 √(162 + 12.82)
✓ 1
A1 20.5 m
✓ 1
Common error: Velocity is found by differentiating position, so position from velocity needs integration with the initial condition applied.
213IBMI3H-3-05r30-07Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium8 marks
Cables must be laid to connect six stages, A to F, at a music festival so that every stage is connected to every other, possibly through other stages. The table shows the length in metres of each possible cable route; a dash means no route is possible.
A
B
C
D
E
F
A
–
45
62
–
–
90
B
45
–
38
71
–
–
C
62
38
–
54
49
–
D
–
71
54
–
33
58
E
–
–
49
33
–
41
F
90
–
–
58
41
–
(a)
Use Kruskal's algorithm to find a minimum spanning tree. List the edges in the order they are selected, and state the total length.
(3)
(b)
Use Prim's algorithm starting at A to find a minimum spanning tree. List the edges in the order they are selected.
(3)
(c)
Cable costs €14.50 per metre. Find the minimum cost of cable needed.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 edges considered in increasing order of length, rejecting any that form a cycle
✓ 1
A1 DE (33), BC (38), EF (41), AB (45), CE (49) selected in this order
✓ 1
A1 total length 206 m
✓ 1
Part (b)
M1 starts at A and repeatedly adds the shortest edge joining the tree to a new vertex
✓ 1
A1 AB (45), BC (38), CE (49)
✓ 1
A1 DE (33), EF (41)
✓ 1
Part (c)
M1 206 × 14.50
✓ 1
A1 €2987.00
✓ 1
Exam tip: In Kruskal's algorithm check each new edge for a cycle; in Prim's algorithm the tree is always a single connected piece.
214IBMI3H-3-05r30-09Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
A charter airline must fly a small aircraft from airport A to each of airports B, C, D and E exactly once and return to A. The table shows the flight times in minutes between each pair of airports.
A
B
C
D
E
A
–
50
80
65
95
B
50
–
45
70
60
C
80
45
–
55
85
D
65
70
55
–
40
E
95
60
85
40
–
(a)
Use the nearest neighbour algorithm, starting at A, to find an upper bound for the minimum total flight time.
(3)
(b)
Use the deleted vertex algorithm, deleting E, to find a lower bound for the minimum total flight time.
(3)
(c)
Hence find the greatest possible amount by which the route in (a) could exceed the minimum total flight time.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 repeatedly moves to the nearest unvisited airport, returning to A at the end
M1 finds a minimum spanning tree for A, B, C and D
✓ 1
A1 BC 45, AB 50, CD 55 with total 150
✓ 1
A1 adds the two shortest edges at E, 40 and 60, to get a lower bound of 250 minutes
✓ 1
Part (c)
A1 285 − 250 = 35 minutes
✓ 1
Exam tip: An upper bound comes from a real route; a lower bound comes from the deleted vertex method. The minimum lies between them.
215IBMI3H-3-05r30-14Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
In a concert hall, a straight cable runs from fixing point A(2, 1, 8) to fixing point B(14, 9, 6), where coordinates are in metres and the third coordinate is height. A microphone hangs at M(6, 3, 2).
(a)
Find a vector equation of the line through A and B.
(2)
(b)
Find AM × AB.
(2)
(c)
Hence find the shortest distance from M to the line through A and B, to 3 significant figures.
(2)
(d)
The point on the line closest to M corresponds to λ = (AM · AB) / |AB|2. Find this value of λ to 3 significant figures, and hence state whether the closest point lies on the cable between A and B.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 direction AB = \(\begin{pmatrix}12\\8\\-2\end{pmatrix}\)
✓ 1
A1 r = \(\begin{pmatrix}2\\1\\8\end{pmatrix}\) + λ\(\begin{pmatrix}12\\8\\-2\end{pmatrix}\) (or equivalent, using B as the point)
✓ 1
Part (b)
M1 AM = \(\begin{pmatrix}4\\2\\-6\end{pmatrix}\) and attempts the vector product
✓ 1
A1 \(\begin{pmatrix}44\\-64\\8\end{pmatrix}\)
✓ 1
Part (c)
M1 distance = |AM × AB| / |AB| = √6096 / √212
✓ 1
A1 5.36 m
✓ 1
Part (d)
A1 λ = 76/212 = 0.358, which lies between 0 and 1, so the closest point is on the cable
✓ 1
Exam tip: The shortest distance from a point to a line equals the magnitude of the vector product of the join vector and the direction vector, divided by the direction vector's magnitude.
216IBMI3H-3-05r31-04Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
Two survey drones fly over a volcanic crater. Relative to a fixed origin, with distances in metres, drone 1 follows the line L1: r = (2, 1, 3) + s(1, 2, −1) and drone 2 follows the line L2: r = (1, 8, −1) + t(3, −3, 2), where s and t are parameters.
(a)
Show that the two flight paths intersect, and find the coordinates of the point of intersection.
(4)
(b)
Find the acute angle between the two flight paths. Give your answer to the nearest 0.1°.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 equate components, e.g. 2 + s = 1 + 3t and 1 + 2s = 8 − 3t
✓ 1
A1 s = 2 and t = 1
✓ 1
R1 check the third component: 3 − 2 = 1 and −1 + 2 = 1, so the equations are consistent
✓ 1
A1 intersection at (4, 5, 1)
✓ 1
Part (b)
M1 scalar product of direction vectors: 3 − 6 − 2 = −5
✓ 1
M1 cos θ = 5 ÷ (√6 × √22)
✓ 1
A1 64.2°
✓ 1
Exam tip: To show lines intersect you must solve two component equations and then verify the third; solving only two proves nothing in three dimensions. Use the absolute value of the scalar product when the question asks for the acute angle.
217IBMI3H-3-05r31-05Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium5 marks
A shipping company serves five ports P, Q, R, S and T. Direct routes between ports are two-way and are shown in the adjacency matrix M below, with rows and columns in the order P, Q, R, S, T. P: 0 1 1 0 1 Q: 1 0 1 1 0 R: 1 1 0 0 1 S: 0 1 0 0 1 T: 1 0 1 1 0
(a)
Write down the number of direct routes from port Q.
(1)
(b)
Use the matrix M3 to find the number of different journeys of exactly three legs that start at S and finish at Q. A journey may revisit a port.
(2)
(c)
Each diagonal entry of M2 is non-zero. Explain what the entry in row P, column P of M2 represents, and state its value.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 3
✓ 1
Part (b)
M1 attempt to find M3 (or row S of M2 multiplied by column Q of M)
✓ 1
A1 6 journeys
✓ 1
Part (c)
A1 the number of two-leg journeys that start and end at P (go out and come back by a direct route)
✓ 1
A1 3, which equals the number of direct routes from P
✓ 1
Exam tip: Entry (i, j) of Mⁿ counts walks of length n from vertex i to vertex j. Name the start and end vertices carefully and keep the row and column the right way round.
218IBMI3H-3-05r31-06Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
A volcano observatory must connect six monitoring stations A to F with cable so that every station is joined to the network. The possible links and their installation costs, in £ thousands, are: AB 14, AC 9, AD 17, BC 11, BE 8, BF 15, CD 13, CF 16, DF 7, EF 12. No other links are possible.
(a)
Use Kruskal's algorithm to find a minimum spanning tree. List the links in the order in which they are chosen.
(3)
(b)
Find the minimum total cost of the network.
(2)
(c)
Explain why link AB is not included in the minimum spanning tree.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 links considered in increasing order of cost, rejecting any that make a cycle
✓ 1
A1 first three chosen: DF (7), BE (8), AC (9)
✓ 1
A1 then BC (11) and EF (12), giving five links in total
✓ 1
Part (b)
M1 add the costs of the chosen links: 7 + 8 + 9 + 11 + 12
✓ 1
A1 £47 thousand
✓ 1
Part (c)
R1 when AB (14) is considered, A and B are already connected (through C, by AC and BC)
✓ 1
R1 adding AB would create a cycle, and a spanning tree of six stations needs exactly five links
✓ 1
Exam tip: In Kruskal's algorithm a link is rejected only if it would complete a cycle. A tree on n vertices has exactly n − 1 edges, so stop as soon as you have them.
219IBMI3H-3-05r31-08Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
A ferry operator based at port A must visit ports B, C, D and E once each and return to A. The distances between ports, in nautical miles, are: AB 12, AC 9, AD 15, AE 11, BC 8, BD 13, BE 10, CD 7, CE 14, DE 6.
(a)
Use the nearest neighbour algorithm, starting at A, to find an upper bound for the length of the shortest tour. State the tour.
(3)
(b)
By deleting port C, find a lower bound for the length of the shortest tour.
(3)
(c)
Write down the interval within which the length of the shortest tour must lie.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 apply nearest neighbour: A to C (9), C to D (7), D to E (6), E to B (10), B to A (12)
✓ 1
A1 tour ACDEBA
✓ 1
A1 upper bound 44 nautical miles
✓ 1
Part (b)
M1 find a minimum spanning tree of A, B, D and E
✓ 1
A1 DE (6) + BE (10) + AE (11) = 27
✓ 1
A1 add the two shortest edges at C, CD (7) and CB (8): 27 + 7 + 8 = 42 nautical miles
✓ 1
Part (c)
A1 between 42 and 44 nautical miles inclusive (ECF from (a) and (b))
✓ 1
Exam tip: A deleted-vertex lower bound uses a minimum spanning tree of what remains plus the two shortest edges at the deleted vertex. A lower bound from any single deletion is valid, so the largest of them gives the best bound.
220IBMI3H-3-05r33-04Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium8 marks
A crop-spraying drone moves over a field. Relative to a fixed origin its velocity, in m s−1, at time t seconds is \(\mathbf{v}=\begin{pmatrix}2t-6\\8-t\end{pmatrix}\) for \(t\ge 0\). When \(t=0\) the drone is at the point with position vector \(\begin{pmatrix}10\\5\end{pmatrix}\) m.
(a)
Find the acceleration of the drone.
(2)
(b)
Find the position vector of the drone when t = 4.
(3)
(c)
Find the minimum speed of the drone, correct to three significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to differentiate the velocity vector with respect to t
✓ 1
A1 (2, −1) m s−2 (constant)
✓ 1
Part (b)
M1 attempt to integrate: \(\mathbf{r}=(t^2-6t+c_1,\ 8t-\tfrac12t^2+c_2)\)
✓ 1
M1 uses the initial position to find c1 = 10 and c2 = 5
✓ 1
A1 (2, 29) m
✓ 1
Part (c)
M1 speed\(^2=(2t-6)^2+(8-t)^2=5t^2-40t+100\)
✓ 1
M1 attempt to find the minimum, e.g. by the graphing facility, or by completing the square to give 5(t − 4)2 + 20
✓ 1
A1 minimum speed \(\sqrt{20}\) = 4.47 m s−1 (at t = 4)
✓ 1
Common error: The speed is the magnitude of the velocity vector, so minimise the speed (or its square), not either component on its own.
221IBMI3H-3-05r33-07Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
A technician must inspect every path linking five stages A, B, C, D and E at a music festival, starting and finishing at the same stage. The path lengths in metres are: AB 120, AC 100, AE 80, BC 90, BD 140, CD 150, DE 110.
(a)
Explain why the technician cannot inspect every path exactly once and return to the start.
(2)
(b)
List the three ways of pairing the four odd vertices, and for each pairing find the least total length of the paths joining the pairs.
(3)
(c)
Hence find the length of the shortest route which inspects every path and returns to the start.
(2)
Show mark scheme
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Mark
Notes
Part (a)
R1 states that an Eulerian circuit needs every vertex to have even degree
✓ 1
A1 vertices A, B, C and D have odd degree 3
✓ 1
Part (b)
M1 attempt to find the shortest path between odd vertices
✓ 1
A1 AB + CD = 120 + 150 = 270 and AD + BC = 190 + 90 = 280 (AD via E)
✓ 1
A1 AC + BD = 100 + 140 = 240
✓ 1
Part (c)
M1 adds the least pairing to the total length of all paths, 790 + 240
✓ 1
A1 1030 m
✓ 1
Common error: Remember to repeat the shortest path between each pair, not just the single edge; a route through another vertex can be shorter than a direct path.
222IBMI3H-3-05r33-12Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
An aircraft has an airspeed of 240 km h−1 and is steered on a heading of bearing 070°. The wind velocity is \(\begin{pmatrix}-20\\35\end{pmatrix}\) km h−1, where the components are east and north respectively.
(a)
Find the components of the velocity of the aircraft in still air, correct to 1 decimal place.
(2)
(b)
Find the speed of the aircraft relative to the ground, correct to three significant figures. Use the unrounded components from part (a).
(2)
(c)
Find the bearing on which the aircraft actually travels, correct to the nearest 0.1°. Use the unrounded components from part (a).
(3)
Show mark scheme
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Notes
Part (a)
M1 east = 240 sin 70°, north = 240 cos 70°
✓ 1
A1 (225.5, 82.1) km h−1
✓ 1
Part (b)
M1 adds the wind: (205.526…, 117.085…) and takes the magnitude
Common error: Bearings are measured clockwise from north, so the east component goes on the numerator of the tangent, not the north component.
223IBMI3H-3-05r33-13Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium8 marks
Three irrigation wells W1(2, 3), W2(10, 5) and W3(5, 11) are marked on a map where one unit represents 1 km. A Voronoi diagram is drawn to show which well is nearest to each point of the district.
(a)
Find the equation of the perpendicular bisector of W1W2, giving your answer in the form y = mx + c.
(3)
(b)
The perpendicular bisector of W1W3 has equation 3x + 8y = 66.5. Find the coordinates of the Voronoi vertex, correct to 3 significant figures.
(3)
(c)
Find the distance from the Voronoi vertex to each well, correct to the nearest 0.1 km. Use the unrounded coordinates of the vertex.
(2)
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Notes
Part (a)
M1 midpoint (6, 4)
✓ 1
A1 gradient of W1W2 is 1/4, so the perpendicular gradient is −4
✓ 1
A1 y = −4x + 28
✓ 1
Part (b)
M1 solves the two bisector equations simultaneously
✓ 1
A1 x = 5.43
✓ 1
A1 y = 6.28
✓ 1
Part (c)
M1 distance from (5.431…, 6.276…) to a well, e.g. W1
✓ 1
A1 4.7 km (the same for all three wells)
✓ 1
Exam tip: The Voronoi vertex is equally far from all three sites, so checking the distance to a second well is a quick test of your coordinates.
224IBMI3H-3-05r33-14Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium8 marks
An aircraft leaves airport A and flies 120 km on a bearing of 070° to a waypoint B. It then flies 85 km on a bearing of 175° to a waypoint C.
(a)
Show that angle ABC = 75°.
(2)
(b)
Find the distance AC, correct to three significant figures.
(3)
(c)
Find the bearing of C from A, correct to the nearest degree.
(3)
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Notes
Part (a)
M1 the bearing of A from B is 250° (back bearing, 70° + 180°)
✓ 1
A1 250° − 175° = 75°, shown
✓ 1
Part (b)
M1 use of the cosine rule, AC2 = 1202 + 852 − 2 × 120 × 85 × cos 75°
✓ 1
A1 AC2 = 16 345.1…
✓ 1
A1 AC = 128 km
✓ 1
Part (c)
M1 use of the sine rule, \(\sin\angle BAC=\frac{85\sin 75^\circ}{AC}\)
✓ 1
A1 angle BAC = 40.0°
✓ 1
A1 bearing 070° + 40° = 110°
✓ 1
Common error: A back bearing differs by 180°; sketching a north line at each point makes the angle inside the triangle much easier to see.
225IBMI3H-3-05r36-02Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium6 marks
Two roof beams of a pavilion meet at the apex A. Relative to A, the other ends of the beams are at B and C, with AB = (5, 2, −1) m and AC = (1, 4, 3) m. A triangular glass panel is fitted between the beams and the line BC.
(a)
Find the angle BAC between the beams, in degrees, correct to 3 significant figures.
(3)
(b)
Find the area of the glass panel ABC, in m2, correct to 3 significant figures.
(3)
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Notes
Part (a)
M1 scalar product AB·AC = 5 + 8 − 3 = 10
✓ 1
M1 use of cos θ = (AB·AC)/(|AB||AC|) with |AB| = √30, |AC| = √26
✓ 1
A1 69.0°
✓ 1
Part (b)
M1 attempt at the vector product AB×AC
✓ 1
A1 (10, −16, 18)
✓ 1
A1 area = ½√680 = 13.0 m2
✓ 1
Exam tip: The area of a triangle with sides given as vectors is half the magnitude of their vector product.
226IBMI3H-3-05r36-03Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
A survey vessel moves so that its position, in km relative to a harbour, at time t hours after departure is given by r(t) = (t2 − 6t + 10, 4 + 2t − 0.1t3), for t ≥ 0.
(a)
Find an expression for the velocity of the vessel at time t.
(2)
(b)
Find the time at which the vessel is moving parallel to the y-axis.
(2)
(c)
Find the magnitude of the acceleration of the vessel when t = 3, in km h−2, correct to 3 significant figures.
(3)
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Notes
Part (a)
M1 differentiate each component with respect to t
✓ 1
A1 v = (2t − 6, 2 − 0.3t2)
✓ 1
Part (b)
M1 set the x-component of velocity equal to zero, 2t − 6 = 0
✓ 1
A1 t = 3 hours
✓ 1
Part (c)
M1 differentiate velocity: a = (2, −0.6t)
✓ 1
M1 substitute t = 3 to get (2, −1.8) and find the magnitude
✓ 1
A1 2.69 km h−2
✓ 1
Common error: Acceleration is the derivative of velocity, so differentiate the velocity vector once more before substituting t.
227IBMI3H-3-05r36-07Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium6 marks
A warden patrols five survey stations A to E in a marine reserve, using the paths below (lengths in km): AB 8, BC 7, CD 9, DE 7, EA 10, BE 12, CE 5, AD 11. The warden must walk along every path at least once, starting and ending at A.
(a)
Explain why the warden cannot walk along every path exactly once and return to A.
(2)
(b)
Use the Chinese postman algorithm to find the minimum total distance the warden must walk.
(4)
Show mark scheme
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Notes
Part (a)
A1 vertices A, B, C, D have odd degree (3 each)
✓ 1
R1 a closed route using every edge exactly once needs all vertices to have even degree
✓ 1
Part (b)
M1 consider the possible pairings of the odd vertices
✓ 1
A1 pairings AB + CD = 17, AD + BC = 18, AC + BD = 31
✓ 1
M1 choose the smallest pairing and add its length to the total of all edges (69)
✓ 1
A1 86 km
✓ 1
Common error: When pairing odd vertices, use the shortest route between them, which may pass through other vertices.
228IBMI3H-3-05r36-08Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium8 marks
A research ship must visit ports A to E once each, returning to its starting port A. The sailing distances between ports are given below in nautical miles (nm). AB 12, AC 19, AD 15, AE 23 BC 14, BD 21, BE 17 CD 11, CE 16 DE 13
(a)
Use the nearest neighbour algorithm, starting at A, to find an upper bound for the length of the shortest tour.
(3)
(b)
Delete vertex A to find a lower bound for the length of the shortest tour.
(4)
(c)
A skipper says a tour of length 62 nm exists. Explain why this is impossible.
(1)
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Notes
Part (a)
M1 choose the nearest unvisited port each time: A→B (12), B→C (14), C→D (11), D→E (13)
✓ 1
M1 return E→A (23)
✓ 1
A1 73 nm
✓ 1
Part (b)
M1 find a minimum spanning tree of B, C, D, E
✓ 1
A1 weight 38 (CD 11, DE 13, BC 14)
✓ 1
M1 add the two shortest edges from A (12 and 15)
✓ 1
A1 65 nm
✓ 1
Part (c)
R1 62 is below the lower bound of 65, so no tour can be that short
✓ 1
Exam tip: The deleted-vertex method needs both a minimum spanning tree of the remaining vertices and the two cheapest edges to the deleted vertex.
229IBMI3H-3-05r36-10Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
On an architect's plan, a transformation T is a horizontal stretch with scale factor 3 (invariant line the y-axis), followed by a rotation of 90° clockwise about the origin.
(a)
Find the matrix that represents T.
(3)
(b)
Find the image of the point (4, 2) under T.
(2)
(c)
An atrium has area 85 m2 on the plan. Find its area after transformation T.
A1 product in the correct order \(\begin{pmatrix}0&1\\-3&0\end{pmatrix}\)
✓ 1
Part (b)
M1 multiply the matrix by the column vector (4, 2)
✓ 1
A1 (2, −12)
✓ 1
Part (c)
M1 area scale factor = |det T| = 3
✓ 1
A1 255 m2
✓ 1
Exam tip: The area scale factor of a transformation matrix is the absolute value of its determinant.
230IBMI3H-3-05r36-12Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium8 marks
A research boat leaves port P and sails 14 km on a bearing of 065° to buoy Q. It then sails 10 km on a bearing of 150° to buoy R.
(a)
Show that angle PQR is 95°.
(2)
(b)
Find the distance PR, in km, correct to 3 significant figures.
(3)
(c)
Find the bearing of P from R, correct to the nearest 0.1°. Use the unrounded value of PR.
(3)
Show mark scheme
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Notes
Part (a)
M1 the bearing of P from Q is 065° + 180° = 245°
✓ 1
A1 245° − 150° = 95°
✓ 1
Part (b)
M1 use of the cosine rule: PR2 = 142 + 102 − 2(14)(10)cos 95°
✓ 1
A1 PR2 = 320.4…
✓ 1
A1 PR = 17.9 km
✓ 1
Part (c)
M1 sine rule to find angle QPR: sin QPR / 10 = sin 95° / PR, giving 33.8°
✓ 1
M1 bearing of R from P = 65° + 33.8° = 98.8°
✓ 1
A1 bearing of P from R = 278.8°
✓ 1
Exam tip: Draw a north line at every point of a bearings diagram and mark the angle between the north line and the path.
231IBMI3H-3-06r36-02Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
A solar panel frame is a parallelogram OABC in space. Relative to corner O, OA = (6, 2, 1) m and OC = (1, 5, 3) m. The z-axis points vertically upwards.
(a)
Find OA × OC.
(2)
(b)
Find the area of the frame, in m2, correct to 3 significant figures.
(2)
(c)
The vector OA × OC is perpendicular to the frame. Find the angle between the frame and the horizontal, correct to 3 significant figures.
(3)
Show mark scheme
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Notes
Part (a)
M1 attempt at the vector product using the component formula
✓ 1
A1 (1, −17, 28)
✓ 1
Part (b)
M1 area = |OA × OC| = √(1 + 289 + 784)
✓ 1
A1 32.8 m2 (ECF from (a))
✓ 1
Part (c)
M1 the angle between the frame and the horizontal equals the angle between its normal and the vertical (0, 0, 1)
✓ 1
M1 cos θ = 28/√1074 (ECF from (a))
✓ 1
A1 31.3°
✓ 1
Common error: The angle between two planes equals the angle between their normals, not the angle between a normal and the plane itself.
232IBMI3H-3-06r36-03Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium8 marks
An inspection drone flies over a wind farm. At time t = 0 s it is at the origin with velocity (3, 1.5) m s−1. Its acceleration is constant and equal to (0.4, −0.2) m s−2.
(a)
Find an expression for the velocity of the drone at time t.
(2)
(b)
Find the speed of the drone when t = 5, in m s−1, correct to 3 significant figures.
(2)
(c)
Find the distance of the drone from the origin when t = 5, in m, correct to 3 significant figures.
(4)
Show mark scheme
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Notes
Part (a)
M1 integrate the acceleration, or use v = u + at
✓ 1
A1 v = (3 + 0.4t, 1.5 − 0.2t)
✓ 1
Part (b)
M1 v(5) = (5, 0.5) and magnitude √(52 + 0.52)
✓ 1
A1 5.02 m s−1
✓ 1
Part (c)
M1 integrate the velocity with respect to time
✓ 1
A1 r = (3t + 0.2t2, 1.5t − 0.1t2), constant of integration zero
✓ 1
M1 substitute t = 5 to get (20, 5) and find the magnitude
✓ 1
A1 20.6 m
✓ 1
Common error: After integrating, use the starting position to fix the constant; here it is zero because the drone starts at the origin.
233IBMI3H-3-06r36-07Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium6 marks
A delivery van must travel along every road in a retail district at least once. The depots are A to F, and the road lengths in km are: AB 9, AC 14, AF 15, BC 7, BD 11, CD 6, CE 8, DE 13, DF 10, EF 12. The van starts at A and finishes at F.
(a)
Write down the vertices of odd degree.
(1)
(b)
Explain why only B and E need to be considered when finding the roads to repeat, and find the shortest route from B to E.
(3)
(c)
Find the minimum distance the van must travel. The total length of all roads is 105 km.
(2)
Show mark scheme
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Notes
Part (a)
A1 A, B, E and F (each has degree 3)
✓ 1
Part (b)
R1 the route starts and ends at the odd vertices A and F, so only the other two odd vertices must be paired
✓ 1
M1 compare routes such as BCE = 15 and BDE = 24
✓ 1
A1 shortest route BCE, 15 km
✓ 1
Part (c)
M1 105 + 15
✓ 1
A1 120 km
✓ 1
Common error: If the route does not return to its start, the two odd vertices at the ends are not paired up; pair the remaining odd vertices instead.
234IBMI3H-3-06r36-10Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium6 marks
A wind turbine blade tip is at (0, 45) m relative to the hub, where the axes are horizontal and vertical through the hub. Each second the blade rotates 150° clockwise about the hub.
(a)
Write down the matrix for a rotation of 150° clockwise about the origin.
(2)
(b)
Find the coordinates of the blade tip after one second, correct to 3 significant figures.
(2)
(c)
Find the smallest number of seconds after which the tip first returns to (0, 45).
(2)
Show mark scheme
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Notes
Part (a)
M1 matrix form \(\begin{pmatrix}\cos 150°&\sin 150°\\-\sin 150°&\cos 150°\end{pmatrix}\)
M1 multiply the matrix by the column vector (0, 45)
✓ 1
A1 (22.5, −39.0)
✓ 1
Part (c)
M1 150n must be a multiple of 360°, so the smallest common multiple of 150 and 360 is 1800
✓ 1
A1 n = 12 seconds
✓ 1
Exam tip: For a clockwise rotation through θ, use the matrix with cos θ on the diagonal and sin θ above the diagonal.
235IBMI3H-3-06r36-12Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium6 marks
In a contact-tracing survey, a drone flies from P to Q, a distance of 12 km, then from Q to R, a distance of 9 km. The angle QPR is 40°. There are two possible positions for R.
(a)
Find the two possible values of angle PRQ, correct to 1 decimal place.
(3)
(b)
For the position of R where angle PRQ is acute, find the distance PR, in km, correct to 3 significant figures. Use the unrounded value of angle PRQ.
(3)
Show mark scheme
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Notes
Part (a)
M1 sine rule: sin PRQ / 12 = sin 40° / 9
✓ 1
A1 PRQ = 59.0°
✓ 1
A1 PRQ = 121.0°
✓ 1
Part (b)
M1 angle PQR = 180° − 40° − 58.99…° = 81.0…°
✓ 1
M1 sine rule: PR / sin 81.0…° = 9 / sin 40°
✓ 1
A1 13.8 km
✓ 1
Common error: When the sine rule is used to find an angle, check whether the obtuse angle 180° − θ is also possible.
236IBMI3H-3-07r37-02Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
Calculator
Two microwave links leave a mast at the origin O. Their directions are given by the vectors a = \(\begin{pmatrix}5\\2\\-1\end{pmatrix}\) and b = \(\begin{pmatrix}2\\6\\3\end{pmatrix}\), where the x-axis points east, the y-axis north and the z-axis vertically up.
(a)
Find a·b.
(2)
(b)
A third link has direction vector p = \(\begin{pmatrix}3\\-2\\m\end{pmatrix}\) and is perpendicular to the link in direction b. Find the value of m.
(2)
(c)
Find the angle between the links in directions a and b. Give your answer in degrees to 3 significant figures.
(3)
Show mark scheme
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Notes
Part (a)
M1 attempts 5×2 + 2×6 + (−1)×3
✓ 1
A1 19
✓ 1
Part (b)
M1 sets b·p = 0, i.e. 6 − 12 + 3m = 0
✓ 1
A1 m = 2
✓ 1
Part (c)
M1 uses cos θ = a·b / (|a||b|)
✓ 1
A1 |a| = √30 and |b| = 7 (ECF from (a) for the scalar product)
✓ 1
A1 θ = 60.3°
✓ 1
Common error: Using the scalar product alone for an angle: the magnitudes of both vectors must be divided out, and the calculator must be in degree mode.
237IBMI3H-3-07r37-03Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
Calculator
A search drone is flown above an avalanche slope. Its velocity, in m s−1, at time t seconds is given by v(t) = \(\begin{pmatrix}4 − 0.5t\\3\\0.2t\end{pmatrix}\) for 0 ≤ t ≤ 20, where the components point east, north and vertically up respectively.
(a)
Find the speed of the drone when t = 6. Give your answer to 3 significant figures.
(3)
(b)
Find the distance of the drone from its starting point when t = 10.
M1 attempts to integrate each component of v(t) from 0 to 10
✓ 1
A1 displacement \(\begin{pmatrix}15\\30\\10\end{pmatrix}\) m
✓ 1
M1 attempts the magnitude √(152 + 302 + 102) (ECF)
✓ 1
A1 35 m
✓ 1
Exam tip: Integrating a velocity vector gives displacement, so using definite limits starting at the start time gives the distance from the starting point directly.
238IBMI3H-3-07r37-09Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
Calculator
A food technologist must visit five tasting labs A, B, C, D and E, starting and ending at lab A and visiting each other lab exactly once. The distances in km between the labs are: AB 12, AC 9, AD 15, AE 11, BC 8, BD 10, BE 14, CD 7, CE 13 and DE 6.
(a)
Use the nearest neighbour algorithm starting at A to find an upper bound for the shortest route. State the route and its length.
(3)
(b)
Use the deleted vertex algorithm, deleting A, to find a lower bound for the shortest route.
(3)
(c)
Hence write down an inequality satisfied by the length L km of the shortest route.
(1)
Show mark scheme
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Notes
Part (a)
M1 applies nearest neighbour from A
✓ 1
A1 route A → C → D → E → B → A
✓ 1
A1 length 9 + 7 + 6 + 14 + 12 = 48 km
✓ 1
Part (b)
M1 finds a minimum spanning tree of B, C, D and E
✓ 1
A1 DE 6, CD 7 and BC 8, total 21
✓ 1
A1 adds the two shortest edges from A, 9 + 11 = 20, giving a lower bound of 41 km
✓ 1
Part (c)
A1 41 ≤ L ≤ 48 (ECF)
✓ 1
Exam tip: An upper bound comes from a real route you have found; a lower bound comes from a spanning-tree argument, so the two bounds are never swapped.
239IBMI3H-3-07r37-13Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium8 marks
Calculator
A food scientist designs a pyramid-shaped cover for a cake. The base is a horizontal rectangle 24 cm by 18 cm and the apex is 12 cm vertically above the centre of the base.
(a)
Show that each sloping edge has length 19.2 cm, correct to 3 significant figures.
(3)
(b)
Find the angle between a sloping edge and the base. Give your answer to 3 significant figures.
(2)
(c)
Find the angle between a triangular face on a 24 cm side and the base. Give your answer to 3 significant figures.
(3)
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Part (a)
M1 finds the base diagonal √(242 + 182) = 30 cm
✓ 1
A1 half-diagonal 15 cm
✓ 1
A1 edge √(152 + 122) = 19.2 cm
✓ 1
Part (b)
M1 uses tan θ = 12/15
✓ 1
A1 38.7°
✓ 1
Part (c)
A1 horizontal distance from centre to the midpoint of the 24 cm side is 9 cm
✓ 1
M1 uses tan φ = 12/9
✓ 1
A1 53.1°
✓ 1
Common error: Using the half-length of a base edge instead of the perpendicular distance from the centre to the edge: the 24 cm side is 9 cm from the centre, not 12 cm.
240IBMI3H-3-05r38-02Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
Two directional antenna beams leave a mast at the origin. Their directions are given by the vectors a = 2i + 3j + 6k and b = 4i − j + 8k. A third beam has direction c = pi + 4j − k, where p is a constant.
(a)
Find a·b.
(2)
(b)
Find the angle between the first two beams, giving your answer to the nearest 0.1°.
(3)
(c)
Beam c is perpendicular to the first beam. Find the value of p.
(2)
Show mark scheme
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Mark
Notes
Part (a)
M1 attempt to sum products of corresponding components
✓ 1
A1 53
✓ 1
Part (b)
M1 use of cos θ = a·b / (|a||b|)
✓ 1
A1 |a| = 7 and |b| = 9
✓ 1
A1 θ = 32.7° (ECF from (a))
✓ 1
Part (c)
M1 set a·c = 0, giving 2p + 12 − 6 = 0
✓ 1
A1 p = −3
✓ 1
Exam tip: Perpendicular vectors have a scalar product of zero, which turns an unknown component into a simple linear equation.
241IBMI3H-3-05r38-07Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
A utility company must connect six hubs A to F with the cheapest possible cable network. The cost of laying cable (in €thousand) between each pair of hubs where a link is possible is shown below. AB 12, AC 9, AF 17, BC 15, BD 7, CD 11, CE 14, DE 6, DF 13, EF 10
(a)
Use Kruskal's algorithm to find the minimum spanning tree. List the edges in the order in which they are accepted.
(3)
(b)
Starting at A, use Prim's algorithm to list the edges in the order in which they are added.
(3)
(c)
Find the minimum total cost of the network.
(1)
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Part (a)
M1 edges considered in increasing order of weight
✓ 1
A1 DE, BD, AC (in this order)
✓ 1
A1 EF, CD (in this order)
✓ 1
Part (b)
M1 at each step the cheapest edge joining the tree to a new vertex
✓ 1
A1 AC, CD, DE
✓ 1
A1 BD, EF
✓ 1
Part (c)
A1 €43 thousand
✓ 1
Exam tip: Kruskal's and Prim's algorithms may add edges in different orders, but with distinct edge weights they always give the same tree and total.
242IBMI3H-3-05r38-08Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
A food research park has five laboratories A to E joined by corridors. The lengths of the corridors, in metres, are AB 7, AC 5, BC 4, BD 11, CD 6, CE 9 and DE 3. A technician must walk along every corridor at least once.
(a)
State which vertices have odd degree.
(2)
(b)
Find the length of the shortest route that begins and ends at the same laboratory and uses every corridor at least once.
(3)
(c)
Find the length of the shortest route that begins at B, ends at D and uses every corridor at least once.
(2)
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Part (a)
A1 B (degree 3)
✓ 1
A1 D (degree 3)
✓ 1
Part (b)
M1 sum of all corridor lengths = 45
✓ 1
M1 shortest path between the two odd vertices, B–C–D = 10 (not the direct 11)
✓ 1
A1 55 m
✓ 1
Part (c)
M1 identification that no corridor needs repeating when starting and ending at the odd vertices
✓ 1
A1 45 m
✓ 1
Exam tip: A route that starts and ends at the two odd vertices needs no repeated edges, so its length is just the total of all edge weights.
243IBMI3H-3-05r39-04Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
A triangular carbon-fibre panel on the floor of a Formula 1 car has vertices A\((10, 20, 0)\), B\((50, 30, 20)\) and C\((20, 60, 30)\), where coordinates are measured in centimetres from a fixed point on the chassis.
M1 area \(=\tfrac12|\overrightarrow{AB}\times\overrightarrow{AC}|\)
✓ 1
A1 \(\tfrac12\sqrt{500^2+1000^2+1500^2}=935\) cm2
✓ 1
Part (c)
M1 mass \(=0.42\times935.41\ldots\)
✓ 1
A1 393 g
✓ 1
Exam tip: The magnitude of the vector product of two sides of a triangle is the area of the parallelogram they span, so halve it for the triangle.
244IBMI3H-3-05r39-07Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium6 marks
The floor plan of a hospital ward is drawn on coordinate axes with units of 1 m. A designer uses the matrix \(\mathbf{M}=\begin{pmatrix}0&-2\\2&0\end{pmatrix}\) to transform the plan to a new scale and orientation for a display board.
(a)
Describe fully the single geometric transformation represented by \(\mathbf{M}\) as a combination of a rotation and an enlargement.
(2)
(b)
Find the image of the point \((3,-1)\).
(2)
(c)
The nurses' station has an area of 7.5 m2 on the original plan. Find its area on the display board.
(2)
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Part (a)
A1 rotation of \(90^\circ\) anticlockwise about the origin
✓ 1
A1 enlargement with scale factor 2, centre the origin
M1 uses \(|\det\mathbf{M}|=4\) as the area scale factor
✓ 1
A1 \(4\times7.5=30\) m2
✓ 1
Exam tip: The determinant of a transformation matrix gives the area scale factor, so its absolute value is used for area and the sign tells you whether orientation is reversed.
245IBMI3H-3-05r39-08Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium8 marks
A track map of a Formula 1 circuit is distorted when it is projected onto a screen. The distortion is a linear transformation represented by the matrix \(\mathbf{T}\). It maps the point \((2,1)\) to \((7,4)\) and the point \((1,1)\) to \((4,3)\).
(a)
Find \(\mathbf{T}\).
(3)
(b)
A region of the track map covers 6.4 cm2 on the original. Find its area after the projection.
(2)
(c)
The pit entry appears at the point \((11,7)\) on the screen. Find the coordinates of the pit entry on the original track map.
(3)
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Part (a)
M1 sets \(\mathbf{T}=\begin{pmatrix}a&b\\c&d\end{pmatrix}\) and forms equations such as \(2a+b=7\), \(a+b=4\)
✓ 1
A1 \(a=3\), \(b=1\)
✓ 1
A1 \(c=1\), \(d=2\), so \(\mathbf{T}=\begin{pmatrix}3&1\\1&2\end{pmatrix}\)
Exam tip: To find an original point from its image, multiply the image by the inverse matrix, not by the matrix itself.
246IBMI3H-3-05r39-10Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
A porter must visit five hospital wards A, B, C, D and E, starting and finishing at A, and visiting each other ward exactly once. The walking times in minutes between wards are: AB 17, AC 13, AD 10, AE 19, BC 27, BD 20, BE 29, CD 23, CE 22, DE 15.
(a)
Use the nearest neighbour algorithm starting at A to find an upper bound for the shortest route, writing down the route.
(3)
(b)
By deleting ward A, find a lower bound for the shortest route.
(3)
(c)
A manager says a route of 78 minutes exists. Explain whether this is possible.
(1)
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Part (a)
M1 at each ward moves to the nearest unvisited ward
✓ 1
A1 route A–D–E–C–B–A
✓ 1
A1 \(10+15+22+27+17=91\) minutes
✓ 1
Part (b)
M1 minimum spanning tree of B, C, D, E
✓ 1
A1 DE 15, BD 20, CE 22, total 57
✓ 1
A1 adds the two shortest edges at A, \(10+13\), to get 80 minutes
✓ 1
Part (c)
R1 not possible, since every route takes at least 80 minutes
✓ 1
Exam tip: A lower bound comes from deleting a vertex and adding its two shortest edges to the minimum spanning tree of the rest; the nearest neighbour route only gives an upper bound.
247IBMI3H-3-05r39-11Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
Relative to a tracking station, the flight path of one bee is the line \(l_1: \mathbf{r}=\begin{pmatrix}2\\1\\3\end{pmatrix}+t\begin{pmatrix}1\\2\\-1\end{pmatrix}\) and the flight path of a second bee is the line \(l_2: \mathbf{r}=\begin{pmatrix}9\\3\\-2\end{pmatrix}+s\begin{pmatrix}-2\\2\\1\end{pmatrix}\), where distances are in metres.
(a)
Show that the two lines intersect and find the coordinates of the point of intersection.
(4)
(b)
Find the acute angle between the two flight paths, giving your answer to the nearest tenth of a degree.
(3)
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Part (a)
M1 equates the components of \(l_1\) and \(l_2\)
✓ 1
A1 solves two equations to get \(t=3\), \(s=2\)
✓ 1
R1 checks the third equation is satisfied
✓ 1
A1 point of intersection \((5,7,0)\)
✓ 1
Part (b)
M1 uses the scalar product of the direction vectors, \(\begin{pmatrix}1\\2\\-1\end{pmatrix}\cdot\begin{pmatrix}-2\\2\\1\end{pmatrix}=1\)
✓ 1
A1 \(\cos\theta=\dfrac{1}{\sqrt6\times3}\)
✓ 1
A1 \(82.2^\circ\)
✓ 1
Common error: When the scalar product of the direction vectors is negative, the angle found is obtuse; take the absolute value to obtain the acute angle between lines.
248IBMI3H-3-05r39-13Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium8 marks
At a Formula 1 circuit, marshal post B is 480 m from post A on a bearing of \(052^\circ\). Post C is 350 m from post B on a bearing of \(125^\circ\).
(a)
Show that angle ABC is \(107^\circ\).
(2)
(b)
Find the distance AC to 3 significant figures.
(3)
(c)
Find the bearing of C from A, giving your answer to the nearest tenth of a degree, as a three-figure bearing.
Exam tip: The magnitude of the vector product is the area of the parallelogram on the two vectors, so the area of the triangle is half of it.
250IBMI3H-3-06r41-05Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium8 marks
The positions of three musicians on a stage plan are D(2, 1), B(4, 1) and S(3, 4), measured in metres from the centre of the stage. For a new layout, the plan is rotated through 90° anticlockwise about the origin, and then reflected in the x-axis.
(a)
Write down the matrix Q for the rotation and the matrix R for the reflection.
(2)
(b)
Find the single matrix T that represents both transformations in this order.
(2)
(c)
Find the new position of the singer S.
(2)
(d)
Find det T, and state what it tells you about the area of the triangle DBS and its orientation.
(2)
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Part (a)
A1 Q = [[0, −1], [1, 0]]
✓ 1
A1 R = [[1, 0], [0, −1]]
✓ 1
Part (b)
M1 T = RQ (reflection matrix on the left)
✓ 1
A1 T = [[0, −1], [−1, 0]]
✓ 1
Part (c)
M1 T(3, 4)T (matrix multiplied by the position vector)
✓ 1
A1 (−4, −3)
✓ 1
Part (d)
A1 det T = (0)(0) − (−1)(−1) = −1
✓ 1
R1 area unchanged (scale factor 1); orientation reversed (negative determinant)
✓ 1
Common error: When two transformations are applied one after the other, the matrix for the first transformation is written on the right of the product.
251IBMI3H-3-06r41-06Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium6 marks
A music festival has five stages A, B, C, D and E. The footpaths between stages are AB, AC, BC, BD, CE and DE, and each path can be walked in either direction.
(a)
Write down the adjacency matrix M for the stages, in the order A, B, C, D, E.
(2)
(b)
Use M3 to find the number of different walks of length 3 from A to E.
M1 evaluating M3 (using a GDC) and reading the entry for row A, column E
✓ 1
A1 2
✓ 1
Part (c)
A1 A–B–C–E
✓ 1
A1 A–B–D–E
✓ 1
Exam tip: The (i, j) entry of Mⁿ counts the walks of length n from vertex i to vertex j, where a walk may repeat vertices and edges.
252IBMI3H-3-06r41-08Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
A farm manager inspects every track on a farm by driving along each track at least once, returning to the starting point. The junctions are A, B, C, D and E, and the tracks (lengths in metres) are: AB 300, AC 200, BC 250, BD 400, CD 350, CE 500, DE 150.
(a)
Show that the graph has exactly two vertices of odd degree, and state them.
(2)
(b)
Find the shortest route between the two odd vertices and its length.
(1)
(c)
Find the length of the shortest inspection route that starts and ends at the same junction.
(2)
(d)
The manager decides instead to start at B and finish at D. Find the length of the shortest route that uses every track at least once.
(2)
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Part (a)
M1 degrees A 2, B 3, C 4, D 3, E 2
✓ 1
A1 odd vertices B and D
✓ 1
Part (b)
A1 B to D directly, 400 m (the other routes, such as B–C–D = 600 m, are longer)
✓ 1
Part (c)
M1 total of all track lengths = 2150 m
✓ 1
A1 2150 + 400 = 2550 m
✓ 1
Part (d)
R1 B and D are the only odd vertices, so an Eulerian trail exists from B to D and no track is repeated
✓ 1
A1 2150 m
✓ 1
Exam tip: In the Chinese postman problem, repeat the shortest path between the odd vertices; if the start and finish are the two odd vertices, no repetition is needed.
253IBMI3H-3-06r41-10Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
An air traffic display shows aircraft positions in kilometres relative to the radar station at the origin. A transformation of the display is a rotation through 90° clockwise about the origin, followed by an enlargement with scale factor 2 and centre the origin.
(a)
Find the single matrix T that represents this transformation.
(2)
(b)
The aircraft P is at (3, −2). Find the position of P after the transformation.
(2)
(c)
After the transformation an aircraft appears at (−8, 10). Find its original position.
Exam tip: To undo a transformation, multiply the image position by the inverse matrix; check by applying T to your answer.
254IBMI3H-3-06r41-13Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
An aircraft flies 180 km from A to B on a bearing of 065°. It then flies 120 km from B to C on a bearing of 150°.
(a)
Show that angle ABC = 95°.
(2)
(b)
Find the distance AC. Give your answer to 3 significant figures.
(2)
(c)
Find the bearing of C from A, to the nearest tenth of a degree. Use the unrounded value of AC.
(3)
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Part (a)
M1 bearing of A from B is 065° + 180° = 245°
✓ 1
A1 245° − 150° = 95°
✓ 1
Part (b)
M1 AC2 = 1802 + 1202 − 2(180)(120) cos 95°
✓ 1
A1 AC = 225 km
✓ 1
Part (c)
M1 sin(∠BAC) / 120 = sin 95° / AC
✓ 1
A1 ∠BAC = 32.1°
✓ 1
A1 bearing = 65° + 32.1° = 097.1°
✓ 1
Common error: The interior angle at B is found from the bearing of A from B (the back bearing), not from the bearing of B from A.
255IBMI3H-3-07r41-02Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium6 marks
Two support cables of a footbridge are attached at the same anchor point. Taking the anchor point as the origin, the cables run in the directions of the vectors \(\mathbf{a}=\begin{pmatrix}4\\-2\\5\end{pmatrix}\) m and \(\mathbf{b}=\begin{pmatrix}1\\6\\-3\end{pmatrix}\) m. A triangular steel gusset plate is to be cut with two of its edges along \(\mathbf{a}\) and \(\mathbf{b}\).
(a)
Find the angle between the two cables. Give your answer to 3 significant figures.
(3)
(b)
Calculate the area of the gusset plate, giving your answer in m2 to 3 significant figures.
(3)
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Part (a)
M1 use of \(\cos\theta=\dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}\)
A1 area \(=\frac12|\mathbf{a}\times\mathbf{b}|=\frac12\sqrt{1541}=19.6\) m2
✓ 1
Exam tip: The area of a triangle with edge vectors a and b is half the magnitude of the vector product; do not forget the factor of one half.
256IBMI3H-3-07r41-03Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium8 marks
Four islands A, B, C and D are linked by footbridges. Each footbridge joins exactly two islands and can be crossed in either direction. The adjacency matrix of the network, with rows and columns in the order A, B, C, D, is \(\mathbf{M}=\begin{pmatrix}0&1&1&0\\1&0&1&1\\1&1&0&1\\0&1&1&0\end{pmatrix}\)
(a)
Write down the degree of each island and hence state, with a reason, whether a walk that crosses every footbridge exactly once and returns to its start is possible.
(3)
(b)
Use \(\mathbf{M}^3\) to find the number of different walks of length 3 from A to B.
(3)
(c)
Verify your answer to (b) by listing all the walks of length 3 from A to B using island letters, for example A–C–D–B.
(2)
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Part (a)
A1 degrees A: 2, B: 3, C: 3, D: 2
✓ 1
R1 two vertices (B and C) have odd degree
✓ 1
A1 so no such closed walk (no Eulerian circuit) is possible
✓ 1
Part (b)
M1 attempt to find \(\mathbf{M}^3\) (or the first row of \(\mathbf{M}^2\) multiplied by \(\mathbf{M}\))
✓ 1
A1 correct entry, row A column B of \(\mathbf{M}^3\)
✓ 1
A1 5 walks
✓ 1
Part (c)
M1 systematic listing of walks of the form A–x–y–B
✓ 1
A1 any complete correct list of five: A–B–A–B, A–C–A–B, A–B–C–B, A–C–D–B, A–B–D–B
✓ 1
Exam tip: A walk of length n corresponds to one entry of the n-th power of the adjacency matrix; read row = start vertex, column = end vertex.
257IBMI3H-3-07r41-07Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
A triangular solar panel is drawn on a design grid with vertices at O(0, 0) and two other points. The panel has area 3 m2. The design is transformed by the matrix \(\mathbf{T}=\begin{pmatrix}0&2\\2&0\end{pmatrix}\), which is a reflection followed by an enlargement, and a vertex P has coordinates (2, 4).
(a)
Describe fully the two transformations that make up T.
(2)
(b)
Find the coordinates of the image of P.
(2)
(c)
Use the determinant of T to find the area of the image of the panel.
M1 attempt at \(\det\mathbf{T}=0\times0-2\times2\)
✓ 1
A1 \(|\det\mathbf{T}|=4\)
✓ 1
A1 area \(=4\times3=12\) m2
✓ 1
Exam tip: The area scale factor of a matrix transformation is the absolute value of its determinant; a negative determinant shows the orientation is reversed.
258IBMI3H-3-07r41-08Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium8 marks
A robotic arm on a production line moves a part by first rotating it \(30^\circ\) anticlockwise about the origin and then enlarging it by scale factor 1.5 about the origin. The combined transformation is represented by the matrix \(\mathbf{M}\). Use unrounded values in your working and give final coordinates to 3 significant figures.
(a)
Write down \(\mathbf{M}\), with entries given to 3 significant figures.
(3)
(b)
Find the image of the point (4, 2).
(2)
(c)
After the transformation a part sits at (6, −3). Find the original coordinates of the part.
M1 use of \(\mathbf{M}^{-1}\) on \(\begin{pmatrix}6\\-3\end{pmatrix}\)
✓ 1
A1 \(\mathbf{M}^{-1}=\frac{1}{1.5}\begin{pmatrix}\cos30^\circ&\sin30^\circ\\-\sin30^\circ&\cos30^\circ\end{pmatrix}\) or equivalent
✓ 1
A1 (2.46, −3.73)
✓ 1
Common error: Matrices for combined transformations are applied right to left; the matrix of the first transformation is the one nearest the point.
259IBMI3H-3-07r41-09Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium6 marks
A small lander moves across an asteroid’s surface. Its position relative to a base station at time \(t\) seconds, \(t \geq 0\), is given by \(\mathbf{r}=\begin{pmatrix}t^3-6t\\4t^2-t\end{pmatrix}\) metres.
(a)
Find the velocity of the lander when \(t=2\).
(2)
(b)
Calculate the speed of the lander when \(t=2\), giving your answer to 3 significant figures.
(2)
(c)
Find the time, for \(t>0\), at which the lander is moving parallel to the \(y\)-axis. Give your answer to 3 significant figures.
(2)
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Part (a)
M1 differentiate each component, \(\mathbf{v}=\begin{pmatrix}3t^2-6\\8t-1\end{pmatrix}\)
✓ 1
A1 \(\begin{pmatrix}6\\15\end{pmatrix}\) m s−1
✓ 1
Part (b)
M1 \(\sqrt{6^2+15^2}\)
✓ 1
A1 \(16.2\) m s−1
✓ 1
Part (c)
M1 set the horizontal component of velocity to zero, \(3t^2-6=0\)
✓ 1
A1 \(t=\sqrt2=1.41\) s
✓ 1
Exam tip: Moving parallel to an axis means the other component of the velocity vector is zero; set that component to zero and solve.
260IBMI3H-3-07r41-13Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium8 marks
Three coffee kiosks are located at A(2, 3), B(8, 3) and C(5, 9), with distances in km. The region closest to each kiosk is shown on a Voronoi diagram.
(a)
Find the equation of the perpendicular bisector of AB.
(2)
(b)
Find the equation of the perpendicular bisector of AC. Give your answer in the form \(y=mx+c\).
(3)
(c)
The two bisectors meet at a vertex V of the diagram. Find the coordinates of V, and the distance from V to kiosk A.
(3)
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Part (a)
M1 midpoint (5, 3) and the line is perpendicular to AB, which is horizontal
✓ 1
A1 \(x=5\)
✓ 1
Part (b)
M1 midpoint of AC, (3.5, 6)
✓ 1
M1 gradient of AC is 2, so the perpendicular gradient is \(-\frac12\)
✓ 1
A1 \(y=-0.5x+7.75\)
✓ 1
Part (c)
M1 substitute \(x=5\) into the bisector of AC
✓ 1
A1 V(5, 5.25)
✓ 1
A1 distance \(=\sqrt{3^2+2.25^2}=3.75\) km
✓ 1
Exam tip: Every point on a Voronoi edge is equidistant from the two sites; the vertex is equidistant from all three, so check by finding its distance to a third site.
261IBMI3H-3-05r44-02Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium6 marks
A triangular shade sail is fixed above the outdoor seating of a shop. Relative to a corner of the shop, the three fixing points are A(1, 0, 2), B(4, 1, 6) and C(2, 5, 3), with all coordinates in metres.
(a)
Write down the vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\).
(2)
(b)
Use a vector product to find the area of the sail, giving your answer to 3 significant figures.
(4)
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Part (a)
A1 \(\overrightarrow{AB}\) = (3, 1, 4)
✓ 1
A1 \(\overrightarrow{AC}\) = (1, 5, 1)
✓ 1
Part (b)
M1 attempt to find \(\overrightarrow{AB}\) × \(\overrightarrow{AC}\) (ECF from (a))
M1 area = ½|\(\overrightarrow{AB}\) × \(\overrightarrow{AC}\)|
✓ 1
A1 area = 11.8 m2
✓ 1
Common error: The vector product is not commutative and the order of subtraction matters: a slip in a single component changes the magnitude, so check each component.
262IBMI3H-3-05r44-03Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
A drone carries test kits across a field hospital site. Using a map with the x-axis pointing east and the y-axis pointing north (distances in metres), its position t seconds after leaving the launch point O is given by r = (2t, 16t − t2), for 0 ≤ t ≤ 16.
(a)
Find the velocity of the drone when t = 3 and hence its speed at that moment, giving the speed to 3 significant figures.
(3)
(b)
Find the value of t when the drone is moving due east.
(2)
(c)
Find the distance of the drone from O at this time, giving your answer to 3 significant figures.
(2)
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Part (a)
M1 differentiate each component of r
✓ 1
A1 v = (2, 16 − 2t) so v(3) = (2, 10)
✓ 1
A1 speed = √104 = 10.2 m s−1
✓ 1
Part (b)
M1 set the north component of velocity to zero: 16 − 2t = 0
✓ 1
A1 t = 8 s
✓ 1
Part (c)
M1 r(8) = (16, 64) (ECF from (b)) and use of magnitude
✓ 1
A1 66.0 m
✓ 1
Exam tip: Moving due east means the velocity has no component in the north direction, not that the position has none.
263IBMI3H-3-05r44-07Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
A courier plans a delivery route along the roads between five junctions A, B, C, D and E. The road lengths in km are AB = 4, BC = 6, CD = 5, DA = 7, AC = 8 and DE = 3. The courier must travel along every road at least once, starting and finishing at A.
(a)
State which junctions have odd degree and explain why the courier cannot use each road exactly once.
(2)
(b)
List the three ways of pairing the odd vertices and find the shortest repeated distance for each pairing.
(3)
(c)
Hence find the length of the shortest route that meets the courier's requirements.
(2)
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Part (a)
A1 A, C, D and E have odd degree (A, C, D have degree 3 and E has degree 1)
✓ 1
R1 an Eulerian circuit exists only if every vertex has even degree, so repeating some roads is unavoidable
✓ 1
Part (b)
A1 AC + DE = 8 + 3 = 11
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A1 AD + CE = 7 + 8 = 15 (shortest C to E is via D)
✓ 1
A1 AE + CD = 10 + 5 = 15 (shortest A to E is via D)
✓ 1
Part (c)
M1 total of all roads (33) added to the smallest pairing (ECF from (b))
✓ 1
A1 44 km
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Exam tip: When pairing odd vertices, use the shortest path between each pair, which may pass through other vertices rather than be a single edge.
264IBMI3H-3-05r44-10Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
Wind-farm planning software maps site coordinates using two successive transformations. First, the map is rotated 90° clockwise about the origin. Then it is stretched by scale factor 3 parallel to the y-axis (the x-coordinates are unchanged).
(a)
Write down the matrix R for the rotation and the matrix S for the stretch.
(2)
(b)
Find the single matrix M that represents the rotation followed by the stretch.
(2)
(c)
A turbine at the point (4, 2) on the original map is transformed by M. Find its image.
(1)
(d)
A turbine appears at (5, −6) on the transformed map. Find its coordinates on the original map.
M1 sine rule to find angle BT1T2: sin α ÷ 11.6 = sin 83° ÷ T1T2
✓ 1
A1 α = 58.8°
✓ 1
M1 bearing of B from T1 is 062° + 180° = 242°, and T2 lies at 242° − α from north
✓ 1
A1 183°
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Exam tip: A back-bearing is the original bearing plus or minus 180°. Sketch a north line at each point before choosing which angle to add or subtract.
266IBMI3H-3-08r44-03Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium8 marks
A theatre uses a flat triangular panel as part of a stage set. Relative to a corner of the stage, the vertices of the panel are A(1, 2, 0), B(5, 2, 1) and C(2, 6, 2), with coordinates in metres and the z-axis vertical.
(a)
Write down the vectors AB and AC.
(2)
(b)
Find AB × AC.
(2)
(c)
Find the area of the panel in m² to 3 significant figures.
(2)
(d)
Find the acute angle between the panel and the horizontal floor (the plane z = 0), in degrees to 3 significant figures.
(2)
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Part (a)
A1 AB = (4, 0, 1)
✓ 1
A1 AC = (1, 4, 2)
✓ 1
Part (b)
M1 attempt to use the vector product formula (correct method with at least two components correct)
✓ 1
A1 (−4, −7, 16)
✓ 1
Part (c)
M1 area = ½|AB × AC| (ECF from (b))
✓ 1
A1 8.96 m² (√321 / 2)
✓ 1
Part (d)
M1 attempt to use the angle between the normal (−4, −7, 16) and the vertical (0, 0, 1), i.e. cos θ = 16/√321 (ECF from (b))
✓ 1
A1 26.7°
✓ 1
Exam tip: The vector product gives a vector perpendicular to the plane of the two vectors; the angle between two planes is the angle between their normals.
267IBMI3H-3-08r44-05Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
Two laser beams in a theatre are modelled by the straight lines L1: r = (1, 2, 3) + s(2, −1, 1) L2: r = (4, −1, 2) + t(1, 1, 3) where coordinates are in metres and s and t are parameters.
(a)
Show that the beams cross, and find the coordinates of the point where they cross.
(4)
(b)
Find the acute angle between the two beams, in degrees to 3 significant figures.
(3)
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Part (a)
M1 equate the components of the two lines: 1 + 2s = 4 + t, 2 − s = −1 + t, 3 + s = 2 + 3t
✓ 1
M1 attempt to solve two of the equations simultaneously
✓ 1
A1 s = 2 and t = 1
✓ 1
R1 the third equation is satisfied (5 = 5), so the lines intersect; point (5, 0, 5)
✓ 1
Part (b)
M1 attempt to use the scalar product of the direction vectors (2, −1, 1) and (1, 1, 3)
✓ 1
A1 cos θ = 4/(√6 × √11)
✓ 1
A1 60.5°
✓ 1
Exam tip: Two lines in 3D can intersect, be parallel or be skew. Solving two component equations is not enough: always check the third equation.
268IBMI3H-3-08r44-07Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
A theatre complex has six buildings, A to F. Electrical cable must be laid so that every building is connected to the network, using as little cable as possible. The possible cable routes and their lengths in metres are: A–B 12, A–C 9, A–D 15, B–C 7, B–E 14, C–D 8, C–F 11, D–E 13, D–F 10, E–F 6.
(a)
Use Kruskal's algorithm to find the minimum spanning tree. List the edges in the order in which they are chosen, and state the first edge that is rejected because it forms a cycle.
(4)
(b)
State the total length of cable required, in metres.
(1)
(c)
The cable costs €18.50 per metre to install. Find the total cost of the cable for this network.
(2)
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Part (a)
M1 attempt to order the edges by length and add the shortest edges that do not form a cycle
✓ 1
A1 E–F, B–C and C–D chosen in that order
✓ 1
A1 A–C (length 9) and D–F (length 10) chosen next
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A1 C–F (length 11) is the first edge rejected, as it forms a cycle C–D–F
✓ 1
Part (b)
A1 40 m
✓ 1
Part (c)
M1 40 × 18.50 (ECF from (b))
✓ 1
A1 €740
✓ 1
Common error: In Kruskal's algorithm an edge is rejected only if it joins two vertices already connected; do not reject an edge just because it is long.
269IBMI3H-3-08r44-13Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
A volcano is modelled as a solid cone with a flat top. The complete cone would have a base of radius 1500 m and a height of 600 m. The top 150 m of this cone is removed to form a flat summit, so that the volcano is a cone with its apex cut off parallel to the base.
(a)
Find the angle between the sloping side of the volcano and the horizontal ground, in degrees to 3 significant figures.
(2)
(b)
Find the radius of the flat summit.
(2)
(c)
Find the volume of the volcano, in m³, giving your answer in the form a × 10k with 1 ≤ a < 10 and a to 3 significant figures.
(3)
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Part (a)
M1 attempt to use tan θ = 600/1500
✓ 1
A1 21.8°
✓ 1
Part (b)
M1 attempt to use similar triangles: radius/1500 = 150/600
✓ 1
A1 375 m
✓ 1
Part (c)
M1 volume of the complete cone: ⅓ × π × 1500² × 600
✓ 1
M1 subtract the volume of the removed cone: ⅓ × π × 375² × 150 (ECF from (b))
✓ 1
A1 1.39 × 109 m³
✓ 1
Common error: The removed top is a smaller cone similar to the whole: its radius is the same fraction of the base radius as its height is of the full height, not the same length as the height.
270IBMI3H-3-08r44-14Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
Three volcano monitoring stations are at S1(2, 6), S2(8, 8) and S3(3, 1) on a map, with distances in km. A Voronoi diagram is constructed using these three stations as sites. An emergency shelter is to be placed at the point that is the same distance from all three stations.
(a)
Find the equation of the perpendicular bisector of S1S3, giving your answer in the form y = mx + c.
(3)
(b)
The perpendicular bisector of S1S2 has equation y = −3x + 22. Find the coordinates of the Voronoi vertex, to 3 significant figures.
(2)
(c)
Find the distance from the shelter to the nearest station, in km to 3 significant figures. Use the unrounded coordinates of the vertex.
(2)
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Part (a)
M1 midpoint (2.5, 3.5) and gradient of S1S3 = −5
✓ 1
M1 perpendicular gradient = 1/5 and attempt to substitute the midpoint
✓ 1
A1 y = 0.2x + 3
✓ 1
Part (b)
M1 equate the two equations: 0.2x + 3 = −3x + 22 (ECF from (a))
✓ 1
A1 (5.94, 4.19)
✓ 1
Part (c)
M1 attempt to use the distance formula between (5.9375, 4.1875) and a station
✓ 1
A1 4.33 km (the three distances are equal)
✓ 1
Exam tip: A Voronoi vertex is where three perpendicular bisectors meet, so it is equidistant from the three nearest sites. Use any one of them to find that distance, then check with a second.
271IBMI3H-3-07r46-03Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium6 marks
A triangular glass panel in a building facade has corners at A(1, 0, 2), B(4, 2, 3) and C(2, 5, 1), where the coordinates are in metres relative to a fixed origin.
(a)
Write down the vectors AB and AC.
(2)
(b)
Find AB × AC.
(2)
(c)
Hence find the area of the panel in m2. Give your answer to 3 significant figures.
(2)
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Part (a)
A1 AB = (3, 2, 1)
✓ 1
A1 AC = (1, 5, −1)
✓ 1
Part (b)
M1 attempt at the vector product, e.g. (2×(−1) − 1×5, 1×1 − 3×(−1), 3×5 − 2×1)
✓ 1
A1 AB × AC = (−7, 4, 13) (ECF from (a) if consistent)
✓ 1
Part (c)
M1 area = ½|AB × AC| = ½√((−7)² + 4² + 13²) (ECF from (b))
✓ 1
A1 area = ½√234 = 7.65 m2
✓ 1
Common error: The magnitude of the vector product gives the area of the parallelogram; remember to halve it for a triangle.
272IBMI3H-3-07r46-05Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium6 marks
Five islands A, B, C, D and E are linked by ferry routes. The direct routes are AB, AC, BC, BD, CE and DE, and each route can be sailed in either direction.
(a)
Write down the adjacency matrix M for the ferry network, using the order A, B, C, D, E.
(2)
(b)
Use M3 to find the number of different journeys of exactly three crossings that start at B and finish at B.
(3)
(c)
List the journeys found in (b), using the letters of the islands.
(1)
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Part (a)
M1 attempt at a 5 × 5 matrix with entries 0 or 1 that is symmetric
✓ 1
A1 rows: A (0 1 1 0 0), B (1 0 1 1 0), C (1 1 0 0 1), D (0 1 0 0 1), E (0 0 1 1 0)
✓ 1
Part (b)
M1 attempt to calculate M3 (accept GDC)
✓ 1
M1 identify the entry in row B, column B
✓ 1
A1 the number of journeys is 2
✓ 1
Part (c)
A1 B–A–C–B and B–C–A–B (both needed; accept B–A–C–B reversed)
✓ 1
Exam tip: The (i, j) entry of the nth power of an adjacency matrix counts walks of length n from vertex i to vertex j; walks may repeat edges.
273IBMI3H-3-07r46-09Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium6 marks
In an architectural design, a decorative floor tile pattern is rotated about the origin O through 35° anticlockwise. The rotation is represented by a 2 × 2 matrix R. A point P on the pattern has coordinates (6, 2), in metres.
(a)
Write down R in terms of cos 35° and sin 35°, and evaluate each entry to 3 significant figures.
(2)
(b)
Find the coordinates of the image of P after the rotation. Give each coordinate to 3 significant figures, using unrounded values of the matrix entries.
(2)
(c)
Show that the distance from O to the image of P is the same as the distance OP, and state this distance to 3 significant figures.
(2)
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Part (a)
A1 R = [[cos 35°, −sin 35°],[sin 35°, cos 35°]]
✓ 1
A1 [[0.819, −0.574],[0.574, 0.819]]
✓ 1
Part (b)
M1 attempt to multiply R by the column vector (6, 2)
✓ 1
A1 (3.77, 5.08)
✓ 1
Part (c)
R1 a rotation about O preserves distances from O (or √(3.768² + 5.079²) = √40, using unrounded values)
✓ 1
A1 OP = √40 = 6.32 m
✓ 1
Exam tip: For calculator work with trigonometric matrices, store the exact values rather than retyping rounded entries.
274IBMI3H-3-07r46-13Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
The glass roof of a museum entrance is a right pyramid VABCD. The base ABCD is a horizontal square of side 6.4 m and the apex V is 4.8 m vertically above the centre M of the base.
(a)
Find the length of the diagonal AC. Give your answer in metres to 3 significant figures.
(2)
(b)
Find the angle between the sloping edge VA and the base. Give your answer in degrees to 3 significant figures.
(3)
(c)
Let N be the midpoint of AB. Find the angle between the triangular face VAB and the base. Give your answer in degrees to 3 significant figures.
(2)
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Part (a)
M1 Pythagoras, AC² = 6.4² + 6.4²
✓ 1
A1 AC = 9.05 m
✓ 1
Part (b)
M1 AM = ½AC = 4.5255…
✓ 1
M1 tan(angle VAM) = 4.8/AM (ECF from (a))
✓ 1
A1 angle = 46.7°
✓ 1
Part (c)
M1 identify angle VNM with MN = 3.2 m, so tan(angle VNM) = 4.8/3.2
✓ 1
A1 angle = 56.3°
✓ 1
Exam tip: To find the angle between a plane and a line or another plane, draw a right-angled triangle that includes the vertical height and a line in the base.
275IBMI3H-3-07r46-14Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
Three delivery depots are located at A(2, 1), B(8, 3) and C(5, 9), where the coordinates are in kilometres. A Voronoi diagram is drawn for the sites A, B and C. A parcel locker is to be placed at the Voronoi vertex, which is equidistant from all three depots.
(a)
Find the equation of the perpendicular bisector of AB, giving your answer in the form y = mx + c.
(3)
(b)
The perpendicular bisector of BC has equation y = 0.5x + 2.75. Find the coordinates of the Voronoi vertex. Give each coordinate to 3 significant figures.
(2)
(c)
Find the distance from the parcel locker to depot A. Give your answer in kilometres to 3 significant figures.
(2)
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Part (a)
M1 midpoint of AB is (5, 2)
✓ 1
M1 gradient of AB is 1/3, so the perpendicular gradient is −3
Exam tip: Every Voronoi vertex is the centre of a circle through three sites, so its distance to those sites is the same.
276IBMI3H-3-06r47-07Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium8 marks
A drug-carrying nanoparticle moves in a blood vessel. Its velocity at time t seconds is \(\mathbf{v}(t)=\begin{pmatrix}6t-t^2\\2e^{-0.5t}\end{pmatrix}\) mm s−1, for \(t\ge0\). When \(t=0\) its position vector is \(\begin{pmatrix}1\\2\end{pmatrix}\) mm.
(a)
Find the acceleration of the nanoparticle when \(t=3\). Give the magnitude to 3 significant figures.
(2)
(b)
Find the speed of the nanoparticle when \(t=3\), to 3 significant figures.
(2)
(c)
Find the position vector of the nanoparticle when \(t=4\). Give each component to 3 significant figures.
A1 \(\mathbf{a}(3)=\begin{pmatrix}0\\-0.223\end{pmatrix}\), magnitude \(0.223\) mm s−2
✓ 1
Part (b)
M1 \(\sqrt{9^2+(2e^{-1.5})^2}\)
✓ 1
A1 \(9.01\) mm s−1
✓ 1
Part (c)
M1 integrate the x-component and use the initial position, \(x=1+3t^2-\frac{t^3}{3}\)
✓ 1
M1 integrate the y-component and use the initial position, \(y=2+4(1-e^{-0.5t})\)
✓ 1
A1 \(x=27.7\)
✓ 1
A1 \(y=5.46\)
✓ 1
Exam tip: When integrating a velocity vector to find position, add a constant vector found from the initial position, not a single constant.
277IBMI3H-3-06r47-08Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
Six villages A, B, C, D, E and F are to be connected to one water supply system using pipes laid along some of the routes below. The cost of each route is in thousands of euros. AB 12, AC 9, BC 7, BD 15, CD 11, CE 14, DE 8, DF 13, EF 10, AF 18.
(a)
Use Kruskal's algorithm to find a minimum spanning tree and its total cost, showing the edges in the order chosen.
(4)
(b)
Route BC becomes unavailable. Find the minimum cost of connecting all six villages using the remaining routes.
(3)
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Part (a)
M1 edges considered in increasing order of cost, rejecting any that make a cycle
✓ 1
A1 BC (7), DE (8), AC (9) chosen
✓ 1
A1 EF (10), CD (11) chosen, and no further edges needed (5 edges for 6 vertices)
✓ 1
A1 total cost €45 000
✓ 1
Part (b)
M1 apply the algorithm again without BC
✓ 1
A1 chosen edges DE, AC, EF, CD, AB
✓ 1
A1 total cost €50 000
✓ 1
Exam tip: A spanning tree on n vertices has exactly \(n-1\) edges, which is a useful check on your list.
278IBMI3H-3-06r47-11Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
A water storage tank is a cuboid with a horizontal rectangular base 12 m by 5 m and height 4 m. A straight pipe runs inside the tank from a corner A on the base to the opposite corner G on the roof, diagonally across the tank.
(a)
Find the length of the pipe, to 3 significant figures.
(2)
(b)
Find the angle the pipe makes with the base of the tank, to 3 significant figures.
(3)
(c)
The tank contains water to a depth of 3 m. A pump removes water at a constant rate of 0.9 m3 per minute. Find the time taken to empty the tank, in minutes.
(2)
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Part (a)
M1 \(\sqrt{12^2+5^2+4^2}\)
✓ 1
A1 \(\sqrt{185}=13.6\) m
✓ 1
Part (b)
M1 base diagonal \(\sqrt{12^2+5^2}=13\) m
✓ 1
M1 \(\tan\theta=\frac{4}{13}\)
✓ 1
A1 \(17.1^\circ\)
✓ 1
Part (c)
M1 volume \(=12\times5\times3=180\) m3
✓ 1
A1 \(\frac{180}{0.9}=200\) minutes
✓ 1
Common error: The angle with the base is found using the diagonal of the base, not one side of it. Sketch the right-angled triangle before using trigonometry.
279IBMI3H-3-06r47-12Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium8 marks
Three wells, A(2, 2), B(10, 4) and C(4, 10), supply a rural area. The coordinates are in km and the area is divided into regions by a Voronoi diagram, so that each point is in the region of its nearest well.
(a)
Find the equation of the perpendicular bisector of AB.
(3)
(b)
The perpendicular bisector of BC is \(y=x\). Find the coordinates of the Voronoi vertex where the three regions meet.
(2)
(c)
A new village is to be built at the Voronoi vertex. Find its distance from well A, to 3 significant figures, and state what this tells you about its distance from B and C.
(3)
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Part (a)
M1 midpoint of AB is (6, 3) and gradient of AB is \(\frac{1}{4}\)
✓ 1
M1 gradient of the perpendicular bisector is \(-4\)
✓ 1
A1 \(y=-4x+27\)
✓ 1
Part (b)
M1 solve \(x=-4x+27\) (ECF from (a))
✓ 1
A1 \((5.4,\,5.4)\)
✓ 1
Part (c)
M1 \(\sqrt{(5.4-2)^2+(5.4-2)^2}\)
✓ 1
A1 \(4.81\) km
✓ 1
R1 the vertex is equidistant from A, B and C, so the distance to each well is also 4.81 km
✓ 1
Exam tip: A Voronoi vertex is equidistant from three sites. Check your vertex by finding its distance from each of the three sites.
280IBMI3H-3-06r48-03Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
Calculator
Two helicopters, A and B, fly on straight paths over flat ground. Relative to a control tower at the origin, with distances in km, their position vectors \(t\) hours after noon are \(\mathbf{r}_A = \binom{0}{2} + t\binom{2}{1}\) and \(\mathbf{r}_B = \binom{10}{-1} + t\binom{-1}{2}\).
(a)
Find the coordinates of the point where the two paths cross.
(4)
(b)
Find, in minutes, the difference between the times at which the two helicopters reach this point.
(2)
(c)
State, with a reason, whether the helicopters collide.
(1)
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Part (a)
M1 equate the position vectors using different parameters, for example \(t\) for A and \(s\) for B
✓ 1
M1 form and attempt to solve \(2t = 10 - s\) and \(2 + t = -1 + 2s\)
✓ 1
A1 \(t = 3.4\) and \(s = 3.2\)
✓ 1
A1 \((6.8, 5.4)\)
✓ 1
Part (b)
M1 attempt to find \(3.4 - 3.2 = 0.2\) hours
✓ 1
A1 \(12\) minutes
✓ 1
Part (c)
R1 no collision, because they reach the crossing point 12 minutes apart (at different times)
✓ 1
Common error: Two paths crossing does not mean a collision; use a separate parameter for each object, and then compare the times.
281IBMI3H-3-06r48-04Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium7 marks
Calculator
A triangular acoustic panel in a recording studio has corners at \(O(0, 0, 0)\), \(P(2, 1, 3)\) and \(Q(1, 4, -2)\), where the coordinates are in metres.
M1 attempt to use \(\frac{1}{2}\left|\overrightarrow{OP}\times\overrightarrow{OQ}\right| = \frac{1}{2}\sqrt{294}\)
✓ 1
A1 \(8.57\) m2
✓ 1
Part (c)
M1 attempt to find \(8.5732\ldots \times 38\)
✓ 1
A1 €326
✓ 1
Exam tip: The magnitude of the vector product is the area of the parallelogram; halve it for the triangle.
282IBMI3H-3-06r48-07Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium6 marks
Calculator
A tanker collects milk from four farms A, B, C and D. The roads, with their lengths in km, are AB 5, BC 6, CD 4, DA 7 and AC 9. The tanker must travel along every road at least once, starting and finishing at A.
(a)
Show that exactly two vertices have odd degree, and state which they are.
(2)
(b)
Find the shortest additional distance that must be repeated so that a closed route is possible.
(2)
(c)
Find the length of the shortest route.
(2)
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Part (a)
M1 degrees found: A 3, B 2, C 3, D 2
✓ 1
A1 the odd vertices are A and C
✓ 1
Part (b)
M1 compare AC (9) with A–B–C (11) and A–D–C (11)
✓ 1
A1 repeat AC, \(9\) km
✓ 1
Part (c)
M1 attempt to add the total of all roads, \(5+6+4+7+9 = 31\), to the repeated distance
✓ 1
A1 \(40\) km
✓ 1
Common error: Only the odd-degree vertices need to be paired; do not repeat edges between even-degree vertices.
283IBMI3H-3-06r48-11Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium8 marks
Calculator
An aircraft flies 180 km from A to B on a bearing of 060°. It then flies 120 km from B to C on a bearing of 140°.
(a)
Show that angle \(ABC\) is \(100°\).
(2)
(b)
Find the distance \(AC\), correct to 3 significant figures.
(3)
(c)
Find the bearing of A from C, correct to the nearest 0.1°. Use the unrounded value of \(AC\).
(3)
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Mark
Notes
Part (a)
M1 use of the bearing of A from B, which is \(240°\) (or co-interior angles)
M1 attempt to use the sine rule, \(\sin\angle ACB = \frac{180\sin 100°}{AC}\)
✓ 1
A1 \(\angle ACB = 49.5°\)
✓ 1
A1 bearing \(= 320° - 49.5° = 270.5°\)
✓ 1
Common error: Draw a north line at each point; the angle at B in the triangle comes from the bearing of A from B, not from the bearing of B from A.
284IBMI3H-3-06r48-12Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium8 marks
Calculator
A music practice room is a cuboid of length 10 m, width 7 m and height 3.5 m. A speaker is in one bottom corner and a microphone is in the opposite top corner. Take the speed of sound as 343 m s−1.
(a)
Find the length of the diagonal of the floor, correct to 3 significant figures.
(2)
(b)
Find the distance from the speaker to the microphone, correct to 3 significant figures.
(2)
(c)
Find the angle between the line from the speaker to the microphone and the floor, correct to 3 significant figures.
(2)
(d)
Find the time taken by sound to travel from the speaker to the microphone, correct to 3 significant figures.
(2)
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Mark
Notes
Part (a)
M1 attempt to use \(\sqrt{10^2 + 7^2}\)
✓ 1
A1 \(12.2\) m
✓ 1
Part (b)
M1 attempt to use \(\sqrt{12.2066^2 + 3.5^2}\) or \(\sqrt{10^2+7^2+3.5^2}\)
✓ 1
A1 \(12.7\) m
✓ 1
Part (c)
M1 attempt to use \(\tan^{-1}\left(\frac{3.5}{12.2066}\right)\)
✓ 1
A1 \(16.0°\)
✓ 1
Part (d)
M1 attempt to use \(\frac{12.698\ldots}{343}\)
✓ 1
A1 \(0.0370\) s
✓ 1
Exam tip: For a space diagonal, find the floor diagonal first, then use a second right-angled triangle with the height.
285IBMI3H-3-06r48-14Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Medium8 marks
Calculator
Three wells A, B and C on a farm are at the points \(A(0, 0)\), \(B(8, 0)\) and \(C(4, 6)\), with coordinates in km. The farm is divided into Voronoi cells, one for each well, showing the area closest to each well.
(a)
Find the equation of the perpendicular bisector of AB.
(1)
(b)
Find the equation of the perpendicular bisector of AC.
(3)
(c)
Find the coordinates of the Voronoi vertex where these two bisectors meet.
(2)
(d)
Find the distance from this vertex to well A, correct to 3 significant figures.
(2)
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Notes
Part (a)
A1 \(x = 4\)
✓ 1
Part (b)
M1 midpoint of AC is \((2, 3)\)
✓ 1
M1 gradient of AC is \(\frac{3}{2}\), so the perpendicular gradient is \(-\frac{2}{3}\)
M1 substitute \(x = 4\) into the equation from (b)
✓ 1
A1 \(\left(4, \frac{5}{3}\right)\)
✓ 1
Part (d)
M1 attempt to use \(\sqrt{4^2 + \left(\frac{5}{3}\right)^2}\)
✓ 1
A1 \(4.33\) km
✓ 1
Exam tip: A Voronoi vertex is the same distance from the three nearest sites, so you can check your vertex by finding its distance to all three.
286IBMI3H-3-03r15-05Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Hard7 marks
Two camera drones fly over a stadium. Relative to a fixed point on the pitch, the position of drone P is \(\mathbf{r}_P = \begin{pmatrix}0\\0\\10\end{pmatrix} + t\begin{pmatrix}3\\2\\1\end{pmatrix}\) and the position of drone Q is \(\mathbf{r}_Q = \begin{pmatrix}30\\10\\0\end{pmatrix} + t\begin{pmatrix}-1\\1\\2\end{pmatrix}\), where distances are in metres and t is the time in seconds, \(0 \leq t \leq 10\).
(a)
Show that \(\overrightarrow{PQ} = \begin{pmatrix}30-4t\\10-t\\t-10\end{pmatrix}\).
(2)
(b)
Hence show that the square of the distance between the drones is \(18t^2 - 280t + 1100\).
(2)
(c)
Find the minimum distance between the drones, and the time at which it occurs. Give both answers to 3 significant figures.
(3)
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Notes
Part (a)
M1 \(\mathbf{r}_Q - \mathbf{r}_P\) attempted
✓ 1
A1 components \(30 - t - 3t\), \(10 + t - 2t\), \(2t - 10 - t\) simplified to the given vector (AG)
✓ 1
Part (b)
M1 \((30-4t)^2 + (10-t)^2 + (t-10)^2\)
✓ 1
A1 expansion and collection of terms to the given result (AG)
✓ 1
Part (c)
M1 use of GDC or of \(t = \tfrac{280}{36}\) to minimise \(18t^2-280t+1100\)
✓ 1
A1 \(t = 7.78\) s
✓ 1
A1 minimum distance \(\sqrt{11.1\ldots} = 3.33\) m
✓ 1
Exam tip: Minimise the squared distance first; the square root is increasing, so the minimum occurs at the same time.
287IBMI3H-3-03r15-09Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Hard8 marks
A rail engineer must visit five depots P, Q, R, S and T, starting and ending at P, visiting each other depot exactly once. The distances in kilometres between depots are given below. PQ 14, PR 22, PS 17, PT 26, QR 12, QS 19, QT 21, RS 15, RT 18, ST 11.
(a)
Use the nearest neighbour algorithm starting at P to find an upper bound for the length of the shortest route.
(3)
(b)
By deleting depot T, find a lower bound for the length of the shortest route.
(4)
(c)
The route P–Q–R–T–S–P has length 72 km. Write down an inequality for the length L of the shortest route using your answers to (a) and (b), and the information in this part, giving the tightest bounds available.
(1)
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Notes
Part (a)
M1 repeatedly moving to the nearest unvisited depot
✓ 1
A1 route P–Q–R–S–T–P
✓ 1
A1 upper bound 78 km
✓ 1
Part (b)
M1 minimum spanning tree of P, Q, R, S
✓ 1
A1 tree edges QR 12, PQ 14, RS 15, length 41
✓ 1
M1 add the two shortest edges from T: 11 and 18
✓ 1
A1 lower bound 70 km
✓ 1
Part (c)
A1 \(70 \leq L \leq 72\) (ECF from (b))
✓ 1
Common error: A lower bound uses a minimum spanning tree of the remaining vertices plus the two shortest edges to the deleted vertex; do not add just one edge.
288IBMI3H-3-03r15-12Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Hard7 marks
A transformation of the plane is represented by a \(2\times2\) matrix M. It maps the point (1, 0) to (3, 1) and the point (0, 1) to (−1, 2). Coordinates are in kilometres on a rail-network map.
(a)
Write down the matrix M.
(2)
(b)
A triangular region has area 12 km2. Find the area of its image.
(3)
(c)
The image of a point X is (7, 0). Find the coordinates of X.
(2)
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Notes
Part (a)
M1 columns of M are the images of (1, 0) and (0, 1)
M1 use of \(\mathbf{M}^{-1} = \frac{1}{7}\begin{pmatrix}2&1\\-1&3\end{pmatrix}\) applied to \(\begin{pmatrix}7\\0\end{pmatrix}\) (or solving simultaneous equations)
✓ 1
A1 \(X = (2, -1)\)
✓ 1
Exam tip: If an image is known, multiply by the inverse matrix on the left of the image vector.
289IBMI3H-3-05r31-14Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Hard8 marks
A tall ash plume rises vertically above a point P on level ground, with top T. Two observers stand at A and B, which are 520 m apart. Angle PAB = 72° and angle PBA = 58°. From A the angle of elevation of T is 28°.
(a)
Find angle APB.
(1)
(b)
Find the distance AP, to the nearest metre.
(2)
(c)
Find the height PT of the plume, to the nearest metre. Use the unrounded value of AP.
(2)
(d)
Find the angle of elevation of T from B. Use unrounded values and give your answer to the nearest 0.1°.
(3)
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Part (a)
A1 180° − 72° − 58° = 50°
✓ 1
Part (b)
M1 sine rule: AP ÷ sin 58° = 520 ÷ sin 50°
✓ 1
A1 AP = 576 m
✓ 1
Part (c)
M1 PT = AP × tan 28°
✓ 1
A1 306 m
✓ 1
Part (d)
M1 sine rule: BP = 520 × sin 72° ÷ sin 50° = 645.6 m
✓ 1
M1 tan(elevation) = PT ÷ BP
✓ 1
A1 25.4°
✓ 1
Exam tip: Work in two stages: find a horizontal distance in the ground triangle, then use a right-angled triangle for the height. Keep unrounded intermediate values so the final angle is not affected by early rounding.
290IBMI3H-3-06r41-12Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Hard8 marks
A drone photographs a rectangular field of area 600 m2. Processing software maps each point (x, y) of the image to a new position using the matrix M = [[k, 2], [1, 3]], where k is a constant.
(a)
Write down an expression for det M in terms of k.
(1)
(b)
The processed image of the field has area 1500 m2. Find the two possible values of k.
(3)
(c)
Take k = 1.5. Find the image of the point (10, 20).
(2)
(d)
Find the value of k for which the transformation is not invertible, and explain what happens to the image of the field.
(2)
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Part (a)
A1 3k − 2
✓ 1
Part (b)
M1 area scale factor = |det M|, so 600 |3k − 2| = 1500
✓ 1
M1 3k − 2 = 2.5 or 3k − 2 = −2.5
✓ 1
A1 k = 1.5 and k = −1/6 (accept −0.167)
✓ 1
Part (c)
M1 M(10, 20)T
✓ 1
A1 (55, 70)
✓ 1
Part (d)
A1 k = 2/3
✓ 1
R1 the area scale factor is zero, so the image collapses to a line segment (or a point), and the original cannot be recovered
✓ 1
Exam tip: The area of the image is the original area multiplied by the absolute value of the determinant; solving |expression| = c gives two cases.
291IBMI3H-3-07r41-05Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Hard7 marks
An inspection team must check five bridge sites A, B, C, D and E. The roads between sites, with their lengths in km, are: AB 5, AC 7, AE 8, BC 4, BD 9, CD 6, CE 10, DE 5. The team must drive along every road at least once, starting and finishing at the same site, and wants the shortest total distance.
(a)
State the vertices of odd degree.
(1)
(b)
Show that the three ways of pairing up the odd vertices need extra distances of 10 km, 26 km and 17 km, and state which pairing should be used. You may use the shortest path between each pair of odd vertices.
(4)
(c)
Calculate the length of the shortest route.
(2)
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Part (a)
A1 A, B, D and E
✓ 1
Part (b)
M1 consider the three pairings AB&DE, AD&BE, AE&BD
✓ 1
A1 shortest paths AB 5, DE 5, AD 13, BE 13, AE 8, BD 9
✓ 1
A1 totals 10, 26 and 17 km
✓ 1
R1 AB and DE pairing chosen as it has the smallest total
✓ 1
Part (c)
M1 sum of all road lengths (54) plus the minimum extra distance
✓ 1
A1 64 km
✓ 1
Exam tip: Check the shortest path between each pair of odd vertices, since the direct road is not always the shortest connection.
292IBMI3H-3-07r41-06Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Hard8 marks
A space probe starts at asteroid A, must visit each of the asteroids B, C, D and E exactly once, and then return to A. Distances between asteroids in millions of km are given. AB 14, AC 22, AD 18, AE 25, BC 12, BD 20, BE 16, CD 15, CE 19, DE 11.
(a)
Use the nearest neighbour algorithm starting at A to find an upper bound for the shortest route. State the route and its length.
(3)
(b)
Use the deleted vertex algorithm, deleting A, to find a lower bound for the shortest route.
(4)
(c)
A mission planner claims that a tour of 68 million km exists. Use your answers to comment on this claim.
(1)
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Part (a)
M1 apply nearest neighbour from A
✓ 1
A1 route A–B–C–D–E–A
✓ 1
A1 length \(14+12+15+11+25=77\) million km
✓ 1
Part (b)
M1 delete A and find a minimum spanning tree on B, C, D, E
✓ 1
A1 tree DE, BC, CD with weight 38
✓ 1
M1 add the two shortest edges from A (AB 14 and AD 18)
✓ 1
A1 lower bound 70 million km
✓ 1
Part (c)
R1 not possible, since the shortest tour is at least 70 million km
✓ 1
Common error: The deleted vertex lower bound adds the two shortest edges at the deleted vertex to the minimum spanning tree of the rest; it is not the tree weight alone.
293IBMI3H-3-07r41-10Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Hard8 marks
A space station S is at the point with position vector \(\begin{pmatrix}10\\4\\0\end{pmatrix}\) km. A probe P moves in a straight line with position vector \(\mathbf{r}=\begin{pmatrix}5\\-2\\8\end{pmatrix}+t\begin{pmatrix}1\\2\\-2\end{pmatrix}\) km, where \(t\) is the time in minutes after the probe passes a reference point.
(a)
Find the speed of the probe in km per minute.
(2)
(b)
Show that the square of the distance between P and S at time \(t\) is \(9t^2-66t+125\).
(3)
(c)
Find the minimum distance between the probe and the station, and the time at which it occurs. Give the time to 3 significant figures.
(3)
Show mark scheme
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Notes
Part (a)
M1 magnitude of the direction vector, \(\sqrt{1^2+2^2+(-2)^2}\)
✓ 1
A1 3 km per minute
✓ 1
Part (b)
M1 form \(\overrightarrow{SP}=\begin{pmatrix}t-5\\2t-6\\8-2t\end{pmatrix}\)
✓ 1
M1 sum of the squares of the components, \((t-5)^2+(2t-6)^2+(8-2t)^2\)
✓ 1
A1 expand and simplify to reach \(9t^2-66t+125\)
✓ 1
Part (c)
M1 minimise the quadratic, for example at \(t=\dfrac{66}{18}\) or by use of a GDC
✓ 1
A1 \(t=3.67\) minutes
✓ 1
A1 minimum distance \(=\sqrt{4}=2\) km
✓ 1
Exam tip: To minimise a distance, minimise its square: the minimum occurs at the same time and avoids differentiating a square root.
294IBMI3H-3-08r44-09Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 1Hard8 marks
A shipping company's vessel must visit five ports, A, B, C, D and E, starting and finishing at A and visiting each of the others exactly once. The sailing times between ports, in hours, are shown in the table. A–B 30, A–C 13, A–D 18, A–E 23, B–C 20, B–D 22, B–E 24, C–D 21, C–E 27, D–E 19.
(a)
Use the nearest neighbour algorithm starting at A to find an upper bound for the shortest time to complete the tour. State the tour and its length.
(3)
(b)
Delete port E and use the deleted vertex algorithm to find a lower bound for the shortest time to complete the tour.
(4)
(c)
Hence find the greatest possible difference, in hours, between the length of the tour from part (a) and the shortest possible tour.
(1)
Show mark scheme
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Notes
Part (a)
M1 attempt to move to the nearest unvisited port each time
✓ 1
A1 tour A–C–B–D–E–A
✓ 1
A1 13 + 20 + 22 + 19 + 23 = 97 hours
✓ 1
Part (b)
M1 attempt to find a minimum spanning tree of A, B, C and D
✓ 1
A1 A–C 13, A–D 18, B–C 20: total 51
✓ 1
M1 add the two shortest edges from E: E–D 19 and E–A 23
✓ 1
A1 lower bound 93 hours
✓ 1
Part (c)
A1 97 − 93 = 4 hours (ECF from (a) and (b))
✓ 1
Common error: The nearest neighbour algorithm gives an upper bound, and the deleted vertex algorithm a lower bound; do not mix them up, and remember the lower bound uses the two shortest edges to the deleted vertex.
295IBMI3H-3-05r38-13Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Easy9 marks
A climber walks 2.4 km from A to B on a bearing of 065°, then 2.0 km from B to C on a bearing of 140°.
(a)
Show that angle ABC = 105°.
(2)
(b)
Find the distance AC, giving your answer to 3 significant figures.
(3)
(c)
Find the bearing of C from A, giving your answer to 1 decimal place. Use unrounded values from earlier working.
Exam tip: Sketch the triangle and mark north at every vertex: the interior angle is found from the bearings, never from the bearings themselves.
296IBMI-032-04-r2-03Graph theory and networksPaper 2Medium13 marks
A gritting lorry treats the roads that connect five junctions, P, Q, R, S and T. The roads and their lengths in km are: PQ 4, PR 6, PS 5, QR 3, QT 7, RS 2, RT 4, ST 8. There are no other roads. The lorry must travel along every road at least once.
(a)
State, with a reason, whether the network has an Eulerian circuit.
(2)
(b)
Find the length of the shortest route from P to T.
(2)
(c)
The lorry is kept at junction R and must start and finish there. Use the Chinese postman algorithm to find the length of the shortest possible route. State which roads must be travelled twice.
(6)
(d)
On a different day, the lorry may start and finish at any two junctions. Find the length of the shortest route that covers every road, and state where it must start and finish.
(3)
Show mark scheme
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Part (a)
A1 for degrees P = 3, Q = 3, R = 4, S = 3, T = 3
✓ 1
R1 for: there is no Eulerian circuit, because some vertices (four of them) have odd degree
✓ 1
Part (b)
M1 for comparing possible routes, e.g. P–R–T, P–Q–T, P–S–R–T
✓ 1
A1 for 10 km via P–R–T
✓ 1
Part (c)
A1 for identifying the odd vertices P, Q, S, T
✓ 1
M1 for considering the three possible pairings
✓ 1
A1 for all three totals correct: PQ + ST = 4 + 6 = 10, PS + QT = 5 + 7 = 12, PT + QS = 10 + 5 = 15
✓ 1
A1 for total length of all roads = 39 km
✓ 1
A1 for minimum route length 39 + 10 = 49 km
✓ 1
A1 for roads repeated: PQ, and RS and RT (the shortest S–T path is S–R–T)
✓ 1
Part (d)
R1 for: start and finish at two of the odd vertices, and repeat the shortest path between the other two
✓ 1
M1 for choosing the smallest single shortest path between two odd vertices, PQ = 4
✓ 1
A1 for 39 + 4 = 43 km, starting at S and finishing at T (or the reverse)
✓ 1
Exam tip: When you pair odd vertices, use the shortest path between them, not just the direct road. Here the shortest path from S to T goes through R and is shorter than road ST.
297IBMI-032-04-r2-05Graph theory and networksPaper 2Medium10 marks
A small website has four pages, W, X, Y and Z. The one-way links between the pages are: W links to X and Y; X links to Z; Y links to W, X and Z; Z links to W. A visitor moves from page to page by clicking a link. On each page, the visitor picks one of that page's links at random, with each link equally likely.
(a)
Write down the transition matrix T. Each column represents the current page and each row the next page, both in the order W, X, Y, Z.
(2)
(b)
A visitor starts on page W. Find the probability that the visitor is on page Z after exactly three clicks.
(2)
(c)
Find the steady-state vector for T, giving each probability to 3 significant figures.
(4)
(d)
The website owner ranks the pages by their long-term popularity. Write down the ranking and interpret the value for page W in context.
M1 for calculating \(T^3\) multiplied by the column vector (1, 0, 0, 0)
✓ 1
A1 for \(\frac16\) ≈ 0.167
✓ 1
Part (c)
M1 for setting up Ts = s with the components summing to 1
✓ 1
M1 for solving the system, or for using a high power of T
✓ 1
A1 for W = 0.333 and Y = 0.167
✓ 1
A1 for X = 0.222 and Z = 0.278
✓ 1
Part (d)
A1 for the order W, Z, X, Y
✓ 1
R1 for: in the long run, about one third of a visitor's clicks land on page W
✓ 1
Exam tip: To check a steady-state vector, multiply it by T on your calculator. You should get the same vector back, and its entries should add up to 1.
298IBMI-032-08-r1-05Vectors and kinematicsPaper 2Medium11 marks
Two delivery drones fly in straight lines at constant velocity above a warehouse yard. Positions are in metres and t is the time in seconds after both drones are launched. Drone P: \(\mathbf{r}_P = \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix} + t\begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}\) Drone Q: \(\mathbf{r}_Q = \begin{pmatrix} -7 \\ 9 \\ 3 \end{pmatrix} + t\begin{pmatrix} 4 \\ -3 \\ 1 \end{pmatrix}\)
(a)
Show that the flight paths of the two drones intersect and find the coordinates of the point of intersection.
(4)
(b)
Explain why the drones do not collide.
(2)
(c)
Find the acute angle between the two flight paths.
(3)
(d)
Find the distance between the drones when t = 2.
(2)
Show mark scheme
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Notes
Part (a)
M1: using different parameters and equating, e.g. 1 + 2λ = −7 + 4μ, 2 − λ = 9 − 3μ
✓ 1
A1: λ = 2
✓ 1
A1: μ = 3
✓ 1
A1: third components both equal 6, so paths meet at (5, 0, 6)
✓ 1
Part (b)
R1: P reaches (5, 0, 6) at t = 2 but Q reaches it at t = 3
✓ 1
A1: they are never at the same point at the same time, so no collision
M1: Q at (1, 3, 5), P at (5, 0, 6), difference \(\begin{pmatrix} 4 \\ -3 \\ 1 \end{pmatrix}\)
✓ 1
A1: \(\sqrt{26}\) = 5.10 m
✓ 1
Common error: Using the same parameter t for both lines when testing whether paths cross. That tests for a collision, not an intersection; use two different parameters first.
299IBMI-032-08-r2-04Vectors and kinematicsPaper 2Medium12 marks
Two delivery drones, P and Q, fly in straight lines at constant velocity above a warehouse. Their position vectors, in metres, at time t seconds are \(\boldsymbol{r}_P = \begin{pmatrix}0\\0\\50\end{pmatrix} + t\begin{pmatrix}4\\3\\1\end{pmatrix}\) and \(\boldsymbol{r}_Q = \begin{pmatrix}52\\18\\63\end{pmatrix} + t\begin{pmatrix}-2\\2\\-0.5\end{pmatrix}\).
(a)
Find the speed of drone P.
(2)
(b)
Using different parameters s and u for the two paths, show that the flight paths intersect, and find the coordinates of the point of intersection.
(5)
(c)
State, with a reason, whether the drones collide.
(2)
(d)
Find the distance between the drones when P passes through the point of intersection.
(3)
Show mark scheme
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Notes
Part (a)
\(\sqrt{4^2+3^2+1^2} = \sqrt{26}\) M1
✓ 1
= 5.10 m s−1 A1
✓ 1
Part (b)
Sets up equations, e.g. 4s = 52 − 2u and 3s = 18 + 2u M1
✓ 1
s = 10 A1
✓ 1
u = 6 A1
✓ 1
Checks third component: 50 + 10 = 60 and 63 − 0.5(6) = 60, so the paths intersect R1
✓ 1
Point (40, 30, 60) A1
✓ 1
Part (c)
P reaches the point at t = 10 s but Q reaches it at t = 6 s R1
✓ 1
The drones are not there at the same time, so they do not collide A1
✓ 1
Part (d)
Substitutes t = 10 into \(\boldsymbol{r}_Q\) M1
✓ 1
Q at (32, 38, 58); P at (40, 30, 60) A1
✓ 1
Distance = \(\sqrt{8^2+8^2+2^2}\) = 11.5 m (11.489…) A1
✓ 1
Common error: Intersecting paths do not mean a collision. A collision needs the same position at the same value of t, so the two parameters must be equal.
300IBMI-032-03-r1-04Vectors and kinematicsPaper 2Medium14 marks
Air traffic controllers model the motion of two aircraft, A and B, flying with constant velocities. Relative to a radar station, their positions t minutes after 14:00 are rA = (−20, 5, 8) + t(4, 1, 0) rB = (14, −19, 3) + t(−2, 6, 1) where distances are in kilometres and the third component is altitude.
(a)
Find the speed of aircraft A in km h−1.
(2)
(b)
Show that the flight paths of the two aircraft intersect, and find the coordinates of the point of intersection.
(5)
(c)
Explain why the aircraft do not collide.
(2)
(d)
Find the distance between the two aircraft at 14:05.
(2)
(e)
Find the angle between the directions of travel of the two aircraft.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1: \(\sqrt{4^2+1^2}=\sqrt{17}\) = 4.12 km per minute
✓ 1
A1: multiplies by 60 to get 247 km h−1
✓ 1
Part (b)
M1: equates the two paths using different parameters, e.g. −20 + 4s = 14 − 2u, 5 + s = −19 + 6u, 8 = 3 + u
✓ 1
A1: u = 5
✓ 1
A1: s = 6
✓ 1
R1: checks the remaining equation: −20 + 24 = 4 and 14 − 10 = 4, so the equations are consistent and the paths intersect
✓ 1
A1: point of intersection (4, 11, 8)
✓ 1
Part (c)
R1: A reaches (4, 11, 8) at t = 6 but B reaches it at t = 5
✓ 1
A1: the aircraft are at the intersection point at different times, so they are never in the same place at the same time
✓ 1
Part (d)
M1: finds positions at t = 5, A at (0, 10, 8) and B at (4, 11, 8), and the magnitude of their difference
✓ 1
A1: \(\sqrt{17}\) = 4.12 km
✓ 1
Part (e)
M1: uses the scalar product of the direction vectors, (4)(−2) + (6) + (0) = −2
✓ 3
M1: substitutes magnitudes into cos θ = \(\frac{-2}{\sqrt{17}\sqrt{41}}\)
✓ 1
A1: θ = 94.3°
✓ 1
Common error: Using the same parameter t for both lines when testing for intersection only checks for a collision. Use different parameters to find where the paths cross, then compare the times.
301IBMI-032-04-r1-03Graph theory and networksPaper 2Medium13 marks
A road-sweeping vehicle must travel along every street in a small town at least once. The street network is modelled by a weighted graph with vertices A, B, C, D and E. The edges and their lengths, in km, are: AB 0.8, AC 1.2, AE 1.5, BC 0.5, BD 1.0, CD 0.7, CE 0.9, DE 0.6
(a)
Explain why the vehicle cannot sweep every street exactly once and return to its starting point.
(2)
(b)
Find the total length of all the streets.
(1)
(c)
The vehicle starts and finishes at its depot at A. By considering all possible pairings of the odd vertices, find which streets must be swept twice and the minimum total distance travelled.
(5)
(d)
Write down one possible route of this minimum length, starting and finishing at A.
(2)
(e)
The council decides that the vehicle may start and finish at different vertices. Find the vertices where it should start and finish, and the new minimum distance.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 vertices A, B, D and E have odd degree (each has degree 3)
✓ 1
R1 a closed route that uses every edge exactly once (an Eulerian circuit) requires every vertex to have even degree
✓ 1
Part (b)
A1 7.2 km
✓ 1
Part (c)
M1 considers all three pairings of A, B, D, E using shortest paths
✓ 1
A1 AB + DE = 0.8 + 0.6 = 1.4
✓ 1
A1 AD + BE = 1.8 + 1.4 = 3.2 (AD via B, BE via C)
✓ 1
A1 AE + BD = 1.5 + 1.0 = 2.5
✓ 1
A1 repeat AB and DE, so the minimum distance is 7.2 + 1.4 = 8.6 km
✓ 1
Part (d)
For example A–B–A–C–B–D–E–D–C–E–A
—
A1 every edge is used, with AB and DE each used twice
✓ 1
A1 the route is a valid closed trail starting and ending at A
✓ 1
Part (e)
M1 recognises that only one pair of odd vertices needs a repeated path, and chooses the pair whose repeat is shortest
✓ 1
A1 start at A and finish at B (or the reverse), repeating DE
✓ 1
A1 7.2 + 0.6 = 7.8 km
✓ 1
Exam tip: With four odd vertices there are exactly three pairings, and you must show all three to earn the method mark. Use the shortest path between each pair of vertices, which is not always the direct edge.
302IBMI-032-04-r1-04Graph theory and networksPaper 2Medium10 marks
A heritage park has four attractions, W, X, Y and Z, joined by footpaths. There are paths between W and X, W and Y, X and Y, and Y and Z only. A visitor at any attraction chooses one of the paths leaving it at random, with each path equally likely, and walks to the next attraction. This is modelled as a random walk with transition matrix T, where the columns represent the current attraction and the rows represent the next attraction, both in the order W, X, Y, Z.
(a)
Write down the matrix T.
(2)
(b)
A visitor starts at Z. Find the probability that the visitor is at W after exactly three moves.
(2)
(c)
Find the steady-state probability vector for this random walk.
(4)
(d)
The park manager wants to open one refreshment kiosk at the attraction that will be visited most often in the long run. State, with a reason, where it should be placed.
(1)
(e)
On a busy day there are 400 visitors, each moving independently in this way. After a long time, estimate the number of visitors at Y.
A1 each column sums to 1 with the correct non-zero positions
✓ 1
A1 all entries correct
✓ 1
Part (b)
M1 calculates T3 applied to (0, 0, 0, 1)T, or traces the paths Z→Y→X→W
✓ 1
A1 \(\frac{1}{6}\) = 0.167
✓ 1
Part (c)
M1 sets up Tv = v, or finds a high power of T
✓ 1
M1 uses the condition that the components sum to 1
✓ 1
A1 W = 0.25, X = 0.25
✓ 1
A1 Y = 0.375, Z = 0.125
✓ 1
Part (d)
R1 at Y, because it has the largest long-run probability (0.375) of being occupied, as it is joined to the most paths
✓ 1
Part (e)
A1 400 × 0.375 = 150 visitors
✓ 1
Common error: Check which way round the matrix is set up. When the columns are 'from', each column must sum to 1. Entering the matrix the other way round gives wrong powers of T.
303IBMI-032-03-r2-04Vectors and kinematicsPaper 2Medium13 marks
A ferry leaves port P at 09:00. Take P as the origin, with the first component of each vector measured east and the second measured north, in kilometres. From 09:00 until 10:30 the ferry has constant velocity \(\begin{pmatrix}6\\8\end{pmatrix}\) km h−1. At 10:30 it changes course, and from then on it has constant velocity \(\begin{pmatrix}2\\10\end{pmatrix}\) km h−1. Harbour H has position vector \(\begin{pmatrix}13\\32\end{pmatrix}\).
(a)
Find the speed of the ferry before 10:30 and the bearing on which it travels, giving the bearing to the nearest 0.1°.
(4)
(b)
Write down the position vector of the ferry at 10:30.
(1)
(c)
Show that the ferry passes through H, and find the time at which it does so.
(3)
(d)
Find the total distance the ferry travels from P to H.
(3)
(e)
Find the angle through which the ferry turns at 10:30.
(2)
Show mark scheme
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Mark
Notes
Part (a)
\(\sqrt{6^2+8^2}\) M1
✓ 1
= 10 km h−1 A1
✓ 1
\(\tan\theta = \frac{6}{8}\), with θ measured clockwise from north M1
Common error: A common slip is to find |PH| as the straight-line distance from P to H. The ferry changes direction at 10:30, so add the lengths of the two stages.
304IBMI3H-3-06-11Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Medium12 marks
A radar station A is 11 km from a lighthouse C. A ship S is detected with angle SAC = 40° and the ship is 8 km from the lighthouse.
(a)
Use the sine rule to find an acute value for angle ASC.
(3)
(b)
Explain why angle ASC could also be obtuse, and find its value.
(2)
(c)
Find the two possible distances AS.
(4)
(d)
A second radar shows that angle ASC is obtuse. State the distance AS.
(1)
(e)
Find the area of triangle ACS for the case in (d).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\dfrac{\sin S}{11}=\dfrac{\sin40^\circ}{8}\)
✓ 1
A1 \(\sin S=0.884\)
✓ 1
A1 \(62.1^\circ\)
✓ 1
Part (b)
A1 \(117.9^\circ\)
✓ 1
R1 \(\sin(180^\circ-x)=\sin x\)
✓ 1
Part (c)
M1 angle ACS \(=180^\circ-40^\circ-S\) for each case (\(77.9^\circ\), \(22.1^\circ\))
✓ 1
M1 sine rule for AS: \(AS=\dfrac{8\sin C}{\sin40^\circ}\)
✓ 1
A1 \(AS=12.2\) km
✓ 1
A1 \(AS=4.68\) km
✓ 1
Part (d)
A1 \(4.68\) km
✓ 1
Part (e)
M1 \(\frac12\times11\times4.68\times\sin40^\circ\) or \(\frac12\times8\times11\times\sin22.1^\circ\)
✓ 1
A1 \(16.6\) km2
✓ 1
Common error: The sine rule can give two triangles when the given angle is not between the known sides; always test the obtuse value.
305IBMI3H-3-06-14Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Medium13 marks
A paper cone is made from a sector of a circle of radius 20 cm and angle 252°. The straight edges of the sector are joined so that the sector forms the curved surface of the cone.
(a)
Find the arc length of the sector.
(2)
(b)
Find the base radius of the cone.
(2)
(c)
Find the vertical height of the cone.
(2)
(d)
Find the volume of the cone.
(2)
(e)
Find the area of the curved surface of the cone.
(2)
(f)
The cone, point downwards, is filled with water to 60% of its height. Find the volume of water.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\dfrac{252}{360}\times2\pi\times20\)
✓ 1
A1 \(88.0\) cm
✓ 1
Part (b)
M1 \(2\pi r=88.0\) (ECF from (a))
✓ 1
A1 \(r=14.0\) cm
✓ 1
Part (c)
M1 \(h=\sqrt{20^2-14^2}\)
✓ 1
A1 \(14.3\) cm
✓ 1
Part (d)
M1 \(\frac13\pi\times14^2\times14.28\)
✓ 1
A1 \(2930\) cm3
✓ 1
Part (e)
M1 \(\pi\times14\times20\)
✓ 1
A1 \(880\) cm2
✓ 1
Part (f)
M1 water forms a similar cone with linear scale factor 0.6
✓ 1
M1 volume scale factor \(0.6^3=0.216\)
✓ 1
A1 \(633\) cm3
✓ 1
Exam tip: For similar solids, lengths scale by k, areas by k² and volumes by k³.
306IBMI3H-3-06b-14Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Medium10 marks
Two observers on level ground stand at A and B, where B is 250 m due east of A. A drone D is observed at a bearing of 035° from A and at a bearing of 323° from B. From A, the angle of elevation of the drone is 24°. Let X be the point on the ground directly below the drone.
(a)
Show that angle XAB = 55° and angle XBA = 53°.
(3)
(b)
Find the horizontal distance AX.
(3)
(c)
Find the height of the drone above the ground.
(2)
(d)
Find the straight-line distance from A to the drone.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use bearings and the east–west line AB
✓ 1
A1 angle XAB = 90° − 35° = 55°
✓ 1
A1 bearing from B to A is 270°, so angle XBA = 323° − 270° = 53°
✓ 1
Part (b)
A1 angle AXB = 180° − 55° − 53° = 72°
✓ 1
M1 attempt to use the sine rule: AX/sin 53° = 250/sin 72°
✓ 1
A1 AX = 210 m
✓ 1
Part (c)
M1 attempt to use tan 24° = height/AX
✓ 1
A1 93.5 m
✓ 1
Part (d)
M1 attempt to use cos 24° = AX/AD, or Pythagoras
✓ 1
A1 230 m
✓ 1
Exam tip: For bearing problems, draw a north line at every point and label each angle before using the sine rule.
307IBMI3H-3-03r6-07Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Medium10 marks
Calculator
A drone is flying over a field. Relative to a fixed origin O, with i pointing east and j north (distances in metres), its velocity at time t seconds, t ≥ 0, is v(t) = (4 − 2t)i + (3 + t)j m s−1. When t = 0 the drone is at the point with position vector 10i + 5j.
(a)
Find the acceleration of the drone.
(2)
(b)
Show that the drone is travelling due north when t = 2, and find its position vector at this time.
(4)
(c)
Find the minimum speed of the drone. Give your answer to 3 significant figures.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 differentiates v with respect to t
✓ 1
A1 a = −2i + j m s−2
✓ 1
Part (b)
A1 v(2) = 0i + 5j, so it is moving due north
✓ 1
M1 integrates the velocity: r = (4t − t2 + c1)i + (3t + ½t2 + c2)j
✓ 1
A1 uses r(0) = 10i + 5j to find c1 = 10 and c2 = 5
✓ 1
A1 r(2) = 14i + 13j
✓ 1
Part (c)
M1 writes (speed)2 = (4 − 2t)2 + (3 + t)2
✓ 1
A1 = 5t2 − 10t + 25
✓ 1
M1 finds the minimum (by differentiating or from the vertex): t = 1
✓ 1
A1 minimum speed = √20 = 4.47 m s−1
✓ 1
Exam tip: To minimise a speed, minimise the square of the speed first; it has the same turning point and avoids a square root.
308IBMI3H-3-04r12-05Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Medium9 marks
Calculator
Two cargo ships follow straight paths. Relative to a lighthouse at the origin, with distances in kilometres, ship P is at \(\mathbf{r}=\begin{pmatrix}2\\1\end{pmatrix}+\lambda\begin{pmatrix}3\\2\end{pmatrix}\) and ship Q is at \(\mathbf{r}=\begin{pmatrix}20\\-5\end{pmatrix}+\mu\begin{pmatrix}-2\\4\end{pmatrix}\), where \(\lambda\) and \(\mu\) are the times in hours after 08:00 for ships P and Q respectively.
(a)
Find the coordinates of the point where the two paths cross.
(4)
(b)
Explain why the ships do not collide, and find how many minutes apart they pass through the crossing point.
(2)
(c)
Find the acute angle between the paths of the two ships. Give your answer in degrees to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 equates the position vectors: \(2+3\lambda=20-2\mu\) and \(1+2\lambda=-5+4\mu\)
✓ 1
A1 solves: \(\lambda=3.75\) and \(\mu=3.375\)
✓ 1
M1 substitutes into either line
✓ 1
A1 \((13.25,\,8.5)\)
✓ 1
Part (b)
R1 the ships reach the crossing point at different times (\(\lambda\neq\mu\)), so they are not there together
✓ 1
A1 difference \(0.375\) hours \(=22.5\) minutes
✓ 1
Part (c)
M1 scalar product of direction vectors \(\begin{pmatrix}3\\2\end{pmatrix}\cdot\begin{pmatrix}-2\\4\end{pmatrix}=2\)
✓ 1
M1 \(\cos\theta=\frac{2}{\sqrt{13}\sqrt{20}}\)
✓ 1
A1 82.9\(^\circ\)
✓ 1
Common error: Two paths crossing does not mean a collision: the two vehicles must also be at the crossing point at the same time, so the parameters are compared separately.
309IBMI3H-3-03r16-05Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Medium9 marks
A hospital cleaner must sweep every corridor in a wing. The wing is modelled as a graph with junctions A to F and corridor lengths in metres: AB 40, AC 55, BC 30, BD 45, CD 35, CE 50, DE 25, DF 60, EF 40.
(a)
Explain why it is impossible to sweep every corridor exactly once and finish at the starting junction.
(2)
(b)
Find the shortest route that sweeps every corridor at least once and returns to its starting junction, and state its length.
(4)
(c)
The cleaner is now allowed to start at B and finish at E. Find the length of the shortest route that sweeps every corridor, and the time it takes at a walking speed of 1.25 m s−1.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 junctions B and E have odd degree (3)
✓ 1
R1 a closed trail using each edge once needs every vertex to have even degree
✓ 1
Part (b)
M1 repeat the shortest path between the two odd vertices B and E
✓ 1
A1 B–D–E = 70 m, compared with B–C–E = 80 m and B–C–D–E = 90 m
✓ 1
A1 total of corridor lengths = 380 m
✓ 1
A1 380 + 70 = 450 m
✓ 1
Part (c)
M1 only B and E are odd, so each corridor can be used exactly once
✓ 1
A1 380 m
✓ 1
A1 380 ÷ 1.25 = 304 s
✓ 1
Exam tip: Count the odd-degree vertices first: with none the route can start anywhere, with exactly two it must start at one and end at the other, otherwise paths must be repeated.
310IBMI3H-3-03r16-12Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Medium9 marks
Three hospitals are modelled as sites A(2, 2), B(8, 2) and C(5, 8) on a map where one unit is 1 km. The Voronoi diagram of the sites is drawn.
(a)
Find the equation of the perpendicular bisector of AB.
(2)
(b)
Find the equation of the perpendicular bisector of AC, giving your answer in the form y = mx + c.
(3)
(c)
Find the coordinates of the Voronoi vertex where the cells of A, B and C meet.
(2)
(d)
A helipad is built at this vertex. Find its distance from hospital A.
(2)
Show mark scheme
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Mark
Notes
Part (a)
M1 midpoint (5, 2) and AB horizontal
✓ 1
A1 x = 5
✓ 1
Part (b)
M1 midpoint of AC is (3.5, 5)
✓ 1
M1 gradient of AC is 2 so perpendicular gradient is −0.5
✓ 1
A1 y = −0.5x + 6.75
✓ 1
Part (c)
M1 substitute x = 5 into the equation from (b)
✓ 1
A1 (5, 4.25)
✓ 1
Part (d)
M1 \(\sqrt{(5-2)^2+(4.25-2)^2}\)
✓ 1
A1 3.75 km
✓ 1
Exam tip: A Voronoi vertex is equidistant from three sites, so after finding it check that its distances to all three sites agree.
311IBMI3H-3-04r16-04Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Medium9 marks
A hospital wants to link six buildings A to F with a single network of data ducts, using the least total length. The possible ducts and their lengths in metres are: AB 90, AC 140, AD 70, BC 80, BD 130, BE 120, CD 110, CF 100, DF 160, EF 60. Installing a duct costs €12.50 per metre.
(a)
Use Prim’s algorithm, starting at A, to find the ducts to be installed in the order they are chosen, and the minimum total length.
(4)
(b)
A new duct DE of length 90 m is proposed. Show that it would be included in the new minimum network and find the new minimum total length.
(3)
(c)
Find the installation cost of the network in part (b).
(2)
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Mark
Notes
Part (a)
M1 start at A and repeatedly add the shortest edge joining the tree to a new vertex
✓ 1
A1 AD (70), AB (90), BC (80)
✓ 1
A1 CF (100), EF (60)
✓ 1
A1 total length 400 m
✓ 1
Part (b)
M1 DE forms a cycle D–A–B–C–F–E whose longest edge is CF (100)
✓ 1
R1 90 < 100, so DE replaces CF
✓ 1
A1 400 + 90 − 100 = 390 m
✓ 1
Part (c)
M1 390 × 12.50
✓ 1
A1 €4875
✓ 1
Exam tip: To test a new edge against a minimum spanning tree, find the cycle it creates and remove the longest edge on that cycle if the new edge is shorter.
312IBMI3H-3-04r16-07Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Medium9 marks
In a car park layout the transformation T is a reflection in the line y = −x followed by an enlargement with scale factor 2, centre the origin. The reflection has matrix \(\begin{pmatrix}0&-1\\-1&0\end{pmatrix}\).
(a)
Find the matrix of T.
(3)
(b)
Triangle XYZ has vertices X(1, 1), Y(4, 1) and Z(4, 3). Find the coordinates of its image under T.
(2)
(c)
Find the coordinates of the point that is mapped by T to (−10, −4).
(2)
(d)
Find the matrix of T applied twice, and describe the single transformation it represents.
M1 multiply the enlargement matrix by the reflection matrix
✓ 1
A1 \(\begin{pmatrix}0&-2\\-2&0\end{pmatrix}\)
✓ 1
Part (b)
M1 multiply the matrix by position vectors
✓ 1
A1 (−2, −2), (−2, −8), (−6, −8)
✓ 1
Part (c)
M1 \(-2y=-10\) and \(-2x=-4\), or use of \(\mathbf{T}^{-1}\)
✓ 1
A1 (2, 5)
✓ 1
Part (d)
A1 \(\begin{pmatrix}4&0\\0&4\end{pmatrix}\)
✓ 1
A1 enlargement, scale factor 4, centre the origin
✓ 1
Common error: Because T is “reflection then enlargement”, the matrices are multiplied with the reflection on the right; here the two commute, but in general the order matters.
313IBMI3H-3-04r23-03Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Medium9 marks
An inspection drone flies near a wind turbine. Its position relative to the base of the turbine at time t seconds (t ≥ 0) is given by the vector r(t) = (2t2, 12t − t2) metres, where the first component is horizontal and the second is vertical.
(a)
Find an expression for the velocity vector v(t).
(2)
(b)
Find the speed of the drone when t = 3, to 3 significant figures.
(2)
(c)
Find the time at which the drone is moving horizontally and its position at that time.
(3)
(d)
Find the magnitude of the drone's acceleration, to 3 significant figures.
Common error: Speed is the magnitude of the velocity vector, not either of its components, and it is never negative.
314IBMI3H-3-04r23-14Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Medium9 marks
Three libraries are located at A(1, 1), B(7, 1) and C(4, 8) on a map where one unit represents 1 km. A Voronoi diagram is drawn with these three libraries as sites.
(a)
Find the equation of the perpendicular bisector of AB.
(2)
(b)
Find the equation of the perpendicular bisector of AC, giving it in the form y = mx + c.
(3)
(c)
Find the coordinates of the vertex where the three cells meet, to 3 significant figures.
(2)
(d)
A new house is at H(4, 5). Show that the nearest library is C, and state the distance.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 midpoint of AB is (4, 1) and AB is horizontal
✓ 1
A1 x = 4
✓ 1
Part (b)
M1 midpoint of AC = (2.5, 4.5)
✓ 1
M1 gradient of AC = 7/3, so perpendicular gradient = −3/7
✓ 1
A1 y = −(3/7)x + 39/7
✓ 1
Part (c)
M1 substitutes x = 4 into the equation from (b)
✓ 1
A1 (4, 3.86)
✓ 1
Part (d)
M1 distances: HA = 5, HB = 5, HC = 3
✓ 1
A1 C is nearest, at 3 km
✓ 1
Exam tip: The vertex of a Voronoi diagram is equidistant from three sites; check it by calculating its distance to each site.
315IBMI3H-3-05r23-07Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Medium10 marks
Calculator
A coffee roaster at A delivers beans to cafés at B, C, D and E, then returns to A. The distances in km between the five locations are AB 12, AC 15, AD 9, AE 20, BC 8, BD 14, BE 11, CD 10, CE 13, DE 7.
(a)
Use the nearest neighbour algorithm starting at A to find an upper bound for the length of the shortest delivery round.
(4)
(b)
By deleting vertex A, find a lower bound for the length of the shortest delivery round.
(4)
(c)
Show that the route A–B–C–E–D–A has length 49 km, and state what this tells you about the shortest possible round.
(2)
Show mark scheme
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Notes
Part (a)
M1 move to the nearest unvisited vertex each time
✓ 1
A1 route A–D–E–B–C–A
✓ 1
M1 add the distances 9 + 7 + 11 + 8 + 15
✓ 1
A1 upper bound 50 km
✓ 1
Part (b)
M1 minimum spanning tree for B, C, D, E using edges DE 7, BC 8, CD 10
✓ 1
A1 weight of tree 25
✓ 1
M1 add the two shortest edges at A, namely AD 9 and AB 12
✓ 1
A1 lower bound 46 km
✓ 1
Part (c)
M1 12 + 8 + 13 + 7 + 9 = 49 km
✓ 1
R1 the shortest round has length between 46 km and 49 km, as 49 is a better upper bound than 50
✓ 1
Common error: A lower bound is found by deleting a vertex and adding two edges back; do not add the two longest edges, and do not forget to use a minimum spanning tree of what remains.
316IBMI3H-3-05r23-12Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Medium11 marks
Calculator
A coffee storage silo is made from a cylinder of radius 1.5 m and height 3.0 m, with an inverted cone of the same radius and height 2.0 m attached below it as a hopper. The silo is full of beans.
(a)
Find the total volume of the silo, correct to 3 significant figures.
(3)
(b)
The bulk density of the beans is 650 kg m−3. Using the unrounded volume, find the mass of beans in the silo, correct to 3 significant figures.
(2)
(c)
Find the angle between the slant side of the cone and the horizontal, correct to 1 decimal place.
(3)
(d)
The curved surfaces of the silo (the cylinder wall and the cone, not the top) are to be painted at a cost of €18.50 per m2. Find the cost, to the nearest cent.
(3)
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Mark
Notes
Part (a)
M1 cylinder volume π×1.52×3.0
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M1 cone volume ⅓×π×1.52×2.0
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A1 25.9 m3
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Part (b)
M1 mass = volume × 650
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A1 16 800 kg
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Part (c)
M1 use of tan θ = 2.0/1.5
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A1 correct angle calculation
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A1 θ = 53.1°
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Part (d)
M1 slant height √(1.52 + 2.02) = 2.5 m
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M1 curved area = 2π×1.5×3.0 + π×1.5×2.5
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A1 cost €741.02
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Common error: The curved surface area of a cone uses the slant height, not the vertical height; find the slant height with Pythagoras first.
317IBMI3H-3-05r23-13Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Medium11 marks
Calculator
Three coffee roasters are located at P(8, 6), Q(1, −1) and R(0, 6), where coordinates are in km. The region served by each roaster is found using a Voronoi diagram.
(a)
Find the equation of the perpendicular bisector of PQ.
(3)
(b)
The perpendicular bisector of PR is the line x = 4. Hence find the coordinates of the Voronoi vertex where the regions of P, Q and R meet.
(3)
(c)
Find the distance from the vertex to each of the three roasters.
(2)
(d)
A café is at C(3, 5). Determine which roaster is closest to the café.
(3)
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Mark
Notes
Part (a)
M1 midpoint of PQ is (4.5, 2.5)
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M1 gradient of PQ = 1 so perpendicular gradient = −1
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A1 y = −x + 7
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Part (b)
R1 the vertex lies on both bisectors
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M1 substitute x = 4 into y = −x + 7
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A1 (4, 3)
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Part (c)
M1 distance from (4, 3) to P, for example √(42 + 32)
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A1 5 km, and the same for Q and R, as expected for a Voronoi vertex
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Part (d)
M1 distances to P, Q and R calculated
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A1 CP = 5.10, CQ = 6.32, CR = 3.16 km
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A1 R is closest
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Exam tip: Every point on a Voronoi edge is the same distance from the two nearest sites, so a vertex is equidistant from three sites; use this as a check on your coordinates.
318IBMI3H-3-05r23-14Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Medium10 marks
Calculator
A bridge has an arch in the shape of an arc of a circle. The arch has a horizontal span of 40 m and a maximum height above the span of 8 m.
(a)
Show that the radius of the circle is 29 m.
(3)
(b)
Find the angle subtended at the centre of the circle by the arch, in radians, correct to 3 significant figures.
(2)
(c)
Using the unrounded angle, find the length of the arch, correct to 3 significant figures.
(2)
(d)
Find the area between the arch and the horizontal span, correct to 3 significant figures, using the unrounded angle.
(3)
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Notes
Part (a)
M1 a right-angled triangle with sides 20, r − 8 and hypotenuse r
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M1 202 + (r − 8)2 = r2
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A1 400 + 64 = 16r, so r = 29
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Part (b)
M1 sin(θ/2) = 20/29
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A1 θ = 1.52 radians
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Part (c)
M1 arc length = rθ
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A1 44.1 m
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Part (d)
M1 sector area ½r2θ
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M1 subtract triangle area ½r2 sin θ
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A1 220 m2
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Common error: Check that your calculator is in radian mode when using arc length = rθ, and keep the unrounded angle for the later parts.
319IBMI3H-3-03r26-06Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Medium10 marks
Calculator
Two fishing boats, A and B, leave at the same time (t = 0, in hours). Relative to a fixed origin, in kilometres, boat A has position \(\mathbf{r}_A=\begin{pmatrix}1\\2\end{pmatrix}+t\begin{pmatrix}3\\1\end{pmatrix}\) and boat B has position \(\mathbf{r}_B=\begin{pmatrix}13\\-4\end{pmatrix}+t\begin{pmatrix}-2\\2\end{pmatrix}\).
(a)
Show that the paths of the two boats cross, and find the coordinates of the crossing point.
(4)
(b)
State whether the boats collide, giving a reason, and find the time difference between the two boats reaching the crossing point.
(2)
(c)
Find the minimum distance between the boats, to 3 significant figures, using unrounded values.
(4)
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Notes
Part (a)
M1 equates the paths using different parameters, \(\lambda\) and \(\mu\)
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A1 \(1+3\lambda=13-2\mu\) and \(2+\lambda=-4+2\mu\)
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A1 \(\lambda=1.5\) and \(\mu=3.75\)
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A1 crossing point \((5.5,\,3.5)\)
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Part (b)
R1 No collision, since A reaches the point at \(t=1.5\) but B reaches it at \(t=3.75\) (different times)
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A1 \(2.25\) hours
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Part (c)
M1 \(\mathbf{r}_B-\mathbf{r}_A=\begin{pmatrix}12-5t\\-6+t\end{pmatrix}\) and squared distance \(26t^2-132t+180\)
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M1 minimises, e.g. \(t=\tfrac{132}{52}\) or uses GDC
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A1 \(t=2.54\) hours (2.5385)
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A1 \(3.53\) km
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Exam tip: Boats collide only if the position vectors are equal at the same value of \(t\); paths crossing is not enough, so use separate parameters to find crossing points.
320IBMI3H-3-06r27-03Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Medium9 marks
A volcanic block is thrown from a vent at the origin O. Air resistance is ignored. The position vector of the block t seconds after leaving the vent is r = (18t, 24t, 35t − 4.9t2) metres, where the first two components are horizontal and the third is vertically upwards. The ground around the vent is horizontal at the level of O.
(a)
Find the velocity vector of the block at time t.
(2)
(b)
Find the speed of the block when t = 3. Give your answer to 3 significant figures.
(2)
(c)
Find the time at which the block reaches its greatest height. Give your answer to 3 significant figures.
(2)
(d)
Find the horizontal distance from the vent at which the block lands. Give your answer to 3 significant figures.
(3)
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Notes
Part (a)
M1 differentiates each component with respect to t
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A1 v = (18, 24, 35 − 9.8t)
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Part (b)
M1 v(3) = (18, 24, 5.6) and takes the magnitude √(182 + 242 + 5.62)
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A1 30.5 m s−1
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Part (c)
M1 vertical component of velocity is zero: 35 − 9.8t = 0
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A1 t = 3.57 s
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Part (d)
M1 35t − 4.9t2 = 0, so t = 35/4.9 (= 7.14 s)
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M1 horizontal distance = √(182 + 242) × t = 30t
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A1 214 m
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Exam tip: To find where a projectile lands, set the vertical component of position equal to the ground level, then use the horizontal components at that time.
321IBMI3H-3-06r27-09Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Medium9 marks
A triangular stage set piece has corners at (1, 1), (4, 1) and (4, 3), in metres. It is rotated 90° anticlockwise about the origin, and the result is then stretched by a factor of 2 in the x-direction.
(a)
Write down the matrix R for the rotation and the matrix S for the stretch.
(2)
(b)
Find the single matrix T that represents the rotation followed by the stretch.
(2)
(c)
Find the coordinates of the corners of the image of the triangle.
(2)
(d)
Find the area of the image, using the determinant of T.
(3)
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Notes
Part (a)
A1 R = ((0, −1), (1, 0))
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A1 S = ((2, 0), (0, 1))
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Part (b)
M1 T = SR (stretch matrix on the left)
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A1 T = ((0, −2), (1, 0))
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Part (c)
M1 multiplies T by the position vectors of the corners
Common error: For "A then B" the product is BA, not AB: the first transformation is the matrix closest to the vector.
322IBMI3H-3-06r27-14Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Medium9 marks
Three coastguard stations are at A(4, 8), B(9, 3) and C(1, −1), where coordinates are in kilometres east and north of a reference point. Each point on the sea is assigned to its nearest station, forming a Voronoi diagram.
(a)
Find the equation of the perpendicular bisector of AB.
(3)
(b)
Find the equation of the perpendicular bisector of BC.
(2)
(c)
Find the coordinates of the Voronoi vertex.
(2)
(d)
A rescue beacon is placed at the Voronoi vertex. Find its distance from each station.
(2)
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Notes
Part (a)
M1 midpoint of AB is (6.5, 5.5)
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A1 gradient of AB is −1, so the perpendicular gradient is 1
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A1 y = x − 1
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Part (b)
M1 midpoint of BC is (5, 1) and gradient of BC is 1/2
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A1 y = −2x + 11
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Part (c)
M1 solves x − 1 = −2x + 11
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A1 (4, 3)
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Part (d)
M1 distance from (4, 3) to A (or to B or C)
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A1 5 km (the same for all three stations, since the vertex is equidistant from them)
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Exam tip: A Voronoi vertex is equidistant from three sites, so it can be found as the intersection of two perpendicular bisectors.
323IBMI3H-3-06r29-03Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Medium11 marks
A robot moves across a warehouse floor. Its velocity at time \(t\) seconds is \(\mathbf{v}=\begin{pmatrix}2t\\3-t\end{pmatrix}\) m s−1, for \(t\ge0\). At \(t=0\) its position is \(\begin{pmatrix}1\\2\end{pmatrix}\) m relative to the origin.
(a)
Find the speed of the robot when \(t=2\).
(2)
(b)
Find the acceleration of the robot.
(2)
(c)
Find the position vector of the robot when \(t=4\).
(3)
(d)
Find the minimum speed of the robot, to 3 significant figures.
Exam tip: Check an inverse or a recovered point by applying the matrix to it again.
326IBMI3H-3-06r29-14Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Medium11 marks
Three fish-processing plants are at A(10, 4), B(2, 0) and C(5, 9), where coordinates are in km. A Voronoi diagram is drawn with these three plants as sites.
(a)
Find the equation of the perpendicular bisector of AB.
(3)
(b)
Find the equation of the perpendicular bisector of AC.
(3)
(c)
Hence find the coordinates of the Voronoi vertex where the edges meet.
(2)
(d)
A new boat landing is at (6, 3). Determine which plant is nearest to it.
(2)
(e)
Find the distance from the Voronoi vertex to plant A.
(1)
Show mark scheme
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Mark
Notes
Part (a)
M1 midpoint \((6,2)\) and gradient of AB \(=\frac12\)
✓ 1
M1 perpendicular gradient \(-2\)
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A1 \(y=-2x+14\)
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Part (b)
M1 midpoint \((7.5,6.5)\) and gradient of AC \(=-1\)
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M1 perpendicular gradient \(1\)
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A1 \(y=x-1\)
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Part (c)
M1 solve \(-2x+14=x-1\)
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A1 (5, 4)
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Part (d)
M1 compare distances, e.g. \(AP^2=17\), \(BP^2=25\), \(CP^2=37\)
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A1 plant A
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Part (e)
A1 5 km
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Exam tip: Every point on a Voronoi edge is equidistant from the two sites it separates, so the vertex is equidistant from all three.
327IBMI3H-3-05r31-09Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Medium10 marks
A theatre designer transforms a triangular set piece with vertices A(1, 1), B(3, 1) and C(3, 2), with coordinates in metres. The piece is first stretched by the matrix S, which doubles all x-coordinates and triples all y-coordinates, and then rotated 90° anticlockwise about the origin by the matrix R.
(a)
Write down the matrices S and R.
(2)
(b)
Find the single matrix T that represents the stretch followed by the rotation.
(2)
(c)
Find the coordinates of the image of vertex C.
(2)
(d)
A lighting point is moved by the combined transformation to (−9, 4). Find its original coordinates.
(2)
(e)
Find the area of the image of triangle ABC.
(2)
Show mark scheme
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Notes
Part (a)
A1 S = [[2, 0], [0, 3]]
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A1 R = [[0, −1], [1, 0]]
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Part (b)
M1 attempt to multiply in the correct order, T = RS
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A1 T = [[0, −3], [2, 0]]
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Part (c)
M1 multiply T by the column vector (3, 2)
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A1 (−6, 6)
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Part (d)
M1 set up −3y = −9 and 2x = 4 (or use T−1)
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A1 (2, 3)
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Part (e)
M1 area scale factor = |det T| = |0 × 0 − (−3)(2)| = 6, with the original area 1 m2
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A1 6 m2
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Exam tip: When one transformation is followed by another, the matrix of the later one multiplies on the left. The area scale factor is the absolute value of the determinant, so it does not depend on the rotation.
328IBMI3H-3-05r31-13Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Medium9 marks
Three weather stations are located at A(2, 1), B(8, 3) and C(4, 9), with coordinates in kilometres. A Voronoi diagram is drawn for the three stations.
(a)
Find the equation of the perpendicular bisector of AB. Give your answer in the form y = mx + c.
(3)
(b)
The perpendicular bisector of AC is y = −0.25x + 5.75. Find the coordinates of the Voronoi vertex where the edges meet, giving your answers to 3 significant figures.
(3)
(c)
A village is at V(5, 6). Determine which station is nearest to V and find the distance to that station. Give the distance to 3 significant figures.
(3)
Show mark scheme
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Notes
Part (a)
M1 midpoint of AB is (5, 2)
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M1 gradient of AB is ⅓, so the perpendicular gradient is −3
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A1 y = −3x + 17
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Part (b)
M1 equate: −3x + 17 = −0.25x + 5.75
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A1 x = 45 ÷ 11 = 4.09
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A1 y = 52 ÷ 11 = 4.73
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Part (c)
M1 calculate distances to the stations: VA = √34, VB = √18, VC = √10
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A1 station C is nearest
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A1 3.16 km
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Exam tip: Every point on a Voronoi edge is equidistant from the two sites it separates, so edges lie along perpendicular bisectors. To find which cell a point is in, compare squared distances; no square roots are needed until the final answer.
329IBMI3H-3-07r37-10Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Medium10 marks
Calculator
The coverage footprint of a small mobile cell is modelled as triangle PQR with P(2, 1), Q(6, 1) and R(2, 4), where coordinates are in km. After an equipment upgrade the footprint is transformed by the matrix M = \(\begin{pmatrix}3&1\\0&2\end{pmatrix}\).
(a)
Find the coordinates of P′, Q′ and R′, the images of P, Q and R.
(3)
(b)
Find the area of the new footprint.
(2)
(c)
A different region has area 90 km2 after the upgrade. Find its area before the upgrade.
(2)
(d)
Find the coordinates of the point that is mapped to (13, 6). Give your answers to 3 significant figures.
(3)
Show mark scheme
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Notes
Part (a)
M1 multiplies M by a position vector
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A1 P′(7, 2) and Q′(19, 2)
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A1 R′(10, 8)
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Part (b)
M1 finds the original area 6 km2 and det M = 6 (accept finding the area of P′Q′R′ directly)
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A1 36 km2
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Part (c)
M1 divides 90 by det M
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A1 15 km2
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Part (d)
M1 finds or uses M−1 or solves 3x + y = 13, 2y = 6
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A1 y = 3
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A1 x = 3.33
✓ 1
Exam tip: The area scale factor of a transformation is the absolute value of the determinant, so it works backwards by dividing.
330IBMI3H-3-07r37-14Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Medium10 marks
Calculator
Three mobile phone towers are at A(2, 3), B(8, 5) and C(5, 10), where coordinates are in km. The region served by each tower is modelled using a Voronoi diagram, each point being served by its nearest tower.
(a)
Find the equation of the perpendicular bisector of AB, in the form y = mx + c.
(3)
(b)
Find the equation of the perpendicular bisector of BC, in the form y = mx + c.
(3)
(c)
Find the coordinates of the Voronoi vertex where these two bisectors meet. Give each coordinate to 3 significant figures.
(2)
(d)
A company wants to place a new tower at this vertex. Find the distance from this vertex to tower A. Use unrounded coordinates from part (c) and give your answer in km to 3 significant figures.
(2)
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Notes
Part (a)
A1 midpoint (5, 4) and gradient of AB is 1/3
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M1 uses the perpendicular gradient −3 through the midpoint
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A1 y = −3x + 19
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Part (b)
A1 midpoint (6.5, 7.5) and gradient of BC is −5/3
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M1 uses the perpendicular gradient 3/5 through the midpoint
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A1 y = 0.6x + 3.6
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Part (c)
M1 equates −3x + 19 = 0.6x + 3.6 (ECF)
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A1 (4.28, 6.17)
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Part (d)
M1 attempts the distance from (4.2778, 6.1667) to (2, 3)
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A1 3.90 km
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Exam tip: A Voronoi vertex is equidistant from three sites, so the distance to any one of them is the same; use this to check your answer against a second tower.
331IBMI3H-3-05r38-03Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Medium9 marks
A triangular face of an ice slope has corners at A(1, 0, 2), B(4, 2, 3) and C(2, 5, 1), where the coordinates are in metres, the x and y axes are horizontal and the z axis is vertical.
(a)
Find the vectors AB and AC.
(2)
(b)
Find AB × AC.
(2)
(c)
Find the area of the face, giving your answer to 3 significant figures.
(2)
(d)
The face is a plane. Find the angle between this plane and the horizontal, to the nearest 0.1°.
(3)
Show mark scheme
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Notes
Part (a)
M1 attempt to subtract coordinates
✓ 1
A1 AB = (3, 2, 1) and AC = (1, 5, −1)
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Part (b)
M1 attempt at the vector product
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A1 (−7, 4, 13) (ECF from (a))
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Part (c)
M1 area = ½|AB × AC| = ½√(49 + 16 + 169)
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A1 7.65 m² (ECF from (b))
✓ 1
Part (d)
M1 identify that the angle between the plane and horizontal equals the angle between the normal and the vertical
✓ 1
M1 cos θ = |13| / √234 (use of scalar product with (0, 0, 1)) (ECF from (b))
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A1 31.8°
✓ 1
Exam tip: A vector product is perpendicular to both original vectors, so it points along the normal to the plane.
332IBMI3H-3-05r38-14Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Medium9 marks
Three mobile masts are located at A(0, 0), B(8, 0) and C(2, 6), where the coordinates are in km. A Voronoi diagram is drawn with these masts as sites.
(a)
Find the equation of the perpendicular bisector of AB.
(2)
(b)
Find the equation of the perpendicular bisector of AC, giving your answer in the form x + ay = b.
(3)
(c)
Find the coordinates of the Voronoi vertex.
(2)
(d)
Find the distance from the Voronoi vertex to each mast, to 3 significant figures.
(2)
Show mark scheme
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Notes
Part (a)
M1 midpoint (4, 0) and the bisector is perpendicular to the x-axis
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A1 x = 4
✓ 1
Part (b)
M1 midpoint (1, 3)
✓ 1
M1 gradient of AC is 3, so perpendicular gradient is −1/3
✓ 1
A1 x + 3y = 10
✓ 1
Part (c)
M1 solve the two bisector equations simultaneously
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A1 (4, 2)
✓ 1
Part (d)
M1 √(4² + 2²)
✓ 1
A1 4.47 km
✓ 1
Exam tip: A Voronoi vertex is equally far from three sites, so its distance to any one of them is the same.
333IBMI3H-3-08r44-01Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Medium10 marks
A cargo ship is travelling in open water. Relative to a lighthouse at the origin O, its position at time t hours after noon is given by r = (−9, 2) + t(4, 3), where distances are in kilometres and the ship moves at constant velocity.
(a)
Find the speed of the ship in km h−1.
(2)
(b)
Find the time at which the ship is closest to the lighthouse.
(3)
(c)
Find the minimum distance between the ship and the lighthouse.
(2)
(d)
The lighthouse can be seen from the ship whenever the distance between them is less than 8 km. Find the length of time, in hours to 3 significant figures, for which the lighthouse can be seen.
(3)
Show mark scheme
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Notes
Part (a)
M1 attempt to find the magnitude of the velocity vector (4, 3)
Exam tip: At the point of closest approach the position vector is perpendicular to the velocity, so their scalar product is zero.
334IBMI3H-3-08r44-04Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Medium9 marks
During a volcanic eruption, a rock is thrown from a vent at the top of a crater rim. Relative to the vent, the position of the rock at time t seconds is r = (22t, 30t − 4.9t²), in metres, where the first component is horizontal and the second is vertical. The rock lands on a flat plain 40 m below the level of the vent.
(a)
Find the velocity vector of the rock at time t.
(2)
(b)
Find the initial speed of the rock in m s−1, to 3 significant figures.
(2)
(c)
Find the time at which the rock lands on the plain, in seconds to 3 significant figures.
(3)
(d)
Find the horizontal distance between the vent and the landing point, in metres to 3 significant figures. Use the unrounded value of t.
(2)
Show mark scheme
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Mark
Notes
Part (a)
A1 horizontal component 22
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A1 vertical component 30 − 9.8t
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Part (b)
M1 attempt to find √(22² + 30²) (ECF from (a))
✓ 1
A1 37.2 m s−1
✓ 1
Part (c)
M1 set 30t − 4.9t² = −40
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M1 attempt to solve the quadratic 4.9t² − 30t − 40 = 0 and select the positive root
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A1 t = 7.25 s
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Part (d)
M1 horizontal distance = 22 × 7.2486…
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A1 159 m
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Common error: A quadratic equation for time usually has two roots; reject the negative one, which corresponds to a time before the motion began.
335IBMI3H-3-06r47-05Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Medium9 marks
In a game, a triangular sprite has vertices P(1, 1), Q(4, 1) and R(1, 3), in pixels. A transformation T is a horizontal stretch with scale factor 2 (the x-coordinate is doubled), followed by an anticlockwise rotation of \(90^\circ\) about the origin.
(a)
Write down the matrices for the stretch and for the rotation, and hence find the single matrix \(\mathbf{T}\) that represents T.
(3)
(b)
Find the coordinates of the image of the sprite, \(P'\), \(Q'\) and \(R'\).
(2)
(c)
The original sprite has area 3 square pixels. Use \(\det\mathbf{T}\) to find the area of the image.
(2)
(d)
A point S is moved by T to the point (−6, 8). Find the coordinates of S.
(2)
Show mark scheme
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Notes
Part (a)
A1 stretch \(\begin{pmatrix}2&0\\0&1\end{pmatrix}\) and rotation \(\begin{pmatrix}0&-1\\1&0\end{pmatrix}\)
✓ 1
M1 multiply with the stretch matrix applied first, i.e. \(\begin{pmatrix}0&-1\\1&0\end{pmatrix}\begin{pmatrix}2&0\\0&1\end{pmatrix}\)
M1 multiply \(\mathbf{T}\) by at least one position vector
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A1 \(P'(-1,2)\), \(Q'(-1,8)\), \(R'(-3,2)\)
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Part (c)
M1 \(\det\mathbf{T}=0\times0-(-1)(2)=2\)
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A1 area \(=2\times3=6\) square pixels
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Part (d)
M1 use \(\mathbf{T}^{-1}=\frac{1}{2}\begin{pmatrix}0&1\\-2&0\end{pmatrix}\) applied to \(\begin{pmatrix}-6\\8\end{pmatrix}\) (or solve \(\mathbf{T}\mathbf{s}=\begin{pmatrix}-6\\8\end{pmatrix}\))
✓ 1
A1 S\((4,6)\)
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Common error: For a sequence of transformations the matrix for the first transformation is written on the right in the product, next to the position vector.
336IBMI3H-3-06r47-06Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Medium9 marks
In a model of a drug molecule, the centres of three atoms have coordinates, in nanometres (nm): carbon C(0.12, 0.30, 0.08), nitrogen N(0.22, 0.38, 0.16) and oxygen O(−0.02, 0.34, 0.11).
(a)
Find the vectors \(\overrightarrow{CN}\) and \(\overrightarrow{CO}\).
(2)
(b)
Find the bond angle \(\angle NCO\), giving your answer to 1 decimal place.
(3)
(c)
Find the vector product \(\overrightarrow{CN}\times\overrightarrow{CO}\).
(2)
(d)
Hence find the area of triangle NCO, giving your answer in nm2 to 3 significant figures.
M1 area \(=\frac12|\overrightarrow{CN}\times\overrightarrow{CO}|\)
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A1 \(0.0104\) nm2
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Exam tip: The magnitude of the vector product of two sides of a triangle gives the area of the parallelogram they span, so halve it for the triangle.
337IBMI-032-08-r1-06Vectors and kinematicsPaper 2Hard12 marks
A coastguard station is at the origin O. At 14:00 a yacht in difficulty is drifting with position vector, in kilometres, \(\mathbf{r}_Y = \begin{pmatrix} 4 \\ 10 \end{pmatrix} + t\begin{pmatrix} 6 \\ -2 \end{pmatrix}\) where t is the time in hours after 14:00 and the unit vectors point east and north. At 14:00 a lifeboat leaves O, travelling in a straight line at a constant speed of 15 km h−1, so as to intercept the yacht.
(a)
Find the speed of the yacht.
(2)
(b)
Find the bearing on which the yacht is drifting.
(2)
(c)
Show that the time of interception, t hours, satisfies \(185t^2 - 8t - 116 = 0\).
(3)
(d)
Find the time at which the lifeboat reaches the yacht, to the nearest minute.
(2)
(e)
Find the bearing on which the lifeboat must travel.
(3)
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Part (a)
M1: \(\sqrt{6^2+(-2)^2}=\sqrt{40}\)
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A1: 6.32 km h−1
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Part (b)
M1: angle below east \(\tan^{-1}\left(\frac{2}{6}\right)=18.4°\)
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A1: bearing 108° (108.4°)
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Part (c)
M1: lifeboat distance 15t equals distance of yacht from O: \((4+6t)^2+(10-2t)^2=(15t)^2\)
Exam tip: Reject the negative root of the quadratic with a reason in context: the interception must happen after the lifeboat sets off.
338IBMI-032-08-r2-05Vectors and kinematicsPaper 2Hard16 marks
A coastguard base is at the origin O. The vector \(\boldsymbol{i}\) points east, \(\boldsymbol{j}\) points north and distances are in kilometres. At time t hours after 12:00, a suspect boat S has position vector \(\boldsymbol{r}_S = \begin{pmatrix}10\\4\end{pmatrix} + t\begin{pmatrix}6\\8\end{pmatrix}\). At 12:30 a patrol boat leaves O. It travels in a straight line at a constant speed of 20 km h−1 and intercepts S at time t = T.
(a)
Find the speed of S and the bearing on which it travels.
(3)
(b)
Find the distance of S from O at 12:30.
(2)
(c)
Show that T satisfies \(300T^2 - 584T - 16 = 0\).
(4)
(d)
Find the time of day at which the interception takes place.
(4)
(e)
Find the bearing on which the patrol boat must travel.
Rejects the negative root (T = −0.027), since the patrol boat leaves at t = 0.5 R1
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Time ≈ 13:58 A1
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Part (e)
Substitutes T into \(\boldsymbol{r}_S\) M1
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Interception point (21.8, 19.8) A1
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Bearing = arctan(21.84/19.79) = 047.8° A1
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Exam tip: The patrol boat starts half an hour late, so the distance it covers is 20(T − 0.5), not 20T. Always check that a root makes sense in context before using it.
339IBMI-032-04-r1-06Graph theory and networksPaper 2Hard14 marks
An engineer must inspect five wind turbines, A, B, C, D and E. She starts and finishes at A and visits each other turbine exactly once. The shortest road distances between the turbines, in km, are: AB 12, AC 18, AD 15, AE 10 BC 9, BD 14, BE 11 CD 8, CE 16 DE 13
(a)
Use the nearest neighbour algorithm, starting at A, to find an upper bound for the length of the engineer's route. State the route.
(3)
(b)
By deleting vertex A, use the deleted vertex algorithm to find a lower bound for the length of the route.
(4)
(c)
Find the lower bound obtained by deleting vertex C instead. Hence state the best interval you can give for the length L km of the optimal route.
(4)
(d)
A colleague suggests the route A–B–C–D–E–A. Find its length, and comment on what this tells you about L, given that all distances are whole numbers of kilometres.
(3)
Show mark scheme
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Notes
Part (a)
M1 at each step chooses the nearest unvisited turbine
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A1 A–E–B–C–D–A
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A1 10 + 11 + 9 + 8 + 15 = 53 km
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Part (b)
M1 finds a minimum spanning tree of B, C, D, E
✓ 1
A1 CD + BC + BE = 8 + 9 + 11 = 28
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M1 adds the two shortest edges from A: AE and AB
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A1 28 + 10 + 12 = 50 km
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Part (c)
A1 minimum spanning tree of A, B, D, E: AE + BE + DE = 10 + 11 + 13 = 34
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A1 two shortest edges from C: CD + BC = 8 + 9 = 17
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A1 lower bound 34 + 17 = 51 km
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R1 the larger lower bound is the better one, so 51 ≤ L ≤ 53
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Part (d)
A1 12 + 9 + 8 + 13 + 10 = 52 km
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R1 this is a valid Hamiltonian cycle shorter than 53, so it gives an improved upper bound: 51 ≤ L ≤ 52
✓ 1
R1 so L is 51 or 52, and this route is at most 1 km longer than the optimal route
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Common error: For a lower bound, choose the largest value from the deleted vertex algorithm. For an upper bound, choose the smallest valid tour length. Mixing these up gives an interval that is either impossible or too wide.
340IBMI-032-03-r2-06Vectors and kinematicsPaper 2Hard14 marks
A golf ball is struck from a point O on level ground. Take horizontal and vertical components in metres, and ignore air resistance. The velocity of the ball \(t\) seconds after it is struck is \(\mathbf{v} = \begin{pmatrix}15\\20-9.8t\end{pmatrix}\) m s−1. A tree of height 17 m stands 40 m horizontally from O.
(a)
Find the initial speed of the ball and the angle its initial velocity makes with the horizontal.
(3)
(b)
Find an expression for the displacement vector \(\mathbf{r}\) of the ball from O at time \(t\).
(3)
(c)
Find the maximum height reached by the ball.
(3)
(d)
Determine whether the ball passes over the tree, stating the clearance or shortfall.
Height \(= 20\left(\frac{8}{3}\right) - 4.9\left(\frac{8}{3}\right)^2\) = 18.5 m (18.489…) A1
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18.5 > 17, so the ball clears the tree by 1.49 m (accept 1.5 m) R1
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Part (e)
\(\mathbf{v} = \begin{pmatrix}15\\-6.13\end{pmatrix}\), so speed \(= \sqrt{15^2+6.133^2}\) M1
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= 16.2 m s−1 A1
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Exam tip: To find when the ball reaches the tree, use the horizontal component, because it is linear in t. Then substitute that time into the vertical component to find the height.
341IBMI-032-03-r1-05Vectors and kinematicsPaper 2Hard13 marks
A coastguard station is at the origin. Position vectors are written as (east component, north component) in kilometres. At time t hours after an alert is raised, a drifting cargo ship has position rS = (10, 4) + t(−2, 5). At t = 0 a rescue boat leaves the station and travels in a straight line at a constant speed of 15 km h−1 so that it meets the ship. Give non-exact answers to three significant figures and bearings to one decimal place.
(a)
Find the speed of the cargo ship.
(2)
(b)
Find the distance and the bearing of the ship from the station when the alert is raised.
(3)
(c)
The rescue boat meets the ship at time T hours. Show that 196T2 = 116.
(3)
(d)
Find the time taken for the rescue boat to reach the ship, giving your answer in minutes.
(2)
(e)
Find the bearing on which the rescue boat must travel.
(3)
Show mark scheme
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Notes
Part (a)
M1: \(\sqrt{(-2)^2+5^2}=\sqrt{29}\)
✓ 1
A1: 5.39 km h−1
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Part (b)
A1: distance \(\sqrt{10^2+4^2}=\sqrt{116}\) = 10.8 km
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M1: uses tan θ = \(\frac{10}{4}\) with the angle measured clockwise from north
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A1: bearing 068.2°
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Part (c)
M1: ship position at time T is (10 − 2T, 4 + 5T)
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M1: sets the distance of this point from the origin equal to the distance travelled by the boat: (10 − 2T)2 + (4 + 5T)2 = (15T)2
✓ 1
A1: expands to 116 + 29T2 = 225T2, with the terms in T cancelling, giving 196T2 = 116
✓ 1
Part (d)
M1: solves to get T = \(\frac{\sqrt{116}}{14}\) = 0.769 h, rejecting the negative root
✓ 1
A1: 46.2 minutes, using the unrounded T (accept 46.1 from 0.769 × 60)
✓ 1
Part (e)
M1: substitutes T into the ship's position to find the meeting point
✓ 1
A1: meeting point (8.46, 7.85)
✓ 1
A1: bearing = arctan(east ÷ north) = 047.2° using unrounded values (accept 047.1° from the rounded coordinates)
✓ 1
Exam tip: The boat should be steered towards where the ship will be, not where it is now. Keep full calculator values until the final answer. To check your result, divide the meeting point by T to get the boat's velocity: its magnitude should be 15.
342IBMI-032-04-r2-06Graph theory and networksPaper 2Hard14 marks
A park ranger must inspect five shelters, A, B, C, D and E. The ranger starts and finishes at A and visits every other shelter. Not every pair of shelters is joined by a direct path. The only paths and their lengths in km are: AB 8, AC 5, BC 4, BD 6, CD 9, CE 7, DE 3. The ranger may pass through a shelter more than once.
(a)
A table of least distances is needed. Find the least distance between each of these pairs: A and D, A and E, B and E.
(3)
(b)
Use the nearest neighbour algorithm on the table of least distances, starting at A, to find an upper bound for the length of the ranger's route. Write down the route the ranger actually follows along the paths.
(4)
(c)
Use the deleted vertex algorithm on the table of least distances to find a lower bound, first by deleting A and then by deleting D. Hence state the best interval for the length L km of the shortest route that you can obtain from your bounds.
(4)
(d)
Find a Hamiltonian cycle in the original path network that is shorter than the route found using the nearest neighbour algorithm. Hence improve the interval for L.
(3)
Show mark scheme
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Notes
Part (a)
A1 for AD = 14 km (via B or via C)
✓ 1
A1 for AE = 12 km (A–C–E)
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A1 for BE = 9 km (B–D–E)
✓ 1
Part (b)
M1 for applying nearest neighbour to the table of least distances
✓ 1
A1 for the cycle A–C–B–D–E–A
✓ 1
A1 for upper bound 5 + 4 + 6 + 3 + 12 = 30 km
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A1 for the actual route A–C–B–D–E–C–A, because E to A goes through C
✓ 1
Part (c)
M1 for the method: find a minimum spanning tree of the remaining vertices, then add the two shortest edges from the deleted vertex
✓ 1
A1 for deleting A: MST DE, BC, BD = 13, plus 5 + 8, giving 26
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A1 for deleting D: MST BC, AC, CE = 16, plus 3 + 6, giving 25
✓ 1
A1 for 26 ≤ L ≤ 30, using the larger lower bound
✓ 1
Part (d)
A1 for A–B–D–E–C–A (or its reverse)
✓ 1
A1 for length 8 + 6 + 3 + 7 + 5 = 29 km
✓ 1
A1 for the improved interval 26 ≤ L ≤ 29
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Exam tip: Always give the best upper bound (the smallest) and the best lower bound (the largest). When an edge in the table of least distances is not a direct path, write out the actual path the ranger follows.
343IBMI3H-3-06-08Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard14 marks
A designer transforms a triangular logo with vertices A(1, 1), B(4, 1) and C(1, 3) using the matrix \(\mathbf{M}=\begin{pmatrix}3&1\\-1&2\end{pmatrix}\), which maps each point (x, y) to its image.
(a)
Find the coordinates of the images A′, B′ and C′ of the vertices.
(3)
(b)
The transformation M is followed by a reflection in the y-axis, with matrix \(\mathbf{Y}=\begin{pmatrix}-1&0\\0&1\end{pmatrix}\). Find the single matrix for the combined transformation.
(2)
(c)
Write down the image of A under the combined transformation.
(2)
(d)
Find \(\mathbf{M}^{-1}\).
(2)
(e)
Under M, a point P maps to (5, −4). Find the coordinates of P.
(2)
(f)
Find the area of triangle A′B′C′.
(3)
Show mark scheme
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Part (a)
M1 multiplies \(\mathbf{M}\) by the position vectors
✓ 1
A1 A′(4, 1) and B′(13, −2)
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A1 C′(6, 5)
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Part (b)
M1 multiplies in the correct order, \(\mathbf{YM}\)
✓ 1
A1 \(\begin{pmatrix}-3&-1\\-1&2\end{pmatrix}\)
✓ 1
Part (c)
M1 applies \(\mathbf{YM}\) to (1, 1) or reflects A′
M1 \(\mathbf{M}^{-1}\begin{pmatrix}5\\-4\end{pmatrix}\) (ECF from (d))
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A1 P(2, −1)
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Part (f)
A1 area of ABC \(=\frac12\times3\times2=3\)
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M1 uses area scale factor \(|\det\mathbf{M}|=7\)
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A1 area \(=21\) square units
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Exam tip: The area of an image equals the area of the original multiplied by the absolute value of the determinant of the transformation matrix.
344IBMI3H-3-06-09Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard15 marks
A courier visits five depots A, B, C, D and E, starting and finishing at A, visiting every depot once. Distances in km: AB 14, AC 9, AD 17, AE 11, BC 6, BD 12, BE 8, CD 10, CE 15, DE 7.
(a)
Use the nearest neighbour algorithm starting at A to find an upper bound for the shortest route.
(4)
(b)
Use the deleted vertex algorithm, deleting A, to find a lower bound.
(5)
(c)
A manager claims that a route of length 40 km exists. Use your answer to (b) to comment on this claim.
(2)
(d)
Find a route that is shorter than your answer to (a), and state its length.
(2)
(e)
The van uses 0.32 litres of fuel per km and fuel costs €1.65 per litre. Find the fuel cost for your route in (d).
(2)
Show mark scheme
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Notes
Part (a)
M1 starts at A and moves to the nearest unvisited depot each time
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A1 route A–C–B–E–D
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M1 returns to A
✓ 1
A1 length 47 km
✓ 1
Part (b)
M1 deletes A, leaving B, C, D, E
✓ 1
M1 finds a minimum spanning tree of B, C, D, E
✓ 1
A1 BC 6, DE 7, BE 8, total 21
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M1 adds the two shortest edges at A (AC 9 and AE 11)
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A1 lower bound 41 km
✓ 1
Part (c)
R1 the lower bound 41 km exceeds 40 km
✓ 1
A1 so the claim cannot be correct
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Part (d)
M1 gives a valid route visiting each depot once, e.g. A–C–B–D–E–A
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A1 length 45 km
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Part (e)
M1 \(45\times0.32=14.4\) litres (ECF from (d))
✓ 1
A1 €23.76
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Common error: The nearest neighbour result is only an upper bound; a shorter route may exist, and the deleted vertex value is only a lower bound.
345IBMI3H-3-06-10Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard15 marks
A triangular solar panel has corners P(0, 0, 3), Q(4, 0, 5) and R(0, 3, 4), where coordinates are in metres and the z-axis is vertical.
(a)
Find the vectors \(\overrightarrow{PQ}\) and \(\overrightarrow{PR}\).
M1 \(\mathbf{n}\cdot\mathbf{r}=\mathbf{n}\cdot\mathbf{p}\) using the normal from (b)
✓ 1
A1 \(3x+2y-6z=-18\) (accept multiples)
✓ 1
Part (e)
M1 angle between normal and \(\mathbf{k}\): \(\cos\theta=\dfrac{12}{14\times1}\)
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A1 \(\cos\theta=\frac67\)
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A1 \(31.0^\circ\)
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Part (f)
M1 \(7\times\cos\theta\) (ECF from (e))
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A1 6 m2
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Part (g)
M1 \(6\times800\times0.19\)
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A1 912 W
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Exam tip: A vector product gives a normal to the plane, and its magnitude gives twice the area of the triangle.
346IBMI3H-3-06b-11Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard13 marks
Two drones P and Q fly near a wind farm. Relative to a control point, with t in seconds and distances in metres, their positions are \(\boldsymbol{r}_P = \begin{pmatrix}0\\0\\50\end{pmatrix} + t\begin{pmatrix}4\\3\\-1\end{pmatrix}\) and \(\boldsymbol{r}_Q = \begin{pmatrix}60\\10\\20\end{pmatrix} + t\begin{pmatrix}-2\\2\\1\end{pmatrix}\).
(a)
Show that the square of the distance between the drones is \(41t^2 - 860t + 4600\).
(3)
(b)
Find the time at which the drones are closest.
(2)
(c)
Find the minimum distance between the drones.
(2)
(d)
Safety rules require the drones to stay at least 12 m apart. Find the interval of time for which this rule is broken.
(3)
(e)
Find the angle between the velocity vectors of the two drones.
M1 differentiate and set equal to zero, or use technology
✓ 1
A1 t = 860/82 = 10.5 s
✓ 1
Part (c)
M1 substitute t = 10.49 into the expression
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A1 √90.2 = 9.50 m
✓ 1
Part (d)
M1 solve 41t2 − 860t + 4600 < 144
✓ 1
A1 critical values t = 9.34 and t = 11.6
✓ 1
A1 9.34 < t < 11.6
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Part (e)
M1 attempt to use the scalar product of (4, 3, −1) and (−2, 2, 1) = −3
✓ 1
A1 |vP| = √26 and |vQ| = 3
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A1 angle = 101°
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Exam tip: Minimising the square of the distance gives the same time as minimising the distance, and avoids the square root.
347IBMI3H-3-06b-12Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard12 marks
The access roads on a wind farm join junctions A to F. The roads and their lengths in km are: AB = 4, AC = 6, BC = 3, BD = 7, CE = 5, DE = 2, DF = 6, EF = 4. An inspection vehicle must travel along every road at least once.
(a)
Show that exactly four junctions have odd degree, and state them.
(2)
(b)
Find the length of the shortest inspection route that starts and finishes at the same junction.
(5)
(c)
The vehicle instead starts at B and ends at E. Find the length of the shortest route.
(3)
(d)
The vehicle averages 35 km h−1 on the closed route from part (b). Find the time taken in minutes.
(2)
Show mark scheme
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Part (a)
M1 attempt to find the degrees
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A1 B, C, D and E have degree 3; A and F have degree 2
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Part (b)
M1 consider the three pairings of B, C, D and E
✓ 1
A1 BC + DE = 3 + 2 = 5
✓ 1
A1 BD + CE = 7 + 5 = 12 and BE + CD = 8 + 7 = 15
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A1 sum of all roads = 37
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A1 shortest closed route = 37 + 5 = 42 km
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Part (c)
M1 B and E stay odd, so pair C and D
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A1 shortest path from C to D is C–E–D = 7 km
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A1 37 + 7 = 44 km
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Part (d)
M1 42/35 hours
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A1 72 minutes
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Common error: Pairing odd vertices using only direct roads; a two-edge path can be shorter than a direct road.
348IBMI3H-3-06b-13Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard12 marks
A transformation of the plane is represented by the matrix M = \(\begin{pmatrix}a&b\\c&d\end{pmatrix}\). It maps A(2, 1) to A′(5, 4) and B(1, 3) to B′(10, 7).
(a)
Find M.
(4)
(b)
Find det M and state what it shows about the effect of M on area.
(2)
(c)
Find the coordinates of the point whose image is (7, 5).
(3)
(d)
Find M2 and hence the image of A under two applications of the transformation.
(3)
Show mark scheme
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Part (a)
M1 form equations 2a + b = 5 and a + 3b = 10 (and similarly for c, d)
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M1 attempt to solve the simultaneous equations
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A1 a = 1, b = 3
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A1 c = 1, d = 2, so M = \(\begin{pmatrix}1&3\\1&2\end{pmatrix}\)
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Part (b)
A1 det M = 2 − 3 = −1
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R1 areas are unchanged in size, but orientation is reversed because the determinant is negative
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Part (c)
M1 attempt to use M−1 = \(\begin{pmatrix}-2&3\\1&-1\end{pmatrix}\)
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A1 correct inverse applied to (7, 5)
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A1 (1, 2)
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Part (d)
A1 M2 = \(\begin{pmatrix}4&9\\3&7\end{pmatrix}\)
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M1 multiply M2 by (2, 1)
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A1 (17, 13)
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Exam tip: If an image is known for two independent points, write one pair of simultaneous equations for each row of the matrix.
349IBMI3H-3-03r6-09Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard12 marks
Calculator
Two aircraft, P and Q, are flying in a region of airspace. Distances are in kilometres relative to a control tower, with i east, j north and k vertically upwards, and time t is in minutes after 14:00. The position vector of P is rP = (−20, 30, 10) + t(6, −4, 0) and the position vector of Q is rQ = (50, −20, 9) + t(−2, 3, 0.1).
(a)
Find the speed of P in km per minute. Give your answer to 3 significant figures.
(2)
(b)
Find the position vector of Q at 14:05.
(2)
(c)
Show that the vector from P to Q at time t is (70 − 8t, −50 + 7t, −1 + 0.1t).
(2)
(d)
Hence find the least distance between the two aircraft, and the time at which it occurs. Give the distance in kilometres to 3 significant figures and the time in minutes after 14:00 to 3 significant figures.
(5)
(e)
Air traffic rules require at least 9 km between aircraft at all times. State whether this rule is broken, giving a reason.
(1)
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Marking point
Mark
Notes
Part (a)
M1 attempts the magnitude of the direction vector (6, −4, 0)
✓ 1
A1 speed = √52 = 7.21 km per minute
✓ 1
Part (b)
M1 substitutes t = 5
✓ 1
A1 rQ = (40, -5, 9.5)
✓ 1
Part (c)
M1 subtracts rP from rQ
✓ 1
A1 collects the components to reach the given result (AG)
M1 finds the minimum of D2 using the vertex, a derivative or a graphing tool
✓ 1
A1 t = 8.05 minutes after 14:00
✓ 1
A1 least distance = 8.47 km
✓ 1
Part (e)
R1 the least distance (8.47 km) is less than 9 km, so the rule is broken (follow through from the previous part)
✓ 1
Exam tip: Minimising the squared distance gives the same time as minimising the distance and keeps the quadratic exact.
350IBMI3H-3-03r6-12Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard10 marks
Calculator
A delivery driver must travel along every one of the tracks on a farm estate at least once. The tracks join five junctions A, B, C, D and E, and their lengths in km are: AB 5, BC 4, CD 6, DE 3, EA 7, AC 8, BD 9. The driver starts and finishes at A.
(a)
Explain why the driver cannot travel along every track exactly once and return to A, and identify the vertices of odd degree.
(2)
(b)
Find the three ways of pairing the odd vertices, and the shortest repeated distance for each pairing.
(3)
(c)
Find the length of the shortest route that uses every track at least once and returns to A.
(2)
(d)
The driver now only needs to start at A and finish at D, with no need to return. Find the length of the shortest route.
(3)
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Notes
Part (a)
R1 a closed route using each edge once needs every vertex to have even degree
✓ 1
A1 vertices A, B, C and D each have degree 3 (odd), so no such route exists
✓ 1
Part (b)
M1 lists the pairings AB + CD, AC + BD and AD + BC
✓ 1
A1 AB + CD = 5 + 6 = 11, and AC + BD = 8 + 9 = 17 (using shortest paths)
✓ 1
A1 AD + BC = 10 (via E) + 4 = 14
✓ 1
Part (c)
M1 adds the total of all track lengths, 42 km, to the smallest pairing total
✓ 1
A1 shortest route = 42 + 11 = 53 km
✓ 1
Part (d)
M1 recognises that only B and C are now odd, so the repeated path joins B and C
✓ 1
A1 shortest path from B to C = 4 km
✓ 1
A1 shortest route = 42 + 4 = 46 km
✓ 1
Common error: Use shortest paths between odd vertices, not just direct edges; here the shortest A to D path goes through E.
351IBMI3H-3-03r6-13Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard10 marks
Calculator
A band plans a tour that visits five cities A, B, C, D and E exactly once each, starting and finishing in A. The direct road distances in km are: AB 120, AC 210, AD 160, AE 300, BC 140, BD 250, BE 190, CD 180, CE 230, DE 100. The tour bus costs €1.85 per km to run.
(a)
Use the nearest neighbour algorithm starting at A to find an upper bound for the length of the tour. State the route.
(3)
(b)
Use the deleted vertex algorithm, deleting E, to find a lower bound for the length of the tour.
(4)
(c)
Hence write down the interval in which the length of the shortest tour lies, and find the greatest possible running cost of the optimal tour using the upper bound.
(3)
Show mark scheme
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Mark
Notes
Part (a)
M1 attempts nearest neighbour from A: A to B, then B to C
✓ 1
A1 route A–B–C–D–E–A
✓ 1
A1 length = 120 + 140 + 180 + 100 + 300 = 840 km
✓ 1
Part (b)
M1 finds a minimum spanning tree of A, B, C and D
✓ 1
A1 edges AB, BC, AD: 120 + 140 + 160 = 420 km
✓ 1
M1 adds the two shortest edges from E: DE and BE
✓ 1
A1 lower bound = 420 + 100 + 190 = 710 km
✓ 1
Part (c)
A1 710 ≤ length ≤ 840 km
✓ 1
M1 multiplies 840 by 1.85
✓ 1
A1 greatest cost = €1554 (accept 1554.00)
✓ 1
Exam tip: Both bounds give a range for the optimal tour, not its exact length: the upper bound comes from any real tour and the lower bound from the deleted vertex method.
352IBMI3H-3-04r9-04Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard11 marks
A tracer particle moves in the horizontal plane of a large food-mixing vat. Positions are in centimetres relative to a sensor at the centre of the vat, with i east and j north. At time t = 0 seconds the particle is at (1, −2), and its velocity is v = (2t − 4)i + (6 − t)j cm/s, for 0 ≤ t ≤ 6.
(a)
Find an expression for the position vector r(t) of the particle.
(4)
(b)
Find the position of the particle at t = 4.
(1)
(c)
The particle is closest to the sensor at time T. Find T and the minimum distance from the sensor, each to 3 significant figures.
(4)
(d)
Find the speed of the particle at time T, using the unrounded value of T, to 3 significant figures.
(2)
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Notes
Part (a)
M1 Attempt to integrate each component of v
✓ 1
A1 r = (t2 − 4t + c1)i + (6t − 0.5t2 + c2)j
✓ 1
M1 Use of r(0) = (1, −2) to find the constants
✓ 1
A1 r = (t2 − 4t + 1)i + (6t − 0.5t2 − 2)j
✓ 1
Part (b)
A1 (1, 14)
✓ 1
Part (c)
M1 Distance = √(x2 + y2) written as a function of t
✓ 1
M1 Use of a graph or minimum function on a calculator
✓ 1
A1 T = 0.323 s
✓ 1
A1 Minimum distance = 0.220 cm
✓ 1
Part (d)
A1 v(T) = (2T − 4, 6 − T) = (−3.35, 5.68)
✓ 1
A1 Speed = 6.59 cm/s
✓ 1
Exam tip: Minimise the distance (or its square) as a function of t; the point of closest approach is generally not where the velocity is zero.
353IBMI3H-3-04r9-10Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard9 marks
A courier based at laboratory A collects food samples from labs B, C, D and E and returns to A, visiting each lab once. The distances in kilometres are: AB 12, AC 9, AD 15, AE 11, BC 8, BD 10, BE 14, CD 13, CE 7, DE 6.
(a)
Use the nearest neighbour algorithm starting at A to find an upper bound for the shortest tour.
(3)
(b)
Use the deleted vertex method, deleting A, to find a lower bound for the shortest tour.
(4)
(c)
Find the percentage by which the upper bound exceeds the lower bound, to 3 significant figures.
(2)
Show mark scheme
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Notes
Part (a)
A1 Route A → C (9) → E (7) → D (6) → B (10)
✓ 1
A1 Return B → A (12)
✓ 1
A1 Upper bound = 44 km
✓ 1
Part (b)
A1 Minimum spanning tree of B, C, D and E uses DE 6, CE 7 and BC 8
✓ 1
A1 MST weight = 21
✓ 1
M1 Add the two shortest edges at A: AC 9 and AE 11
✓ 1
A1 Lower bound = 41 km
✓ 1
Part (c)
M1 (44 − 41) ÷ 41 × 100
✓ 1
A1 = 7.32%
✓ 1
Exam tip: The deleted vertex method gives a lower bound only: the true optimum lies between the two bounds, not necessarily at either.
354IBMI3H-3-04r9-12Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard12 marks
On a map grid (units of kilometres), a linear transformation represented by the matrix \(M=\begin{pmatrix}a&b\\c&d\end{pmatrix}\) is used to model how the layout of an antenna array is distorted. It maps the point (1, 2) to (4, 5) and the point (3, 1) to (7, 0).
(a)
Find a, b, c and d.
(4)
(b)
A region of area 5 km2 is transformed by M. Find the area of the image.
(2)
(c)
Find the point whose image under M is (7, −7).
(2)
(d)
The image from M is then reflected in the x-axis. Find the single matrix for the combined transformation, and the distance from the origin to the final image of (4, −1), to 3 significant figures.
(4)
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Part (a)
M1 Set up a + 2b = 4, 3a + b = 7, c + 2d = 5 and 3c + d = 0
✓ 1
M1 Solve the pairs of simultaneous equations
✓ 1
A1 a = 2, b = 1
✓ 1
A1 c = −1, d = 3
✓ 1
Part (b)
A1 det M = 2 × 3 − 1 × (−1) = 7
✓ 1
A1 Image area = 35 km2
✓ 1
Part (c)
M1 Use of M−1 = (1/7)[[3, −1], [1, 2]]
✓ 1
A1 Multiply to obtain (4, −1)
✓ 1
Part (d)
M1 Combined matrix is the reflection matrix multiplied on the left of M, with reflection matrix [[1, 0], [0, −1]]
✓ 1
A1 Combined matrix = [[2, 1], [1, −3]]
✓ 1
A1 Image of (4, −1) = (7, 7)
✓ 1
A1 Distance = 7√2 = 9.90 km
✓ 1
Common error: For a transformation applied first and one applied second, the second matrix goes on the left of the product.
355IBMI3H-3-04r12-02Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard10 marks
Calculator
Two ships move at constant velocity in open water. Relative to a harbour at the origin, with distances in kilometres and time t in hours after noon, the position of ship A is \(\mathbf{r}_A=\begin{pmatrix}2\\5\end{pmatrix}+t\begin{pmatrix}4\\-3\end{pmatrix}\) and the position of ship B is \(\mathbf{r}_B=\begin{pmatrix}14\\-1\end{pmatrix}+t\begin{pmatrix}-2\\5\end{pmatrix}\).
(a)
Show that the vector from A to B at time t is \(\begin{pmatrix}12-6t\\8t-6\end{pmatrix}\).
(2)
(b)
Find the time at which the ships are closest together, and the distance between them at that time.
(4)
(c)
The ships' captains must keep at least 8 km apart. Find the length of time, in hours, for which the ships are less than 8 km apart. Give your answer to 3 significant figures.
M1 minimises \(D^2\) (or \(D\)): derivative zero or GDC minimum
✓ 1
A1 \(t=1.2\) hours (1:12 pm)
✓ 1
A1 minimum distance \(\sqrt{36}=6.00\) km
✓ 1
Part (c)
M1 forms \(100t^2-240t+180<64\), that is \(100t^2-240t+116<0\)
✓ 1
A1 roots \(t=0.6708\) and \(t=1.7292\) (hours)
✓ 1
M1 duration is difference of roots
✓ 1
A1 1.06 hours
✓ 1
Exam tip: To find a closest approach, work with the squared distance: it has its minimum at the same time and avoids the square root.
356IBMI3H-3-04r12-10Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard9 marks
A touring theatre company must visit five towns A, B, C, D and E, starting and finishing in town A and visiting every other town exactly once. The travel costs, in pounds, between towns are:
A
B
C
D
E
A
–
40
55
70
35
B
40
–
30
65
50
C
55
30
–
45
60
D
70
65
45
–
25
E
35
50
60
25
–
(a)
Use the nearest neighbour algorithm starting at A to find an upper bound for the minimum cost of the tour. State the route and its cost.
(3)
(b)
By deleting town E, find a lower bound for the minimum cost of the tour.
(4)
(c)
Hence state the minimum cost of the tour, and give a reason for your answer.
(2)
Show mark scheme
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Mark
Notes
Part (a)
M1 at each town moves to the cheapest unvisited town
✓ 1
A1 route A–E–D–C–B–A
✓ 1
A1 cost \(35+25+45+30+40=175\), so the upper bound is £175
✓ 1
Part (b)
M1 minimum spanning tree of \(A\), \(B\), \(C\), \(D\) found by Kruskal's or Prim's algorithm
✓ 1
A1 tree uses BC (30), AB (40), CD (45), total 115
✓ 1
M1 adds the two cheapest links to E: DE (25) and AE (35)
✓ 1
A1 lower bound \(115+25+35=175\)
✓ 1
Part (c)
R1 the upper bound equals the lower bound, so no tour can cost less than 175 and the tour found costs exactly 175
✓ 1
A1 minimum cost £175
✓ 1
Exam tip: A lower bound from a deleted vertex is only a bound on the cost; it can show a tour is optimal only if a tour of exactly that cost has been found.
357IBMI3H-3-04r12-12Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard9 marks
Calculator
On a navigation chart, positions are given relative to a buoy at the origin. A reflection R in the line through the origin that makes an angle of \(30^\circ\) with the positive x-axis is applied to a map.
(a)
Show that the matrix of R is \(\begin{pmatrix}\frac12&\frac{\sqrt3}{2}\\\frac{\sqrt3}{2}&-\frac12\end{pmatrix}\).
(2)
(b)
The map is then enlarged by scale factor 3 about the origin. Find the matrix N of the combined transformation (reflection followed by enlargement), calculate \(\det\mathbf{N}\) and state what it shows about the effect on areas and on orientation.
(3)
(c)
After the reflection R, a shear with matrix \(\begin{pmatrix}1&k\\0&1\end{pmatrix}\) is applied. The point \(P(4,0)\) is mapped by the reflection and then the shear to a point with x-coordinate 10. Find the value of \(k\), giving your answer to 3 significant figures.
(4)
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Part (a)
M1 uses the reflection matrix \(\begin{pmatrix}\cos2\theta&\sin2\theta\\\sin2\theta&-\cos2\theta\end{pmatrix}\) with \(\theta=30^\circ\) (or tracks the images of \((1,0)\) and \((0,1)\))
✓ 1
A1 \(\cos60^\circ=\frac12\) and \(\sin60^\circ=\frac{\sqrt3}{2}\) giving the given matrix (AG)
R1 areas are multiplied by \(|{-9}|=9\) and the negative sign shows orientation is reversed
✓ 1
Part (c)
M1 image of \(P\) under R: \((2,\,2\sqrt3)\)
✓ 1
M1 shear: \(x'=2+k\times2\sqrt3\)
✓ 1
M1 sets \(2+2\sqrt3k=10\)
✓ 1
A1 \(k=\frac{4}{\sqrt3}=2.31\)
✓ 1
Common error: Apply the transformations in the stated order: the matrix of 'R then shear' is \(\mathbf{S}\mathbf{R}\), not \(\mathbf{R}\mathbf{S}\).
358IBMI3H-3-03r16-06Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard11 marks
A Formula 1 team’s truck leaves the factory A, must visit four storage depots B, C, D and E exactly once each, and then returns to A. The distances in kilometres between sites are: AB 14, AC 22, AD 18, AE 29, BC 26, BD 11, BE 24, CD 20, CE 15, DE 23.
(a)
Use the nearest neighbour algorithm starting at A to find an upper bound for the shortest possible length of the journey.
(3)
(b)
By deleting vertex A, find a lower bound for the length of the journey.
(5)
(c)
The tour A–B–D–E–C–A is found. Calculate its length and use it together with your answer to (b) to write down an interval that contains the shortest possible length of the journey.
(3)
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Part (a)
M1 move each time to the nearest unvisited vertex
✓ 1
A1 A–B–D–C–E–A
✓ 1
A1 14 + 11 + 20 + 15 + 29 = 89 km
✓ 1
Part (b)
M1 find a minimum spanning tree of the graph on B, C, D, E
✓ 1
A1 edges BD (11), CE (15) and CD (20)
✓ 1
A1 weight 46 km
✓ 1
M1 add the two shortest edges at A (AB 14 and AD 18)
✓ 1
A1 lower bound 78 km
✓ 1
Part (c)
M1 14 + 11 + 23 + 15 + 22
✓ 1
A1 85 km
✓ 1
A1 shortest length is between 78 km and 85 km
✓ 1
Common error: A lower bound from deleting a vertex uses the two shortest edges at that vertex; using only one, or using the nearest-neighbour tour length, does not give a valid lower bound.
359IBMI3H-3-03r16-08Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard10 marks
The tessellated cells of a honeycomb are modelled in a plane. A transformation T is an enlargement, centre the origin with scale factor k > 0, followed by an anticlockwise rotation of 60° about the origin. T maps the point (4, 0) to the point \((3,\,3\sqrt{3})\).
(a)
Show that k = 1.5.
(3)
(b)
Write down the matrix T that represents the transformation, giving exact entries.
(2)
(c)
Find the coordinates of the point that is mapped by T to \((-3\sqrt{3},\,3)\).
(3)
(d)
A region of area 8 cm2 is transformed by T. Find the area of its image.
(2)
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Notes
Part (a)
M1 image of (4, 0) is \((4k\cos60^\circ,\,4k\sin60^\circ)\)
M1 area scale factor \(|\det\mathbf{T}|=\frac94\) (or \(k^2\))
✓ 1
A1 18 cm2
✓ 1
Exam tip: The area scale factor of any matrix transformation is the absolute value of its determinant, so a rotation never changes area and an enlargement multiplies it by the square of the scale factor.
360IBMI3H-3-04r16-05Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard10 marks
A Formula 1 circuit has six garages A to F joined by service roads. A road inspector must drive along every road. The road lengths in metres are: AB 110, AE 120, AF 180, BC 190, BE 70, CD 200, CF 130, DF 140, EF 50. The total length of all the roads is 1190 m.
(a)
Write down the vertices of odd degree.
(2)
(b)
The inspector must start and finish at the same garage. Find the length of the shortest route that covers every road at least once.
(5)
(c)
The inspector may now start and finish at different garages. Find the length of the shortest route, and state where it starts and finishes.
(3)
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Part (a)
M1 attempt to find degrees (A 3, B 3, C 3, D 2, E 3, F 4)
✓ 1
A1 A, B, C and E
✓ 1
Part (b)
M1 consider the three ways of pairing the four odd vertices
✓ 1
A1 AB + CE = 110 + 180 = 290 (CE via F)
✓ 1
A1 AC + BE = 300 + 70 = 370 and AE + BC = 120 + 190 = 310
✓ 1
M1 repeat the pairing of least total length
✓ 1
A1 1190 + 290 = 1480 m
✓ 1
Part (c)
M1 start and finish at two odd vertices, repeating the shortest path between the other two
✓ 1
A1 repeat B–E (70), start and finish at A and C
✓ 1
A1 1190 + 70 = 1260 m
✓ 1
Exam tip: For four odd vertices there are three pairings to compare; each pair must be joined by its shortest path, which is not always a single direct road.
361IBMI3H-3-04r16-06Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard11 marks
A courier must deliver medicines from a central hospital C to four other hospitals A, B, D and E, visiting each exactly once before returning to C. The distances in kilometres are: AB 19, AC 10, AD 12, AE 23, BC 16, BD 21, BE 13, CD 14, CE 9, DE 17.
(a)
Use the nearest neighbour algorithm starting at C to find an upper bound for the shortest possible journey, giving the route.
(3)
(b)
By deleting C, find a lower bound for the length of the journey.
(5)
(c)
Show that the route C–A–D–B–E–C has length 65 km. What does this tell you about the shortest possible journey?
(3)
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Notes
Part (a)
M1 go to the nearest unvisited hospital each time
✓ 1
A1 C–E–B–A–D–C
✓ 1
A1 9 + 13 + 19 + 12 + 14 = 67 km
✓ 1
Part (b)
M1 minimum spanning tree of the graph on A, B, D, E
✓ 1
A1 edges AD (12), BE (13), DE (17)
✓ 1
A1 weight 42 km
✓ 1
M1 add the two shortest edges at C (CE 9 and AC 10)
✓ 1
A1 lower bound 61 km
✓ 1
Part (c)
M1 10 + 12 + 21 + 13 + 9
✓ 1
A1 65 km
✓ 1
A1 the shortest journey is between 61 km and 65 km
✓ 1
Exam tip: A tour you can write down gives an upper bound and a deleted-vertex calculation gives a lower bound; the shortest tour lies in the interval between your best bounds.
362IBMI3H-3-04r16-08Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard10 marks
A logo is transformed by the matrix M that represents a horizontal stretch with scale factor k (k > 0), centre the y-axis, followed by a reflection in the x-axis. A triangle of area 6 cm2 is mapped by M to a triangle of area 21 cm2.
(a)
Write down the matrix M in terms of k.
(2)
(b)
Find the value of k.
(3)
(c)
Find the coordinates of the point that is mapped by M to (14, 5).
(3)
(d)
A circular badge of radius 2 cm is transformed by M. Find the area of its image, in cm2 to 3 significant figures.
(2)
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Notes
Part (a)
M1 stretch matrix \(\begin{pmatrix}k&0\\0&1\end{pmatrix}\) and reflection matrix \(\begin{pmatrix}1&0\\0&-1\end{pmatrix}\)
M1 use of \(\mathbf{M}^{-1}\) or equations \(3.5x=14\), \(-y=5\)
✓ 1
A1 x = 4
✓ 1
A1 y = −5, so the point is (4, −5)
✓ 1
Part (d)
M1 \(\pi(2)^2\times3.5\)
✓ 1
A1 44.0 cm2
✓ 1
Exam tip: A negative determinant signals a reflection (orientation reversed), but the area scale factor is always the absolute value of the determinant.
363IBMI3H-3-06r20-12Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard9 marks
Calculator
A touring sports team must visit five towns A, B, C, D and E, starting and finishing at A and visiting each other town exactly once. The distances between towns, in kilometres, are: AB 34, AC 52, AD 41, AE 60, BC 29, BD 47, BE 55, CD 38, CE 44, DE 31.
(a)
Use the nearest neighbour algorithm starting at A to find an upper bound for the shortest tour. State the tour and its length.
(3)
(b)
By deleting town A, find a lower bound for the shortest tour.
(4)
(c)
Show that the tour ABCEDA has length 179 km. State what this tells you about the length of the shortest tour.
(2)
Show mark scheme
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Notes
Part (a)
M1 repeatedly move to the nearest unvisited town: A to B (34), B to C (29), C to D (38), D to E (31)
✓ 1
A1 return E to A (60), tour ABCDEA
✓ 1
A1 length \(=192\) km, an upper bound
✓ 1
Part (b)
M1 minimum spanning tree of B, C, D, E
✓ 1
A1 edges BC 29, DE 31, CD 38, total 98
✓ 1
M1 add the two shortest edges from A: 34 and 41
✓ 1
A1 lower bound \(=98+34+41=173\) km
✓ 1
Part (c)
A1 \(34+29+44+31+41=179\) km
✓ 1
R1 the shortest tour is at most 179 km, so it lies between 173 km and 179 km
✓ 1
Exam tip: The nearest neighbour answer gives an upper bound, and deleting a vertex gives a lower bound. A tour you find by inspection can tighten the upper bound.
364IBMI3H-3-06r20-13Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard9 marks
Calculator
A tracking drone films a cycling time trial. Relative to its launch point, its velocity at time \(t\) seconds is \(\mathbf{v}=\begin{pmatrix}4t\\12-3t\end{pmatrix}\) m s−1 for \(0\le t\le4\), with the components giving east and north. The drone is at the launch point when \(t=0\).
(a)
Find the acceleration of the drone.
(1)
(b)
Find the position vector of the drone at time \(t\).
(2)
(c)
Find the minimum speed of the drone, and the time at which it occurs.
(4)
(d)
Find the position of the drone at the instant of minimum speed, to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(\begin{pmatrix}4\\-3\end{pmatrix}\) m s−2
✓ 1
Part (b)
M1 integrate each component of \(\mathbf{v}\)
✓ 1
A1 \(\mathbf{r}=\begin{pmatrix}2t^2\\12t-1.5t^2\end{pmatrix}\) (with constants of integration zero)
✓ 1
Part (c)
M1 speed \(=\sqrt{16t^2+(12-3t)^2}\)
✓ 1
M1 minimise the speed (or its square \(25t^2-72t+144\)) by differentiating or by using technology
✓ 1
A1 \(t=1.44\) s
✓ 1
A1 minimum speed \(=9.6\) m s−1
✓ 1
Part (d)
M1 substitute \(t=1.44\) into the position vector from part (b)
✓ 1
A1 \((4.15,\ 14.2)\) m
✓ 1
Common error: The minimum of the speed occurs at the same time as the minimum of the speed squared, which is a quadratic here. Do not confuse minimum speed with zero velocity: the velocity never vanishes in this motion.
365IBMI3H-3-06r20-14Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard10 marks
Calculator
A space probe moves along the line \(L:\ \mathbf{r}=\begin{pmatrix}2\\-1\\4\end{pmatrix}+\lambda\begin{pmatrix}1\\2\\2\end{pmatrix}\), where positions are in thousands of kilometres relative to a tracking station and \(\lambda\) is the time in hours. A fixed relay satellite sits at \(P(6,3,1)\). Let \(A(2,-1,4)\) be the probe's position at \(\lambda=0\) and let \(\mathbf{d}=\begin{pmatrix}1\\2\\2\end{pmatrix}\).
(a)
Find \(\overrightarrow{AP}\) and calculate \(\overrightarrow{AP}\times\mathbf{d}\).
(3)
(b)
Hence find the shortest distance from P to the line \(L\), to 3 significant figures.
(2)
(c)
Find the position of the probe when it is closest to P.
(3)
(d)
Find the time at which the probe is closest to P, in minutes, and the speed of the probe in km h−1.
M1 the closest point satisfies \(\left(\overrightarrow{AP}-\lambda\mathbf{d}\right)\cdot\mathbf{d}=0\)
✓ 1
A1 \(6-9\lambda=0\), so \(\lambda=\frac23\)
✓ 1
A1 \(\left(\frac{8}{3},\ \frac{1}{3},\ \frac{16}{3}\right)\), i.e. \((2.67,\ 0.333,\ 5.33)\)
✓ 1
Part (d)
A1 time \(=\frac23\) hour \(=40\) minutes
✓ 1
A1 speed \(=|\mathbf{d}|\times1000=3000\) km h−1
✓ 1
Exam tip: There are two routes to the closest distance: the vector product formula, or finding where the line from P meets the path at right angles. Using both is a good check.
366IBMI3H-3-04r23-04Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard12 marks
Two drones, A and B, inspect a wind farm. Relative to the control hub, in metres, their position vectors at time t seconds (t ≥ 0) are rA = (10 + 3t, 40 − 2t) and rB = (70 − 2t, 10 + 3t).
(a)
Write down the velocity of drone A and find its speed, to 3 significant figures.
(2)
(b)
Find the vector d(t) = rB − rA, the displacement from A to B.
(2)
(c)
Show that the square of the distance between the drones is |d|2 = 50t2 − 900t + 4500.
(3)
(d)
The drones are closest when the velocity of B relative to A is perpendicular to d(t). Use a scalar product to find the time at which they are closest.
(3)
(e)
Find the minimum distance between the drones, to 3 significant figures.
A1 collects terms to reach 50t2 − 900t + 4500 (no incorrect working)
✓ 1
Part (d)
A1 relative velocity of B relative to A = (−2, 3) − (3, −2) = (−5, 5)
✓ 1
M1 sets (60 − 5t)(−5) + (5t − 30)(5) = 0
✓ 1
A1 50t − 450 = 0, so t = 9 s
✓ 1
Part (e)
M1 substitutes t = 9: d(9) = (15, 15) or |d|2 = 450
✓ 1
A1 √450 = 21.2 m
✓ 1
Exam tip: To find a minimum distance between moving objects, either minimise |d|² or use the perpendicularity of relative velocity and displacement; both give the same time.
367IBMI3H-3-04r23-09Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard13 marks
A maintenance crew must visit five turbines A, B, C, D and E, starting and finishing at A, visiting each turbine once. The distances between turbines, in km, are: AB 12, AC 9, AD 15, AE 11, BC 8, BD 10, BE 14, CD 13, CE 7, DE 6.
(a)
Use the nearest neighbour algorithm starting at A to find a tour and its length.
(3)
(b)
State what this tour gives for the length of the shortest possible tour, and explain why.
(2)
(c)
Find a lower bound by deleting vertex E, finding a minimum spanning tree of the remaining vertices and adding the two shortest edges from E.
(4)
(d)
Repeat the lower bound calculation by deleting vertex C instead. State the best lower bound you now have and the interval in which the shortest tour length lies.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 at each stage moves to the nearest unvisited turbine
✓ 1
A1 tour A–C–E–D–B–A
✓ 1
A1 length 9 + 7 + 6 + 10 + 12 = 44 km
✓ 1
Part (b)
A1 an upper bound of 44 km
✓ 1
R1 the tour is a valid route, so the shortest tour cannot be longer than it
✓ 1
Part (c)
M1 minimum spanning tree of A, B, C, D using BC 8, AC 9, BD 10
✓ 1
A1 length 27 km
✓ 1
M1 adds the two shortest edges at E: ED 6 and EC 7
✓ 1
A1 lower bound 27 + 13 = 40 km
✓ 1
Part (d)
M1 minimum spanning tree of A, B, D, E using DE 6, BD 10, AE 11 = 27 km
✓ 1
M1 adds the two shortest edges at C: CE 7 and CB 8, giving 27 + 15
✓ 1
A1 lower bound 42 km, which is greater than 40 km
✓ 1
A1 the shortest tour length is between 42 km and 44 km
✓ 1
Exam tip: Different deleted vertices give different lower bounds; always quote the largest lower bound and the smallest upper bound.
368IBMI3H-3-04r23-11Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard12 marks
The blades of a wind turbine rotate clockwise at a constant 15 revolutions per minute. Take the hub as the origin, with x horizontal and y vertical (in metres). At time t = 0 the tip of blade 1 is at (0, 40). The matrix for a clockwise rotation through an angle θ about the origin is (cos θ, sin θ; −sin θ, cos θ), meaning first row cos θ, sin θ and second row −sin θ, cos θ. The hub is 85 m above the ground.
(a)
Find the angle through which the blades have turned when t = 2.5 s.
(2)
(b)
Write down the rotation matrix for this angle, with entries to 3 significant figures.
(2)
(c)
Hence find the coordinates of the tip of blade 1 when t = 2.5 s, to 3 significant figures.
(2)
(d)
Find the height of this tip above the ground, to 3 significant figures.
(2)
(e)
Blade 2 is 120° further round clockwise from blade 1 at all times. Find the coordinates of the tip of blade 2 when t = 2.5 s, to 3 significant figures.
(2)
(f)
Find the distance between the tips of the two blades when t = 2.5 s, to 3 significant figures.
(2)
Show mark scheme
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Mark
Notes
Part (a)
M1 15 rev/min = 0.25 rev/s = 90° per second
✓ 1
A1 90 × 2.5 = 225°
✓ 1
Part (b)
A1 (cos 225°, sin 225°; −sin 225°, cos 225°)
✓ 1
A1 (−0.707, −0.707; 0.707, −0.707)
✓ 1
Part (c)
M1 multiplies the matrix by the column vector (0, 40)
✓ 1
A1 (−28.3, −28.3)
✓ 1
Part (d)
M1 adds the hub height: 85 + (−28.28…)
✓ 1
A1 56.7 m
✓ 1
Part (e)
M1 uses a total rotation of 225° + 120° = 345° applied to its starting point (0, 40)
✓ 1
A1 (−10.4, 38.6)
✓ 1
Part (f)
M1 finds the difference of the position vectors, or uses 2 × 40 × sin 60° for the chord
✓ 1
A1 69.3 m
✓ 1
Common error: A clockwise rotation matrix has the opposite signs of sin θ to the anticlockwise matrix; test it on (0, 40) with 90° to check the tip ends at (40, 0).
369IBMI3H-3-05r23-03Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard13 marks
Calculator
Two survey probes, A and B, move through space at constant velocity. Relative to a station, their positions after t hours are rA = (0, 10, 4) + t(3, −2, 1) and rB = (12, 0, 10) + t(−1, 2, −1), with all distances in thousands of kilometres.
(a)
Show that the position vector of B relative to A at time t is (12 − 4t, −10 + 4t, 6 − 2t).
(2)
(b)
Find the time at which the probes are closest together, and the least distance between them, each correct to 3 significant figures.
(4)
(c)
Find the angle between the directions of travel of the two probes, correct to 1 decimal place.
(3)
(d)
Show that the paths followed by the two probes do not intersect.
(4)
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Notes
Part (a)
M1 rB − rA
✓ 1
A1 (12 − 0 − 3t − t, 0 − 10 + 2t + 2t, 10 − 4 − t − t), which simplifies to the given vector
M1 minimise (set derivative to zero, or use the minimum of the graph)
✓ 1
A1 t = 2.78 hours
✓ 1
A1 minimum distance = 1.49 thousand km
✓ 1
Part (c)
M1 scalar product of direction vectors (3, −2, 1)·(−1, 2, −1) = −8
✓ 1
M1 substitute into cos θ = (a·b)/(|a||b|) with |a| = √14 and |b| = √6
✓ 1
A1 θ = 150.8°
✓ 1
Part (d)
M1 equate the paths using different parameters: 3s = 12 − u, 10 − 2s = 2u, 4 + s = 10 − u
✓ 1
M1 attempt to solve two of the equations
✓ 1
A1 the second equation gives u = 5 − s and the third gives u = 6 − s
✓ 1
R1 these are inconsistent, so there is no common point and the paths do not intersect
✓ 1
Common error: When testing whether two lines meet, use a separate parameter for each line; using the same t only tests whether the probes collide.
370IBMI3H-3-05r23-10Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard10 marks
Calculator
An imaging system on a space probe maps a grid of points using a linear transformation represented by the matrix T = \(\begin{pmatrix}a & b\\c & d\end{pmatrix}\). The transformation maps the point (2, 1) to (5, 6) and the point (1, 3) to (0, 8).
(a)
Find the values of a, b, c and d.
(4)
(b)
A region of the grid has area 7.5 cm2. Find the area of its image under the transformation.
(2)
(c)
Find T−1.
(2)
(d)
Find the coordinates of the point that is mapped to (10, 4).
(2)
Show mark scheme
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Mark
Notes
Part (a)
M1 form equations such as 2a + b = 5 and a + 3b = 0
✓ 1
M1 form equations for the second row, 2c + d = 6 and c + 3d = 8
✓ 1
A1 a = 3, b = −1
✓ 1
A1 c = 2, d = 2
✓ 1
Part (b)
M1 determinant of T = 3×2 − (−1)×2 = 8
✓ 1
A1 area 60 cm2
✓ 1
Part (c)
M1 use of 1/det × swap leading diagonal and change signs on the other diagonal
✓ 1
A1 (1/8) × the matrix with rows (2, 1) and (−2, 3)
✓ 1
Part (d)
M1 multiply T−1 by the column vector (10, 4)
✓ 1
A1 (3, −1)
✓ 1
Exam tip: The columns of a transformation matrix are the images of the unit vectors; if you are given other points instead, form a matrix equation and use an inverse.
371IBMI3H-3-03r24-06Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard9 marks
A warehouse robot moves in a plane so that its velocity at time \(t\) seconds is \(\mathbf{v}=\begin{pmatrix}4\cos 2t\\3\sin t\end{pmatrix}\) m s−1. At \(t=0\) its position vector is \(\begin{pmatrix}1\\2\end{pmatrix}\) m. In this question all angles are in radians.
(a)
Find the acceleration of the robot at time \(t\).
(2)
(b)
Find the position vector of the robot at time \(t\).
(5)
(c)
Find the distance of the robot from its starting point at \(t=3\), giving your answer to 3 significant figures.
M1 use the position at \(t=0\) to find the constants
✓ 1
A1 \(\mathbf{r}=\begin{pmatrix}1+2\sin 2t\\5-3\cos t\end{pmatrix}\) m
✓ 1
Part (c)
M1 displacement \(=\mathbf{r}(3)-\mathbf{r}(0)=\begin{pmatrix}2\sin 6\\3-3\cos 3\end{pmatrix}\) and its magnitude (ECF from (b))
✓ 1
A1 6.00 m
✓ 1
Exam tip: Set your calculator to radians before evaluating trigonometric functions of time in kinematics questions.
372IBMI3H-3-03r24-10Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard9 marks
A fashion label delivers samples from its studio A to four boutiques B, C, D and E, and then returns to A, visiting each boutique exactly once. The distances in km between the sites are: AB 12, AC 15, AD 20, AE 9, BC 8, BD 14, BE 17, CD 11, CE 13, DE 10.
(a)
Use the nearest neighbour algorithm starting at A to find an upper bound for the length of the shortest delivery route. State the route.
(3)
(b)
By deleting vertex A, find a lower bound for the length of the shortest route.
(4)
(c)
Hence state the minimum length of the route, and find the fuel cost for one delivery round at €0.45 per km.
(2)
Show mark scheme
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Mark
Notes
Part (a)
M1 attempt at nearest neighbour from A
✓ 1
A1 route A–E–D–C–B–A
✓ 1
A1 upper bound \(9+10+11+8+12=50\) km
✓ 1
Part (b)
M1 minimum spanning tree of B, C, D, E
✓ 1
A1 BC 8, DE 10, CD 11, total 29
✓ 1
M1 add the two shortest edges from A
✓ 1
A1 \(29+9+12=50\) km
✓ 1
Part (c)
R1 upper bound equals lower bound, so the minimum length is 50 km
✓ 1
A1 €22.50
✓ 1
Exam tip: If the upper and lower bounds are equal, the tour is optimal and you should say so explicitly as your conclusion.
373IBMI3H-3-03r26-04Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard11 marks
Calculator
A drone flies along a straight line. Its position, in metres relative to a fixed origin, is \(\mathbf{r}=\begin{pmatrix}1\\2\\0\end{pmatrix}+\lambda\begin{pmatrix}2\\-1\\2\end{pmatrix}\), where \(\lambda\) is the time in seconds. A sensor is placed at the point P(4, 2, 5).
(a)
Write down the speed of the drone.
(2)
(b)
Find the value of \(\lambda\) at which the drone is closest to P, and the minimum distance, to 3 significant figures.
(5)
(c)
The sensor can detect the drone when it is within 3.5 m of P. Find the length of time, in seconds, for which the drone is detected. Give your answer to 3 significant figures, using unrounded values throughout.
(4)
Show mark scheme
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Notes
Part (a)
M1 magnitude of direction vector, \(\sqrt{2^2+(-1)^2+2^2}\)
✓ 1
A1 \(3\) m s−1
✓ 1
Part (b)
M1 vector from drone to P: \(\begin{pmatrix}3-2\lambda\\ \lambda\\5-2\lambda\end{pmatrix}\)
✓ 1
M1 sets scalar product with direction vector equal to zero: \(2(3-2\lambda)-\lambda+2(5-2\lambda)=0\)
✓ 1
A1 \(16-9\lambda=0\) so \(\lambda=\tfrac{16}{9}\) (1.78)
✓ 1
M1 magnitude of the vector at this \(\lambda\) (unrounded), \(\sqrt{(-\tfrac{5}{9})^2+(\tfrac{16}{9})^2+(\tfrac{13}{9})^2}\)
M1 solves, \(\lambda=0.915\) and \(\lambda=2.64\) (unrounded \(0.9153\) and \(2.6402\))
✓ 1
A1 difference \(\dfrac{\sqrt{241}}{9}=1.72\) s
✓ 1
Exam tip: For the closest point on a line, set the scalar product of the direction vector with the vector to the external point equal to zero.
374IBMI3H-3-03r26-10Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard9 marks
Calculator
A research vessel visits five fish farms A, B, C, D and E, starting and ending at A and visiting each other farm exactly once. The distances in kilometres between farms are: AB 14, AC 20, AD 11, AE 17, BC 9, BD 16, BE 12, CD 13, CE 8, DE 15.
(a)
Use the nearest neighbour algorithm starting at A to find an upper bound for the shortest route. State the route and its length.
(3)
(b)
Use the deleted vertex method, deleting A, to find a lower bound for the shortest route.
(4)
(c)
A manager claims that a route of length 54 km exists. Use your answer to (b) to comment.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 chooses nearest unvisited farm each time, starting AD (11)
✓ 1
A1 route A–D–C–E–B–A
✓ 1
A1 length \(11+13+8+12+14=58\) km
✓ 1
Part (b)
M1 finds a minimum spanning tree of B, C, D, E
✓ 1
A1 edges CE (8), BC (9), CD (13), total \(30\)
✓ 1
M1 adds the two shortest edges from A: AD (11) and AB (14)
✓ 1
A1 lower bound \(30+11+14=55\) km
✓ 1
Part (c)
M1 compares the claim with the lower bound
✓ 1
R1 the claim is false, because every route has length at least \(55\) km
✓ 1
Exam tip: The nearest neighbour length is an upper bound, and the deleted vertex length is a lower bound; the shortest route lies between them.
375IBMI3H-3-06r27-08Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard10 marks
A theatre company tours five towns A, B, C, D and E, starting and finishing at A and visiting each other town exactly once. The distances between towns, in km, are shown in the table.
A
B
C
D
E
A
–
11
19
23
15
B
11
–
20
25
30
C
19
20
–
27
8
D
23
25
27
–
29
E
15
30
8
29
–
(a)
Use the nearest neighbour algorithm, starting at A, to find a tour and state its length.
(3)
(b)
Delete vertex A. Find a minimum spanning tree for the remaining four towns, and use it to find a lower bound for the length of the shortest tour.
(4)
(c)
Write down the best interval that these results give for the length of the shortest tour.
(1)
(d)
Show that the tour A → B → D → C → E → A gives a better upper bound than the one found in part (a).
(2)
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Notes
Part (a)
M1 from A goes to the nearest unvisited town each time
✓ 1
A1 A → B → C → E → D → A
✓ 1
A1 11 + 20 + 8 + 29 + 23 = 91 km
✓ 1
Part (b)
M1 builds a minimum spanning tree on B, C, D, E (e.g. Kruskal)
✓ 1
A1 CE (8), BC (20), BD (25), total 53
✓ 1
M1 adds the two shortest edges at A: AB (11) and AE (15)
✓ 1
A1 lower bound 53 + 26 = 79 km
✓ 1
Part (c)
A1 79 ≤ shortest tour ≤ 91 (ECF from (a) and (b))
✓ 1
Part (d)
M1 11 + 25 + 27 + 8 + 15
✓ 1
A1 86 km, which is less than 91 km
✓ 1
Exam tip: A lower bound uses a minimum spanning tree of the remaining vertices plus the two shortest edges to the deleted vertex; an upper bound is the length of any actual tour.
376IBMI3H-3-06r27-10Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard9 marks
A projector transforms points on a flat screen using a linear transformation with matrix M. It maps the point (1, 2) to (4, 5) and the point (3, 1) to (7, 0).
(a)
Find the matrix M.
(3)
(b)
Find the determinant of M and state what it tells you about the effect of the transformation on areas.
(2)
(c)
The point (7, 7) is the image of a point X under the transformation. Find the coordinates of X.
(3)
(d)
A region of area 4.5 m2 on the screen is transformed. Find the area of its image.
(1)
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Mark
Notes
Part (a)
M1 forms M((1, 3), (2, 1)) = ((4, 7), (5, 0)) and uses M = ((4, 7), (5, 0))((1, 3), (2, 1))−1 (or equivalent simultaneous equations)
✓ 1
A1 ((1, 3), (2, 1))−1 = −(1/5)((1, −3), (−2, 1))
✓ 1
A1 M = ((2, 1), (−1, 3))
✓ 1
Part (b)
A1 det M = 2 × 3 − 1 × (−1) = 7
✓ 1
R1 areas are multiplied by 7 (and orientation is preserved since the determinant is positive)
✓ 1
Part (c)
M1 uses M−1
✓ 1
A1 M−1 = (1/7)((3, −1), (1, 2))
✓ 1
A1 X = (2, 3)
✓ 1
Part (d)
A1 4.5 × 7 = 31.5 m2 (ECF from (b))
✓ 1
Exam tip: If you know where two independent vectors go, write the images as the columns of one matrix and the originals as the columns of another, then multiply by an inverse.
377IBMI3H-3-06r27-11Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard10 marks
Two ships, A and B, sail in open water. Relative to a lighthouse, the position vectors, in km, t hours after noon are rA = (−10, 20) + t(8, −6) and rB = (30, −5) + t(−4, 10). The first component is east and the second north.
(a)
Find the vector d = rB − rA in terms of t.
(2)
(b)
Show that the square of the distance between the ships is 400t2 − 1760t + 2225.
(2)
(c)
Find the time at which the ships are closest together, and their least distance apart.
(3)
(d)
A ship is in danger if the other is within 20 km. Find the length of time, in hours, for which the ships are within 20 km of each other. Give your answer to 3 significant figures.
Common error: Minimise the square of the distance to avoid a square root, but remember to take the square root when stating the actual distance.
378IBMI3H-3-06r29-04Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard11 marks
Two fishing boats move at constant velocity. Distances are in km relative to a harbour and time \(t\) is in hours after noon. Boat A has position \(\mathbf{r}_A=\begin{pmatrix}2\\1\end{pmatrix}+t\begin{pmatrix}3\\4\end{pmatrix}\) and boat B has position \(\mathbf{r}_B=\begin{pmatrix}14\\-3\end{pmatrix}+t\begin{pmatrix}-1\\6\end{pmatrix}\). The boats can communicate by radio when they are within 1.5 km of each other.
(a)
Find the speed of boat A.
(2)
(b)
Show that \(\overrightarrow{AB}=\begin{pmatrix}12-4t\\2t-4\end{pmatrix}\), where \(\overrightarrow{AB}\) is the vector from A to B.
(2)
(c)
Show that the square of the distance between the boats is \(20t^2-112t+160\).
(2)
(d)
Find the minimum distance between the boats, to 3 significant figures.
(4)
(e)
Hence state whether the boats can communicate by radio at any time.
(1)
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Part (a)
M1 \(\sqrt{3^2+4^2}\)
✓ 1
A1 5 km h−1
✓ 1
Part (b)
M1 \(\mathbf{r}_B-\mathbf{r}_A\)
✓ 1
A1 correct simplification to the given vector (AG)
✓ 1
Part (c)
M1 \((12-4t)^2+(2t-4)^2\)
✓ 1
A1 expansion and simplification to \(20t^2-112t+160\) (AG)
✓ 1
Part (d)
M1 differentiate \(20t^2-112t+160\) and set to zero
✓ 1
A1 \(t=2.8\)
✓ 1
M1 substitute to find \(\text{distance}^2=3.2\)
✓ 1
A1 1.79 km
✓ 1
Part (e)
R1 minimum distance 1.79 km is greater than 1.5 km, so they cannot
✓ 1
Exam tip: To find a closest approach, minimise the squared distance, which is a quadratic in \(t\).
379IBMI3H-3-06r29-07Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard10 marks
A warehouse robot must inspect every aisle. The aisles join six junctions A to F, with lengths in metres: AB 6, AC 5, AE 10, BC 4, BD 7, CD 3, DE 6, DF 9, EF 4.
(a)
Find the degree of each vertex and state which vertices have odd degree.
(2)
(b)
The robot must start and finish at the same junction and traverse every aisle at least once. Find the pairing of odd vertices that gives the shortest repeated distance, showing the three possible pairings.
(3)
(c)
Hence find the length of the shortest route.
(2)
(d)
The robot may instead start and finish at different junctions. Find the minimum length of a route, and state where it starts and finishes.
(3)
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Part (a)
A1 degrees A3, B3, C3, D4, E3, F2
✓ 1
A1 odd vertices A, B, C, E
✓ 1
Part (b)
M1 consider all three pairings of the odd vertices
✓ 1
A1 AB+CE \(=6+9=15\) and AC+BE \(=5+13=18\)
✓ 1
A1 AE+BC \(=10+4=14\) is the least
✓ 1
Part (c)
M1 total of all aisles \(=54\) plus repeated distance
✓ 1
A1 68 m
✓ 1
Part (d)
M1 leave two odd vertices as the start and end, and repeat only the path between the other two
✓ 1
A1 repeat BC (4 m), so length \(54+4=58\) m
✓ 1
A1 start and finish at A and E (in either order)
✓ 1
Exam tip: For a route that starts and ends at different vertices, the start and end must be the two odd vertices you do not pair up.
380IBMI3H-3-06r29-08Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard9 marks
A fisheries inspector must visit five ports A, B, C, D and E, starting and finishing at A, visiting each port exactly once. The distances between ports, in km, are: AB 12, AC 18, AD 25, AE 15, BC 9, BD 20, BE 14, CD 11, CE 22, DE 16.
(a)
Use the nearest-neighbour algorithm starting at A to find an upper bound for the shortest tour.
(3)
(b)
By deleting vertex E, find a lower bound for the length of the shortest tour.
(4)
(c)
Write down the interval in which the length of the shortest tour lies, and explain why the lower bound is valid.
(2)
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Part (a)
M1 apply the algorithm from A
✓ 1
A1 A–B–C–D–E–A
✓ 1
A1 63 km
✓ 1
Part (b)
M1 minimum spanning tree of A, B, C and D
✓ 1
A1 BC, CD, AB with total 32
✓ 1
A1 two shortest edges from E: EB 14 and EA 15, total 29
✓ 1
A1 61 km
✓ 1
Part (c)
A1 \(61\le\) shortest tour \(\le63\)
✓ 1
R1 any tour minus its two edges at E is a spanning path of the other vertices, so it is at least the minimum spanning tree plus the two shortest edges at E
✓ 1
Common error: A nearest-neighbour tour gives an upper bound only; it is not guaranteed to be the shortest tour.
381IBMI3H-3-05r30-11Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard13 marks
Two aircraft, P and Q, fly at constant velocity in an air-traffic region. Relative to a control tower at the origin, with distances in km and time t in hours after 12:00, the position vectors are p = \(\begin{pmatrix}0\\0\\10\end{pmatrix}\) + t\(\begin{pmatrix}240\\180\\-6\end{pmatrix}\) and q = \(\begin{pmatrix}100\\12\\8.5\end{pmatrix}\) + t\(\begin{pmatrix}-200\\150\\0\end{pmatrix}\). The third component is altitude. Safety rules require the distance between the aircraft to be at least 8 km at all times.
(a)
Show that the position vector of Q relative to P is \(\begin{pmatrix}100 − 440t\\12 − 30t\\−1.5 + 6t\end{pmatrix}\).
(2)
(b)
Hence show that the square of the distance between the aircraft is D2 = 194 536t2 − 88 738t + 10 146.25.
(3)
(c)
Find the time at which the aircraft are closest together. Give your answer in minutes after 12:00, to 3 significant figures.
(2)
(d)
Find the minimum distance between the aircraft, to 3 significant figures.
(2)
(e)
State, with a reason, whether the safety rule is broken.
(1)
(f)
Find the length of time, in seconds to 3 significant figures, for which the distance between the aircraft is less than 8 km.
M1 minimises the quadratic, for example by t = 88 738 / (2 × 194 536) or by differentiating
✓ 1
A1 t = 0.22808 h = 13.7 minutes
✓ 1
Part (d)
M1 substitutes the unrounded t into D2 to get 26.74
✓ 1
A1 D = 5.17 km
✓ 1
Part (e)
R1 yes, because the minimum distance 5.17 km is less than 8 km
✓ 1
Part (f)
M1 sets D2 = 64 to give 194 536t2 − 88 738t + 10 082.25 = 0
✓ 1
A1 roots t = 0.2142 and t = 0.2419
✓ 1
A1 difference 0.02768 h = 99.6 seconds
✓ 1
Common error: Minimise the square of the distance rather than the distance itself; it has the same minimum point and avoids a square root.
382IBMI3H-3-05r30-12Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard11 marks
A laser light show at a music venue moves a spot of light on a wall. Positions on the wall are given as coordinates (x, y) in metres. A light point is first reflected in the line y = (tan 25°)x and then rotated by 40° anticlockwise about the origin. In this question, give all matrix entries to 3 significant figures.
(a)
Write down the matrix R for a rotation of 40° anticlockwise about the origin.
(2)
(b)
Write down the matrix F for reflection in the line y = (tan 25°)x.
(2)
(c)
Find the single matrix that represents the reflection followed by the rotation.
(3)
(d)
Describe fully the single geometric transformation represented by the matrix in (c).
(2)
(e)
A light point starts at (6, −2). Find the distance between its starting position and its final position, to 3 significant figures.
M1 final position (−2, 6) from the matrix in (c), then distance √(82 + 82)
✓ 1
A1 11.3 m
✓ 1
Common error: When the first transformation is applied first, its matrix is on the right in the product, so write RF, not FR.
383IBMI3H-3-05r30-13Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard12 marks
A park ranger must patrol every one of the paths in a nature reserve. The paths join six junctions, A to F, and their lengths in kilometres are: AB 8, AC 6, BC 5, BD 9, CE 7, DE 4, DF 6, EF 11, CD 3, AF 12. The paths meet only at the junctions.
(a)
State which junctions have an odd number of paths meeting at them.
(2)
(b)
The ranger wants to walk every path at least once, starting and finishing at the same junction. Find the three possible pairings of the odd junctions, the shortest distance between the junctions in each pair, and the total for each pairing.
(4)
(c)
Hence find the length of the shortest route in which the ranger starts and finishes at the same junction.
(2)
(d)
The ranger is now allowed to start and finish at different junctions. State the junctions where the ranger should start and finish, and find the length of the shortest route that walks every path at least once.
(4)
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Part (a)
M1 degrees found: A 3, B 3, C 4, D 4, E 3, F 3
✓ 1
A1 A, B, E and F
✓ 1
Part (b)
M1 lists pairings AB & EF, AE & BF, AF & BE
✓ 1
A1 shortest distances AB 8, EF 10 (via D), AE 13 (via C), BF 14 (via C and D), AF 12, BE 12 (via C)
✓ 1
A1 pairing totals 18, 27 and 24
✓ 1
A1 the minimum is 18, from AB and EF
✓ 1
Part (c)
M1 total length of all paths = 71 km added to the minimum pairing total
✓ 1
A1 71 + 18 = 89 km
✓ 1
Part (d)
M1 start and finish at two of the odd junctions, with the other two odd junctions paired
✓ 1
M1 chooses the pair with the shortest connection, which is AB with length 8 km
✓ 1
A1 starts at E and finishes at F, or starts at F and finishes at E
✓ 1
A1 71 + 8 = 79 km
✓ 1
Exam tip: With two odd vertices left as the start and end, the other odd vertices must still be paired by repeated paths, so choose the cheapest pair to repeat.
384IBMI3H-3-05r31-07Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard10 marks
A volcanologist inspects every path in a trail network connecting six lookout points A to F. The paths and their lengths in kilometres are: AB 4, AC 6, AF 9, BC 3, BD 7, CD 5, CE 8, DE 4, DF 6, EF 5. The total length of all the paths is 57 km. She must walk along every path at least once and return to her starting point A.
(a)
Write down the vertices of odd degree.
(2)
(b)
Show that the shortest inspection route that starts and finishes at A has length 66 km.
(4)
(c)
State which paths are walked twice in this route.
(1)
(d)
Another plan lets her start at A but finish at a different lookout point. Find the shortest possible route length for this plan and state where it finishes.
(3)
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Part (a)
A1 A, B, E and F
✓ 1
R1 their degrees are all 3 (the others, C and D, have degree 4)
✓ 1
Part (b)
M1 consider the three ways of pairing the odd vertices
✓ 1
A1 AB + EF = 4 + 5 = 9; AE + BF = 14 + 13 = 27; AF + BE = 9 + 11 = 20
✓ 1
M1 repeat the pairing with the least total, 9, and add to the total length: 57 + 9
✓ 1
A1 66 km (AG)
✓ 1
Part (c)
A1 AB and EF
✓ 1
Part (d)
M1 the route must finish at an odd vertex; the remaining two odd vertices must be paired by their shortest connection
✓ 1
A1 finish at B, with EF (5) repeated
✓ 1
A1 57 + 5 = 62 km
✓ 1
Exam tip: For a route that starts and ends at the same vertex, odd vertices must be paired and the shortest connecting paths repeated. Find the shortest path between each pair using any route in the network, not only a direct edge.
385IBMI3H-3-05r31-10Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard10 marks
During an eruption a rock is thrown from a vent at the origin. Taking x as horizontal distance and y as height, both in metres, its position at time t seconds is r = (20t, 30t − 4.9t2) for 0 ≤ t ≤ 6.1.
(a)
Find the velocity vector of the rock at time t.
(2)
(b)
Find the speed of the rock at the instant it leaves the vent. Give your answer to 3 significant figures.
(2)
(c)
Find the two times at which the speed of the rock is 32 m s−1. Give your answers to 3 significant figures.
(4)
(d)
At the later time, find the angle below the horizontal at which the rock is moving. Use the unrounded value of t. Give your answer to the nearest 0.1°.
(2)
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Part (a)
M1 differentiate each component with respect to t
✓ 1
A1 v = (20, 30 − 9.8t)
✓ 1
Part (b)
M1 √(202 + 302)
✓ 1
A1 36.1 m s−1
✓ 1
Part (c)
M1 202 + (30 − 9.8t)2 = 322
✓ 1
M1 (30 − 9.8t)2 = 624, so 30 − 9.8t = ±√624
✓ 1
A1 t = 0.512 s
✓ 1
A1 t = 5.61 s
✓ 1
Part (d)
M1 vertical component is −√624, so tan θ = √624 ÷ 20
✓ 1
A1 51.3°
✓ 1
Exam tip: Velocity is the derivative of the position vector, and speed is its magnitude. A quadratic equation in the speed gives two times, one on the way up and one on the way down; check both lie in the allowed time interval.
386IBMI3H-3-05r33-08Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard9 marks
A band plans a tour that visits five cities A, B, C, D and E once each and returns to the start. The distances in km between the cities are: AB 140, AC 210, AD 320, AE 180, BC 95, BD 250, BE 130, CD 160, CE 220, DE 270.
(a)
Starting at A, use the nearest neighbour algorithm to find an upper bound for the length of the tour. State the route.
(3)
(b)
By deleting vertex A, find a lower bound for the length of the tour.
(4)
(c)
The band’s manager claims that a tour of length 700 km exists. Use your answers to comment on this claim.
(2)
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Part (a)
M1 attempt at the nearest neighbour algorithm from A
M1 attempt to find the minimum spanning tree of B, C, D and E
✓ 1
A1 tree BC, BE, CD of weight 385
✓ 1
M1 adds the two shortest edges from A
✓ 1
A1 385 + 140 + 180 = 705 km
✓ 1
Part (c)
M1 compares 700 with the lower bound
✓ 1
R1 700 is less than the lower bound of 705, so no tour can be this short and the claim is false
✓ 1
Exam tip: Every tour must be at least as long as the lower bound, and the best tour is no longer than the upper bound, so the shortest tour lies between the two.
387IBMI3H-3-05r33-10Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard9 marks
A flight display team uses a linear transformation with matrix M to relocate points on a map. It maps the point (2, 1) to (5, 1) and the point (1, 3) to (5, 8).
(a)
Find the matrix M.
(4)
(b)
A triangle with vertices (0, 0), (3, 0) and (0, 4) is transformed by M. Find the area of the image triangle.
(2)
(c)
A point R is mapped by M to the point (4, −9). Find the coordinates of R.
(3)
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Part (a)
M1 forms \(\mathbf{M}\begin{pmatrix}2&1\\1&3\end{pmatrix}=\begin{pmatrix}5&5\\1&8\end{pmatrix}\)
✓ 1
M1 multiplies on the right by the inverse of \(\begin{pmatrix}2&1\\1&3\end{pmatrix}\)
Exam tip: To undo a transformation, multiply the image point by the inverse matrix; check by applying M to your answer.
388IBMI3H-3-05r33-11Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard11 marks
Two light aircraft, A and B, fly at the same altitude. Relative to a control tower at the origin, their position vectors in km at time t hours after 10:00 are \(\mathbf{r}_A=\begin{pmatrix}-120\\50\end{pmatrix}+t\begin{pmatrix}360\\-60\end{pmatrix}\) and \(\mathbf{r}_B=\begin{pmatrix}100\\-70\end{pmatrix}+t\begin{pmatrix}-240\\240\end{pmatrix}\). Aircraft must remain at least 10 km apart.
(a)
Find the speed of aircraft A, correct to the nearest km h−1.
(2)
(b)
Show that the vector from B to A at time t is \(\begin{pmatrix}-220+600t\\120-300t\end{pmatrix}\) km.
(2)
(c)
Find the least distance between the aircraft and the time, in minutes after 10:00, at which this occurs. Give the distance to the nearest 0.1 km and the time to the nearest 0.1 minute.
(4)
(d)
Find, to the nearest second, the length of time for which the aircraft are less than 10 km apart.
M1 attempt to minimise, e.g. t = 336 000 ÷ 900 000 or by the graphing facility
✓ 1
A1 t = 0.3733 h = 22.4 minutes after 10:00
✓ 1
A1 least distance 8.9 km
✓ 1
Part (d)
M1 solves 450 000t2 − 336 000t + 62 800 = 100
✓ 1
A1 t = 0.36667 h and t = 0.38 h (22.0 and 22.8 minutes)
✓ 1
A1 difference 0.01333 h = 48 seconds
✓ 1
Exam tip: Work with the square of the distance: it is a quadratic in t, so its minimum and the times it crosses a given value are easy to find without square roots.
389IBMI3H-3-05r36-04Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard9 marks
Two ships leave positions at 09:00. Relative to a harbour (units km), ship A starts at (2, 10) and has constant velocity (5, −2) km h−1. Ship B starts at (20, −2) and has constant velocity (−4, 3) km h−1. Let t be the time in hours after 09:00.
(a)
Show that the vector from A to B at time t is (18 − 9t, −12 + 5t).
(2)
(b)
Find the time, in hours after 09:00, at which the ships are closest together. Give your answer correct to 3 significant figures.
(4)
(c)
Find the minimum distance between the ships, in km, correct to 3 significant figures, using the unrounded value of t from (b). State, with a reason, whether the ships ever come within 2 km of each other.
(3)
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Part (a)
M1 write both position vectors: rA = (2 + 5t, 10 − 2t) and rB = (20 − 4t, −2 + 3t)
✓ 1
A1 subtract, rB − rA = (18 − 9t, −12 + 5t), with the stated result obtained
M1 minimise: differentiate and set to zero, or use the vertex of the quadratic or a GDC minimum
✓ 1
A1 t = 444/212 (= 2.0943…)
✓ 1
A1 t = 2.09 hours
✓ 1
Part (c)
M1 substitute t = 444/212 into the distance expression
✓ 1
A1 1.75 km
✓ 1
R1 yes, since 1.75 < 2
✓ 1
Exam tip: Minimising the squared distance gives the same time as minimising the distance and avoids a square root.
390IBMI3H-3-05r36-11Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard10 marks
A triangular marine enclosure of area 12 m2 is mapped by the matrix M = \(\begin{pmatrix}k&2\\1&k\end{pmatrix}\), where k is a positive constant, onto a triangle of area 60 m2.
(a)
Write down an expression for det M.
(1)
(b)
Find the exact value of k.
(4)
(c)
Find M−1 in terms of √7.
(2)
(d)
A point P is mapped to (7, 3). Find the coordinates of P, and the distance OP from the origin, correct to 3 significant figures.
(3)
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Part (a)
A1 det M = k2 − 2
✓ 1
Part (b)
M1 area scale factor = |det M| = 60/12 = 5
✓ 1
M1 form |k2 − 2| = 5, so k2 − 2 = 5 or k2 − 2 = −5
✓ 1
A1 k2 = 7 (the equation k2 = −3 has no real solution)
R1 check the remaining equation holds (for example 1 + 2 = 9 − 6), so the point of intersection is (6, 3, 6)
✓ 1
Part (b)
M1 scalar product of direction vectors = 2 − 2 + 6 = 6
✓ 1
M1 magnitudes √14 and 3 in cos θ = |d1·d2|/(|d1||d2|)
✓ 1
A1 57.7°
✓ 1
Part (c)
M1 |2(2, 1, 3)| = 2√14, or distance between (2, 1, 0) and (6, 3, 6)
✓ 1
A1 7.48 m
✓ 1
Exam tip: Substitute values into the equation you did not use to solve; if it fails, the lines do not intersect.
393IBMI3H-3-06r36-11Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard10 marks
A map of an outbreak region is transformed by a matrix M. The point A(1, 2) is mapped to A′(4, 5) and the point B(3, 1) is mapped to B′(7, 0).
(a)
Find M.
(4)
(b)
The point C′(10, −1) is the image of a point C. Find the exact coordinates of C.
(3)
(c)
Triangle OAB has vertices at the origin O, A and B. Find the area of its image under M.
(3)
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Part (a)
M1 set up M\(\begin{pmatrix}1&3\\2&1\end{pmatrix}\) = \(\begin{pmatrix}4&7\\5&0\end{pmatrix}\), or two pairs of simultaneous equations
✓ 1
M1 multiply by the inverse of \(\begin{pmatrix}1&3\\2&1\end{pmatrix}\) (determinant −5) or solve the equations
✓ 1
A1 first row (2, 1)
✓ 1
A1 second row (−1, 3)
✓ 1
Part (b)
M1 det M = 7 and M−1 = \(\frac{1}{7}\begin{pmatrix}3&-1\\1&2\end{pmatrix}\)
✓ 1
M1 multiply M−1 by the column vector (10, −1)
✓ 1
A1 C = (31/7, 8/7)
✓ 1
Part (c)
M1 area of OAB = ½|1 × 1 − 3 × 2| = 2.5
✓ 1
M1 area scale factor = |det M| = 7
✓ 1
A1 17.5 square units
✓ 1
Exam tip: Two points and their images give four equations, which is enough to find the four entries of a 2×2 matrix.
394IBMI3H-3-06r36-14Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard10 marks
Three retail stores are at A(0, 0), B(8, 0) and C(7, 7), with coordinates in km. A Voronoi diagram is drawn for the stores. Customers go to their nearest store.
(a)
Find the equation of the perpendicular bisector of AB.
(2)
(b)
Find the equation of the perpendicular bisector of AC.
(3)
(c)
Hence find the coordinates of the vertex V where the three cells meet.
(2)
(d)
A fourth store D opens at (4, −3). Find the distance VD, and explain why V is still a vertex of the Voronoi diagram for the four stores.
(3)
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Part (a)
M1 midpoint of AB is (4, 0) and AB is horizontal
✓ 1
A1 x = 4
✓ 1
Part (b)
M1 midpoint of AC = (3.5, 3.5) and gradient of AC = 1
✓ 1
M1 perpendicular gradient −1
✓ 1
A1 y = −x + 7
✓ 1
Part (c)
M1 substitute x = 4 into y = −x + 7
✓ 1
A1 V = (4, 3)
✓ 1
Part (d)
M1 VD = √((4 − 4)2 + (3 + 3)2)
✓ 1
A1 6 km
✓ 1
R1 V is 5 km from each of A, B and C, and 6 km from D, so D is not as close, and V still has three equally near nearest sites
✓ 1
Common error: A Voronoi vertex is only valid if no other site is closer to it than the three sites that define it.
395IBMI3H-3-07r37-04Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard12 marks
Calculator
Two relay drones fly above a valley carrying telecoms equipment. Relative to a ground station at the origin, the position vectors of the drones t minutes after launch are rA = \(\begin{pmatrix}1\\2\\7\end{pmatrix}\) + t\(\begin{pmatrix}2\\1\\1\end{pmatrix}\) and rB = \(\begin{pmatrix}8\\4\\3\end{pmatrix}\) + t\(\begin{pmatrix}-1\\0\\2\end{pmatrix}\). One unit on each axis represents 100 m.
(a)
Show that the directions of travel of the two drones are perpendicular.
(3)
(b)
Show that the two flight paths intersect and find the coordinates of the point of intersection.
(4)
(c)
Explain why the drones do not collide.
(2)
(d)
Find the distance between the drones when t = 2. Give your answer in metres, correct to 3 significant figures.
(3)
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Part (a)
M1 attempts the scalar product of the direction vectors, 2×(−1) + 1×0 + 1×2
✓ 1
A1 = 0
✓ 1
R1 so the directions are perpendicular (angle 90°)
✓ 1
Part (b)
M1 equates components of rA and rB using different parameters, e.g. t for A and s for B
✓ 1
A1 from the y-components, t = 2, and from the z-components, s = 3
✓ 1
R1 checks the x-components: 1 + 4 = 5 and 8 − 3 = 5, so the paths meet
✓ 1
A1 point of intersection (5, 4, 9)
✓ 1
Part (c)
R1 drone A is at the intersection point when t = 2
✓ 1
R1 drone B is at the intersection point when t = 3, a different time, so they are never there together
✓ 1
Part (d)
M1 finds both positions at t = 2, (5, 4, 9) and (6, 4, 7)
✓ 1
M1 attempts the magnitude of the difference, √(12 + 02 + 22) = √5 units
✓ 1
A1 224 m
✓ 1
Common error: Using one parameter for both lines when testing for an intersection forces both drones to be at the same point at the same time, which is a different question.
396IBMI3H-3-07r37-11Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard12 marks
Calculator
A rotating antenna sweeps a beam. Its movement is modelled using matrices acting on position vectors, with coordinates in km. Let R be the matrix for an anticlockwise rotation of 50° about the origin and let F = \(\begin{pmatrix}-1&0\\0&1\end{pmatrix}\) represent a reflection in the y-axis.
(a)
Write down R, giving each entry to 3 significant figures.
(2)
(b)
Find the image of the point (8, 3) under the rotation R. Use unrounded values of R and give your answer to 3 significant figures.
(2)
(c)
The point (8, 3) is first reflected using F and then rotated using R. Find the single matrix N that represents this combined transformation, giving each entry to 3 significant figures.
(3)
(d)
Show that the determinant of N equals −1 and state what this tells you about the transformation.
(2)
(e)
The transformation N leaves every point on one line through the origin unchanged. Find the equation of this line in the form y = mx, giving m to 3 significant figures.
(3)
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Part (a)
M1 uses the rotation matrix \(\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}\) with θ = 50°
Common error: Multiplying the matrices in the wrong order: for 'first F, then R' the matrix product is RF, with the later transformation written on the left.
397IBMI3H-3-05r38-05Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard12 marks
A weather balloon is released from the origin. For 0 ≤ t ≤ 4, its position in metres is given by r(t) = (3t2, 20t − t3), where t is the time in seconds, the first component is horizontal and the second is vertical.
(a)
Find the velocity of the balloon when t = 2, and hence its speed. Give the speed to 3 significant figures.
(3)
(b)
Find the value of t at which the balloon is moving horizontally.
(2)
(c)
Find the magnitude of the acceleration when t = 3.
(3)
(d)
Find the minimum speed of the balloon for 0 ≤ t ≤ 4, giving your answer to 3 significant figures.
(4)
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Notes
Part (a)
M1 differentiate r(t) to obtain v = (6t, 20 − 3t2)
✓ 1
A1 v(2) = (12, 8) m s−1
✓ 1
A1 speed = √208 = 14.4 m s−1
✓ 1
Part (b)
M1 set the vertical component of velocity to zero, 20 − 3t2 = 0
✓ 1
A1 t = 2.58 s
✓ 1
Part (c)
M1 differentiate velocity to obtain a = (6, −6t)
✓ 1
A1 a(3) = (6, −18)
✓ 1
A1 magnitude = √360 = 19.0 m s−2
✓ 1
Part (d)
M1 speed squared = 36t2 + (20 − 3t2)2
✓ 1
M1 minimise (differentiate and set equal to zero, or use a graphing method)
✓ 1
A1 t2 = 14/3 (t = 2.16 s)
✓ 1
A1 14.3 m s−1
✓ 1
Exam tip: Minimising the speed is easier if you minimise the speed squared, which has the same turning point and avoids the square root.
398IBMI3H-3-05r38-09Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard10 marks
Five mountain huts A to E must all be visited by a guide, who starts and ends at A and visits every other hut exactly once. The walking times (in minutes) between huts are shown in the table (symmetric). AB 17, AC 19, AD 13, AE 33 BC 22, BD 25, BE 14 CD 12, CE 32 DE 10
(a)
Use the nearest neighbour algorithm starting at A to find an upper bound for the shortest route. State the route and its total time.
(3)
(b)
Delete vertex A and use the deleted vertex algorithm to find a lower bound.
(4)
(c)
A route A–B–E–D–C–A is found. State its total time.
(1)
(d)
Find, to 3 significant figures, the percentage by which the nearest neighbour route exceeds the time of the route in (c).
(2)
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Part (a)
M1 repeatedly move to the nearest unvisited hut
✓ 1
A1 route A–D–E–B–C–A
✓ 1
A1 78 minutes
✓ 1
Part (b)
M1 find a minimum spanning tree of B, C, D, E
✓ 1
A1 DE 10, CD 12, BE 14, total 36
✓ 1
M1 add the two smallest edges from A (13 and 17)
✓ 1
A1 lower bound 66 minutes
✓ 1
Part (c)
A1 72 minutes
✓ 1
Part (d)
M1 (78 − 72) / 72 × 100
✓ 1
A1 8.33%
✓ 1
Exam tip: The shortest route always lies between the lower and upper bounds, so a route inside that interval is possible but may still not be the best.
399IBMI3H-3-05r38-12Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard10 marks
A sensor calibration applies transformation A to every point of a map, and then transformation M to the result. The matrices are M = ((2, 1), (1, 1)) and the combined transformation is T = MA = ((2, 7), (1, 5)) (rows listed in each case).
(a)
Find det M and M−1.
(2)
(b)
Find the matrix A.
(3)
(c)
A point (x, y) is mapped by A to (10, 12). Find x and y.
(3)
(d)
A triangular region on the map has area 8 cm2. Find its area after transformation T.
(2)
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Part (a)
A1 det M = 1
✓ 1
A1 M−1 = ((1, −1), (−1, 2))
✓ 1
Part (b)
M1 A = M−1T (correct order)
✓ 1
M1 attempt at matrix multiplication
✓ 1
A1 A = ((1, 2), (0, 3))
✓ 1
Part (c)
M1 set up x + 2y = 10 and 3y = 12
✓ 1
A1 y = 4
✓ 1
A1 x = 2
✓ 1
Part (d)
M1 det T = 2 × 5 − 7 × 1 = 3
✓ 1
A1 24 cm2
✓ 1
Common error: To undo the left-hand matrix in a product, multiply both sides by its inverse on the left, not the right.
400IBMI3H-3-05r39-09Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard12 marks
A monitoring drone circles a hive. Relative to a control point, its position vector is \(\mathbf{r}(t)\) metres at time \(t\) seconds, where \(t\geq0\). Its velocity is \(\mathbf{v}(t)=\begin{pmatrix}4\cos(0.5t)\\-3\sin(0.5t)\end{pmatrix}\) m s−1, and its initial position is \(\begin{pmatrix}10\\5\end{pmatrix}\). Angles are in radians.
(a)
Find the magnitude of the acceleration of the drone when \(t=2\), giving your answer to 3 significant figures.
(3)
(b)
Show that the speed \(s\) of the drone satisfies \(s^2=9+7\cos^2(0.5t)\), and hence find the minimum speed of the drone and the first time at which it occurs.
(3)
(c)
Find an expression for \(\mathbf{r}(t)\).
(4)
(d)
Find the straight-line distance between the drone's initial position and its position at \(t=\pi\).
A1 \(\mathbf{r}(\pi)=\begin{pmatrix}18\\-1\end{pmatrix}\) (ECF from (c))
✓ 1
A1 distance \(=\sqrt{8^2+6^2}=10\) m
✓ 1
Common error: Calculators must be in radian mode when differentiating or evaluating sin and cos of a time variable such as 0.5t.
401IBMI3H-3-06r41-02Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard12 marks
Two aircraft fly at constant velocity through the same region of airspace. Relative to an air traffic control tower at the origin, with components in kilometres (east, north) and time t in hours, their position vectors are aircraft A: rA = (−80, 40) + t(240, −60) aircraft B: rB = (50, −30) + t(−100, 150).
(a)
Find the speed of aircraft A, to 3 significant figures.
(2)
(b)
Show that the vector from A to B at time t is (130 − 340t, −70 + 210t).
(2)
(c)
Hence show that the square of the distance between the aircraft is D2 = 159 700t2 − 117 800t + 21 800.
(3)
(d)
Find the time at which the aircraft are closest together, and the minimum distance between them. Give the time to 3 significant figures and the distance to 3 significant figures.
(4)
(e)
The minimum safe separation is 10 km. State, with a reason, whether the aircraft stay at least this far apart.
M1 minimising D2 (vertex at t = 117 800 / (2 × 159 700), or derivative equal to zero, or GDC)
✓ 1
A1 t = 0.369 hours
✓ 1
M1 substituting the unrounded t into D2 and taking the square root
✓ 1
A1 minimum distance = 8.76 km
✓ 1
Part (e)
R1 No, since the minimum distance (8.76 km) is less than 10 km
✓ 1
Exam tip: To find the closest approach, minimise the square of the distance; it is a quadratic in t and avoids differentiating a square root.
402IBMI3H-3-06r41-09Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard10 marks
A regional airline serves five airports A, B, C, D and E. The distances in kilometres between airports are: AB 340, AC 520, AD 410, AE 290, BC 380, BD 450, BE 310, CD 270, CE 430, DE 360. An aircraft must start at A, visit every other airport exactly once and return to A.
(a)
Use the nearest neighbour algorithm starting at A to find an upper bound for the length of the shortest route.
(3)
(b)
By deleting airport D, find a lower bound for the length of the shortest route.
(4)
(c)
Find a route shorter than your answer to part (a) and state the length of this route. Hence write down an interval in which the length of the shortest route lies.
(2)
(d)
Explain why the value found in part (b) cannot be larger than the length of the shortest route.
(1)
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Part (a)
M1 choosing the nearest unvisited airport at each step
✓ 1
A1 A–E–B–C–D–A
✓ 1
A1 290 + 310 + 380 + 270 + 410 = 1660 km
✓ 1
Part (b)
M1 minimum spanning tree for A, B, C, E
✓ 1
A1 MST length 290 + 310 + 380 = 980 km
✓ 1
M1 adding the two shortest edges from D (270 and 360)
✓ 1
A1 lower bound 1610 km
✓ 1
Part (c)
A1 for example A–B–C–D–E–A with length 340 + 380 + 270 + 360 + 290 = 1640 km
✓ 1
A1 shortest route is between 1610 km and 1640 km
✓ 1
Part (d)
R1 removing D from the shortest route leaves a spanning path on the other vertices, which is at least as long as the MST, and the two edges at D are at least as long as the two shortest edges at D
✓ 1
Common error: The lower bound uses a minimum spanning tree of the remaining vertices plus the two shortest edges at the deleted vertex; do not add the shortest edge twice or only one edge.
403IBMI3H-3-05r44-04Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard12 marks
Two service boats leave different points of an offshore wind farm at noon. Relative to a fixed origin, their position vectors (in km) t hours after noon are rA = (2, 1) + t(12, 16) and rB = (14, −3) + t(−4, 20).
(a)
Show that boat A travels at a constant speed of 20 km h−1.
(2)
(b)
Find an expression for the vector \(\overrightarrow{AB}\) from boat A to boat B at time t.
(2)
(c)
Hence show that the square of the distance between the boats is 272t2 − 416t + 160.
(3)
(d)
Find the time, in minutes after noon, at which the boats are closest together. Give your answer to 3 significant figures.
(3)
(e)
Find the least distance between the boats, giving your answer in km to 3 significant figures. Use the unrounded value of t from part (d).
M1 expand both brackets: 144 − 384t + 256t2 and 16t2 − 32t + 16
✓ 1
A1 collect terms to reach 272t2 − 416t + 160 (AG)
✓ 1
Part (d)
M1 minimise 272t2 − 416t + 160 (vertex t = 416/544, or GDC minimum)
✓ 1
A1 t = 0.765 hours (0.7647…)
✓ 1
A1 45.9 minutes
✓ 1
Part (e)
M1 substitute t = 416/544 into 272t2 − 416t + 160 and take the square root
✓ 1
A1 0.970 km
✓ 1
Exam tip: Minimising the square of the distance gives the same time as minimising the distance itself and avoids differentiating a square root.
404IBMI3H-3-05r44-08Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard10 marks
A delivery van starts at depot A, must visit stores B, C, D and E once each and then return to A. The distances in km between the locations are: AB 12, AC 9, AD 15, AE 8, BC 10, BD 7, BE 14, CD 11, CE 6, DE 13.
(a)
Use the nearest-neighbour algorithm starting at A to find an upper bound for the shortest route. State the route and its length.
(3)
(b)
Use the deleted vertex method, deleting A, to find a lower bound for the shortest route.
(4)
(c)
Using your answers to (a) and (b), find the greatest possible percentage by which the route in (a) could exceed the shortest possible route.
(3)
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Part (a)
M1 at each step move to the nearest unvisited vertex
✓ 1
A1 route A → E → C → B → D → A
✓ 1
A1 upper bound 8 + 6 + 10 + 7 + 15 = 46 km
✓ 1
Part (b)
M1 minimum spanning tree of B, C, D and E
✓ 1
A1 edges CE (6), BD (7), BC (10) giving weight 23
✓ 1
M1 add the two shortest edges at A: AE (8) and AC (9)
✓ 1
A1 lower bound 23 + 17 = 40 km
✓ 1
Part (c)
A1 shortest route L satisfies 40 ≤ L ≤ 46 (ECF)
✓ 1
M1 (46 − 40) ÷ 40 × 100 (ECF)
✓ 1
A1 15%
✓ 1
Common error: The nearest-neighbour route gives an upper bound only, and the deleted-vertex method gives a lower bound only; neither is guaranteed to be the optimal route.
405IBMI3H-3-05r44-11Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard11 marks
A mapping app applies a linear transformation represented by the matrix T to coordinates on a plan. It maps the point P(2, 1) to P′(7, 4) and the point Q(1, 3) to Q′(11, 7).
(a)
Find the matrix T.
(5)
(b)
A rectangular park on the original plan has area 35 hectares. Find its area on the transformed plan, justifying your answer.
(2)
(c)
The line y = x on the original plan is transformed by T. Find the equation of its image.
A1 area is unchanged at 35 hectares because the area scale factor is 1
✓ 1
Part (c)
M1 a general point (t, t) is transformed by T
✓ 1
A1 image point (5t, 3t)
✓ 1
M1 eliminate t from x = 5t, y = 3t
✓ 1
A1 y = 0.6x (or y = ⅗x)
✓ 1
Exam tip: Because the images are the columns of the right-hand matrix, the inverse must multiply on the right of the images, not the left.
406IBMI3H-3-05r44-14Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard10 marks
A retail chain has stores at A(2, 1), B(8, 3) and C(4, 8) on a map where one unit is 1 km. The company uses a Voronoi diagram to decide which store serves each customer.
(a)
Find the equation of the perpendicular bisector of AB, giving your answer in the form y = mx + c.
(3)
(b)
The perpendicular bisector of AC has equation 4x + 14y = 75. Find the coordinates of the Voronoi vertex where the bisectors of AB and AC meet, to 3 significant figures.
(3)
(c)
A customer is at P(6, 6). Determine which store is nearest to P.
(2)
(d)
The company plans a fourth store at the Voronoi vertex found in (b). Find its distance from store A, giving your answer to 3 significant figures. Use unrounded coordinates.
(2)
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Part (a)
A1 midpoint of AB = (5, 2)
✓ 1
M1 gradient of AB = 1/3, so the perpendicular gradient is −3
✓ 1
A1 y = −3x + 17
✓ 1
Part (b)
M1 solve y = −3x + 17 and 4x + 14y = 75 simultaneously (ECF from (a))
Exam tip: A Voronoi vertex is equidistant from three sites, so you can check your coordinates by confirming the distances to all three stores agree.
407IBMI3H-3-08r44-08Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard9 marks
A volcano park has six viewpoints, A to F, joined by trails. The lengths of the trails, in km, are: A–B 4, A–C 7, A–F 10, B–C 5, B–D 6, C–D 3, C–E 8, D–E 5, D–F 9, E–F 4. A warden must walk along every trail at least once, starting and finishing at viewpoint A, while walking as little as possible. The total length of all the trails is 61 km.
(a)
State which vertices have odd degree and explain why the warden has to walk along some trails more than once.
(2)
(b)
Find the shortest distance between each pair of the odd vertices that are needed to complete the three possible pairings, and the total for each pairing.
(3)
(c)
Hence find the minimum distance the warden must walk.
(2)
(d)
The warden walks at an average speed of 4.6 km h−1. Find the time taken, in hours to 3 significant figures.
(2)
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Notes
Part (a)
A1 A, B, E and F have odd degree (3); C and D have degree 4
✓ 1
R1 a closed route using every edge once requires all vertices to have even degree, so some edges must be repeated
✓ 1
Part (b)
M1 attempt to consider all three pairings of the odd vertices: AB + EF, AE + BF and AF + BE
✓ 1
A1 AB + EF = 4 + 4 = 8
✓ 1
A1 AE + BF = 14 + 14 = 28 and AF + BE = 10 + 11 = 21 (using AE = A–F–E, BF = B–A–F, AF direct, BE = B–D–E)
✓ 1
Part (c)
M1 61 + the smallest pairing total
✓ 1
A1 69 km
✓ 1
Part (d)
M1 69 ÷ 4.6 (ECF from (c))
✓ 1
A1 15.0 hours
✓ 1
Exam tip: In the Chinese postman problem, pair up the odd vertices in every possible way and use the shortest path between each pair; the cheapest pairing gives the repeated edges.
408IBMI3H-3-08r44-10Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard11 marks
A theatre designer plans the movement of a triangular set piece on a stage grid, with distances in metres. The set piece has corners A(1, 1), B(4, 1) and C(2, 3). Three transformations are applied in order: P: reflection in the y-axis Q: anticlockwise rotation of 90° about the origin S: a stretch with factor 2 in the x-direction and factor 1.5 in the y-direction. The single matrix M represents the combination of P, then Q, then S.
(a)
Find M.
(3)
(b)
Find the coordinates of the images of A, B and C after the combined transformation.
(2)
(c)
Find M−1 and state what it represents in this context.
(3)
(d)
Find the area of the image of the set piece, in m².
(3)
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Part (a)
M1 attempt to multiply the matrices in the order SQP (the first transformation applied is on the right)
✓ 1
A1 QP = rows (0, −1), (−1, 0)
✓ 1
A1 M = rows (0, −2), (−1.5, 0)
✓ 1
Part (b)
A1 A′(−2, −1.5) and B′(−2, −6)
✓ 1
A1 C′(−6, −3)
✓ 1
Part (c)
M1 attempt to find the inverse using det M = −3 (or technology)
✓ 1
A1 M−1 = rows (0, −2/3), (−1/2, 0)
✓ 1
R1 it moves the set piece back from its final position to its original position
✓ 1
Part (d)
A1 area of triangle ABC = 3 m²
✓ 1
M1 multiply by |det M| (ECF from (a))
✓ 1
A1 9 m²
✓ 1
Exam tip: For a sequence of transformations, the matrix of the first one goes on the right in the product; the area scale factor is the absolute value of the determinant.
409IBMI3H-3-07r46-07Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard11 marks
A marine park has six jetties A, B, C, D, E and F joined by walkways. A ranger wants to inspect every walkway at least once, starting and finishing at jetty A. The lengths of the walkways are given in hundreds of metres: AB 5, AC 7, AF 9, BC 4, BD 8, CD 6, CE 10, DE 5, DF 7, EF 6.
(a)
Explain why it is impossible for the ranger to inspect every walkway exactly once and return to A.
(2)
(b)
The ranger must repeat some walkways. By considering each of the three ways of pairing the four odd vertices, find the least total length of walkway that must be repeated.
(4)
(c)
Find the minimum length of a route that inspects every walkway and returns to A. Give your answer in metres.
(3)
(d)
The ranger walks at a constant 1.2 m s−1. Find the time taken for the route in (c), in minutes, to 3 significant figures. Use the value from (c).
(2)
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Part (a)
A1 the vertices A, B, E and F have odd degree (each has degree 3)
✓ 1
R1 a closed route using every edge exactly once needs every vertex to have even degree
✓ 1
Part (b)
M1 identify the three pairings AB+EF, AE+BF and AF+BE and attempt shortest distances between each pair
✓ 1
A1 AB+EF: 5 + 6 = 11
✓ 1
A1 AE+BF: 15 (via F) + 14 (via A) = 29 and AF+BE: 9 + 13 (via D) = 22
✓ 1
A1 least repeated length is 11, by repeating AB and EF
✓ 1
Part (c)
M1 total of all walkway lengths = 67
✓ 1
M1 67 + 11 (ECF from (b))
✓ 1
A1 78 hundred metres = 7800 m
✓ 1
Part (d)
M1 7800 ÷ 1.2 = 6500 s (ECF from (c))
✓ 1
A1 108 minutes (108.33…)
✓ 1
Exam tip: Check your pairings systematically: with four odd vertices there are exactly three pairings, and the shortest path between two odd vertices may pass through other vertices.
410IBMI3H-3-07r46-08Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard10 marks
A courier company has five depots A, B, C, D and E. A van must visit every depot exactly once and return to its starting depot. The distances in kilometres between each pair of depots are: AB 14, AC 21, AD 16, AE 13, BC 20, BD 27, BE 10, CD 12, CE 17, DE 22.
(a)
Use the nearest neighbour algorithm, starting at A, to find an upper bound for the shortest route. State the route and its length.
(3)
(b)
Use the deleted vertex algorithm, deleting A, to find a lower bound for the shortest route.
(4)
(c)
Write down an interval in which the length of the shortest route must lie.
(1)
(d)
The van costs £1.35 per kilometre to run. Find the greatest possible cost of the shortest route according to your interval. Give your answer to the nearest penny.
(2)
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Notes
Part (a)
M1 attempt to move to the nearest unvisited depot at each stage
✓ 1
A1 route A → E → B → C → D → A
✓ 1
A1 length 13 + 10 + 20 + 12 + 16 = 71 km
✓ 1
Part (b)
M1 attempt at a minimum spanning tree on B, C, D, E
✓ 1
A1 edges BE (10), CD (12), CE (17), total 39
✓ 1
M1 add the two shortest edges from A, AE (13) and AB (14)
✓ 1
A1 lower bound 39 + 27 = 66 km
✓ 1
Part (c)
A1 66 ≤ length ≤ 71 (km) (ECF from (a) and (b))
✓ 1
Part (d)
M1 71 × 1.35 (ECF from (c))
✓ 1
A1 £95.85
✓ 1
Common error: The nearest neighbour tour gives an upper bound, not necessarily the shortest tour, while the deleted vertex method gives a lower bound; the true minimum lies between them.
411IBMI3H-3-07r46-10Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard10 marks
Two research vessels are at sea. At time t hours after noon, vessel A has position vector rA = (2 + 3t)i + (1 + 4t)j and vessel B has position vector rB = (14 − t)i + (−3 + 5t)j, with distances in kilometres relative to a fixed buoy.
(a)
Show that the position of B relative to A is given by (12 − 4t)i + (t − 4)j.
(2)
(b)
Show that the square of the distance between the vessels is 17t2 − 104t + 160 km2.
(2)
(c)
Find the minimum distance between the vessels, and the value of t at which it occurs. Give both answers to 3 significant figures.
(3)
(d)
The vessels must not be closer than 3 km. Find the length of time, in hours, for which they are less than 3 km apart. Give your answer to 3 significant figures.
M1 attempt to minimise the quadratic, e.g. t = 104/(2×17) or differentiation or GDC
✓ 1
A1 t = 3.06 hours
✓ 1
A1 minimum distance = 0.970 km (√0.941…)
✓ 1
Part (d)
M1 set 17t² − 104t + 160 = 9
✓ 1
A1 t = 2.37 and t = 3.75 (2.3703… and 3.7473…)
✓ 1
A1 duration = 1.38 hours (use unrounded roots)
✓ 1
Exam tip: Working with the square of the distance avoids a square root and leads to a quadratic that can be minimised or solved directly.
412IBMI3H-3-06r47-09Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard10 marks
An engineer must inspect every pipe in a water network and return to the starting junction. The network has junctions A to F and the pipe lengths in km are: AB 4, AC 6, AF 8, BC 3, BD 5, CD 7, CE 4, DE 6, DF 3, EF 5.
(a)
Find the degree of each junction, and hence explain why the engineer cannot inspect every pipe exactly once and return to the start.
(2)
(b)
Using the Chinese postman algorithm, find the minimum total length of repeated pipes needed for a route that returns to the start.
(3)
(c)
Hence find the shortest length of a route that inspects every pipe and returns to its starting junction.
(2)
(d)
The engineer now starts at A and finishes at F instead of returning to A. Find the length of the shortest route that inspects every pipe.
(3)
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Part (a)
A1 degrees: A 3, B 3, C 4, D 4, E 3, F 3
✓ 1
R1 four junctions (A, B, E, F) have odd degree, so no Eulerian circuit exists and some pipes must be repeated
✓ 1
Part (b)
M1 consider the three ways of pairing the odd junctions: AB+EF, AE+BF, AF+BE
M1 total of pipe lengths \(=51\) km plus the repeated 9 km
✓ 1
A1 60 km
✓ 1
Part (d)
M1 only the odd junctions B and E need to be paired as A and F are the start and end
✓ 1
A1 shortest path BE \(=7\) km (B–C–E)
✓ 1
A1 \(51+7=58\) km
✓ 1
Common error: Pair up odd vertices using shortest paths through the network, not just direct edges, and check all possible pairings before choosing the smallest.
413IBMI3H-3-06r47-13Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard11 marks
In a game, two drones fly in straight lines. Their positions, in metres, at time t seconds are \(\mathbf{r}_A=\begin{pmatrix}2\\-1\\3\end{pmatrix}+t\begin{pmatrix}1\\2\\-1\end{pmatrix}\) and \(\mathbf{r}_B=\begin{pmatrix}8\\-4\\9\end{pmatrix}+t\begin{pmatrix}-1\\3\\-2\end{pmatrix}\). A proximity alert sounds whenever the drones are less than 3 m apart.
(a)
Show that \(\mathbf{r}_B-\mathbf{r}_A=\begin{pmatrix}6-2t\\-3+t\\6-t\end{pmatrix}\) and that the square of the distance between the drones is \(d^2=6t^2-42t+81\).
(2)
(b)
Find the time at which the drones are closest together, and the minimum distance between them. Give the distance to 3 significant figures.
(3)
(c)
Find the time interval during which the proximity alert sounds.
(3)
(d)
Find the acute angle between the paths of the two drones, to 3 significant figures.
(3)
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Mark
Notes
Part (a)
M1 subtract the position vectors
✓ 1
A1 \((6-2t)^2+(t-3)^2+(6-t)^2=6t^2-42t+81\) (AG)
✓ 1
Part (b)
M1 use differentiation, \(12t-42=0\), or completing the square
✓ 1
A1 \(t=3.5\) s
✓ 1
A1 \(d^2=7.5\) so minimum distance \(=2.74\) m
✓ 1
Part (c)
M1 set \(6t^2-42t+81<9\)
✓ 1
M1 \(t^2-7t+12<0\) and factorise or solve
✓ 1
A1 \(3
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Part (d)
M1 scalar product of direction vectors \(=-1+6+2=7\)
✓ 1
M1 \(\cos\theta=\frac{7}{\sqrt6\sqrt{14}}\)
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A1 \(40.2^\circ\)
✓ 1
Exam tip: The distance between two moving points is the magnitude of the difference of their position vectors; work with \(d^2\) so you can avoid the square root until the end.
414IBMI3H-3-06r48-08Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard10 marks
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A band plans a tour starting and finishing at city A, visiting each of B, C, D and E exactly once. The distances in km are: AB 120, AC 200, AD 150, AE 90, BC 80, BD 160, BE 110, CD 100, CE 190, DE 140.
(a)
Use the nearest-neighbour algorithm, starting at A, to find a tour and its length.
(3)
(b)
State what the answer to (a) tells you about the length of the shortest possible tour.
(1)
(c)
Delete vertex A and find a minimum spanning tree of the remaining four cities. State its total length.
(3)
(d)
Hence find a lower bound for the length of the shortest tour, by adding the two shortest edges at A to your answer to (c).
(2)
(e)
Write down an inequality for the length \(L\) of the shortest tour.
(1)
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Notes
Part (a)
M1 at each step move to the nearest city not yet visited
✓ 1
A1 A–E–B–C–D–A
✓ 1
A1 \(90 + 110 + 80 + 100 + 150 = 530\) km
✓ 1
Part (b)
R1 the shortest tour is at most \(530\) km (this is an upper bound)
✓ 1
Part (c)
M1 attempt to build a minimum spanning tree on B, C, D, E
✓ 1
A1 BC, CD and BE chosen
✓ 1
A1 \(80 + 100 + 110 = 290\) km
✓ 1
Part (d)
M1 the two shortest edges at A, namely AE 90 and AB 120
✓ 1
A1 \(290 + 90 + 120 = 500\) km
✓ 1
Part (e)
A1 \(500 \le L \le 530\) (ECF from (a) and (d))
✓ 1
Exam tip: The lower bound comes from deleting a vertex; the upper bound comes from any complete tour. Together they trap the true minimum.
415IBMI3H-3-06r48-10Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard10 marks
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An aerial photograph of a triangular plot is distorted by the transformation with matrix \(T = \begin{pmatrix} 2 & 1 \\ 1 & 3 \end{pmatrix}\). On the plan, the plot has corners \(O(0, 0)\), \(B(4, 0)\) and \(C(0, 3)\), with coordinates in hundreds of metres.
(a)
Find the images \(B'\) and \(C'\) of \(B\) and \(C\) under \(T\).
(2)
(b)
Find the area of triangle \(OB'C'\).
(2)
(c)
Show that the area of triangle \(OB'C'\) is \(|\det T|\) times the area of triangle \(OBC\).
(2)
(d)
Find \(T^{-1}\).
(2)
(e)
A point \(P\) is mapped by \(T\) to the point \((13, 17)\). Find the coordinates of \(P\).
(2)
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Part (a)
A1 \(B' = (8, 4)\)
✓ 1
A1 \(C' = (3, 9)\)
✓ 1
Part (b)
M1 attempt to use \(\frac{1}{2}\left|8\times9 - 3\times4\right|\)
✓ 1
A1 \(30\) square units
✓ 1
Part (c)
M1 \(\det T = 2\times3 - 1\times1 = 5\) and area of \(OBC\) is \(6\)
✓ 1
R1 \(30 = 5 \times 6\), so the area scale factor equals \(|\det T|\)
✓ 1
Part (d)
M1 attempt to use \(\frac{1}{\det T}\) with the entries rearranged
Exam tip: The determinant of a transformation matrix gives the area scale factor; a negative determinant also means orientation is reversed.
416IBMI3H-3-06r48-13Vectors, scalar and vector products and vector kinematics; graph theory: adjacency matrices, minimum spanning trees, Chinese postman, travelling salesman; matrix transformationsPaper 2Hard9 marks
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An agricultural drone moves over a field so that its position, in metres relative to a base station, is \(\mathbf{r}(t) = \begin{pmatrix} 2t^2 \\ 3t - t^2 \end{pmatrix}\), where \(t\) is the time in seconds after take-off. The \(x\)-axis points east and the \(y\)-axis points north.
(a)
Find the velocity vector of the drone at time \(t\).
(2)
(b)
Find the speed of the drone when \(t = 2\), correct to 3 significant figures.
(2)
(c)
Find the acceleration vector and state whether it changes with time.
(1)
(d)
Find the time at which the drone is moving due east.
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