Edexcel International A Level Maths · P4 · Pure Mathematics 4
Vectors in two and three dimensions
28 free, high-quality past-paper-style questions on Vectors in two and three dimensions for Edexcel IAL Maths, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Maths notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.28 new this month · 251 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
1IALM-P4-8-B-01P4 · A2Easy6 marks
The vectors a and b are given by a = 4i − 2j + 4k and b = i + 3j − k.
Common error: Sign slips when subtracting. Work out 2a = 8i − 4j + 8k and 3b = 3i + 9j − 3k first: the k component of the difference is 8 − (−3) = 11, not 5.
In a plane, relative to a fixed origin O, the points P and Q have position vectors −3i + 4j and 5i − 2j respectively, where i and j are perpendicular unit vectors.
Figure: the points P and Q relative to O
(a)
Find PQ.
(1)
(b)
Find the distance PQ.
(2)
(c)
Find the angle between PQ and the unit vector i, giving your answer in degrees to 1 decimal place.
(2)
(d)
The point R has position vector 13i − 8j. Show that P, Q and R lie on a straight line.
(2)
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Answer
Mark
Notes
Part (a)
8i − 6j
B1
Part (b)
√(82 + 62)
M1
= 10
A1
Part (c)
tan−1(6/8) or cos−1(8/10)
M1
36.9° (below i)
A1
Part (d)
PR = 16i − 12j = 2PQ (or QR = 8i − 6j = PQ)
M1
the vectors are parallel and share the point P, so P, Q and R are collinear
A1
Exam tip: Collinearity needs two things in writing: one vector is a multiple of the other, and the two vectors share a point.
Two vectors are defined by u = 3i − 4j + 12k and v = 2i + 2j − k.
(a)
Find u·v.
(1)
(b)
Find the angle between u and v, giving your answer in degrees to 1 decimal place.
(3)
(c)
Find the value of the constant λ for which u + λv is perpendicular to v.
(2)
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Answer
Mark
Notes
Part (a)
6 − 8 − 12 = −14
B1
Part (b)
|u| = 13 and |v| = 3
B1
cos θ = −14/(13 × 3) = −14/39
M1
θ = 111.0°
A1
Part (c)
(u + λv)·v = −14 + 9λ = 0
M1
λ = 14/9
A1
Exam tip: A negative scalar product means an obtuse angle between the vectors. Keep it: the question asks for the angle between the vectors, not between two lines.
Figure 1 shows the quadrilateral ABCD drawn on a coordinate grid of unit squares. Each vertex lies at a grid point. The unit vectors i and j are in the directions of the positive x- and y-axes.
(a)
Use Figure 1 to find the vectors AB and DC.
(2)
(b)
Hence show that ABCD is a trapezium, and find the ratio AB : DC.
(2)
(c)
Find the exact length of AD.
(2)
(d)
Find the unit vector in the direction of BC.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
from the grid A(−2, 1), B(4, 3), C(6, −2), D(−3, −5); AB = 6i + 2j
B1
DC = 9i + 3j
B1
Part (b)
DC = 3/2 AB, so AB is parallel to DC (and AD = −i − 6j is not parallel to BC = 2i − 5j)
M1
AB : DC = 2 : 3
A1
Part (c)
AD = −i − 6j
M1
AD = √(1 + 36) = √37
A1
Part (d)
BC = 2i − 5j, so the unit vector is (1/√29)(2i − 5j)
B1
Common error: Finding BC as b − c. The vector from B to C is c − b: end point minus start point, so (6 − 4)i + (−2 − 3)j = 2i − 5j.
Figure 1 shows a storage locker modelled as a cuboid OABCDEFG. Relative to the origin O, the edges OA, OC and OD lie along the positive x-, y- and z-axes, and the vertices E, F and G are directly above A, B and C. The dimensions, in metres, are shown in Figure 1.
(a)
Write down the position vectors of B and F.
(2)
(b)
Find the length of the diagonal OF.
(2)
(c)
Find the acute angle between the diagonals OF and AG, giving your answer in degrees to 1 decimal place.
(3)
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Answer
Mark
Notes
Part (a)
OB = 4i + 4j
B1
OF = 4i + 4j + 7k
B1
Part (b)
√(42 + 42 + 72) = √81
M1
= 9 m
A1
Part (c)
AG = −4i + 4j + 7k
B1
OF·AG = −16 + 16 + 49 = 49 and cos θ = 49/(9 × 9)
M1
θ = 52.8°
A1
Exam tip: Build each position vector from the dimensions: G is above C, so it is 0 m along x, 4 m along y and 7 m up, giving (0, 4, 7).
Figure 1 shows two vectors p and q drawn on squared paper. The unit vectors i and j are one square to the right and one square up, as shown.
(a)
Write p and q in terms of i and j.
(2)
(b)
Find p·q.
(1)
(c)
Find the angle between p and q, giving your answer in degrees to 1 decimal place.
(2)
(d)
Find the vector of magnitude 10 that is perpendicular to p and has a positive i component.
(2)
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Answer
Mark
Notes
Part (a)
p = 3i + 4j
B1
q = −2i + 5j
B1
Part (b)
3 × (−2) + 4 × 5 = 14
B1
Part (c)
|p| = 5, |q| = √29, cos θ = 14/(5√29)
M1
θ = 58.7°
A1
Part (d)
4i − 3j (or a multiple) is perpendicular, since (3i + 4j)·(4i − 3j) = 12 − 12 = 0
M1
scaled to magnitude 10: 8i − 6j
A1
Exam tip: A vector does not need to start at the origin. Count squares from the tail to the head of the arrow: q goes 2 squares left and 5 up, so its i component is −2.
Figure 1 shows the straight lines L1 and L2 on a coordinate grid of unit squares. L1 passes through the grid points A and B, and L2 passes through the grid points C and D. The unit vectors i and j are in the directions of the positive x- and y-axes.
(a)
Find a vector equation of L1, using the parameter s.
(2)
(b)
Find a vector equation of L2, using the parameter t.
(1)
(c)
Show that L1 and L2 are perpendicular.
(1)
(d)
Find the coordinates of the point where L1 and L2 meet.
(3)
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Answer
Mark
Notes
Part (a)
AB = 4i − 2j (or a multiple)
M1
r = (−2i + 3j) + s(2i − j) or equivalent
A1
Part (b)
r = (−i − 2j) + t(i + 2j) or equivalent
B1
Part (c)
(2i − j)·(i + 2j) = 2 − 2 = 0, so the lines are perpendicular
B1
Part (d)
−2 + 2s = −1 + t and 3 − s = −2 + 2t
M1
s = 7/5 (t = 9/5)
A1
(0.8, 1.6)
A1
Common error: Reading the crossing point off the grid. The lines do not cross at a grid point, so the point must come from solving the two component equations.
In a model of a factory, two straight pipes are represented by the lines l1 and l2, where l1: r = (2i + 3j − k) + λ(i − j + 2k) l2: r = (−4i + 3j + 2k) + μ(3i − j + k) The unit of length is the metre. The pipes are modelled as lines with no thickness.
(a)
Show that the lines intersect, and find the coordinates of the point of intersection.
(5)
(b)
Find the acute angle between the pipes, in degrees to 1 decimal place.
(3)
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Answer
Mark
Notes
Part (a)
equating two components, e.g. 2 + λ = −4 + 3μ and 3 − λ = 3 − μ
M1
solving simultaneously
M1
λ = 3, μ = 3
A1
third component: −1 + 2(3) = 5 and 2 + 3 = 5, consistent, so the lines intersect
A1
point (5, 0, 5)
A1
Part (b)
(i − j + 2k)·(3i − j + k) = 3 + 1 + 2 = 6
M1
cos θ = 6/(√6 × √11)
M1
θ = 42.4°
A1
Exam tip: Use two of the component equations to find both parameters, then show the third one working with both values. Without that check the lines could be skew.
Two drones fly along straight paths. Relative to a fixed origin on level ground, with distances in metres, the paths are modelled by path 1: r = (i + 2k) + s(2i + j − k) path 2: r = (4i + 3j) + t(i − j + 2k) Each drone is modelled as a particle.
(a)
Explain why the paths are not parallel.
(1)
(b)
Show that the two paths do not meet.
(4)
(c)
Find the acute angle between the directions of the two paths, in degrees to 1 decimal place.
(3)
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Answer
Mark
Notes
Part (a)
2i + j − k is not a scalar multiple of i − j + 2k (e.g. the i components need a factor 2 but the j components need −1)
B1
Part (b)
1 + 2s = 4 + t and s = 3 − t
M1
s = 2, t = 1
A1
k components: 2 − s = 0 and 2t = 2
M1
0 ≠ 2, so the paths do not intersect (and with part (a) they are skew)
A1
Part (c)
2 × 1 + 1 × (−1) + (−1) × 2 = −1
M1
cos θ = −1/(√6 × √6) = −1/6, θ = 99.6°
M1
acute angle = 80.4°
A1
Common error: Giving 99.6° as the angle between two lines. The scalar product was negative, so subtract from 180° for the acute angle.
A straight overhead cable is modelled as the line l with equation r = (−i + 2j) + t(2i − j + 2k), relative to a fixed origin, with distances in metres. A lamp is fixed at the point P(5, 5, 0). The point F lies on l.
Figure: the cable l, the lamp P and a point F on l; schematic, not to scale
(a)
Write down PF in terms of t.
(1)
(b)
Find the coordinates of the point on the cable that is nearest to P.
(3)
(c)
Find the shortest distance from P to the cable.
(2)
(d)
The point Q is such that the nearest point of the cable is the midpoint of PQ. Find the coordinates of Q.
(2)
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Answer
Mark
Notes
Part (a)
(2t − 6)i + (−3 − t)j + 2tk
B1
Part (b)
2(2t − 6) − (−3 − t) + 2(2t) = 0
M1
9t − 9 = 0, so t = 1
A1
(1, 1, 2)
A1
Part (c)
PF = −4i − 4j + 2k
M1
√(16 + 16 + 4) = 6 m
A1
Part (d)
OQ = 2(1, 1, 2) − (5, 5, 0)
M1
(−3, −3, 4)
A1
Exam tip: The nearest point is the foot of the perpendicular: set PF·d = 0 using the direction vector of the line.
A flat triangular solar panel has corners at A(2, 1, 0), B(6, 1, 3) and C(2, 5, 3), relative to a fixed origin, with distances in metres. The panel is modelled as a triangle of negligible thickness.
(a)
Find AB and AC.
(2)
(b)
Find the size of angle BAC, in degrees to 1 decimal place.
(3)
(c)
Find the area of the panel, giving your answer to 3 significant figures.
(3)
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Answer
Mark
Notes
Part (a)
AB = 4i + 3k
B1
AC = 4j + 3k
B1
Part (b)
AB·AC = 0 + 0 + 9 = 9
M1
|AB| = |AC| = 5, cos θ = 9/25
M1
θ = 68.9°
A1
Part (c)
sin θ = √(1 − (9/25)2) = √544/25, or sin 68.9°
M1
½ × 5 × 5 × sin θ
M1
= 2√34 = 11.7 m2
A1
Common error: Using cos θ in the area formula. The area of a triangle is ½ab sin C, with the angle between the two sides used.
Figure 1 shows a vertical mast OA standing on horizontal ground, which is the plane z = 0. A straight wire runs from the top A of the mast, through a bracket at B, to a peg at G on the ground. The coordinates of A and B are shown, relative to the origin O at the foot of the mast, with distances in metres. The wire is modelled as a straight line.
(a)
Find a vector equation of the line of the wire.
(2)
(b)
Find the coordinates of G.
(2)
(c)
Find the length of the wire AG, giving your answer to 3 significant figures.
(2)
(d)
Find the angle between the wire and the ground, giving your answer in degrees to 1 decimal place.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
AB = 2i + 3j − 3k
M1
r = 9k + t(2i + 3j − 3k) or equivalent
A1
Part (b)
9 − 3t = 0, so t = 3
M1
G(6, 9, 0)
A1
Part (c)
AG = 6i + 9j − 9k
M1
√198 = 14.1 m
A1
Part (d)
GA·GO = 36 + 81 + 0 = 117 and cos θ = 117/(√198 × √117), or sin θ = 9/√198
M1
θ = 39.8°
A1
Exam tip: The angle between the wire and the ground is the angle between the wire and the line OG directly beneath it on the ground. Equivalently, use the height of the mast: sin θ = 9/|AG|.
Two straight rods AB and CD are fixed in a frame. Figure 1 shows the coordinates of their end points, relative to a fixed origin O. The rods are modelled as line segments lying on the lines AB and CD.
(a)
Find vector equations of the lines AB and CD.
(2)
(b)
Show that the lines AB and CD are skew.
(4)
(c)
The end D is moved vertically to D′(4, 4, h), with C fixed, so that the rods touch. Find the value of h and the coordinates of the point where the rods touch.
(4)
Show mark scheme
Answer
Mark
Notes
Part (a)
AB: r = (4i + k) + s(−2i + 2j + k) or equivalent
B1
CD: r = 2k + t(2i + 2j + k) or equivalent
B1
Part (b)
−2i + 2j + k is not a multiple of 2i + 2j + k, so the lines are not parallel
B1
4 − 2s = 2t and 2s = 2t
M1
s = 1, t = 1
A1
z: 1 + 1 = 2 but 2 + 1 = 3; 2 ≠ 3 so the lines do not meet, and they are not parallel, so they are skew
A1
Part (c)
line CD′: r = 2k + u(4i + 4j + (h − 2)k)
M1
4 − 2s = 4u and 2s = 4u give s = 1, u = 1/2
A1
z: 1 + 1 = 2 + (h − 2)/2, so h = 2
A1
point (2, 2, 2)
A1
Common error: Stopping at “not parallel”. In three dimensions, lines that are not parallel may still miss each other; only the third component equation decides whether they meet.
Figure 1 shows a flat metal plate ABCD in the shape of a parallelogram, with C opposite A. The coordinates of A, B and D are shown, relative to a fixed origin, with distances in centimetres. The plate is modelled as a parallelogram of negligible thickness.
(a)
Find the coordinates of C.
(2)
(b)
Show that ABCD is a rhombus.
(2)
(c)
Show that the diagonals AC and BD are perpendicular.
(2)
(d)
Find the exact area of the plate.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
OC = OB + AD
M1
C(3, 5, 3)
A1
Part (b)
AB = i + 2j + 2k and AD = 2i + 2j − k
M1
|AB| = |AD| = 3, so the parallelogram has equal adjacent sides: a rhombus
A1
Part (c)
AC = 3i + 4j + k and BD = i − 3k
M1
AC·BD = 3 + 0 − 3 = 0, so perpendicular
A1
Part (d)
AB·AD = 2 + 4 − 2 = 4, cos BAD = 4/9
M1
sin BAD = √(1 − 16/81) = √65/9
M1
area = 3 × 3 × √65/9 = √65 cm2 (or ½ × √26 × √10)
A1
Exam tip: The area of a parallelogram is |AB||AD| sin θ, twice the area of triangle ABD, so there is no ½. For a rhombus, ½ × (product of the diagonals) gives the same value and makes a quick check.
Common error: Using 120° for the angle between a and b. The scalar product needs the angle between the vectors when they start from the same point; here they run head to tail, so the angle is 180° − 120° = 60°.
Figure 1 shows a map on a grid of 1 km squares, with the unit vectors i and j (each 1 km) pointing due east and due north. A ship sails on a straight course through the points P and Q. A lighthouse is at L. The points P, Q and L are at grid points. The ship is modelled as a particle.
(a)
Find a vector equation of the ship's course.
(2)
(b)
Find the acute angle between the ship's course and the direction due north, giving your answer in degrees to 1 decimal place.
(2)
(c)
The light from the lighthouse can be seen from any point within 5 km of L. The position of the ship is r = (−4i − j) + t(2i + j), so that t = 0 at P and t = 3 at Q. Show that the light can be seen from the ship when 5t2 − 24t + 28 ≤ 0.
(3)
(d)
Find the coordinates of the points where the light first becomes visible from the ship and where it stops being visible.
(2)
(e)
Find the exact distance the ship travels while the light can be seen.
(14/5 − 2) × |2i + j| = 0.8√5 = 4√5/5 km (≈ 1.79 km)
B1
Exam tip: Work with the squared distance. The condition “distance from L ≤ 5” becomes (distance)2 ≤ 25, which gives a quadratic in the parameter with no square roots; its two roots are the boundary points.
Figure 1 shows a pyramid VOABC. The base OABC is a square of side 4 units in the plane z = 0, with O at the origin, A on the positive x-axis and C on the positive y-axis. The vertex V is vertically above the centre N of the base, and NV = 6 units.
(a)
Write down the position vectors of B and V.
(2)
(b)
Find the exact length of the edge VA.
(2)
(c)
Find the size of angle AVC, giving your answer in degrees to 1 decimal place.
(3)
(d)
Find the angle between the edge VA and the diagonal AC of the base, giving your answer in degrees to 1 decimal place.
Exam tip: For an angle at a vertex, both vectors must start at that vertex: for angle AVC use VA and VC. Using AV with VC gives the supplementary angle.
With respect to a fixed origin O, the lines l1 and l2 are given by l1: r = (3i − j + 2k) + λ(i + 2j − k) l2: r = (7i + 2j + pk) + μ(2i − j + 3k) where p is a constant. The lines intersect at the point X.
(a)
Find the value of p.
(4)
(b)
Find the coordinates of X.
(1)
(c)
Find the acute angle between l1 and l2, in degrees to 1 decimal place.
(3)
(d)
The points Y on l2 satisfy XY = 2√14. Find the coordinates of the two possible positions of Y.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
3 + λ = 7 + 2μ and −1 + 2λ = 2 − μ
M1
solving simultaneously
M1
λ = 2, μ = −1
A1
k: 2 − 2 = p − 3, so p = 3
A1
Part (b)
(5, 3, 0)
B1
Part (c)
1 × 2 + 2 × (−1) + (−1) × 3 = −3
M1
cos θ = −3/(√6 × √14), θ = 109.1°
M1
acute angle 70.9°
A1
Part (d)
XY = |μ + 1| × √14, so |μ + 1| = 2
M1
μ = 1 or μ = −3
A1
(9, 1, 6) and (1, 5, −6)
A1
Exam tip: When a line contains an unknown constant, use the two component equations that do not involve it to find the parameters; the third equation then gives the constant.
A rectangular glass panel PQSR is held in a frame. Relative to a fixed origin, the corners P, Q and R have coordinates (1, 2, −1), (3, 3, 1) and (−3, 6, 1) respectively, with distances in metres. The panel is modelled as a rectangle of negligible thickness.
(a)
Show that PQ is perpendicular to PR.
(2)
(b)
Find the area of triangle PQR.
(2)
(c)
Use a scalar product to find the size of angle PRQ, in degrees to 1 decimal place.
(3)
(d)
Find the coordinates of S.
(2)
(e)
The point F lies on the diagonal QR and PF is perpendicular to QR. Find the coordinates of F and the exact length PF.
(4)
Show mark scheme
Answer
Mark
Notes
Part (a)
PQ = 2i + j + 2k and PR = −4i + 4j + 2k
B1
PQ·PR = −8 + 4 + 4 = 0, so perpendicular
B1
Part (b)
|PQ| = 3 and |PR| = 6
M1
area = ½ × 3 × 6 = 9 m2
A1
Part (c)
RP = 4i − 4j − 2k and RQ = 6i − 3j
M1
RP·RQ = 24 + 12 = 36 and cos θ = 36/(6 × 3√5)
M1
θ = 26.6°
A1
Part (d)
OS = OQ + PR (or OR + PQ)
M1
(−1, 7, 3)
A1
Part (e)
line QR: r = (3i + 3j + k) + t(−2i + j), so PF = (2 − 2t)i + (1 + t)j + 2k
M1
−2(2 − 2t) + (1 + t) = 0, i.e. 5t − 3 = 0
M1
t = 3/5, F(9/5, 18/5, 1)
A1
PF = √(16/25 + 64/25 + 4) = 6√5/5 m
A1
Exam tip: Check part (e) against part (b): the area of triangle PQR is also ½ × QR × PF = ½ × 3√5 × 6√5/5 = 9.
In a laboratory, a laser beam travels in a straight line from a source at A(−4, 6, 9) through the point B(−1, 4, 3). The floor of the laboratory is the plane z = 0 and distances are in metres. The beam is modelled as a straight line and the source as a point.
(a)
Find a vector equation of the path of the beam.
(2)
(b)
Find the coordinates of the point where the beam meets the floor.
(2)
(c)
Find the acute angle between the beam and the vertical, in degrees to 1 decimal place.
(2)
(d)
A sensor is at the point C(1, 7, 3). Find the coordinates of the point F on the path of the beam that is closest to the sensor.
(4)
(e)
Find the shortest distance from the sensor to the beam, giving your answer to 3 significant figures.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
AB = 3i − 2j − 6k
M1
r = (−4i + 6j + 9k) + t(3i − 2j − 6k) or equivalent
A1
Part (b)
9 − 6t = 0, t = 3/2
M1
(0.5, 3, 0)
A1
Part (c)
(3i − 2j − 6k)·k = −6, |3i − 2j − 6k| = 7, so the acute angle has cos θ = 6/7
M1
θ = 31.0°
A1
Part (d)
CF = (−5 + 3t)i + (−1 − 2t)j + (6 − 6t)k
M1
3(−5 + 3t) − 2(−1 − 2t) − 6(6 − 6t) = 0
M1
49t − 49 = 0, t = 1
A1
F(−1, 4, 3)
A1
Part (e)
CF = −2i − 3j
M1
√13 = 3.61 m
A1
Exam tip: The vertical is the direction k. The angle with the floor would be 90° minus the angle with the vertical — read which one is asked for.
Figure 1 shows a cube OABCDEFG of edge 6 cm. Relative to the origin O, the edges OA, OC and OD lie along the positive x-, y- and z-axes, and E, F and G are directly above A, B and C. The points P, Q and R are the midpoints of the edges AE, BC and DG respectively.
(a)
Find the position vectors of P, Q and R.
(3)
(b)
Show that triangle PQR is equilateral, and find the exact length of its sides.
(3)
(c)
Show that the diagonal OF is perpendicular to both PQ and QR.
(2)
(d)
Find the exact area of triangle PQR.
(2)
(e)
Find the angle between OP and OF, giving your answer in degrees to 1 decimal place.
Figure 1 shows the straight line L through the points P and Q, and the point A, which is not on L. Coordinates are shown relative to a fixed origin O. A square ABCD is formed so that its diagonal BD lies along L, with B nearer to Q than D is. The diagonals of the square meet at M.
(a)
Find a vector equation of L.
(2)
(b)
Explain why M is the foot of the perpendicular from A to L, and find the coordinates of M.
(4)
(c)
Find the coordinates of B and D.
(3)
(d)
Find the coordinates of C.
(1)
(e)
Find the area of the square ABCD.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
PQ = 4i + 8j + 8k
M1
r = (i + j + k) + t(i + 2j + 2k) or equivalent
A1
Part (b)
the diagonals of a square meet at right angles, so AM is perpendicular to L
the diagonals are equal and bisect each other, so MB = MD = 3 = |i + 2j + 2k|, giving t = 2 ± 1
M1
B(4, 7, 7) and D(2, 3, 3)
A1
Part (d)
OC = 2OM − OA: C(1, 4, 7)
B1
Part (e)
diagonal AC = 2 × 3 = 6 (or side AB = 3√2)
M1
area = ½ × 6 × 6 = 18
A1
Exam tip: Turn the shape facts into vector facts before calculating: the diagonals of a square are equal, bisect each other and meet at right angles. Each fact gives one of the steps here.
Figure 1 shows a straight rail through the points R and S, and two fixed points A and B. Coordinates are shown relative to a fixed origin, with distances in metres. A slider P moves along the rail and is joined to A and to B by taut cables. The rail is modelled as a straight line, the slider as a particle and the cables as straight line segments.
(a)
Find a vector equation of the rail, in the form r = a + tb, where t = 0 at R.
(2)
(b)
Find PA and PB in terms of t.
(2)
(c)
Show that the two cables are perpendicular when t2 − 3t + 2 = 0.
(3)
(d)
Find the coordinates of the two positions of P for which the cables are perpendicular.
(2)
(e)
For the position of P further from R, find the area of triangle APB.
Figure 1 shows triangle OAB on a coordinate grid of unit squares, with O at the origin and A and B at grid points. OA = a and OB = b. The point P lies on OA and the point Q lies on OB, dividing them in the ratios shown. The lines AQ and BP meet at X.
(a)
Find AQ and BP in terms of a and b.
(2)
(b)
By writing OX in two different ways, find OX in terms of a and b.
(5)
(c)
Find the ratio AX : XQ.
(1)
(d)
Use Figure 1 to write a and b in terms of i and j, and hence find the coordinates of X.
(2)
(e)
Find the acute angle between the lines AQ and BP, giving your answer in degrees to 1 decimal place.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
AQ = ½b − a
B1
BP = ¾a − b
B1
Part (b)
OX = a + λ(½b − a) = (1 − λ)a + ½λb
M1
OX = b + μ(¾a − b) = ¾μa + (1 − μ)b
M1
a and b are not parallel, so compare coefficients: 1 − λ = ¾μ and ½λ = 1 − μ
M1
λ = 2/5, μ = 4/5
A1
OX = (3/5)a + (1/5)b
A1
Part (c)
λ = 2/5, so AX : XQ = 2 : 3
B1
Part (d)
a = 8i, b = 2i + 6j
B1
X(5.2, 1.2)
A1
Part (e)
AQ = −7i + 3j, BP = 4i − 6j, AQ·BP = −28 − 18 = −46, cos θ = −46/(√58 × √52)
M1
θ = 146.9°, so the acute angle is 33.1°
A1
Exam tip: Comparing coefficients of a and b is valid only because they are not parallel. Say so in your working before you equate them.
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