443 free, high-quality past-paper-style questions on Geometry and trigonometry for IB Maths AA HL, each with a full mark scheme. Covers: Three-dimensional geometry: distance, solids and angles; Applications of trigonometry: elevation, depression and bearings; Radians, arc length and sector area; and more. For when you have finished the past papers, or want to practise one topic at a time.Topic 3 at Standard Level →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.3,476 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
1AAHL3-q00013.1 Three-dimensional geometry: distance, solids and anglesPaper 1Easy9 marks
The points \(A\) and \(B\) have coordinates \((1,\,-2,\,4)\) and \((k,\,2,\,0)\) respectively. The distance \(AB\) is \(6\) units.
(a)
Find the possible values of \(k\).
(3)
(b)
It is now given that \(k=3\). The point \(C\) lies on the \(z\)-axis and has coordinates \((0,\,0,\,c)\). The angle \(A\hat{C}B\) is a right angle. Find the possible values of \(c\).
(3)
(c)
Let \(M\) be the midpoint of \([AB]\). Show that \(MC=3\) for both values of \(c\).
(2)
(d)
Explain why the result in part (c) could have been predicted without calculation.
\([AB]\) is a diameter of the circle through \(A\), \(B\) and \(C\) (angle in a semicircle), so \(M\) is its centre and \(MC=\dfrac{1}{2}AB=3\)
R1
Accept: the midpoint of the hypotenuse is equidistant from the three vertices.
Answers: (a) \(k=-1\) or \(k=3\) · (b) \(c=2\pm\sqrt5\)
2AAHL3-q00023.1 Three-dimensional geometry: distance, solids and anglesPaper 1Easy8 marks
A regular tetrahedron \(ABCD\) has all six edges of length \(6\) cm. The face \(BCD\) is horizontal and \(A\) is vertically above the point \(G\), the centre of the equilateral triangle \(BCD\), so that \(GB=GC=GD\) and \(B\hat{G}C=C\hat{G}D=D\hat{G}B=\dfrac{2\pi}{3}\).
Plan view of the face BCD (A is vertically above G). Diagram not to scale
(a)
Show that \(BG=2\sqrt3\) cm.
(2)
(b)
Find the exact height \(AG\) of the tetrahedron.
(2)
(c)
Find the exact volume of the tetrahedron.
(3)
(d)
Write down the exact value of the cosine of the angle between the edge \([AB]\) and the face \(BCD\).
(1)
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Working
Mark
Notes
(a)
cosine rule in triangle \(BGC\) with \(BG=GC=x\): \(36=x^{2}+x^{2}-2x^{2}\cos\dfrac{2\pi}{3}\)
M1
Or: the median of \(BCD\) has length \(3\sqrt3\) and \(BG=\tfrac{2}{3}\) of it.
\(3x^{2}=36\), so \(x=\sqrt{12}\)
A1
\(BG=2\sqrt3\)
AG
(b)
\(AG^{2}=6^{2}-\left(2\sqrt3\right)^{2}=24\)
(M1)
\(AG=2\sqrt6\) cm
A1
Accept \(\sqrt{24}\).
(c)
area of \(BCD=\dfrac{1}{2}(6)(6)\sin\dfrac{\pi}{3}=9\sqrt3\)
3AAHL3-q00083.3 Applications of trigonometry: elevation, depression and bearingsPaper 1Easy8 marks
An orienteer runs \(6\) km from checkpoint \(A\) on a bearing of \(060^{\circ}\) to checkpoint \(B\). She then runs \(6\sqrt3\) km on a bearing of \(150^{\circ}\) to checkpoint \(C\).
(a)
Draw a labelled diagram and show that \(A\hat{B}C=90^{\circ}\).
(2)
(b)
Find the distance \(AC\).
(2)
(c)
Find the bearing of \(C\) from \(A\).
(2)
(d)
The point \(D\) is due east of \(A\) and due north of \(C\). Find the exact distances \(AD\) and \(DC\).
(2)
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Mark
Notes
(a)
diagram with north lines at \(A\) and \(B\); the bearing of \(A\) from \(B\) is \(240^{\circ}\)
M1
\(240^{\circ}-150^{\circ}=90^{\circ}\)
A1
\(A\hat{B}C=90^{\circ}\)
AG
(b)
\(AC^{2}=6^{2}+\left(6\sqrt3\right)^{2}=144\)
(M1)
\(12\) km
A1
(c)
\(\tan B\hat{A}C=\dfrac{6\sqrt3}{6}=\sqrt3\), so \(B\hat{A}C=60^{\circ}\)
(M1)
\(120^{\circ}\)
A1
(d)
\(AD=12\cos30^{\circ}=6\sqrt3\) km
A1
\(DC=12\sin30^{\circ}=6\) km
A1
Answers: (b) 12 km · (c) 120° · (d) \(AD=6\sqrt3\) km, \(DC=6\) km
Model sketch
Model diagram (not to scale)
4AAHL3-q00123.4 Radians, arc length and sector areaPaper 1Easy8 marks
A bend of a running track is a semicircle. The track has five lanes, each \(1.2\) m wide. A runner in lane \(n\) runs along a semicircular arc of radius \((34.8+1.2n)\) m, so that the radius for lane \(1\) is \(36\) m. All arcs have the same centre \(O\) and every runner finishes the bend on the same straight line through \(O\).
Lanes 1 (inside) to 5 (outside) on the semicircular bend. Diagram not to scale
(a)
Find the exact length of the bend for a runner in lane \(1\) and for a runner in lane \(5\).
(2)
(b)
Show that the bend in lane \(n+1\) is \(1.2\pi\) m longer than the bend in lane \(n\).
(2)
(c)
The runner in lane \(5\) starts further round the bend so that she runs the same distance on the bend as the runner in lane \(1\). Find the angle, in radians, between \([OS_{1}]\) and \([OS_{5}]\), where \(S_{1}\) and \(S_{5}\) are the starting points of the runners in lanes \(1\) and \(5\).
(3)
(d)
Express your answer to part (c) in degrees, as a fraction in its simplest form.
(1)
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Mark
Notes
(a)
lane \(1\): \(36\pi\) m
A1
lane \(5\): radius \(40.8\), length \(40.8\pi\) m
A1
Accept \(\dfrac{204\pi}{5}\).
(b)
\(\pi(34.8+1.2(n+1))-\pi(34.8+1.2n)\)
M1
\(=1.2\pi\), independent of \(n\)
A1
bend in lane \(n+1\) is \(1.2\pi\) m longer
AG
(c)
extra distance in lane \(5\) \(=40.8\pi-36\pi=4.8\pi\)
(M1)
Or the lane-5 runner covers \(36\pi\) on radius \(40.8\): angle \(\dfrac{36\pi}{40.8}=\dfrac{15\pi}{17}\).
\(\theta=\dfrac{s}{r}=\dfrac{4.8\pi}{40.8}\)
(M1)
Or \(\pi-\dfrac{15\pi}{17}\).
\(\dfrac{2\pi}{17}\)
A1
(d)
\(\left(\dfrac{360}{17}\right)^{\circ}\)
A1
Accept \(21\tfrac{3}{17}^{\circ}\). FT from (c).
Answers: (a) \(36\pi\) m and \(40.8\pi\) m · (c) \(\dfrac{2\pi}{17}\) · (d) \(\dfrac{360}{17}^{\circ}\)
5AAHL3-q00203.6 Pythagorean and double angle identitiesPaper 1Easy7 marks
(a)
Show that \(\dfrac{1-\cos2x}{\sin2x}\equiv\tan x\), for \(\sin2x\neq0\).
(3)
(b)
Hence find the exact value of \(\tan\dfrac{\pi}{8}\), in the form \(\sqrt{a}+b\), where \(a,b\in\mathbb{Z}\).
(2)
(c)
Hence find the exact value of \(\tan\dfrac{3\pi}{8}\), in the same form.
(2)
Show mark scheme
Working
Mark
Notes
(a)
substituting \(\cos2x=1-2\sin^{2}x\) and \(\sin2x=2\sin x\cos x\)
14AAHL3-q00563.12 Vectors: concepts and algebraPaper 1Easy7 marks
The vectors \(\boldsymbol{u}=3\boldsymbol{i}+p\boldsymbol{j}-6\boldsymbol{k}\) and \(\boldsymbol{v}=-2\boldsymbol{i}+4\boldsymbol{j}+q\boldsymbol{k}\) are parallel.
(a)
Find the value of \(p\) and the value of \(q\).
(4)
(b)
Show that \(|\boldsymbol{u}|=9\).
(1)
(c)
Show that \(\dfrac{\boldsymbol{u}}{|\boldsymbol{u}|}=-\dfrac{\boldsymbol{v}}{|\boldsymbol{v}|}\) and interpret this result.
\(\boldsymbol{u}\) and \(\boldsymbol{v}\) are parallel but in opposite directions
A1
Answers: (a) \(p=-6,\ q=4\)
15AAHL3-q00593.12 Vectors: concepts and algebraPaper 1Easy8 marks
The diagram shows a cuboid OABCDEFG, where O is the origin, \(\overrightarrow{OA}=4\boldsymbol{i}\), \(\overrightarrow{OC}=3\boldsymbol{j}\) and \(\overrightarrow{OD}=2\boldsymbol{k}\). The point M is the midpoint of [EF].
Diagram not to scale
(a)
Write down the position vectors of B, F and G.
(3)
(b)
Find \(\overrightarrow{AG}\).
(1)
(c)
Show that \(\left|\overrightarrow{AG}\right|=\left|\overrightarrow{OF}\right|\).
(2)
(d)
Find the exact value of \(\left|\overrightarrow{OM}\right|\).
16AAHL3-q00633.13 Scalar productPaper 1Easy7 marks
Let \(\boldsymbol{a}=2\boldsymbol{i}-\boldsymbol{j}+2\boldsymbol{k}\) and \(\boldsymbol{b}=\boldsymbol{i}+4\boldsymbol{j}-8\boldsymbol{k}\). The angle between \(\boldsymbol{a}\) and \(\boldsymbol{b}\) is \(\theta\).
(a)
Find \(\boldsymbol{a}\cdot\boldsymbol{b}\).
(1)
(b)
Find the exact value of \(\cos\theta\).
(3)
(c)
Hence find the exact area of the parallelogram that has \(\boldsymbol{a}\) and \(\boldsymbol{b}\) as adjacent sides.
(3)
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Working
Mark
Notes
(a)
\(2-4-16=-18\)
A1
(b)
\(|\boldsymbol{a}|=\sqrt{4+1+4}=3\) and \(|\boldsymbol{b}|=\sqrt{1+16+64}=9\)
Show that \(L_{1}\) and \(L_{3}\) are parallel but not coincident.
(2)
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Working
Mark
Notes
(a)
\(\begin{pmatrix}-4\\2\\-6\end{pmatrix}=-2\begin{pmatrix}2\\-1\\3\end{pmatrix}\), so the lines are parallel
R1
testing whether \((5,0,6)\) lies on \(L_{1}\): \(1+2\lambda=5\Rightarrow\lambda=2\)
M1
\(\lambda=2\) gives \(y=2-2=0\) and \(z=6\), so \((5,0,6)\) is on \(L_{1}\)
A1
parallel with a common point, so coincident
AG
(b)
testing \((1,0,0)\) on \(L_{1}\): \(x=1\) gives \(\lambda=0\), but then \(y=2\neq0\)
M1
same direction as \(L_{1}\) (parallel) but no common point, so not coincident
R1
Answers: (a) coincident · (b) parallel, distinct
21AAHL3-q00783.16 Vector productPaper 1Easy6 marks
Let \(\boldsymbol{a}=2\boldsymbol{i}-\boldsymbol{j}+3\boldsymbol{k}\), \(\boldsymbol{b}=\boldsymbol{i}+4\boldsymbol{j}-2\boldsymbol{k}\) and \(\boldsymbol{c}=-4\boldsymbol{i}+2\boldsymbol{j}-6\boldsymbol{k}\).
(a)
Find \(\boldsymbol{a}\times\boldsymbol{b}\).
(2)
(b)
Verify that \(\boldsymbol{a}\times\boldsymbol{b}\) is perpendicular to \(\boldsymbol{a}\).
(1)
(c)
Write down \(\boldsymbol{b}\times\boldsymbol{a}\).
(1)
(d)
Find \(\boldsymbol{a}\times\boldsymbol{c}\) and hence state the geometrical relationship between \(\boldsymbol{a}\) and \(\boldsymbol{c}\).
23AAHL3-q00873.18 Intersections and angles of lines and planesPaper 1Easy7 marks
The plane \(\Pi\) has equation \(2x-y+3z=15\). The line \(L\) has equation \(\boldsymbol{r}=\begin{pmatrix}1\\2\\-1\end{pmatrix}+t\begin{pmatrix}1\\-1\\2\end{pmatrix}\).
(a)
Find the coordinates of the point where \(L\) meets \(\Pi\).
(3)
(b)(i)
The line \(M\) has equation \(\boldsymbol{r}=\begin{pmatrix}0\\0\\c\end{pmatrix}+s\begin{pmatrix}1\\2\\0\end{pmatrix}\), where \(c\in\mathbb{R}\).
Show that \(M\) is parallel to \(\Pi\) for all values of \(c\).
(1)
(b)(ii)
Find the value of \(c\) for which \(M\) lies in \(\Pi\).
(2)
(b)(iii)
Write down the number of points of intersection of \(M\) and \(\Pi\) when \(c\) does not take this value.
(1)
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Working
Mark
Notes
(a)
\(2(1+t)-(2-t)+3(-1+2t)=15\)
M1
\(9t-3=15\Rightarrow t=2\)
A1
\((3,0,3)\)
A1
(b)(i)
\(\begin{pmatrix}1\\2\\0\end{pmatrix}\cdot\begin{pmatrix}2\\-1\\3\end{pmatrix}=2-2+0=0\), so the direction of \(M\) is perpendicular to the normal
24IBMA2H-049-05-013.4 Radians, arc length and sector areaPaper 1Easy5 marks
No calculator
A bus windscreen wiper arm pivots at a fixed point O. The rubber blade lies along the arm, from 12 cm to 54 cm from O. On each sweep the arm turns through an angle of \(\frac{2\pi}{3}\) radians.
Blade shown in red. Diagram not to scale
(a)
Find the length of the arc traced by the tip of the blade during one sweep. Give your answer in terms of π.
(2)
(b)
Find the area of windscreen cleaned during one sweep. Give your answer in terms of π.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of \(s=r\theta\) with \(r=54\) and \(\theta=\frac{2\pi}{3}\)
✓ 1
A1 \(36\pi\) cm
✓ 1
Part (b)
M1 use of \(\frac{1}{2}r^2\theta\) for a sector
✓ 1
M1 outer sector minus inner sector: \(\frac{1}{2}(54^2-12^2)\times\frac{2\pi}{3}\)
✓ 1
A1 \(924\pi\) cm2
✓ 1
Common error: The blade starts 12 cm from the pivot, so the cleaned region is an annular sector. Using the full sector of radius 54 cm gives 972π, which is too large.
25IBMA2H-049-05-053.15 Intersecting, parallel and skew linesPaper 1Easy4 marks
No calculator
In an alignment rig, two laser beams are modelled by the lines \(L_1:\ \boldsymbol{r}=\begin{pmatrix}1\\2\\-1\end{pmatrix}+\lambda\begin{pmatrix}2\\-1\\3\end{pmatrix}\) and \(L_2:\ \boldsymbol{r}=\begin{pmatrix}5\\0\\5\end{pmatrix}+\mu\begin{pmatrix}-4\\2\\-6\end{pmatrix}\).
(a)
Show that \(L_1\) and \(L_2\) are parallel.
(1)
(b)
Determine whether the beams are distinct parallel lines or the same line. Justify your answer.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \((-4,2,-6)=-2(2,-1,3)\), so the direction vectors are parallel
✓ 1
Part (b)
M1 consider the vector from \((1,2,-1)\) to \((5,0,5)\), or substitute \((5,0,5)\) into \(L_1\)
✓ 1
A1 \((4,-2,6)=2(2,-1,3)\), so \(\lambda=2\) gives \((5,0,5)\)
✓ 1
R1 a point of \(L_2\) lies on \(L_1\) and the lines are parallel, so they are the same line (coincident)
✓ 1
Exam tip: Parallel direction vectors alone do not show that lines are distinct. Test whether a point on one line lies on the other before you conclude.
26IBMA2H-049-05-123.17 Equations of a planePaper 1Easy5 marks
No calculator
A delivery drone flies in a straight line. Its position, in metres, \(t\) seconds after a timer starts is given by \(\boldsymbol{r}=\begin{pmatrix}2\\-1\\5\end{pmatrix}+t\begin{pmatrix}3\\4\\0\end{pmatrix}\).
(a)
Show that the drone passes through the point \(A(14,15,5)\), and state the value of \(t\) when it does.
(2)
(b)
Show that the drone does not pass through the point \(B(8,7,6)\).
(1)
(c)
Find the distance travelled by the drone between \(t=0\) and reaching A.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 solves one component equation for \(t\), e.g. \(2+3t=14 \Rightarrow t=4\)
✓ 1
A1 checks the other components with \(t=4\): \(-1+4(4)=15\) and \(z=5\), so A lies on the path, reached at \(t=4\)
✓ 1
Part (b)
R1 the \(z\)-coordinate is always 5 but B has \(z=6\), so B is not on the path (or: \(t=2\) from \(x\) gives \((8,7,5)\neq B\))
✓ 1
Part (c)
M1 finds \(4\left|\begin{pmatrix}3\\4\\0\end{pmatrix}\right|=4\times5\), or \(\left|\begin{pmatrix}12\\16\\0\end{pmatrix}\right|\)
✓ 1
A1 \(20\) m
✓ 1
Common error: To show that a point lies on a line, the same value of t must satisfy every component. Checking only the x-coordinate is not enough.
27IBMA2H-049-06-013.13 Scalar productPaper 1Easy5 marks
No calculator
An underwater survey robot is launched from point A(2, −1, 3) and travels in a straight line in the direction of the vector i + 2j − 2k. Distances are in metres.
(a)
Write down a vector equation for the path of the robot.
(1)
(b)
Show that the point P(5, 5, −3) lies on the path of the robot.
(2)
(c)
Find the distance AP.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 r = 2i − j + 3k + λ(i + 2j − 2k)
✓ 1
Part (b)
M1 set one component equal to its coordinate, e.g. 2 + λ = 5, giving λ = 3
✓ 1
A1 check the other two components: −1 + 2(3) = 5 and 3 − 2(3) = −3, so P lies on the path (AG)
Exam tip: In a 'show that' question, find λ from one component and then check every remaining component explicitly; one matching coordinate is not enough.
28IBMA2H-049-06-023.17 Equations of a planePaper 1Easy4 marks
No calculator
A flat solar panel lies in a plane Π that passes through the point Q(1, 2, 3) and is perpendicular to the vector n = 2i − j + 2k.
(a)
Find the Cartesian equation of Π.
(2)
(b)
A corner of the panel is at R(p, 4, 1). Find the value of p.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use r·n = a·n, with a·n = 2×1 − 2 + 2×3
✓ 1
A1 2x − y + 2z = 6
✓ 1
Part (b)
M1 substitute the coordinates of R into the equation of Π: 2p − 4 + 2 = 6
✓ 1
A1 p = 4
✓ 1
Common error: Writing 2x − y + 2z = 0 forgets that the plane passes through Q, not the origin. The constant must be a·n.
29IBMA2H-049-06-043.18 Intersections and angles of lines and planesPaper 1Easy6 marks
No calculator
In a radiotherapy test, a narrow beam leaves a source S(1, 0, 2) and travels along the line r = i + 2k + t(2i + j − k). A flat detector plate lies in the plane x + 2y + z = 12. Distances are in centimetres.
(a)
Find the coordinates of the point P where the beam meets the detector plate.
(4)
(b)
Find the exact distance SP.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 substitute the line into the plane: (1 + 2t) + 2t + (2 − t) = 12
✓ 1
A1 3 + 3t = 12, t = 3
✓ 1
M1 substitute t = 3 into the line
✓ 1
A1 P(7, 3, −1)
✓ 1
Part (b)
M1 \(\vec{SP}\) = 6i + 3j − 3k
✓ 1
A1 SP = √54 = 3√6 cm
✓ 1
Common error: Stopping at t = 3 does not answer the question: the point of intersection is a set of coordinates, found by substituting t back into the line.
30IBMA2H-049-00-013.13 Scalar productPaper 1Easy4 marks
No calculator
A robotic arm has two straight links. The first link makes an acute angle \(\alpha\) with the horizontal, where \(\tan\alpha=\frac{1}{2}\). At the joint the second link turns upwards through a further acute angle \(\beta\), so that it makes an angle \(\alpha+\beta\) with the horizontal, where \(\tan(\alpha+\beta)=3\).
(a)
Find the exact value of \(\tan\beta\).
(3)
(b)
Hence write down the value of \(\beta\) in radians.
Exam tip: When the angle you want is a difference of two known angles, write it as that difference before choosing the identity. Here \(\beta=(\alpha+\beta)-\alpha\) leads straight to the tan subtraction formula.
31IBMA2H-049-00-033.9 Reciprocal and inverse trigonometric functionsPaper 1Easy5 marks
No calculator
The hinge of a folding chair opens to an obtuse angle \(\theta\), where \(\frac{\pi}{2}<\theta<\pi\) and \(\sin\theta=\frac{2}{3}\).
Common error: Taking the positive root of tan²θ = 8 without checking the quadrant loses the R mark. In the second quadrant sine is positive but cosine and tangent are negative.
34IBMA2H-049-01-053.6 Pythagorean and double angle identitiesPaper 1Easy6 marks
No calculator
(a)
Show that \(\frac{1-\cos2\theta}{\sin2\theta} \equiv \tan\theta\), for \(\sin2\theta \neq 0\).
(3)
(b)
Hence find the exact value of \(\tan\frac{\pi}{8}\), giving your answer in the form \(\sqrt{a}+b\), where \(a, b \in \mathbb{Z}\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses \(\cos2\theta = 1 - 2\sin^2\theta\) so that the numerator is \(2\sin^2\theta\)
Exam tip: Pick the version of cos 2θ that cancels the 1 in the numerator. Using 1 − 2sin²θ leaves a single term, which then simplifies directly.
35IBMA2H-049-02-013.12 Vectors: concepts and algebraPaper 1Easy5 marks
No calculator
A delivery drone flies in a straight line from a loading point A to a drop-off point B. Relative to a fixed origin on the ground, with distances in metres, the position vectors are \(\vec{OA}=\begin{pmatrix}1\\2\\0\end{pmatrix}\) and \(\vec{OB}=\begin{pmatrix}7\\-1\\6\end{pmatrix}\).
(a)
Find \(\vec{AB}\).
(1)
(b)
Find a unit vector in the direction of \(\vec{AB}\).
(2)
(c)
The drone flies at a constant speed of 4.5 m s−1. Find the time it takes to fly from A to B.
A1 unit vector \(\frac{1}{9}\begin{pmatrix}6\\-3\\6\end{pmatrix}=\begin{pmatrix}2/3\\-1/3\\2/3\end{pmatrix}\)
✓ 1
Part (c)
M1 distance divided by speed: \(\frac{9}{4.5}\)
✓ 1
A1 \(t=2\) s
✓ 1
Common error: \(\vec{AB}\) is \(\vec{OB}-\vec{OA}\), not \(\vec{OA}-\vec{OB}\). Reversing the order gives a vector of the right length pointing the wrong way, which loses the mark for the unit vector.
36IBMA2H-049-02-023.7 Circular functions and their transformationsPaper 1Easy5 marks
No calculator
A wearable monitor models the arterial blood pressure of a patient at rest by \(P(t)=95+25\sin(2.5\pi t)\), where \(P\) is measured in mmHg and \(t\) is the time in seconds. Each complete cycle of the model corresponds to one heartbeat.
(a)
Write down the maximum and minimum values of \(P\).
(2)
(b)
Find the period of \(P\).
(2)
(c)
Hence find the patient's heart rate in beats per minute.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 maximum 120 mmHg
✓ 1
A1 minimum 70 mmHg
✓ 1
Part (b)
M1 period \(=\frac{2\pi}{2.5\pi}\)
✓ 1
A1 0.8 s
✓ 1
Part (c)
A1 \(\frac{60}{0.8}=75\) beats per minute
✓ 1
Exam tip: For \(a\sin(bt)+d\) the period is \(\frac{2\pi}{b}\). Here \(b\) already contains \(\pi\), so the \(\pi\) cancels and the period is a simple decimal.
37IBMA2H-049-02-033.17 Equations of a planePaper 1Easy5 marks
No calculator
On a communications satellite, the main antenna points in the direction \(\boldsymbol{u}=\begin{pmatrix}2\\1\\-2\end{pmatrix}\). The edge of a solar panel lies in the direction \(\boldsymbol{w}=\begin{pmatrix}p\\4\\1\end{pmatrix}\), where \(p\in\mathbb{R}\).
(a)
The panel edge is perpendicular to the antenna direction. Find the value of \(p\).
(2)
(b)
A sensor points in the direction \(\boldsymbol{v}=\begin{pmatrix}1\\1\\0\end{pmatrix}\). Find the exact angle, in radians, between \(\boldsymbol{u}\) and \(\boldsymbol{v}\).
Exam tip: Perpendicular vectors have a scalar product of zero, so 'perpendicular' in a question is a signal to form an equation from \(\boldsymbol{u}\cdot\boldsymbol{w}=0\).
38IBMA2H-049-02-043.7 Circular functions and their transformationsPaper 1Easy7 marks
No calculator
The brightness of a pulsing LED indicator light is modelled by \(L(x)=4\sin\left(2x-\frac{\pi}{3}\right)+5\), \(x\ge 0\), where \(x\) is the time in milliseconds and \(L\) is measured in arbitrary units.
(a)
Write down the period of \(L\).
(1)
(b)
Find the range of \(L\).
(2)
(c)
The graph of \(y=L(x)\) can be obtained from \(y=\sin x\) by a horizontal stretch followed by a horizontal translation (together with vertical transformations). State the scale factor of the horizontal stretch and the horizontal translation.
(2)
(d)
Find the smallest positive value of \(x\) at which the brightness is a maximum.
A1 translation by \(\frac{\pi}{6}\) to the right, i.e. since \(2x-\frac{\pi}{3}=2\left(x-\frac{\pi}{6}\right)\)
✓ 1
Part (d)
M1 sets \(2x-\frac{\pi}{3}=\frac{\pi}{2}\)
✓ 1
A1 \(x=\frac{5\pi}{12}\) ms
✓ 1
Common error: The horizontal shift is not \(\frac{\pi}{3}\). Factorise first: \(2x-\frac{\pi}{3}=2\left(x-\frac{\pi}{6}\right)\), so after stretching by factor \(\frac{1}{2}\) the translation is \(\frac{\pi}{6}\).
39IBMA2H-049-07-013.2 Right-angled and non-right-angled trianglesPaper 1Easy4 marks
No calculator
In a CAD model of a bicycle frame, two tubes leave the same joint O. They are represented by the vectors a = 2i − j + 3k and b = i + 4j − 2k, with lengths in centimetres. A third tube joins the free ends of the first two tubes to form a triangle.
(a)
Find a × b.
(2)
(b)
Hence find the exact area of the triangle formed by the three tubes.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 correct use of the component (determinant) formula, e.g. i: (−1)(−2) − (3)(4) ; A1 a × b = −10i + 7j + 9k
✓ 2
Part (b)
M1 |a × b| = \(\sqrt{100+49+81}=\sqrt{230}\) and area = ½|a × b| ; A1 area = \(\frac{\sqrt{230}}{2}\) cm²
✓ 2
Common error: The j-component of a vector product is a₃b₁ − a₁b₃. If you expand a determinant, remember the minus sign on the middle term, then check your answer is perpendicular to both vectors using the scalar product.
40IBMA2H-049-07-033.8 Solving trigonometric equationsPaper 1Easy5 marks
No calculator
A triangular sail on a model yacht has sides of length 5 m, 7 m and 8 m. Let θ be the angle between the 5 m side and the 8 m side.
M1 area = ½ × 5 × 8 × sin(\(\frac{\pi}{3}\)) ; A1 area = \(10\sqrt{3}\) m²
✓ 2
Exam tip: In a 'show that' question, write the exact fraction before simplifying and state why only one angle is possible. Simply writing the target answer earns nothing.
41IBMA2H-049-07-043.18 Intersections and angles of lines and planesPaper 1Easy4 marks
No calculator
A flat solar panel on a rooftop contains the direction vectors u = 3i + k and v = i + 2j − k. A light sensor is to be mounted along a line perpendicular to the panel.
(a)
Find u × v.
(2)
(b)
Hence find a unit vector in the direction of the sensor's line.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 correct use of the component formula, e.g. i component: 0×(−1) − 1×2 = −2 ; A1 u × v = −2i + 4j + 6k
Common error: Giving the vector product itself as the 'unit vector' without dividing by its magnitude loses the final mark. Either sign gives a valid perpendicular direction.
42IBMA2H-049-03-013.10 Compound and double angle identitiesPaper 1Easy4 marks
No calculator
A robotic welding arm has two segments. Their directions are given by the vectors a = (2, −1, 3) and b = (k, 4, −2), where k ∈ ℝ. For a particular weld, the two segments must be perpendicular.
(a)
Find the value of k.
(2)
(b)
For this value of k, find |b|, giving your answer in the form p√q, where p, q ∈ ℤ.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sets a·b = 0, giving 2k − 4 − 6 = 0
✓ 1
A1 k = 5
✓ 1
Part (b)
M1 |b| = √(5² + 4² + (−2)²) = √45
✓ 1
A1 3√5
✓ 1
Common error: Perpendicular means the scalar product is zero, not that the vectors' components multiply to zero individually. Write out all three products and add them.
43IBMA2H-049-03-033.8 Solving trigonometric equationsPaper 1Easy5 marks
No calculator
The tilt angle x radians of a satellite dish mounting satisfies the equation \(2\sin^2 x = 3\cos x\).
(a)
Show that the equation can be written as \(2\cos^2 x + 3\cos x - 2 = 0\).
A1 rearranges correctly to 2cos²x + 3cos x − 2 = 0
✓ 1
Part (b)
M1 factorises: (2cos x − 1)(cos x + 2) = 0
✓ 1
R1 rejects cos x = −2 because −1 ≤ cos x ≤ 1
✓ 1
A1 x = −π/3 and x = π/3
✓ 1
Common error: Cosine is an even function, so in an interval symmetric about zero a solution x = π/3 always comes with x = −π/3. Missing the negative solution loses a mark.
44IBMA2H-049-04-013.13 Scalar productPaper 1Easy4 marks
No calculator
A glass atrium roof has two sloping panels. The panels lie in the planes Π1: \(x + z = 5\) and Π2: \(y + z = 3\), where distances are in metres.
(a)
Find the acute angle between the two panels. Give your answer in radians.
(4)
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Marking point
Mark
Notes
Part (a)
M1 normals identified as n1 = i + k and n2 = j + k
✓ 1
M1 use of \(\cos\theta = \frac{|\boldsymbol{n}_1\cdot\boldsymbol{n}_2|}{|\boldsymbol{n}_1||\boldsymbol{n}_2|}\)
Exam tip: The angle between two planes is the angle between their normal vectors. Take the modulus of the dot product so that you get the acute angle directly.
45IBMA2H-049-04-023.13 Scalar productPaper 1Easy4 marks
No calculator
In a laboratory, a laser beam travels along the line r = (2i − j) + t(i + 2j + 3k). A flat mirror lies in the plane \(2x - y + z = 11\).
(a)
Find the coordinates of the point where the beam strikes the mirror.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 substitute the general point (2 + t, −1 + 2t, 3t) into the plane equation
Common error: Substituting only the direction vector, or only the position vector, into the plane equation loses the method mark. Use the full general point of the line.
46IBMA2H-049-04-033.11 Symmetry properties of trigonometric graphsPaper 1Easy5 marks
No calculator
A robotic arm joint is set at an acute angle α, where \(\sin\alpha = \frac{3}{5}\). The control software needs several related trigonometric values.
(a)
Find the exact value of \(\cos(-\alpha)\).
(2)
(b)
Find the exact value of \(\sin(\pi + \alpha)\).
(1)
(c)
Find the exact value of \(\tan(\pi - \alpha)\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\cos(-\alpha) = \cos\alpha\) and use of \(\cos^2\alpha = 1 - \sin^2\alpha\)
✓ 1
A1 \(\frac{4}{5}\)
✓ 1
Part (b)
A1 \(\sin(\pi+\alpha) = -\sin\alpha = -\frac35\)
✓ 1
Part (c)
M1 \(\tan(\pi-\alpha) = -\tan\alpha\)
✓ 1
A1 \(-\frac{3}{4}\)
✓ 1
Exam tip: Picture the unit circle: π − α reflects the point in the y-axis and π + α rotates it half a turn. Decide the sign of each ratio from the quadrant before writing a value.
47IBMA3H-3-19r5-013.14 Vector equation of a linePaper 1Easy5 marks
No calculator
Two straight irrigation pipes on a farm run in the directions of the vectors \(\mathbf{p}=\begin{pmatrix}2\\-1\\3\end{pmatrix}\) and \(\mathbf{q}=\begin{pmatrix}k\\4\\2\end{pmatrix}\), where \(k\) is a constant. The pipes are perpendicular.
(a)
Find the value of \(k\).
(3)
(b)
For this value of \(k\), find the exact length of \(\mathbf{q}\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses \(\mathbf{p}\cdot\mathbf{q}=0\)
✓ 1
A1 \(2k-4+6=0\)
✓ 1
A1 \(k=-1\)
✓ 1
Part (b)
M1 \(|\mathbf{q}|=\sqrt{(-1)^2+4^2+2^2}\)
✓ 1
A1 \(\sqrt{21}\) (=4.5826 to 5 s.f.)
✓ 1
Common error: Setting the vector product, or the product of the magnitudes, to zero instead of the scalar product.
48IBMA3H-3-19r5-033.10 Compound and double angle identitiesPaper 1Easy6 marks
No calculator
The angle \(\theta\) is acute and \(\cos 2\theta=\dfrac{7}{25}\).
(a)
Find the exact value of \(\sin\theta\).
(3)
(b)
Hence find the exact value of \(\tan2\theta\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses \(\cos2\theta=1-2\sin^2\theta\)
✓ 1
A1 \(\sin^2\theta=\dfrac{9}{25}\)
✓ 1
A1 \(\sin\theta=\dfrac35\) (positive since acute)
✓ 1
Part (b)
M1 \(\cos\theta=\dfrac45\) (Pythagoras or identity)
A1 \(\tan2\theta=\dfrac{24/25}{7/25}=\dfrac{24}{7}\) (=3.4286 to 5 s.f.)
✓ 1
Exam tip: When the double-angle value is given, choose the form of \(\cos2\theta\) that contains only the function you need.
49IBMA3H-3-18r8-023.7 Circular functions and their transformationsPaper 1Easy5 marks
No calculator
The graph of \(y=\sin x\) is transformed into the graph of \(y=2\sin\!\left(x-\frac{\pi}{4}\right)+3\) to model the strength of a mountain-rescue beacon signal.
(a)
Describe a sequence of transformations that maps \(y=\sin x\) onto the new graph.
(3)
(b)
The point \(\left(\frac{\pi}{2},1\right)\) lies on \(y=\sin x\). Find the coordinates of its image.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 vertical stretch with scale factor 2
✓ 1
A1 horizontal translation by \(\frac{\pi}{4}\) to the right
✓ 1
A1 vertical translation of 3 upwards
✓ 1
Part (b)
M1 apply the same transformations to the point
✓ 1
A1 \(\left(\frac{3\pi}{4},5\right)\)
✓ 1
Common error: A translation inside the bracket moves the graph the opposite way to the sign: \((x-\frac{\pi}{4})\) shifts it to the right.
50IBMA3H-3-19r10-013.11 Symmetry properties of trigonometric graphsPaper 1Easy5 marks
No calculator
The vibration of a bridge deck is monitored by a sensor whose signal strength, \(S\) units, at angle setting \(x\) radians is modelled by \(S = 2\sin x\).
(a)
Find all values of \(x\) in the interval \(0 \le x \le 2\pi\) for which \(S = \sqrt{3}\).
(3)
(b)
Hence, or otherwise, find all values of \(x\) in the interval \(0 \le x \le \pi\) for which \(2\sin 2x=\sqrt3\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\sin x=\frac{\sqrt3}{2}\)
✓ 1
A1 \(x=\frac{\pi}{3}\)
✓ 1
A1 \(x=\frac{2\pi}{3}\)
✓ 1
Part (b)
M1 \(2x=\frac{\pi}{3}\) or \(\frac{2\pi}{3}\), recognising that \(0\le 2x\le 2\pi\)
✓ 1
A1 \(x=\frac{\pi}{6}\) and \(x=\frac{\pi}{3}\)
✓ 1
Common error: A sine equation with a positive right-hand side always has a second solution \(\pi - x\) in \([0, 2\pi]\); stopping at the first one loses marks.
51IBMA3H-3-19r10-073.17 Equations of a planePaper 1Easy4 marks
No calculator
A space probe moves in a straight line through the points A\((2,-1,4)\) and B\((5,3,-2)\), with coordinates in km relative to a tracking station.
(a)
Find a vector equation of the line \(L\) through A and B.
(2)
(b)
A relay beacon is at C\((8,7,-8)\). Determine whether C lies on \(L\).
M1 equate components, e.g. \(2+3\lambda=8\) gives \(\lambda=2\)
✓ 1
A1 \(-1+4\times2=7\) and \(4-6\times2=-8\), so C lies on \(L\)
✓ 1
Common error: After finding \(\lambda\) from one component you must check the other two; matching a single coordinate proves nothing.
52IBMA3H-3-19r10-103.17 Equations of a planePaper 1Easy5 marks
No calculator
The flat solar-panel array of a space probe lies in a plane \(\Pi\) which has normal vector \(\begin{pmatrix}2\\-1\\3\end{pmatrix}\) and contains the point A\((1,4,2)\). Coordinates are in metres.
(a)
Find the Cartesian equation of \(\Pi\).
(2)
(b)
A boom lies along the line \(\mathbf r=\begin{pmatrix}0\\-1\\2\end{pmatrix}+\lambda\begin{pmatrix}1\\2\\1\end{pmatrix}\). Find the coordinates of the point where the boom meets \(\Pi\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(2x-y+3z=2\times1-4+3\times2\)
✓ 1
A1 \(2x-y+3z=4\)
✓ 1
Part (b)
M1 substitute \(x=\lambda\), \(y=-1+2\lambda\), \(z=2+\lambda\) into the plane equation
✓ 1
A1 \(3\lambda+7=4\), so \(\lambda=-1\)
✓ 1
A1 \((-1,-3,1)\)
✓ 1
Exam tip: Check your intersection point by substituting it back into the plane equation: here \(2\times(-1)-(-3)+3\times1=4\).
53IBMA3H-3-18r11-033.10 Compound and double angle identitiesPaper 1Easy5 marks
No calculator
A library door sensor sweeps a beam through an angle \(\theta\) (in radians), and the lateral offset of the beam is \(2\sin\theta\) metres for \(0\leq\theta\leq 2\pi\). The offset is \(\sqrt{3}\) m.
(a)
Find the two possible values of \(\theta\).
(3)
(b)
Find the exact value of \(\cos2\theta\) for the smaller value of \(\theta\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\sin\theta=\frac{\sqrt3}{2}\)
✓ 1
A1 \(\theta=\frac{\pi}{3}\)
✓ 1
A1 \(\theta=\frac{2\pi}{3}\)
✓ 1
Part (b)
M1 \(\cos2\theta=1-2\sin^2\theta\) with \(\sin\theta=\frac{\sqrt3}{2}\)
✓ 1
A1 \(1-2\left(\frac34\right)=-\frac12\)
✓ 1
Exam tip: Learn the exact values of sine and cosine at \(\frac{\pi}{6}\), \(\frac{\pi}{4}\) and \(\frac{\pi}{3}\); many no-calculator equations rely on them.
54IBMA3H-3-18r11-083.9 Reciprocal and inverse trigonometric functionsPaper 1Easy4 marks
No calculator
A surveyor measures the angle of elevation \(\theta\) of a turbine hub, where \(\theta\) is acute and \(\sin\theta=\frac{5}{13}\).
Exam tip: Reciprocal functions are found by inverting the matching primary ratio: \(\sec=1/\cos\), \(\csc=1/\sin\), \(\cot=1/\tan\).
55IBMA3H-3-18r12-083.17 Equations of a planePaper 1Easy5 marks
No calculator
Two light beams from a telescope are described by the direction vectors \(\mathbf{a} = \begin{pmatrix}2\\-1\\3\end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix}1\\4\\2\end{pmatrix}\). A third beam has direction \(\mathbf{c} = \begin{pmatrix}k\\4\\2\end{pmatrix}\).
(a)
Find \(\mathbf{a}\cdot\mathbf{b}\).
(2)
(b)
State, with a reason, whether the angle between \(\mathbf{a}\) and \(\mathbf{b}\) is acute or obtuse.
(1)
(c)
Given that \(\mathbf{a}\) is perpendicular to \(\mathbf{c}\), find the value of \(k\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(2\times1 + (-1)\times4 + 3\times2\)
✓ 1
A1 4
✓ 1
Part (b)
R1 acute, because \(\mathbf{a}\cdot\mathbf{b} > 0\) so \(\cos\theta > 0\)
✓ 1
Part (c)
M1 \(\mathbf{a}\cdot\mathbf{c} = 2k - 4 + 6 = 0\)
✓ 1
A1 \(k = -1\)
✓ 1
Common error: Perpendicular vectors have scalar product zero; a scalar product of zero does not require either vector to be the zero vector.
56IBMA3H-3-19r12-043.10 Compound and double angle identitiesPaper 1Easy6 marks
No calculator
The angles \(A\) and \(B\) are both acute, with \(\sin A=\frac35\) and \(\tan B=\frac{5}{12}\).
(a)
Find the exact values of \(\cos A\), \(\sin B\) and \(\cos B\).
(2)
(b)
Find the exact value of \(\cos(A+B)\).
(2)
(c)
Find the exact value of \(\sin(A-B)\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(\cos A=\frac45\)
✓ 1
A1 \(\sin B=\frac{5}{13}\) and \(\cos B=\frac{12}{13}\) (from a triangle with sides 5, 12, 13)
M1 \(1-2\sin^{2}\theta\) or \(\cos^{2}\theta-\sin^{2}\theta\)
✓ 1
A1 \(\frac{7}{25}\) (\(=0.28\))
✓ 1
Exam tip: Three forms of \(\cos2\theta\) exist; pick the one that uses the value you already know.
58IBMA3H-3-18r15-023.8 Solving trigonometric equationsPaper 1Easy5 marks
No calculator
In a model of a Formula 1 car's suspension, the displacements (in cm) of two linked dampers at time \(t\) seconds are \(y_1=\sin2t\) and \(y_2=\sin t\), for \(0\le t\le2\pi\). Find all values of \(t\) for which the displacements are equal.
(a)
Solve \(\sin2t=\sin t\) for \(0\le t\le2\pi\), giving exact answers.
(5)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(2\sin t\cos t=\sin t\)
✓ 1
M1 \(\sin t(2\cos t-1)=0\)
✓ 1
A1 \(\sin t=0\) gives \(t=0,\pi,2\pi\)
✓ 1
A1 \(\cos t=\frac12\) gives \(t=\frac\pi3\)
✓ 1
A1 and \(t=\frac{5\pi}3\)
✓ 1
Common error: Dividing both sides by \(\sin t\) loses the solutions where \(\sin t=0\); factorise instead.
59IBMA3H-3-18r15-083.15 Intersecting, parallel and skew linesPaper 1Easy4 marks
No calculator
Two bees fly in directions given by \(\mathbf u=\begin{pmatrix}2\\k\\-1\end{pmatrix}\) and \(\mathbf v=\begin{pmatrix}3\\-2\\k\end{pmatrix}\), where \(k\) is a constant.
(a)
Find the value of \(k\) for which the bees' directions are perpendicular.
(2)
(b)
The vector \(\mathbf u\) is parallel to \(\mathbf w=\begin{pmatrix}-6\\-9\\3\end{pmatrix}\). Find the value of \(k\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\mathbf u\cdot\mathbf v=6-2k-k=0\)
✓ 1
A1 \(k=2\)
✓ 1
Part (b)
M1 \(\mathbf w=-3\mathbf u\) so \(-3k=-9\) (or equate ratios of components)
✓ 1
A1 \(k=3\)
✓ 1
Common error: Perpendicular vectors have scalar product zero; parallel vectors are scalar multiples of each other, not equal scalar products.
60IBMA3H-3-18r15-123.17 Equations of a planePaper 1Easy5 marks
No calculator
The tray of a medical scanner lies in a plane that passes through \(A(1,2,3)\) and is parallel to the vectors \(\mathbf a=\begin{pmatrix}1\\0\\2\end{pmatrix}\) and \(\mathbf b=\begin{pmatrix}0\\1\\-1\end{pmatrix}\).
(a)
Find a vector normal to the plane.
(2)
(b)
Find the Cartesian equation of the plane.
(2)
(c)
Determine whether the point \(B(2,4,3)\) lies on the plane.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\mathbf a\times\mathbf b\)
✓ 1
A1 \(\begin{pmatrix}-2\\1\\1\end{pmatrix}\) (or a multiple)
✓ 1
Part (b)
M1 \(-2\times1+1\times2+1\times3=3\) for the constant
✓ 1
A1 \(-2x+y+z=3\) (or \(2x-y-z=-3\))
✓ 1
Part (c)
R1 \(-4+4+3=3\) so \(B\) lies on the plane
✓ 1
Exam tip: Substituting a known point is the quickest way to find the constant in a Cartesian plane equation.
61IBMA3H-3-18r17-083.9 Reciprocal and inverse trigonometric functionsPaper 1Easy5 marks
No calculator
The angle θ satisfies cos θ = −5/13, where π/2 < θ < π.
(a)
Find the exact value of sin θ.
(2)
(b)
Write down the exact values of sec θ and cosec θ.
(2)
(c)
Find the exact value of cot θ.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses sin2θ = 1 − 25/169 = 144/169
✓ 1
A1 sin θ = 12/13 (positive in the second quadrant)
✓ 1
Part (b)
A1 sec θ = −13/5
✓ 1
A1 cosec θ = 13/12
✓ 1
Part (c)
A1 cot θ = cos θ / sin θ = −5/12
✓ 1
Common error: Check the sign of each ratio from the quadrant: in the second quadrant only sin and cosec are positive.
62IBMA3H-3-19r18-013.10 Compound and double angle identitiesPaper 1Easy6 marks
No calculator
A theatre technician tilts a spotlight so that it makes an acute angle \(\theta\) with the horizontal stage, where \(\sin\theta=\frac{5}{13}\).
(a)
Find the exact value of \(\cos\theta\).
(2)
(b)
Find the exact value of \(\sin 2\theta\).
(2)
(c)
Hence find the exact value of \(\tan 2\theta\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of \(\sin^2\theta+\cos^2\theta=1\)
✓ 1
A1 \(\cos\theta=\frac{12}{13}\) (positive since \(\theta\) is acute)
Common error: Writing \(\sin 2\theta=2\sin\theta\) loses the factor \(\cos\theta\); always apply the double angle identity in full.
63IBMA3H-3-19r18-033.12 Vectors: concepts and algebraPaper 1Easy5 marks
No calculator
Two gas sensors on a volcano are at points \(A(2,-1,3)\) and \(B(5,3,15)\), with coordinates in hundreds of metres relative to a reference point at the base station. Position vectors are taken from the base station.
(a)
Find \(\overrightarrow{AB}\).
(1)
(b)
A third sensor \(P\) lies on \([AB]\) with \(AP:PB=1:3\). Find the coordinates of \(P\).
(2)
(c)
Find the distance between the sensors \(A\) and \(B\), in metres.
Common error: The ratio \(AP:PB=1:3\) means \(P\) is one quarter of the way along \(AB\), not one third.
64IBMA3H-3-19r18-043.17 Equations of a planePaper 1Easy5 marks
No calculator
A theatre cable has vector equation \(\boldsymbol{r}=\begin{pmatrix}1\\2\\0\end{pmatrix}+t\begin{pmatrix}2\\1\\2\end{pmatrix}\), where coordinates are in metres and \(t\in\mathbb{R}\).
(a)
Write down the coordinates of the point on the cable when \(t=3\).
(1)
(b)
Show that the point \(Q(-3,0,-4)\) lies on the cable.
(2)
(c)
Find the length of cable between the points where \(t=0\) and \(t=3\).
Common error: \(\sec x\) is \(\frac{1}{\cos x}\), not \(\cos^{-1}x\); the reciprocal and the inverse function are different things.
69IBMA3H-3-19r25-023.10 Compound and double angle identitiesPaper 1Easy6 marks
No calculator
Exact values of tangents of multiples of \(\frac{\pi}{12}\) can be found using compound angle identities.
(a)
Use the identity for \(\tan(A-B)\) with \(A=\frac{\pi}{4}\) and \(B=\frac{\pi}{6}\) to show that \(\tan\frac{\pi}{12}=2-\sqrt{3}\).
(4)
(b)
Hence find the exact value of \(\tan\frac{5\pi}{12}\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 \(\tan\frac{\pi}{4}=1\) and \(\tan\frac{\pi}{6}=\frac{1}{\sqrt3}\)
✓ 1
M1 substitute into \(\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}\) to get \(\frac{1-\frac{1}{\sqrt3}}{1+\frac{1}{\sqrt3}}=\frac{\sqrt3-1}{\sqrt3+1}\)
✓ 1
M1 rationalise by multiplying numerator and denominator by \(\sqrt3-1\) to get \(\frac{4-2\sqrt3}{2}\)
✓ 1
A1 \(\tan\frac{\pi}{12}=2-\sqrt3\) (AG, all steps shown). Accept a valid use of \(\tan(\frac{\pi}{4}+\ldots)\) or \(\tan(\frac{\pi}{3}-\frac{\pi}{4})\) if fully shown.
✓ 1
Part (b)
M1 use \(\tan\frac{5\pi}{12}=\tan\left(\frac{\pi}{2}-\frac{\pi}{12}\right)=\frac{1}{\tan(\pi/12)}\)
✓ 1
A1 \(\frac{1}{2-\sqrt3}=2+\sqrt3\) (rationalised)
✓ 1
Exam tip: In a 'show that' question the final line must match the printed result exactly, and every intermediate rationalising step must be visible.
70IBMA3H-3-18r32-023.2 Right-angled and non-right-angled trianglesPaper 1Easy4 marks
No calculator
A retailer packs poster tubes in a cuboid carton measuring 30 cm by 40 cm by 120 cm. A tube is modelled as a straight line segment.
Diagram not to scale. The tube joins opposite corners
(a)
Find the length of the longest tube that can fit inside the carton, touching opposite corners.
(3)
(b)
Determine whether a tube of length 1.28 m can fit in the carton.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(30^{2}+40^{2}+120^{2}\)
✓ 1
A1 16900
✓ 1
A1 130 cm
✓ 1
Part (b)
R1 yes, since 128 cm < 130 cm
✓ 1
Exam tip: The space diagonal of a cuboid is \(\sqrt{l^{2}+w^{2}+h^{2}}\): apply Pythagoras twice, or once in three dimensions.
71IBMA3H-3-18r32-043.17 Equations of a planePaper 1Easy5 marks
No calculator
Two rows of solar panels have directions given by the vectors \(\mathbf{a}=\begin{pmatrix}2\\-1\\3\end{pmatrix}\) and \(\mathbf{b}=\begin{pmatrix}4\\5\\-1\end{pmatrix}\).
(a)
Find \(\mathbf{a}\cdot\mathbf{b}\).
(2)
(b)
State what your answer to part (a) tells you about the two rows.
(1)
(c)
A third row has direction \(\mathbf{c}=\begin{pmatrix}k\\2\\1\end{pmatrix}\). Given that \(\mathbf{c}\) is perpendicular to \(\mathbf{a}\), find the value of \(k\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(2\times4+(-1)\times5+3\times(-1)\)
✓ 1
A1 0
✓ 1
Part (b)
R1 the rows are perpendicular
✓ 1
Part (c)
M1 \(2k-2+3=0\)
✓ 1
A1 \(k=-\frac12\)
✓ 1
Exam tip: Two non-zero vectors are perpendicular exactly when their scalar product is zero; set up the equation before solving for the unknown.
72IBMA3H-3-19r35-023.12 Vectors: concepts and algebraPaper 1Easy5 marks
No calculator
In a library atrium, a book lift runs along a straight track from A to B. Relative to an origin O at floor level, with units in metres, \(\overrightarrow{OA}=\mathbf{i}+2\mathbf{j}+\mathbf{k}\) and \(\overrightarrow{OB}=5\mathbf{i}-2\mathbf{j}+3\mathbf{k}\).
(a)
Find \(\overrightarrow{AB}\).
(1)
(b)
The lift stops at the point P on AB where AP : PB = 1 : 2. Find the position vector of P.
Common error: Forgetting that cosine is negative for angles between π/2 and π gives sec θ and cot θ the wrong sign.
74IBMA3H-3-20r40-033.7 Circular functions and their transformationsPaper 1Easy5 marks
No calculator
A Formula 1 engineer models the vertical movement of a front wing on a test rig by \(f(t)=4\cos(2t)-1\), where \(f\) is the displacement in millimetres from a reference level and \(t\) is the time in seconds. Angles are in radians.
(a)
Write down the amplitude, the period and the equation of the principal axis of the graph of \(y=f(t)\).
(3)
(b)
Find the exact displacement when \(t=\frac{\pi}{6}\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 amplitude = 4
✓ 1
A1 period = \(\pi\)
✓ 1
A1 principal axis \(y=-1\)
✓ 1
Part (b)
M1 \(4\cos\frac{\pi}{3}-1\)
✓ 1
A1 1 mm
✓ 1
Exam tip: The period of \(a\cos(bt)+d\) is \(\frac{2\pi}{b}\), and the principal axis is the horizontal line \(y=d\), halfway between the maximum and minimum values.
75IBMA3H-3-20r40-083.17 Equations of a planePaper 1Easy6 marks
No calculator
A beekeeper tracks a swarm flying in a straight line from a hive at A(1, 2, 3) to a tree at B(4, −2, 15). Coordinates are in metres relative to a fixed point, and the third coordinate is the height above the ground.
(a)
Find the vector \(\overrightarrow{AB}\).
(1)
(b)
Find the distance from A to B.
(2)
(c)
Write down a vector equation of the line along which the swarm flies.
(1)
(d)
Determine whether the swarm passes through the point C(2.5, 0, 9).
A1 \(\boldsymbol{r}=\begin{pmatrix}1\\2\\3\end{pmatrix}+\lambda\begin{pmatrix}3\\-4\\12\end{pmatrix}\) (or any equivalent form using B as the point)
✓ 1
Part (d)
M1 finds \(\lambda\) from one coordinate, e.g. \(1+3\lambda=2.5\) so \(\lambda=0.5\)
✓ 1
A1 checks the other coordinates (\(2-4(0.5)=0\) and \(3+12(0.5)=9\)), so C lies on the line
✓ 1
Exam tip: To test whether a point lies on a line, solve for \(\lambda\) from one coordinate and then check that the same \(\lambda\) satisfies the other two. Never stop after checking just one.
76IBMA3H-3-20r40-093.9 Reciprocal and inverse trigonometric functionsPaper 1Easy4 marks
No calculator
The angle \(\theta\) satisfies \(\sec\theta=-\frac{5}{3}\) and \(\frac{\pi}{2}<\theta<\pi\).
(a)
Write down the value of \(\cos\theta\).
(1)
(b)
Find the value of \(\sin\theta\).
(2)
(c)
Find the value of \(\cot\theta\).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(\cos\theta=-\frac{3}{5}\)
✓ 1
Part (b)
M1 uses \(\sin^2\theta=1-\cos^2\theta=\frac{16}{25}\) (or a right-angled triangle with sides 3, 4, 5)
✓ 1
A1 \(\sin\theta=\frac{4}{5}\) (positive in the second quadrant)
✓ 1
Part (c)
A1 \(\cot\theta=\frac{\cos\theta}{\sin\theta}=-\frac{3}{4}\) (ECF from (a) and (b))
✓ 1
Common error: \(\sec\theta\) is the reciprocal of \(\cos\theta\), not of \(\sin\theta\). Convert to cosine first, then use the quadrant to fix the sign of the sine.
77IBMA3H-3-20r41-023.8 Solving trigonometric equationsPaper 1Easy5 marks
No calculator
The angle θ is acute and sin θ = 5/13.
(a)
Find the exact value of cos θ.
(2)
(b)
Write down the value of sin(π − θ).
(1)
(c)
Write down the value of cos(π + θ).
(1)
(d)
Find the exact value of tan(−θ).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of sin2θ + cos2θ = 1 (or a right-angled triangle with sides 5, 12, 13)
✓ 1
A1 cos θ = 12/13 (positive since θ is acute)
✓ 1
Part (b)
A1 5/13
✓ 1
Part (c)
A1 −12/13
✓ 1
Part (d)
A1 −5/12
✓ 1
Exam tip: Sketch the unit circle and mark θ, π − θ and π + θ; the signs of sine and cosine can then be read off rather than remembered.
78IBMA3H-3-20r41-083.12 Vectors: concepts and algebraPaper 1Easy7 marks
No calculator
A survey drone flies in a straight line. Its position, in metres relative to a base point on a wind farm, is given by r = (−20, 35, 12) + t(2, −3, 6), where t is the time in seconds after take-off and the third component is the height above the ground.
(a)
Find the speed of the drone.
(2)
(b)
The top of a turbine is at the point T(−8, 17, 48). Show that the drone passes through T.
(2)
(c)
Find the time at which the drone is at a height of 60 m, and its position at that time.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 magnitude of the direction vector: √(22 + (−3)2 + 62)
✓ 1
A1 speed = 7 m s−1
✓ 1
Part (b)
M1 −20 + 2t = −8 gives t = 6
✓ 1
R1 check: y = 35 − 18 = 17 and z = 12 + 36 = 48, so the drone is at T when t = 6
✓ 1
Part (c)
M1 12 + 6t = 60
✓ 1
A1 t = 8 s
✓ 1
A1 position (−4, 11, 60)
✓ 1
Exam tip: To show that a point lies on a line, find the parameter from one component and then check that it gives the other two components.
79IBMA3H-3-18r42-013.17 Equations of a planePaper 1Easy4 marks
No calculator
A delivery drone carrying vaccines flies along a straight path through the point \(A(1,2,4)\) with direction vector \(\begin{pmatrix}2\\-1\\3\end{pmatrix}\). Coordinates are in metres.
(a)
Write down a vector equation of the path of the drone.
(1)
(b)
A clinic has a landing pad at \(C(5,0,10)\). Show that the pad lies on the path of the drone.
A1 checks the other two: \(2-2=0\) and \(4+6=10\), so \(C\) lies on the path
✓ 1
Exam tip: To show a point lies on a line, find the parameter from one component and verify that it satisfies the other two.
80IBMA3H-3-18r42-043.14 Vector equation of a linePaper 1Easy6 marks
No calculator
A survey balloon is released near a wind farm. Its position, in metres, at time \(t\) seconds is given by \(\mathbf r=\begin{pmatrix}2\\-4\\10\end{pmatrix}+t\begin{pmatrix}6\\3\\-2\end{pmatrix}\), where the third coordinate is the height above the ground.
(a)
Find the speed of the balloon.
(2)
(b)
Find the time at which the balloon reaches the ground.
(2)
(c)
Find the distance travelled by the balloon from its release until it reaches the ground.
M1 distance \(=\) speed \(\times\) time \(=7\times5\) (ECF from (a) and (b))
✓ 1
A1 \(35\) m
✓ 1
Common error: The speed is the magnitude of the velocity vector, not the sum of its components.
81IBMA3H-3-18r42-093.4 Radians, arc length and sector areaPaper 1Easy4 marks
No calculator
A solar farm occupies a sector of a circle with centre at a substation \(O\). The radius is 60 m and the angle of the sector is \(\dfrac{\pi}{3}\) radians. Give exact answers.
Diagram not to scale
(a)
Find the length of the curved fence along the arc of the sector.
Common error: The formulas s = rθ and A = ½r²θ only work when θ is in radians.
82IBMA3H-3-19r42-023.11 Symmetry properties of trigonometric graphsPaper 1Easy5 marks
No calculator
Exact trigonometric values are required throughout this question.
(a)
Write down the exact value of \(\sin\frac{5\pi}{6}\).
(1)
(b)
Solve \(2\cos2x+1=0\) for \(0\le x\le2\pi\), giving your answers as multiples of \(\pi\).
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(\frac12\)
✓ 1
Part (b)
M1 \(\cos2x=-\frac12\)
✓ 1
A1 \(2x=\frac{2\pi}{3}\) (or \(\frac{4\pi}{3}\)) as a first solution
✓ 1
M1 uses symmetry and period to find further values of \(2x\) in \([0,4\pi]\): \(\frac{2\pi}{3},\frac{4\pi}{3},\frac{8\pi}{3},\frac{10\pi}{3}\)
✓ 1
A1 \(x=\frac{\pi}{3},\frac{2\pi}{3},\frac{4\pi}{3},\frac{5\pi}{3}\) (all four, no extras)
✓ 1
Exam tip: When the argument is \(2x\), the interval for \(2x\) is twice as long as the interval for \(x\); list every solution there before dividing by 2.
83IBMA3H-3-19r43-013.9 Reciprocal and inverse trigonometric functionsPaper 1Easy5 marks
No calculator
An archaeologist models the slope of a stone ramp by an acute angle \(\theta\), where \(\sin\theta=\frac{5}{13}\).
(a)
Find the exact value of \(\cos\theta\).
(2)
(b)
Find the exact value of \(\sin2\theta\).
(1)
(c)
Find the exact value of \(\cos2\theta\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of \(\cos^2\theta=1-\sin^2\theta\), or a right-angled triangle with sides 5, 12, 13
✓ 1
A1 \(\cos\theta=\frac{12}{13}\) (positive since \(\theta\) is acute)
Common error: When \(\sin\theta\) is given, remember the sign of \(\cos\theta\) comes from the stated quadrant; an acute angle gives a positive value.
84IBMA3H-3-19r43-023.8 Solving trigonometric equationsPaper 1Easy6 marks
No calculator
The heights, relative to the crank axle, of two points on a cyclist's pedals are modelled by \(y_1=\cos2x\) and \(y_2=\sin x\), where \(x\) is the angle turned in radians and \(0\le x\le2\pi\). The two points are level when \(y_1=y_2\).
(a)
Show that the points are level when \(2\sin^2x+\sin x-1=0\).
(2)
(b)
Hence find all values of \(x\) for which the points are level. Give exact answers.
M1 attempt to factorise or solve the quadratic in \(\sin x\)
✓ 1
A1 \((2\sin x-1)(\sin x+1)=0\), so \(\sin x=\frac12\) or \(\sin x=-1\)
✓ 1
A1 \(x=\frac{\pi}{6}\) and \(x=\frac{5\pi}{6}\)
✓ 1
A1 \(x=\frac{3\pi}{2}\)
✓ 1
Exam tip: After substituting a double angle identity, treat the result as a quadratic in a single trigonometric ratio, then list every solution in the stated interval.
85AAHL3-q00053.2 Right-angled and non-right-angled trianglesPaper 1Medium10 marks
In triangle \(ABC\), \(BC=7\) cm, \(AB+AC=13\) cm and \(B\hat{A}C=\dfrac{\pi}{3}\). It is also given that \(AB\gt AC\).
Diagram not to scale
(a)
Show that \(AB\times AC=40\).
(3)
(b)
Find \(AB\) and \(AC\).
(3)
(c)
Find the exact area of triangle \(ABC\).
(2)
(d)
A circle with centre \(I\) and radius \(\rho\) cm lies inside the triangle and touches all three sides. By considering the areas of triangles \(IBC\), \(ICA\) and \(IAB\), find the exact value of \(\rho\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
cosine rule with \(AB=c\), \(AC=b\): \(49=b^{2}+c^{2}-2bc\cos\dfrac{\pi}{3}=b^{2}+c^{2}-bc\)
M1
using \(b^{2}+c^{2}=(b+c)^{2}-2bc\)
M1
\(49=169-3bc\)
A1
\(bc=40\)
AG
(b)
\(AB\) and \(AC\) are the roots of \(x^{2}-13x+40=0\)
Each triangle has height ρ, as the radius to a point of contact is perpendicular to the side.
\(\rho=\sqrt3\)
A1
Answers: (b) \(AB=8\) cm, \(AC=5\) cm · (c) \(10\sqrt3\text{ cm}^{2}\) · (d) \(\rho=\sqrt3\)
86AAHL3-q00133.4 Radians, arc length and sector areaPaper 1Medium9 marks
A sector of a circle has radius \(r\) cm and angle \(\theta\) radians at the centre, where \(0\lt\theta\lt 2\pi\). The perimeter of the sector is \(20\) cm.
Diagram not to scale
(a)
Show that \(\theta=\dfrac{20-2r}{r}\).
(2)
(b)
Show that the area of the sector, \(A\text{ cm}^{2}\), is given by \(A=10r-r^{2}\).
(2)
(c)
Find the maximum area of the sector and the value of \(\theta\) at which it occurs.
(3)
(d)
The sector is required to be no larger than a semicircle, so that \(\theta\le\pi\). Find the set of possible values of \(r\).
Upper bound from \(\theta\gt0\). Award A1 only for both bounds.
Answers: (c) 25 cm² when \(\theta=2\) · (d) \(\dfrac{20}{\pi+2}\le r\lt10\)
87AAHL3-q00173.5 The unit circle, exact values and the ambiguous casePaper 1Medium9 marks
The diagram shows part of the unit circle with centre \(O\). The point \(A\) is \((1,\,0)\) and the point \(P(\cos\theta,\,\sin\theta)\) lies on the circle, where \(0\lt\theta\lt\dfrac{\pi}{2}\). The tangent to the circle at \(P\) meets the \(x\)-axis at \(T\).
Diagram not to scale
(a)
Explain why \(O\hat{P}T=\dfrac{\pi}{2}\), and hence show that \(OT=\dfrac{1}{\cos\theta}\).
(2)
(b)
Show that \(PT=\tan\theta\).
(1)
(c)
By considering the areas of triangle \(OAP\), sector \(OAP\) and triangle \(OPT\), show that \(\sin\theta\lt\theta\lt\tan\theta\).
(4)
(d)
By taking \(\theta=\dfrac{\pi}{6}\), show that \(3\lt\pi\lt2\sqrt3\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
a tangent to a circle is perpendicular to the radius \([OP]\) at the point of contact
R1
in right-angled triangle \(OPT\): \(\cos\theta=\dfrac{OP}{OT}=\dfrac{1}{OT}\)
A1
\(OT=\dfrac{1}{\cos\theta}\)
AG
(b)
\(\tan\theta=\dfrac{PT}{OP}=\dfrac{PT}{1}\)
A1
\(PT=\tan\theta\)
AG
(c)
area of triangle \(OAP=\dfrac{1}{2}(1)(1)\sin\theta=\dfrac{1}{2}\sin\theta\)
A1
area of sector \(OAP=\dfrac{1}{2}(1)^{2}\theta=\dfrac{1}{2}\theta\)
A1
area of triangle \(OPT=\dfrac{1}{2}(OP)(PT)=\dfrac{1}{2}\tan\theta\)
A1
triangle \(OAP\) lies inside sector \(OAP\), which lies inside triangle \(OPT\), so \(\dfrac{1}{2}\sin\theta\lt\dfrac{1}{2}\theta\lt\dfrac{1}{2}\tan\theta\)
R1
\(\sin\theta\lt\theta\lt\tan\theta\)
AG
(d)
\(\sin\dfrac{\pi}{6}=\dfrac{1}{2}\) and \(\tan\dfrac{\pi}{6}=\dfrac{1}{\sqrt3}\), so \(\dfrac{1}{2}\lt\dfrac{\pi}{6}\lt\dfrac{1}{\sqrt3}\)
88AAHL3-q00183.5 The unit circle, exact values and the ambiguous casePaper 1Medium10 marks
A regular twelve-sided polygon \(A_{0}A_{1}A_{2}\ldots A_{11}\) is inscribed in a circle with centre \(O(0,\,0)\) and radius \(2\). The vertex \(A_{k}\) has coordinates \(\left(2\cos\dfrac{k\pi}{6},\ 2\sin\dfrac{k\pi}{6}\right)\), for \(k=0,1,\ldots,11\).
Regular 12-sided polygon inscribed in a circle of radius 2
(a)
Write down the exact coordinates of \(A_{5}\) and of \(A_{8}\).
(2)
(b)
Find the exact distance \(A_{5}A_{8}\).
(2)
(c)
Find the exact area of the polygon.
(2)
(d)
Show that \(A_{0}A_{1}=\sqrt{8-4\sqrt3}\).
(2)
(e)
Show that \(A_{0}A_{1}=\sqrt6-\sqrt2\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(A_{5}\left(-\sqrt3,\,1\right)\)
A1
\(A_{8}\left(-1,\,-\sqrt3\right)\)
A1
(b)
\(\sqrt{\left(\sqrt3-1\right)^{2}+\left(1+\sqrt3\right)^{2}}\), or \(A_{5}\hat{O}A_{8}=\dfrac{\pi}{2}\) so \(\sqrt{2^{2}+2^{2}}\)
dividing numerator and denominator by \(\cos^{2}x\): \(\dfrac{1-\frac{\sin^{2}x}{\cos^{2}x}}{1+\frac{\sin^{2}x}{\cos^{2}x}}\)
A1
\(\cos2x\equiv\dfrac{1-\tan^{2}x}{1+\tan^{2}x}\)
AG
(c)
\(\sin2x=\dfrac{4}{5}\)
A1
\(\cos2x=-\dfrac{3}{5}\)
A1
(d)
both expressions depend only on \(\tan x\) (the two possible angles differ by \(\pi\), so the values of \(2x\) differ by \(2\pi\))
R1
(e)
\(\dfrac{2t}{1+t^{2}}=\dfrac{3}{5}\), where \(t=\tan x\)
M1
\(3t^{2}-10t+3=0\), so \((3t-1)(t-3)=0\)
A1
\(\tan x=3\) or \(\tan x=\dfrac{1}{3}\)
A1
Both values required.
Answers: (c) \(\sin2x=\dfrac{4}{5}\), \(\cos2x=-\dfrac{3}{5}\) · (e) \(\tan x=3\) or \(\dfrac{1}{3}\)
90AAHL3-q00233.7 Circular functions and their transformationsPaper 1Medium11 marks
Let \(f(x)=a\sin\big(b(x+c)\big)+d\), where \(a,b\gt0\) and \(0\lt c\lt\pi\). The function \(f\) has maximum value \(7\), minimum value \(-1\) and period \(\pi\). It is also given that \(f(0)=5\) and that \(f\) is increasing at \(x=0\).
(a)
Write down the value of \(a\) and the value of \(d\).
(2)
(b)
Write down the value of \(b\).
(1)
(c)
Show that \(c=\dfrac{\pi}{12}\).
(3)
(d)
Find the smallest positive value of \(x\) at which \(f\) has a minimum.
(2)
(e)
Solve \(f(x)=1\), for \(0\le x\le\pi\).
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(a=4\)
A1
\(d=3\)
A1
(b)
\(b=2\)
A1
(c)
\(4\sin2c+3=5\), so \(\sin2c=\dfrac{1}{2}\)
M1
\(2c=\dfrac{\pi}{6}\) or \(\dfrac{5\pi}{6}\), so \(c=\dfrac{\pi}{12}\) or \(\dfrac{5\pi}{12}\)
A1
if \(c=\dfrac{5\pi}{12}\), \(f\) has a maximum at \(x=-\dfrac{\pi}{6}\) and its next minimum at \(x=\dfrac{\pi}{3}\), so \(f\) is decreasing at \(x=0\); reject
R1
Or: for \(c=\dfrac{\pi}{12}\) the argument \(2x+\dfrac{\pi}{6}\) is in \(\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)\) near \(x=0\), where sine increases.
91AAHL3-q00243.7 Circular functions and their transformationsPaper 1Medium9 marks
Let \(g(x)=3\cos\left(2x-\dfrac{\pi}{3}\right)-1\), for \(x\in\mathbb{R}\).
(a)
Write \(g(x)\) in the form \(3\cos\big(2(x-p)\big)-1\).
(1)
(b)
Describe a sequence of transformations that maps the graph of \(y=\cos x\) onto the graph of \(y=g(x)\).
(3)
(c)
The point \((\pi,\,-1)\) is a minimum point on the graph of \(y=\cos x\). Find the coordinates of its image on the graph of \(y=g(x)\).
(2)
(d)
The graph of \(y=g(x)\) is translated by the vector \(\begin{pmatrix}k\\0\end{pmatrix}\), where \(k\gt0\), onto the graph of \(y=3\sin2x-1\). Find the smallest possible value of \(k\).
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(p=\dfrac{\pi}{6}\)
A1
(b)
horizontal stretch with scale factor \(\dfrac{1}{2}\) (parallel to the \(x\)-axis), then a horizontal translation of \(\dfrac{\pi}{6}\) to the right
A1
Or translation \(\dfrac{\pi}{3}\) to the right followed by the horizontal stretch.
vertical stretch with scale factor \(3\)
A1
vertical translation of \(1\) down, performed after the vertical stretch
A1
Accept a single translation by \(\begin{pmatrix}\pi/6\\-1\end{pmatrix}\) after both stretches.
95AAHL3-q00353.9 Reciprocal and inverse trigonometric functionsPaper 1Medium6 marks
Consider the equation \(2\tan^{2}x=3\sec x\), for \(0\le x\le2\pi\).
(a)
Show that the equation can be written as \(2\sec^{2}x-3\sec x-2=0\).
(2)
(b)
Hence solve the equation.
(4)
Show mark scheme
Working
Mark
Notes
(a)
use of \(\tan^{2}x=\sec^{2}x-1\)
M1
\(2(\sec^{2}x-1)=3\sec x\)
A1
\(2\sec^{2}x-3\sec x-2=0\)
AG
(b)
\((\sec x-2)(2\sec x+1)=0\Rightarrow\sec x=2\) or \(\sec x=-\dfrac{1}{2}\)
A1
\(\sec x=-\dfrac12\Rightarrow\cos x=-2\), which is impossible as \(|\cos x|\le1\)
R1
\(x=\dfrac{\pi}{3}\)
A1
\(x=\dfrac{5\pi}{3}\)
A1
Answers: (b) \(x=\frac{\pi}{3},\ \frac{5\pi}{3}\)
96AAHL3-q00413.10 Compound and double angle identitiesPaper 1Medium9 marks
The angles \(A\) and \(B\) are such that \(\sin A=\dfrac{7}{25}\), where \(\dfrac{\pi}{2}\lt A\lt\pi\), and \(\tan B=\dfrac{15}{8}\), where \(0\lt B\lt\dfrac{\pi}{2}\).
(a)
Find the exact values of \(\cos A\), \(\sin B\) and \(\cos B\).
(2)
(b)
Find the exact value of \(\sin(A-B)\).
(2)
(c)
Find the exact value of \(\tan(A+B)\).
(3)
(d)
Hence, or otherwise, determine the quadrant in which the angle \(A+B\) lies. Justify your answer.
98AAHL3-q00443.10 Compound and double angle identitiesPaper 1Medium7 marks
The line \(L_{1}\) has equation \(y=2x+1\) and the line \(L_{2}\) has equation \(y=\dfrac{1}{3}x-4\). The lines make angles \(\alpha\) and \(\beta\) respectively with the positive \(x\)-axis, where \(0\lt\beta\lt\alpha\lt\dfrac{\pi}{2}\).
L₁: y = 2x + 1, L₂: y = x/3 − 4. Diagram not to scale
(a)
Show that the acute angle between \(L_{1}\) and \(L_{2}\) is \(\dfrac{\pi}{4}\).
(3)
(b)
The line \(L_{3}\) has equation \(y=kx\), where \(k\neq-\dfrac{1}{2}\). The acute angle between \(L_{1}\) and \(L_{3}\) is \(\dfrac{\pi}{4}\). Find the possible values of \(k\).
midpoint of [AC] \(=\left(\dfrac72,\dfrac52,\dfrac52\right)\)
M1
midpoint of [BD] \(=\left(\dfrac72,\dfrac52,\dfrac52\right)\), the same point
A1
FT their D.
(c)
\(\overrightarrow{EB}=\overrightarrow{AC}\), so \(\overrightarrow{OE}=\overrightarrow{OB}-\overrightarrow{AC}\)
M1
E\((-1,\ -2,\ -3)\)
A1
Answers: (a) D\((3,6,3)\) · (c) E\((-1,-2,-3)\)
103AAHL3-q00583.12 Vectors: concepts and algebraPaper 1Medium7 marks
In triangle OAB, \(\overrightarrow{OA}=\boldsymbol{a}\) and \(\overrightarrow{OB}=\boldsymbol{b}\). The points L, M and N are the midpoints of [OA], [AB] and [OB] respectively. The point G lies on [OM] such that \(\overrightarrow{OG}=\dfrac23\overrightarrow{OM}\).
Diagram not to scale
(a)
Find \(\overrightarrow{OM}\) in terms of \(\boldsymbol{a}\) and \(\boldsymbol{b}\).
(1)
(b)
Find \(\overrightarrow{OG}\) in terms of \(\boldsymbol{a}\) and \(\boldsymbol{b}\).
(1)
(c)
Show that \(\overrightarrow{AG}=\dfrac23\overrightarrow{AN}\).
(3)
(d)
Hence prove that the three medians of triangle OAB pass through a single point.
105AAHL3-q00653.13 Scalar productPaper 1Medium9 marks
The vectors \(\boldsymbol{a}\) and \(\boldsymbol{b}\) are such that \(|\boldsymbol{a}|=3\), \(|\boldsymbol{b}|=5\) and \(|\boldsymbol{a}-\boldsymbol{b}|=7\).
(a)
Show that \(\boldsymbol{a}\cdot\boldsymbol{b}=-\dfrac{15}{2}\).
(3)
(b)
Find the angle between \(\boldsymbol{a}\) and \(\boldsymbol{b}\).
(2)
(c)
Find the value of \(k\) for which \(\boldsymbol{a}+k\boldsymbol{b}\) is perpendicular to \(\boldsymbol{a}\).
106AAHL3-q00663.13 Scalar productPaper 1Medium6 marks
The points A, B and P lie on a circle with centre O, where [AB] is a diameter and P is distinct from A and B. Let \(\overrightarrow{OA}=\boldsymbol{a}\) and \(\overrightarrow{OP}=\boldsymbol{p}\).
Diagram not to scale
(a)
Write down \(\overrightarrow{OB}\) in terms of \(\boldsymbol{a}\).
(1)
(b)
Find \(\overrightarrow{AP}\) and \(\overrightarrow{BP}\) in terms of \(\boldsymbol{a}\) and \(\boldsymbol{p}\).
(2)
(c)
Hence prove that \(\mathrm{A}\hat{\mathrm{P}}\mathrm{B}\) is a right angle.
Must see the cross terms cancel, using \(\boldsymbol{p}\cdot\boldsymbol{a}=\boldsymbol{a}\cdot\boldsymbol{p}\).
\(|\boldsymbol{p}|=|\boldsymbol{a}|\) (both are radii), so \(\overrightarrow{AP}\cdot\overrightarrow{BP}=0\); since P is distinct from A and B both vectors are non-zero, so they are perpendicular
R1
Must refer to equal radii and to the vectors being non-zero.
107AAHL3-q00673.13 Scalar productPaper 1Medium8 marks
The vector \(\boldsymbol{v}=2\boldsymbol{i}-3\boldsymbol{j}+6\boldsymbol{k}\) makes angles \(\alpha\), \(\beta\) and \(\gamma\) with the positive \(x\)-, \(y\)- and \(z\)-axes respectively.
(a)
Find \(|\boldsymbol{v}|\).
(1)
(b)
Find the values of \(\cos\alpha\), \(\cos\beta\) and \(\cos\gamma\).
(2)
(c)
Show that, for any non-zero vector \(\boldsymbol{w}=w_{1}\boldsymbol{i}+w_{2}\boldsymbol{j}+w_{3}\boldsymbol{k}\) making angles \(\alpha\), \(\beta\) and \(\gamma\) with the positive \(x\)-, \(y\)- and \(z\)-axes, \(\cos^{2}\alpha+\cos^{2}\beta+\cos^{2}\gamma=1\).
(2)
(d)
A unit vector \(\boldsymbol{u}\) makes an angle of \(\dfrac{\pi}{3}\) with the positive \(x\)-axis, an angle of \(\dfrac{\pi}{4}\) with the positive \(y\)-axis and an acute angle \(\gamma\) with the positive \(z\)-axis. Find \(\gamma\) and write \(\boldsymbol{u}\) in component form.
108AAHL3-q00713.14 Vector equation of a linePaper 1Medium6 marks
The line \(L_{1}\) has vector equation \(\boldsymbol{r}=\begin{pmatrix}1\\0\\-2\end{pmatrix}+\lambda\begin{pmatrix}2\\-1\\2\end{pmatrix}\) and the line \(L_{2}\) has Cartesian equations \(x-3=\dfrac{y+1}{-4}=\dfrac{z}{-8}\).
(a)
Find the exact value of the cosine of the acute angle between \(L_{1}\) and \(L_{2}\).
(4)
(b)
The line \(L_{3}\) has direction vector \(\begin{pmatrix}p\\2\\p-1\end{pmatrix}\) and is perpendicular to \(L_{1}\). Find the value of \(p\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
direction of \(L_{2}\): \(\begin{pmatrix}1\\-4\\-8\end{pmatrix}\)
Award A0 for \(-\tfrac{10}{27}\) (this is the cosine of the obtuse angle).
(b)
\(2p-2+2(p-1)=0\)
M1
\(p=1\)
A1
Answers: (a) \(\tfrac{10}{27}\) · (b) \(p=1\)
109AAHL3-q00753.15 Intersecting, parallel and skew linesPaper 1Medium8 marks
The lines \(L_{1}\) and \(L_{2}\) have equations \(\boldsymbol{r}=\begin{pmatrix}1\\-2\\3\end{pmatrix}+\lambda\begin{pmatrix}1\\2\\-1\end{pmatrix}\) and \(\boldsymbol{r}=\begin{pmatrix}4\\1\\-1\end{pmatrix}+\mu\begin{pmatrix}-1\\1\\2\end{pmatrix}\) respectively.
(a)
Show that \(L_{1}\) and \(L_{2}\) intersect, and find the coordinates of their point of intersection.
(4)
(b)
Find a vector that is perpendicular to both \(L_{1}\) and \(L_{2}\).
(2)
(c)
Hence find the Cartesian equation of the plane that contains \(L_{1}\) and \(L_{2}\).
110AAHL3-q00763.15 Intersecting, parallel and skew linesPaper 1Medium8 marks
The lines \(L_{1}\) and \(L_{2}\) have equations \(\boldsymbol{r}=\begin{pmatrix}2\\0\\1\end{pmatrix}+\lambda\begin{pmatrix}1\\-1\\2\end{pmatrix}\) and \(\boldsymbol{r}=\begin{pmatrix}0\\-1\\4\end{pmatrix}+\mu\begin{pmatrix}2\\1\\1\end{pmatrix}\) respectively.
(a)
Show that \(L_{1}\) and \(L_{2}\) are not parallel.
(1)
(b)
Show that \(L_{1}\) and \(L_{2}\) are skew.
(4)
(c)
The line \(L_{3}\) passes through the point \((0,-1,4)\) and has direction vector \(\begin{pmatrix}2\\1\\p\end{pmatrix}\). Given that \(L_{3}\) intersects \(L_{1}\), find the value of \(p\) and the coordinates of the point of intersection.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(\begin{pmatrix}2\\1\\1\end{pmatrix}\neq k\begin{pmatrix}1\\-1\\2\end{pmatrix}\) for any \(k\) (eg \(x\) gives \(k=2\), \(y\) gives \(k=-1\))
112AAHL3-q00843.17 Equations of a planePaper 1Medium8 marks
The plane \(\Pi_{1}\) has vector equation \(\boldsymbol{r}=\begin{pmatrix}1\\2\\-1\end{pmatrix}+\lambda\begin{pmatrix}1\\0\\2\end{pmatrix}+\mu\begin{pmatrix}0\\1\\-1\end{pmatrix}\).
(a)
Find a vector normal to \(\Pi_{1}\).
(2)
(b)
Hence show that the Cartesian equation of \(\Pi_{1}\) is \(2x-y-z=1\).
(2)
(c)
The point P\((3,5,0)\) lies in \(\Pi_{1}\). Find the values of \(\lambda\) and \(\mu\) that give P.
(2)
(d)
The plane \(\Pi_{2}\) has Cartesian equation \(x+2y-4z=8\). Find a vector equation of \(\Pi_{2}\) in the form \(\boldsymbol{r}=\boldsymbol{a}+s\boldsymbol{b}+t\boldsymbol{c}\).
a point on \(\Pi_{2}\), eg \(\boldsymbol{a}=\begin{pmatrix}8\\0\\0\end{pmatrix}\)
A1
two non-parallel vectors perpendicular to \(\begin{pmatrix}1\\2\\-4\end{pmatrix}\), eg \(\boldsymbol{r}=\begin{pmatrix}8\\0\\0\end{pmatrix}+s\begin{pmatrix}2\\-1\\0\end{pmatrix}+t\begin{pmatrix}4\\0\\1\end{pmatrix}\)
A1
Accept any correct equation; check that each direction has zero scalar product with the normal.
113AAHL3-q00853.17 Equations of a planePaper 1Medium8 marks
The line \(L\) has equation \(\boldsymbol{r}=\begin{pmatrix}1\\-1\\2\end{pmatrix}+\lambda\begin{pmatrix}2\\1\\-1\end{pmatrix}\) and the point P has coordinates \((3,2,4)\).
(a)
Show that P does not lie on \(L\).
(2)
(b)
Find the Cartesian equation of the plane \(\Pi\) that contains \(L\) and P.
(4)
(c)
The line \(M\) has equation \(\boldsymbol{r}=\begin{pmatrix}0\\0\\a\end{pmatrix}+\mu\begin{pmatrix}4\\4\\1\end{pmatrix}\). Given that \(M\) lies in \(\Pi\), find the value of \(a\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(1+2\lambda=3\Rightarrow\lambda=1\)
M1
then \(y=-1+1=0\neq2\), so P is not on \(L\)
R1
(b)
vector from \((1,-1,2)\) to P: \(\begin{pmatrix}2\\3\\2\end{pmatrix}\)
114AAHL3-q00863.17 Equations of a planePaper 1Medium7 marks
The points A and B have position vectors \(\boldsymbol{a}=\begin{pmatrix}1\\-2\\3\end{pmatrix}\) and \(\boldsymbol{b}=\begin{pmatrix}5\\4\\-1\end{pmatrix}\). The set of all points that are the same distance from A as from B is a plane \(\Pi\).
(a)
Show that the position vector \(\boldsymbol{r}\) of any point on \(\Pi\) satisfies \(\boldsymbol{r}\cdot(\boldsymbol{b}-\boldsymbol{a})=\dfrac12\left(|\boldsymbol{b}|^{2}-|\boldsymbol{a}|^{2}\right)\).
(3)
(b)
Hence find the Cartesian equation of \(\Pi\) in the form \(px+qy+rz=s\), where \(p,q,r,s\in\mathbb{Z}\).
(3)
(c)
Find the coordinates of the point where \(\Pi\) meets the \(y\)-axis.
116IBMA2H-049-05-033.4 Radians, arc length and sector areaPaper 1Medium7 marks
No calculator
A flexible solar panel for a camping backpack is cut in the shape of a sector of a circle. The sector has radius \(r\) cm and angle \(\theta\) radians. Its perimeter is 20 cm and its area is 16 cm2.
Diagram not to scale
(a)
Write down two equations in \(r\) and \(\theta\).
(2)
(b)
Hence find the value of \(r\) and the value of \(\theta\), justifying any value you reject.
(5)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(2r+r\theta=20\)
✓ 1
A1 \(\frac{1}{2}r^2\theta=16\)
✓ 1
Part (b)
M1 substitute \(r\theta=20-2r\) into the area equation: \(\frac{1}{2}r(20-2r)=16\)
✓ 1
A1 \(r^2-10r+16=0\)
✓ 1
R1 \(r=2\) gives \(\theta=8>2\pi\), which is impossible for a sector, so reject
✓ 1
A1 \(r=8\)
✓ 1
A1 \(\theta=\frac{1}{2}\)
✓ 1
Exam tip: Always check both solutions of the quadratic against the context. A sector angle must lie between 0 and 2π. Giving a clear reason for rejecting a root earns the reasoning mark.
117IBMA2H-049-05-063.15 Intersecting, parallel and skew linesPaper 1Medium6 marks
No calculator
Two straight tunnels are planned beneath a city. Their centre lines are modelled by \(L_1:\ \boldsymbol{r}=\begin{pmatrix}2\\1\\0\end{pmatrix}+\lambda\begin{pmatrix}1\\-1\\2\end{pmatrix}\) and \(L_2:\ \boldsymbol{r}=\begin{pmatrix}k\\5\\-1\end{pmatrix}+\mu\begin{pmatrix}2\\1\\-1\end{pmatrix}\), where \(k\) is a constant. The engineers need the tunnels to meet at a junction.
(a)
Find the value of \(k\) for which the lines intersect.
M1 solve the \(y\) and \(z\) equations simultaneously
✓ 1
A1 \(\lambda=3,\ \mu=-7\)
✓ 1
A1 \(k=19\)
✓ 1
Part (b)
M1 substitute \(\lambda=3\) into \(L_1\) (or \(\mu=-7\) into \(L_2\))
✓ 1
A1 \((5,-2,6)\)
✓ 1
Exam tip: Use the two equations that do not contain k to find λ and μ. Then substitute into the third equation to find k.
118IBMA2H-049-05-103.15 Intersecting, parallel and skew linesPaper 1Medium13 marks
No calculator
In a radiotherapy treatment, two radiation beams must meet exactly at a tumour T. Relative to a fixed origin, with units in centimetres, beam 1 is modelled by \(\boldsymbol{r}=\begin{pmatrix}-4\\2\\7\end{pmatrix}+\lambda\begin{pmatrix}2\\0\\-1\end{pmatrix}\). Beam 2 passes through the points \(E(5,-7,10)\) and \(F(4,-4,8)\).
(a)
Find a vector equation for beam 2.
(2)
(b)
Show that the beams intersect, and find the coordinates of T.
(5)
(c)
Show that the beams are perpendicular.
(2)
(d)
A third beam, which is wrongly aligned, follows the line \(\boldsymbol{r}=t\begin{pmatrix}1\\1\\1\end{pmatrix}\). Show that this beam and beam 1 are skew.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 direction \(\overrightarrow{EF}=(-1,3,-2)\)
✓ 1
A1 \(\boldsymbol{r}=(5,-7,10)+\mu(-1,3,-2)\) or equivalent
R1 the \(x\)-equation gives \(6\neq2\), so there is no intersection and the lines are skew
✓ 1
Common error: Finding values of λ and μ from two equations does not prove that the lines intersect. You must show that the third equation is also satisfied.
119IBMA2H-049-06-053.15 Intersecting, parallel and skew linesPaper 1Medium7 marks
No calculator
In a chemical plant, two straight pipes are modelled by the lines L1: r = i + 2j + s(i − j + 2k) L2: r = 3i + pk + t(j + k) where p is a constant. The pipes are to be joined where they meet.
(a)
Find the value of p for which the lines intersect.
(4)
(b)
Find the coordinates of the point of intersection.
(1)
(c)
Find the exact value of the cosine of the acute angle between the pipes.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 equate components: 1 + s = 3, 2 − s = t, 2s = p + t
✓ 1
M1 solve the x-equation: s = 2
✓ 1
A1 t = 0
✓ 1
A1 p = 4
✓ 1
Part (b)
A1 (3, 0, 4)
✓ 1
Part (c)
M1 (i − j + 2k)·(j + k) = 1 with magnitudes √6 and √2
✓ 1
A1 cos θ = 1/√12 = √3/6
✓ 1
Exam tip: Use different parameters (s and t) for the two lines. Using the same letter wrongly assumes that the pipes reach the meeting point at the same parameter value. Here that assumption leads to a contradiction.
120IBMA2H-049-06-063.17 Equations of a planePaper 1Medium6 marks
No calculator
A triangular shade sail has corners at A(1, 1, 1), B(0, 3, 1) and C(0, 1, 4), with coordinates in metres.
(a)
Find \(\vec{AB}\) × \(\vec{AC}\).
(2)
(b)
Hence find the Cartesian equation of the plane containing the sail.
(2)
(c)
Find the area of the sail.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\vec{AB}\) = −i + 2j, \(\vec{AC}\) = −i + 3k and attempt at vector product
✓ 1
A1 6i + 3j + 2k
✓ 1
Part (b)
M1 uses normal (6, 3, 2) and substitutes a point on the plane, e.g. A: 6×1 + 3×1 + 2×1 = 11
Exam tip: One vector product gives both the normal to the plane and twice the triangle's area, so compute it carefully once and reuse it.
121IBMA2H-049-06-083.18 Intersections and angles of lines and planesPaper 1Medium7 marks
No calculator
The two panels of a folding phone lie in the planes Π1: x + y + z = 6 and Π2: x − y + 2z = 3. The hinge lies along the line of intersection of the planes.
(a)
Find the vector product of the normals of Π1 and Π2.
(2)
(b)
Find the point where the hinge line meets the plane y = 0.
(2)
(c)
Hence write down a vector equation of the hinge line.
(1)
(d)
Find the exact value of the cosine of the acute angle between the panels.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt (i + j + k) × (i − j + 2k)
✓ 1
A1 3i − j − 2k
✓ 1
Part (b)
M1 solve x + z = 6 and x + 2z = 3 simultaneously
✓ 1
A1 (9, 0, −3)
✓ 1
Part (c)
A1 r = 9i − 3k + t(3i − j − 2k) or equivalent, e.g. any scalar multiple of the direction vector or any correct point on the line; must include 'r ='
✓ 1
Part (d)
M1 cos θ = |1 − 1 + 2|/(√3 √6)
✓ 1
A1 cos θ = 2/√18 = √2/3
✓ 1
Exam tip: The direction of a line of intersection is perpendicular to both normals, which is exactly what the vector product of the normals gives.
122IBMA2H-049-06-103.18 Intersections and angles of lines and planesPaper 1Medium7 marks
No calculator
A lidar unit at P(3, 5, 7) is calibrated using a flat mirror lying in the plane x + 2y + 2z = 9. Distances are in metres.
(a)
Write down a vector equation of the line through P perpendicular to the mirror.
(1)
(b)
Find the coordinates of F, the foot of the perpendicular from P to the mirror.
(3)
(c)
Find the coordinates of P′, the reflection of P in the mirror.
Exam tip: Multiply the interval by 3 before listing solutions for \(3x\). Solving only in \([0,\pi]\) loses half of the answers.
124IBMA2H-049-00-093.10 Compound and double angle identitiesPaper 1Medium10 marks
No calculator
In a cinema, the bottom edge of the screen is 2 m above a viewer's eye level and the top edge is 8 m above eye level. The viewer sits a horizontal distance \(d\) metres from the screen, where \(d>0\). The angle subtended at the viewer's eye by the screen is \(\theta\).
Side view. Diagram not to scale
(a)
Show that \(\tan\theta=\frac{6d}{d^2+16}\).
(3)
(b)
Find the values of \(d\) for which \(\tan\theta=\frac{3}{5}\).
(3)
(c)
By considering \((d-4)^2\), find the maximum value of \(\tan\theta\) and the value of \(d\) at which it occurs.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\theta=A-B\), where \(A\) and \(B\) are the angles of elevation of the top and bottom edges, with \(\tan A=\frac{8}{d}\) and \(\tan B=\frac{2}{d}\)
A1 multiplies numerator and denominator by \(d^2\) to obtain \(\frac{6d}{d^2+16}\)
✓ 1
Part (b)
M1 \(30d=3d^2+48\)
✓ 1
A1 \(d^2-10d+16=0\)
✓ 1
A1 \(d=2\) or \(d=8\)
✓ 1
Part (c)
M1 \((d-4)^2\ge0\)
✓ 1
A1 \(d^2+16\ge8d\)
✓ 1
R1 since \(d>0\) and \(d^2+16>0\), \(\tan\theta=\frac{6d}{d^2+16}\le\frac{6d}{8d}=\frac{3}{4}\)
✓ 1
A1 maximum \(\tan\theta=\frac{3}{4}\), with equality when \(d=4\)
✓ 1
Exam tip: When a question asks you to consider a given expression, use that inequality argument instead of calculus. State when equality holds, because that gives the value of \(d\).
125IBMA2H-049-01-033.10 Compound and double angle identitiesPaper 1Medium6 marks
No calculator
Consider the equation \(\arctan(2x) + \arctan(x) = \frac{\pi}{4}\).
(a)
Show that \(2x^2 + 3x - 1 = 0\).
(3)
(b)
Hence solve the equation, justifying your answer.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 takes tangent of both sides and uses the compound angle formula \(\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}\)
✓ 1
A1 \(\frac{2x + x}{1 - 2x^2} = 1\)
✓ 1
A1 \(3x = 1 - 2x^2 \Rightarrow 2x^2+3x-1=0\)
✓ 1
Part (b)
A1 \(x = \frac{-3 \pm \sqrt{17}}{4}\)
✓ 1
R1 when \(x = \frac{-3-\sqrt{17}}{4} < 0\) both arctan terms are negative, so their sum cannot equal \(\frac{\pi}{4}\): reject
✓ 1
A1 \(x = \frac{\sqrt{17}-3}{4}\) only
✓ 1
Common error: Taking tan of both sides can introduce extra solutions, because tan has period π. Check that each root gives an angle sum of π/4, not −3π/4.
126IBMA2H-049-01-063.6 Pythagorean and double angle identitiesPaper 1Medium6 marks
No calculator
This question is about the expression \(8\sin x\cos x\cos2x\).
(a)
Show that \(8\sin x\cos x\cos2x \equiv 2\sin4x\).
(2)
(b)
Hence solve \(8\sin x\cos x\cos2x = 1\) for \(0 \le x \le \pi\), giving your answers as exact values.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of \(2\sin x\cos x = \sin2x\) to give \(8\sin x\cos x\cos 2x = 4\sin2x\cos2x\)
Common error: Since \(0 \le x \le \pi\), the angle \(4x\) runs from \(0\) to \(4\pi\). Find every solution of \(\sin4x = \frac12\) in that wider interval. Stopping after the first cycle gives only two of the four answers.
127IBMA2H-049-01-083.9 Reciprocal and inverse trigonometric functionsPaper 1Medium7 marks
No calculator
(a)
Show that \(\sec\theta - \cos\theta \equiv \sin\theta\tan\theta\), where \(\cos\theta \ne 0\).
Common error: Dividing both sides by sin θ loses the solutions 0, π and 2π. Factorise instead. These values are valid because cos θ ≠ 0 at each of them.
128IBMA2H-049-02-073.16 Vector productPaper 1Medium8 marks
No calculator
An engineer is checking the frame of a solar panel. Three corners of the frame have coordinates A(1, 0, 2), B(3, −2, 3) and C(5, 1, 5), in metres. The frame ABCD is designed to be a parallelogram.
R1 \(=0\), so AB is perpendicular to BC, and a parallelogram with a right angle is a rectangle
✓ 1
Part (c)
M1 \(\vec{OD}=\vec{OA}+\vec{BC}\)
✓ 1
A1 D(3, 3, 4)
✓ 1
Part (d)
M1 \(|\vec{AB}|=3\) and \(|\vec{BC}|=\sqrt{17}\) (or \(|\vec{AB}\times\vec{BC}|=\sqrt{153}\))
✓ 1
A1 area \(=3\sqrt{17}\) m2
✓ 1
Common error: For parallelogram ABCD the fourth vertex satisfies \(\vec{AD}=\vec{BC}\). Using \(\vec{OD}=\vec{OA}+\vec{AB}+\vec{AC}\) or mixing up the order of the letters gives a point that makes a different quadrilateral.
129IBMA2H-049-02-083.11 Symmetry properties of trigonometric graphsPaper 1Medium8 marks
No calculator
Let \(f(x)=2\cos(2x)+1\) for \(0\le x\le 2\pi\).
(a)
Find the range of \(f\).
(2)
(b)
Solve \(f(x)=0\).
(4)
(c)
Find the value of \(k\) for which the equation \(f(x)=k\) has exactly two solutions in the given domain.
R1 \(f\) has minima at \(x=\frac{\pi}{2},\frac{3\pi}{2}\) only, while \(k=3\) gives three solutions (\(0,\pi,2\pi\)) and \(-1<k<3\) gives four
✓ 1
A1 \(k=-1\)
✓ 1
Exam tip: When the argument is \(2x\), double the interval before solving: \(0\le x\le 2\pi\) becomes \(0\le 2x\le 4\pi\). Then halve every solution at the end.
130IBMA2H-049-07-063.16 Vector productPaper 1Medium8 marks
No calculator
A triangular roof truss has vertices A(1, 0, 2), B(3, 1, 0) and C(0, 2, 3), with coordinates in metres.
(a)
Find the vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\).
M1 |\(\overrightarrow{AB}\times\overrightarrow{AC}\)| = \(\sqrt{50}=5\sqrt2\) ; A1 area = \(\frac{5\sqrt2}{2}\) m²
✓ 2
Part (d)
M1 distance = \(\frac{|\overrightarrow{AB}\times\overrightarrow{AC}|}{|\overrightarrow{AB}|}\) with |\(\overrightarrow{AB}\)| = 3, or area = ½ × 3 × h ; A1 \(\frac{5\sqrt2}{3}\) m
✓ 2
Exam tip: 'Hence' signals that you should reuse the area: the triangle's area equals ½ × base AB × height, so the height drops out with almost no extra work.
131IBMA2H-049-07-093.2 Right-angled and non-right-angled trianglesPaper 1Medium7 marks
No calculator
A steel bracket is in the shape of triangle ABC, with angle BAC = \(\frac{\pi}{6}\), angle ABC = \(\frac{\pi}{4}\) and BC = 6 cm.
Diagram not to scale
(a)
Find the exact length of AC.
(3)
(b)
Show that \(\sin\left(\frac{7\pi}{12}\right)=\frac{\sqrt6+\sqrt2}{4}\).
(2)
(c)
Hence find the exact area of the bracket, giving your answer in the form \(p(1+\sqrt3)\), where p ∈ ℤ.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sine rule \(\frac{AC}{\sin(\pi/4)}=\frac{6}{\sin(\pi/6)}\) ; A1 substitutes \(\frac{\sqrt2}{2}\) and \(\frac12\) ; A1 AC = \(6\sqrt2\) cm
M1 angle ACB = \(\frac{7\pi}{12}\) and area = ½ × 6 × \(6\sqrt2\) × \(\frac{\sqrt6+\sqrt2}{4}\) ; A1 area = \(9(1+\sqrt3)\) cm²
✓ 2
Exam tip: Find angle ACB first: π − π/6 − π/4 = 7π/12. This is why part (b) is there. When expanding, use √2 × √6 = 2√3.
132IBMA2H-049-03-063.6 Pythagorean and double angle identitiesPaper 1Medium7 marks
No calculator
The combined output of two alternating signals is modelled by y = sin x + √3 cos x, where x is measured in radians.
(a)
Express y in the form R sin(x + α), where R > 0 and 0 < α < π/2.
(3)
(b)
Hence solve sin x + √3 cos x = 1 for 0 ≤ x ≤ 2π.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 expands R sin(x + α) = R sin x cos α + R cos x sin α and compares: R cos α = 1, R sin α = √3
✓ 1
A1 R = 2
✓ 1
A1 α = π/3
✓ 1
Part (b)
M1 sin(x + π/3) = 1/2
✓ 1
M1 considers π/3 ≤ x + π/3 ≤ 7π/3, giving x + π/3 = 5π/6 or 13π/6
✓ 1
A1 x = π/2
✓ 1
A1 x = 11π/6
✓ 1
Exam tip: Shift the interval for the compound angle before solving. Here x + π/3 runs from π/3 to 7π/3, so the solution 13π/6 is easy to miss if you only look in [0, 2π].
133IBMA2H-049-03-083.8 Solving trigonometric equationsPaper 1Medium7 marks
No calculator
The light intensity on a sensor is modelled using an angle x radians which satisfies \(\cos 2x + 5\sin x = 3\).
A1 obtains 2sin²x − 5 sin x + 2 = 0, so (2 sin x − 1)(sin x − 2) = 0
✓ 1
R1 rejects sin x = 2 since |sin x| ≤ 1
✓ 1
A1 x = π/6
✓ 1
A1 x = 5π/6
✓ 1
Part (b)
M1 recognises x = 2θ with 0 ≤ 2θ ≤ 2π, so 2θ = π/6 or 5π/6
✓ 1
A1 θ = π/12 and θ = 5π/12
✓ 1
Common error: Choose the version of cos 2x that matches the other term. Using cos²x − sin²x or 2cos²x − 1 leaves a mixture of sine and cosine that cannot be factorised.
134IBMA2H-049-04-053.17 Equations of a planePaper 1Medium10 marks
No calculator
A folded architectural canopy is made from two flat sheets lying in the planes Π1: \(x + y - z = 2\) Π2: \(2x - y + z = 7\). The sheets meet along a straight crease. A lamp is placed at the origin O. Distances are in metres.
(a)
Find a vector equation of the crease.
(4)
(b)
Show that the two sheets are perpendicular.
(2)
(c)
Find the exact shortest distance from the lamp to the crease.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to eliminate a variable, e.g. adding the equations
✓ 1
A1 \(3x = 9\) so \(x = 3\)
✓ 1
M1 introduce a parameter, e.g. \(z = \lambda\), giving \(y = \lambda - 1\)
A1 distance \(\sqrt{9 + \frac14 + \frac14} = \frac{\sqrt{38}}{2}\) m
✓ 1
Exam tip: There are many correct forms of the crease. Check your answer by substituting the general point into both plane equations; both should hold for every λ.
135IBMA2H-049-04-063.18 Intersections and angles of lines and planesPaper 1Medium11 marks
No calculator
In a motion-capture studio, three flat laser sheets are modelled by the planes Π1: \(x + y + z = 6\) Π2: \(2x - y + z = 3\) Π3: \(x + 2y + kz = 2\), where \(k \in \mathbb{R}\).
(a)
Show that the line of intersection of Π1 and Π2 can be written as r = (−3i + 9k) + t(2i + j − 3k).
(3)
(b)
For \(k = -1\), find the coordinates of the point where all three sheets meet.
(3)
(c)
Find the value of k for which the three planes do not meet at a unique point.
(2)
(d)
For this value of k, describe the geometrical arrangement of the three planes, justifying your answer.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 eliminate a variable, e.g. Π2 − Π1 gives \(x - 2y = -3\)
✓ 1
A1 with \(y = t\): \(x = 2t - 3\)
✓ 1
A1 \(z = 6 - x - y = 9 - 3t\), giving the required line
✓ 1
(Alternative: M1 direction from the cross product of the normals, (1,1,1)×(2,−1,1) = (2,1,−3); A1 the point (−3, 0, 9) lies on both planes; A1 conclusion.)
—
Part (b)
M1 substitute (2t − 3, t, 9 − 3t) into \(x + 2y - z = 2\), or another valid method for solving the system
✓ 1
A1 \(7t - 12 = 2\), so \(t = 2\)
✓ 1
A1 point (1, 2, 3)
✓ 1
Part (c)
M1 general substitution giving \((4 - 3k)t + 9k - 3 = 2\), OR determinant of the coefficient matrix \(= 4 - 3k\)
✓ 1
A1 no unique solution when \(4 - 3k = 0\), so \(k = \frac43\)
✓ 1
Part (d)
M1 with \(k = \frac43\), substituting the line into Π3 gives \(0t + 9 = 2\), or equivalent row reduction
✓ 1
R1 this is impossible, so the three planes have no common point
✓ 1
R1 the normals (1,1,1), (2,−1,1) and (1,2,\(\frac43\)) are pairwise non-parallel, so no two planes are parallel; the planes form a triangular prism
✓ 1
Exam tip: When there is no unique solution, check the constant term: a contradiction means no common point, while an identity means the planes share a whole line.
136IBMA2H-049-04-073.8 Solving trigonometric equationsPaper 1Medium7 marks
No calculator
The height, h metres, of the tip of a wind-turbine blade above the ground is modelled by \(h(t) = 30 + 20\sin\left(\frac{\pi t}{3}\right)\), \(t \ge 0\), where t is the time in seconds.
(a)
Show that \(h(3 - t) = h(t)\).
(2)
(b)
Given that \(h(0.5) = 40\), write down another value of t, with \(0 \le t \le 3\), for which \(h(t) = 40\).
(1)
(c)
Show that \(h(t + 3) = 60 - h(t)\).
(2)
(d)
Hence find the two values of t, with \(3 \le t \le 6\), for which the blade tip is 20 m above the ground.
M1 use \(h(t+3) = 60 - 40 = 20\) for \(t = 0.5\) and \(t = 2.5\)
✓ 1
A1 \(t = 3.5\) and \(t = 5.5\)
✓ 1
Exam tip: "Hence" means use the symmetry results you have just proved. Re-solving the equation from scratch may not earn the method mark.
137IBMA2H-049-04-093.17 Equations of a planePaper 1Medium10 marks
No calculator
A straight guide rail L1 is fixed onto a ramp. The ramp lies in the plane \(2x + y - 3z = d\) and the rail has equation r = (i + 2j − k) + t(ai + j + k), where \(a, d \in \mathbb{R}\). The rail lies entirely in the plane of the ramp.
(a)
Find the value of a.
(2)
(b)
Find the value of d.
(2)
(c)
A support cable L2 has equation r = k + s(i − j + k). Find the exact value of the sine of the acute angle between L2 and the ramp.
(3)
(d)
Show that L1 and L2 are skew.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 direction perpendicular to normal: \(2a + 1 - 3 = 0\)
✓ 1
A1 \(a = 1\)
✓ 1
Part (b)
M1 substitute (1, 2, −1) into the plane
✓ 1
A1 \(d = 2 + 2 + 3 = 7\)
✓ 1
Part (c)
M1 use \(\sin\phi = \frac{|\boldsymbol{n}\cdot\boldsymbol{d}|}{|\boldsymbol{n}||\boldsymbol{d}|}\)
R1 directions (1, 1, 1) and (1, −1, 1) are not scalar multiples, so the lines are not parallel
✓ 1
M1 equate components: \(1 + t = s\), \(2 + t = -s\), \(-1 + t = 1 + s\)
✓ 1
A1 first two give \(t = -\frac32\), \(s = -\frac12\), but then \(-\frac52 \ne \frac12\), so no intersection; hence skew
✓ 1
Exam tip: A line lies in a plane only if two conditions hold: its direction is perpendicular to the normal and one point of the line satisfies the plane equation.
138IBMA2H-049-04-103.11 Symmetry properties of trigonometric graphsPaper 1Medium6 marks
No calculator
In an audio circuit, two signals are modelled by \(V_1 = \cos 3x\) and \(V_2 = \sin x\), where x is a phase angle in radians.
(a)
Use a symmetry property of trigonometric graphs to write \(\sin x\) in terms of a cosine.
(1)
(b)
Hence solve \(\cos 3x = \sin x\) for \(0 \le x \le \pi\).
(5)
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Marking point
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Notes
Part (a)
A1 \(\sin x = \cos\left(\frac{\pi}{2} - x\right)\)
A1 all three solutions \(\frac{\pi}{8}, \frac{5\pi}{8}, \frac{3\pi}{4}\) and no others
✓ 1
Common error: Equating the angles only as 3x = π/2 − x misses the solutions from cos A = cos B when A = −B + 2kπ, because cosine is an even function.
139IBMA3H-3-18r8-073.6 Pythagorean and double angle identitiesPaper 1Medium5 marks
No calculator
A telecoms engineer models a test signal by \(f(x)=a\sin(bx)+d\) with \(a>0\) and \(b>0\). The maximum value of \(f\) is 7, the minimum value is \(-1\), and the period is 8.
(a)
Find the values of \(a\) and \(d\).
(2)
(b)
Find the exact value of \(b\).
(1)
(c)
Find the exact value of \(f(1)\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(a=\frac{7-(-1)}{2}\) or \(d=\frac{7+(-1)}{2}\)
✓ 1
A1 \(a=4,\ d=3\)
✓ 1
Part (b)
A1 \(b=\frac{2\pi}{8}=\frac{\pi}{4}\)
✓ 1
Part (c)
M1 \(f\left(1\right)=4\sin\frac{\pi}{4}+3\)
✓ 1
A1 \(3+2\sqrt2\) (\(\approx5.828\))
✓ 1
Exam tip: Maximum and minimum values give the amplitude (half the difference) and the midline (the mean) directly.
140IBMA3H-3-19r10-033.8 Solving trigonometric equationsPaper 1Medium6 marks
No calculator
The output of a space probe's spin sensor is modelled by \(f(\theta)=2\sin^2\theta+3\cos\theta\), where \(\theta\) is in radians and \(0\le\theta\le2\pi\).
(a)
Show that the equation \(f(\theta)=3\) can be written as \(2\cos^2\theta-3\cos\theta+1=0\).
(2)
(b)
Hence solve \(f(\theta)=3\) for \(0\le\theta\le2\pi\).
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use \(\sin^2\theta=1-\cos^2\theta\)
✓ 1
A1 \(2-2\cos^2\theta+3\cos\theta=3\), rearranged to the given form
✓ 1
Part (b)
M1 factorise \((2\cos\theta-1)(\cos\theta-1)=0\)
✓ 1
A1 \(\cos\theta=\frac12\) or \(\cos\theta=1\)
✓ 1
A1 \(\theta=\frac{\pi}{3},\ \frac{5\pi}{3}\)
✓ 1
A1 \(\theta=0,\ 2\pi\)
✓ 1
Common error: When a quadratic in \(\cos\theta\) gives \(\cos\theta=1\), do not forget both end-points \(0\) and \(2\pi\) if the interval is closed.
141IBMA3H-3-19r10-093.15 Intersecting, parallel and skew linesPaper 1Medium7 marks
No calculator
Two cables on a bridge are modelled by the lines \(L_1:\ \mathbf r=\begin{pmatrix}1\\2\\3\end{pmatrix}+s\begin{pmatrix}1\\-1\\2\end{pmatrix}\) and \(L_2:\ \mathbf r=\begin{pmatrix}0\\1\\k\end{pmatrix}+t\begin{pmatrix}2\\1\\-1\end{pmatrix}\), where \(k\) is a constant.
(a)
Find the value of \(k\) for which the two lines intersect.
(5)
(b)
Given that \(k=2\), explain why the lines are skew.
M1 substitute into the third equation: \(3+\frac23=k-\frac23\)
✓ 1
A1 \(k=\frac{13}{3}\) (\(\approx4.3333\))
✓ 1
Part (b)
R1 the direction vectors are not parallel (neither is a multiple of the other)
✓ 1
R1 intersection requires \(k=\frac{13}{3}\) and \(2\ne\frac{13}{3}\), so the lines do not meet; hence skew
✓ 1
Common error: Lines that are not parallel and do not meet are skew; showing only that the direction vectors differ is not enough.
142IBMA3H-3-19r10-113.17 Equations of a planePaper 1Medium7 marks
No calculator
Three bolts on a bridge roof panel are at A\((1,0,2)\), B\((3,1,1)\) and C\((2,2,4)\), with distances in metres from a fixed origin O. The panel lies in a plane \(\Pi\).
(a)
Find the Cartesian equation of \(\Pi\).
(4)
(b)
Find the shortest distance from O to \(\Pi\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\overrightarrow{AB}=(2,1,-1)\) and \(\overrightarrow{AC}=(1,2,2)\)
✓ 1
M1 attempt at the vector product \(\overrightarrow{AB}\times\overrightarrow{AC}\)
✓ 1
A1 normal vector \((4,-5,3)\)
✓ 1
A1 \(4x-5y+3z=10\)
✓ 1
Part (b)
M1 line through O perpendicular to \(\Pi\): \(\mathbf r=\mu(4,-5,3)\)
✓ 1
M1 substitute into the plane equation: \(50\mu=10\), so \(\mu=\frac15\)
✓ 1
A1 distance \(=\frac15\sqrt{50}=\sqrt2\) m (\(\approx1.4142\))
✓ 1
Common error: The shortest distance is the length of the vector from O to the foot of the perpendicular, \(|\mu|\,|\mathbf n|\), not the value of \(\mu\) itself.
143IBMA3H-3-18r11-093.9 Reciprocal and inverse trigonometric functionsPaper 1Medium7 marks
No calculator
A lane-line sweeper in a swimming pool turns through an angle \(f(x)=3\arccos\left(\frac{x}{2}\right)\) radians when it is a lateral distance \(x\) metres from the centre line.
(a)
State the domain and range of \(f\).
(2)
(b)
Find \(f^{-1}(x)\) and state its domain.
(3)
(c)
Find the lateral distance at which the sweeper has turned through \(2\pi\) radians.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 domain \(-2\leq x\leq 2\)
✓ 1
A1 range \(0\leq f(x)\leq 3\pi\)
✓ 1
Part (b)
M1 \(y=3\arccos\left(\frac{x}{2}\right)\) rearranged to \(\cos\left(\frac y3\right)=\frac x2\)
Common error: The range of \(\arccos\) is \([0,\pi]\), so its domain and range are swapped with those of \(\cos\) on \([0,\pi]\); do not give the range as \([-1,1]\).
144IBMA3H-3-18r11-103.9 Reciprocal and inverse trigonometric functionsPaper 1Medium7 marks
No calculator
The glare factor from a library skylight is modelled by \(G(\theta)=\csc\theta-\sin\theta\) for \(0<\theta<\pi\).
Exam tip: When a compound-angle calculation gives a tangent value, state the interval the angle lies in before choosing between the possible angles.
146IBMA3H-3-18r11-123.9 Reciprocal and inverse trigonometric functionsPaper 1Medium8 marks
No calculator
A wind-farm yaw controller produces a signal that is zero when \(2\tan^2\theta+3\sec\theta=0\), for \(0\leq\theta\leq2\pi\).
(a)
Show that the equation can be written as \(2\sec^2\theta+3\sec\theta-2=0\).
(2)
(b)
Solve for \(\sec\theta\).
(2)
(c)
Explain why one of these values gives no solution.
(1)
(d)
Find all values of \(\theta\) in the given domain.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use \(\tan^2\theta=\sec^2\theta-1\)
✓ 1
A1 \(2\sec^2\theta-2+3\sec\theta=0\), which rearranges to the required form
✓ 1
Part (b)
M1 \((2\sec\theta-1)(\sec\theta+2)=0\)
✓ 1
A1 \(\sec\theta=\frac12\) or \(\sec\theta=-2\)
✓ 1
Part (c)
R1 \(|\sec\theta|\geq1\) for all \(\theta\), so \(\sec\theta=\frac12\) is impossible
✓ 1
Part (d)
M1 \(\cos\theta=-\frac12\)
✓ 1
A1 \(\theta=\frac{2\pi}{3}\)
✓ 1
A1 \(\theta=\frac{4\pi}{3}\)
✓ 1
Common error: The range of \(\sec\theta\) excludes the interval \((-1,1)\); a value such as \(\frac12\) cannot occur.
147IBMA3H-3-18r12-073.8 Solving trigonometric equationsPaper 1Medium7 marks
No calculator
In a model of an athlete's stride, the angle \(x\) radians satisfies \(2\sin^2 x + 3\cos x = 3\), where \(0 \le x \le 2\pi\).
(a)
Show that the equation can be written as \(2\cos^2 x - 3\cos x + 1 = 0\).
(2)
(b)
Hence find the possible values of \(\cos x\).
(2)
(c)
Hence solve the equation, giving your answers as exact values.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 replace \(\sin^2 x\) with \(1 - \cos^2 x\)
✓ 1
A1 \(2 - 2\cos^2x + 3\cos x = 3\), which rearranges to \(2\cos^2 x - 3\cos x + 1 = 0\) (AG)
✓ 1
Part (b)
M1 factorise: \((2\cos x - 1)(\cos x - 1) = 0\)
✓ 1
A1 \(\cos x = \frac{1}{2}\) or \(\cos x = 1\)
✓ 1
Part (c)
A1 \(x = \frac{\pi}{3}\)
✓ 1
A1 \(x = \frac{5\pi}{3}\)
✓ 1
A1 \(x = 0\) and \(x = 2\pi\)
✓ 1
Common error: Both ends of the interval count: \(\cos x = 1\) gives \(x = 0\) and \(x = 2\pi\) when the domain is \(0 \le x \le 2\pi\).
148IBMA3H-3-18r12-133.18 Intersections and angles of lines and planesPaper 1Medium6 marks
No calculator
A straight rail track is modelled by the line \(L\): \(\mathbf{r} = \begin{pmatrix}2\\1\\-1\end{pmatrix} + t\begin{pmatrix}1\\2\\2\end{pmatrix}\), where distances are in hundreds of metres. A sensor is at \(Q(5, 4, -1)\). The point \(N\) on \(L\) is the closest point of the track to \(Q\).
(a)
Find \(\overrightarrow{NQ}\) in terms of \(t\), where \(N\) has parameter \(t\), and use a scalar product to find the value of \(t\) at \(N\).
(3)
(b)
Write down the coordinates of \(N\).
(1)
(c)
Find the shortest distance from the sensor to the track, in metres.
(2)
Show mark scheme
Marking point
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Notes
Part (a)
M1 \(\overrightarrow{NQ} = (3-t,\ 3-2t,\ -2t)\)
✓ 1
M1 \(\overrightarrow{NQ}\) is perpendicular to the direction, so \((3-t) + 2(3-2t) + 2(-2t) = 0\)
Exam tip: The shortest distance from a point to a line always meets the line at a right angle, so set the scalar product of the connecting vector and the direction to zero.
149IBMA3H-3-19r12-063.8 Solving trigonometric equationsPaper 1Medium7 marks
No calculator
Consider the equation \(\cos2x=3\sin x-1\).
(a)
Show that the equation can be written as \(2\sin^2x+3\sin x-2=0\).
(2)
(b)
Hence find all solutions of the equation in the interval \(0\le x\le2\pi\). Give exact answers.
M1 factorises or solves the quadratic in \(\sin x\): \((2\sin x-1)(\sin x+2)=0\)
✓ 1
A1 \(\sin x=\frac12\) or \(\sin x=-2\)
✓ 1
R1 rejects \(\sin x=-2\) because \(-1\le\sin x\le1\)
✓ 1
A1 \(x=\frac{\pi}{6}\) (0.5236)
✓ 1
A1 \(x=\frac{5\pi}{6}\)
✓ 1
Common error: Always check each root of a quadratic in \(\sin x\) against \(-1\le\sin x\le1\); a root outside this range gives no solutions.
150IBMA3H-3-19r12-073.15 Intersecting, parallel and skew linesPaper 1Medium7 marks
No calculator
Two straight cables supporting a solar array are modelled by \(L_1:\ \mathbf r=\begin{pmatrix}1\\2\\-3\end{pmatrix}+\lambda\begin{pmatrix}1\\-1\\4\end{pmatrix}\) and \(L_2:\ \mathbf r=\begin{pmatrix}8\\k\\7\end{pmatrix}+\mu\begin{pmatrix}2\\1\\-1\end{pmatrix}\), where \(k\) is a constant and distances are in metres.
(a)
The two cables intersect. Find the value of \(k\) and the coordinates of the point of intersection.
(5)
(b)
A third cable passes through the point of intersection and is perpendicular to both \(L_1\) and \(L_2\). Find a vector equation of this cable.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 equates \(x\) and \(z\) components: \(1+\lambda=8+2\mu\) and \(-3+4\lambda=7-\mu\) and solves
✓ 1
A1 \(\lambda=3\), \(\mu=-2\)
✓ 1
M1 equates \(y\) components: \(2-\lambda=k+\mu\), so \(-1=k-2\)
✓ 1
A1 \(k=1\)
✓ 1
A1 point \((4,-1,9)\)
✓ 1
Part (b)
M1 vector product of direction vectors: \((1,-1,4)\times(2,1,-1)=(-3,9,3)\)
✓ 1
A1 \(\mathbf r=\begin{pmatrix}4\\-1\\9\end{pmatrix}+s\begin{pmatrix}-1\\3\\1\end{pmatrix}\) (any non-zero multiple of the direction)
✓ 1
Exam tip: Use two components to find the parameters, then use the third component to find \(k\) or to check that the lines really meet.
151IBMA3H-3-19r12-083.7 Circular functions and their transformationsPaper 1Medium7 marks
No calculator
Let \(g(x)=4\sin\left(2x-\frac{\pi}{3}\right)\).
(a)
The graph of \(y=\sin x\) is transformed onto the graph of \(y=g(x)\) using three transformations. Describe each transformation.
(3)
(b)
Write down the range of \(g\).
(1)
(c)
Solve \(g(x)=2\sqrt3\) for \(0\le x\le\pi\). Give exact answers.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 vertical stretch (parallel to the \(y\)-axis) with scale factor 4
✓ 1
A1 horizontal stretch (parallel to the \(x\)-axis) with scale factor \(\frac12\)
✓ 1
A1 horizontal translation by \(\frac{\pi}{6}\) to the right, i.e. \(\begin{pmatrix}\pi/6\\0\end{pmatrix}\), since \(\sin(2x-\frac\pi3)=\sin\left(2\left(x-\frac\pi6\right)\right)\)
✓ 1
Accept any equivalent order if the translation is adjusted accordingly.
A1 \(2x-\frac\pi3=\frac\pi3\), so \(x=\frac\pi3\) (1.0472)
✓ 1
A1 \(2x-\frac\pi3=\frac{2\pi}{3}\), so \(x=\frac\pi2\)
✓ 1
Common error: A factor inside the brackets acts on \(x\) the 'wrong way round': \(\sin(2x-\frac\pi3)\) is a stretch of scale factor \(\frac12\), and the shift is \(\frac\pi6\), not \(\frac\pi3\).
152IBMA3H-3-18r15-033.6 Pythagorean and double angle identitiesPaper 1Medium5 marks
No calculator
A beekeeper's model for the rate of bee arrivals at a hive involves the expression below. Work with \(x\) in radians.
(a)
Show that \(\dfrac{1-\cos2x}{\sin2x}=\tan x\).
(3)
(b)
Hence solve \(\dfrac{1-\cos2x}{\sin2x}=\sqrt3\) for \(0<x<\pi\).
Exam tip: In a 'show that' question, start from one side only and finish with exactly the expression given.
153IBMA3H-3-18r15-043.11 Symmetry properties of trigonometric graphsPaper 1Medium6 marks
No calculator
The number of flights per minute made by bees from a hive is modelled by \(f(t)=\cos2t+3\sin t\), where \(t\) is in radians and \(0\le t\le2\pi\).
(a)
Find all values of \(t\) for which \(f(t)=2\), giving exact answers.
(6)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\cos2t=1-2\sin^{2}t\)
✓ 1
A1 \(2\sin^{2}t-3\sin t+1=0\)
✓ 1
M1 \((2\sin t-1)(\sin t-1)=0\)
✓ 1
A1 \(\sin t=\frac12\) or \(\sin t=1\)
✓ 1
A1 \(t=\frac\pi6,\frac{5\pi}6\)
✓ 1
A1 \(t=\frac\pi2\)
✓ 1
Common error: After substituting for \(\cos2t\), check that the equation is a quadratic in a single trigonometric function before factorising.
154IBMA3H-3-18r15-063.8 Solving trigonometric equationsPaper 1Medium6 marks
No calculator
A hospital monitor displays a signal whose strength is proportional to \(4\sin x\cos x\cos2x\), with \(x\) in radians.
(a)
Show that \(4\sin x\cos x\cos2x=\sin4x\).
(3)
(b)
Hence solve \(4\sin x\cos x\cos2x=\frac12\) for \(0\le x\le\frac\pi2\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(2\sin x\cos x=\sin2x\) so expression is \(2\sin2x\cos2x\)
✓ 1
M1 \(\sin4x=\sin(2\times2x)=2\sin2x\cos2x\)
✓ 1
A1 conclusion reached with no errors
✓ 1
Part (b)
M1 \(\sin4x=\frac12\) with \(0\le4x\le2\pi\)
✓ 1
A1 \(4x=\frac\pi6\) so \(x=\frac{\pi}{24}\)
✓ 1
A1 \(4x=\frac{5\pi}6\) so \(x=\frac{5\pi}{24}\)
✓ 1
Common error: Write down the interval for \(4x\) before solving, otherwise solutions are missed or extra ones are included.
155IBMA3H-3-18r17-093.9 Reciprocal and inverse trigonometric functionsPaper 1Medium7 marks
No calculator
Consider the equation cot2x + cosec x = 1, for 0 < x < 2π.
(a)
Show that the equation can be written as cosec2x + cosec x − 2 = 0.
(2)
(b)
Hence solve the equation, giving your answers as exact multiples of π.
(4)
(c)
Find the sum of all the solutions found in part (b).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses the identity 1 + cot2x = cosec2x, so cot2x = cosec2x − 1
✓ 1
A1 substitutes and rearranges to the given equation (AG)
✓ 1
Part (b)
M1 factorises: (cosec x + 2)(cosec x − 1) = 0
✓ 1
A1 cosec x = 1 or cosec x = −2, so sin x = 1 or sin x = −1/2
✓ 1
A1 x = π/2
✓ 1
A1 x = 7π/6 and x = 11π/6
✓ 1
Part (c)
A1 π/2 + 7π/6 + 11π/6 = 7π/2 (ECF from (b))
✓ 1
Exam tip: Convert cosec x values back to sin x before using the unit circle, and list the solutions in each quadrant where sin x has the required sign.
156IBMA3H-3-18r17-103.9 Reciprocal and inverse trigonometric functionsPaper 1Medium7 marks
No calculator
The function f is defined by f(x) = 2 arcsin(x − 1) + π, for the largest possible domain.
(a)
Write down the domain of f.
(1)
(b)
Write down the range of f.
(1)
(c)
Find an expression for the inverse function f−1(x).
(3)
(d)
Find the exact value of f−1(4π/3).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 0 ≤ x ≤ 2
✓ 1
Part (b)
A1 0 ≤ f(x) ≤ 2π
✓ 1
Part (c)
M1 sets y = 2 arcsin(x − 1) + π and rearranges: arcsin(x − 1) = (y − π)/2
✓ 1
M1 takes sine of both sides: x − 1 = sin((y − π)/2)
Exam tip: Arcsine has range [−π/2, π/2]; use this to find the range of any transformed arcsine, and recall that the domain of the inverse is the range of the original.
157IBMA3H-3-19r18-053.10 Compound and double angle identitiesPaper 1Medium8 marks
No calculator
The height of a counterweight above the stage floor is modelled by \(h=\cos2t+3\sin t\) metres, where \(t\) is the time in seconds and \(0\le t\le 2\pi\).
A1 \((2\sin t-1)(\sin t-1)=0\), so \(\sin t=\frac12\) or \(\sin t=1\)
✓ 1
A1 \(t=\frac{\pi}{6},\frac{5\pi}{6}\)
✓ 1
A1 \(t=\frac{\pi}{2}\)
✓ 1
Part (c)
M1 treat as quadratic in \(s=\sin t\) with vertex at \(s=\frac34\) (valid since \(|s|\le1\))
✓ 1
A1 \(h_{max}=1+\frac94-\frac98=\frac{17}{8}=2.125\) m
✓ 1
Common error: The equation \(\sin t=1\) gives only one solution in \([0,2\pi]\), not two, so count the solutions of each case carefully.
158IBMA3H-3-19r18-143.12 Vectors: concepts and algebraPaper 1Medium11 marks
No calculator
A stage canopy is a flat parallelogram \(ABCD\) with \(A(0,0,4)\), \(B(6,0,6)\) and \(D(2,3,3)\), in metres.
(a)
Find the coordinates of \(C\).
(2)
(b)
Show that the diagonals \(AC\) and \(BD\) have the same midpoint.
(3)
(c)
The point \(E\) lies on \([AC]\) with \(AE:EC=1:3\). Find the coordinates of \(E\).
(2)
(d)
Find angle \(BAD\) to the nearest tenth of a degree. Leave your working in exact form until the final answer (a calculator is needed only for the last step).
Common error: Use \(\overrightarrow{AB}\) and \(\overrightarrow{AD}\), both starting at \(A\), for angle \(BAD\); mixing in \(\overrightarrow{BA}\) gives the supplementary angle.
159IBMA3H-3-18r22-123.6 Pythagorean and double angle identitiesPaper 1Medium6 marks
No calculator
Consider the identity \(\frac{1-\cos2\theta}{\sin2\theta}\equiv\tan\theta\).
(a)
Prove the identity.
(3)
(b)
Hence find the exact value of \(\tan\frac{\pi}{8}\).
Exam tip: When asked to 'hence' find an exact value, choose the angle that makes the double angle a standard value.
160IBMA3H-3-18r23-033.8 Solving trigonometric equationsPaper 1Medium6 marks
No calculator
The outputs of two sensors on a wind turbine are modelled by \(V_1=\cos 2x\) and \(V_2=\sin x\) volts, where \(x\) is the rotor angle in radians, \(0\le x\le 2\pi\). Find all values of \(x\) for which the two outputs are equal.
(a)
Find all values of \(x\) for which \(V_1=V_2\).
(6)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use \(\cos 2x=1-2\sin^2x\)
✓ 1
A1 \(2\sin^2x+\sin x-1=0\)
✓ 1
M1 factorise or solve the quadratic in \(\sin x\)
✓ 1
A1 \(\sin x=\frac12\) or \(\sin x=-1\)
✓ 1
A1 \(x=\frac{\pi}{6}\) and \(x=\frac{5\pi}{6}\)
✓ 1
A1 \(x=\frac{3\pi}{2}\)
✓ 1
Exam tip: Choose the form of \(\cos 2x\) that leaves a single trigonometric function, here \(1-2\sin^2x\), so the equation becomes a quadratic.
161IBMA3H-3-18r23-073.9 Reciprocal and inverse trigonometric functionsPaper 1Medium7 marks
No calculator
A diver enters the water at an acute angle \(\theta\) to the surface, where \(\sin\theta=\frac{5}{13}\).
(a)
Find the exact value of \(\cos\theta\).
(2)
(b)
Find the exact value of \(\sin 2\theta\).
(2)
(c)
Find the exact value of \(\cos 2\theta\).
(2)
(d)
Hence write down the exact value of \(\tan 2\theta\).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use \(\cos^2\theta=1-\sin^2\theta\), with \(\cos\theta>0\) since \(\theta\) is acute
✓ 1
A1 \(\cos\theta=\frac{12}{13}\)
✓ 1
Part (b)
M1 \(\sin2\theta=2\sin\theta\cos\theta\)
✓ 1
A1 \(\frac{120}{169}\)
✓ 1
Part (c)
M1 use \(\cos 2\theta=1-2\sin^2\theta\) (or \(\cos^2\theta-\sin^2\theta\))
✓ 1
A1 \(\frac{119}{169}\)
✓ 1
Part (d)
A1 \(\frac{120}{119}\)
✓ 1
Exam tip: Find the missing ratio first (here \(\cos\theta\)) and note its sign from the quadrant; every double-angle value then follows.
162IBMA3H-3-18r23-093.9 Reciprocal and inverse trigonometric functionsPaper 1Medium6 marks
No calculator
The function \(f\) is defined by \(f(x)=3\arcsin\left(\frac{x}{2}\right)\).
(a)
Find the largest possible domain of \(f\).
(2)
(b)
Find the range of \(f\).
(2)
(c)
Find the exact value of \(f(\sqrt2)\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(-1\le\frac{x}{2}\le1\)
✓ 1
A1 \(-2\le x\le2\)
✓ 1
Part (b)
M1 range of \(\arcsin\) is \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\), multiplied by 3
✓ 1
A1 \(-\frac{3\pi}{2}\le f(x)\le\frac{3\pi}{2}\)
✓ 1
Part (c)
M1 \(\arcsin\frac{\sqrt2}{2}=\frac{\pi}{4}\)
✓ 1
A1 \(\frac{3\pi}{4}\)
✓ 1
Exam tip: The argument of \(\arcsin\) must lie in \([-1,1]\); use this to get the domain, and the output range \([-\frac{\pi}{2},\frac{\pi}{2}]\) to get the range.
163IBMA3H-3-18r23-103.10 Compound and double angle identitiesPaper 1Medium7 marks
No calculator
Two sensors on a wind farm record phase angles \(A\) and \(B\), where \(\sin A=\frac35\) with \(A\) acute, and \(\cos B=-\frac{12}{13}\) with \(B\) obtuse.
(a)
Find the exact values of \(\cos A\) and \(\sin B\).
(2)
(b)
Find the exact value of \(\sin(A+B)\).
(3)
(c)
Find the exact value of \(\cos(A-B)\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use \(\sin^2\theta+\cos^2\theta=1\), noting \(\cos A>0\) and \(\sin B>0\)
✓ 1
A1 \(\cos A=\frac45\) and \(\sin B=\frac{5}{13}\)
✓ 1
Part (b)
M1 \(\sin(A+B)=\sin A\cos B+\cos A\sin B\)
✓ 1
A1 \(\frac35\times\left(-\frac{12}{13}\right)+\frac45\times\frac{5}{13}\) (ECF from (a))
✓ 1
A1 \(-\frac{16}{65}\)
✓ 1
Part (c)
M1 \(\cos(A-B)=\cos A\cos B+\sin A\sin B\)
✓ 1
A1 \(-\frac{33}{65}\)
✓ 1
Common error: \(\cos(A-B)\) has a plus sign between the products, and an obtuse angle has a negative cosine but a positive sine.
164IBMA3H-3-19r25-063.9 Reciprocal and inverse trigonometric functionsPaper 1Medium6 marks
No calculator
Solve the equation \(3\tan^2x+5\sec x+1=0\) for \(0\le x\le 2\pi\).
(a)
Solve the equation, giving exact values.
(6)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use \(\tan^2x=\sec^2x-1\)
✓ 1
A1 \(3\sec^2x+5\sec x-2=0\)
✓ 1
M1 factorise or solve: \((3\sec x-1)(\sec x+2)=0\)
✓ 1
R1 reject \(\sec x=\frac13\) because \(|\sec x|\ge1\)
✓ 1
M1 \(\sec x=-2\) so \(\cos x=-\frac12\)
✓ 1
A1 \(x=\frac{2\pi}{3},\ \frac{4\pi}{3}\)
✓ 1
Common error: Keeping the root sec x = 1/3 as valid: because cos x lies between −1 and 1, sec x can never lie strictly between −1 and 1.
165IBMA3H-3-19r35-143.9 Reciprocal and inverse trigonometric functionsPaper 1Medium6 marks
No calculator
Consider the equation \(\tan^2x=\sec x+1\) for \(0\le x\le 2\pi\).
(a)
Show that the equation can be written as \(\sec^2x-\sec x-2=0\).
(2)
(b)
Hence find all the solutions of the equation, giving exact values.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of \(\tan^2x=\sec^2x-1\)
✓ 1
A1 \(\sec^2x-1=\sec x+1\), which rearranges to \(\sec^2x-\sec x-2=0\) (AG)
✓ 1
Part (b)
M1 factorise: \((\sec x-2)(\sec x+1)=0\)
✓ 1
A1 \(\sec x=2\) or \(\sec x=-1\)
✓ 1
A1 from \(\cos x=\tfrac12\): \(x=\dfrac{\pi}{3},\ \dfrac{5\pi}{3}\)
✓ 1
A1 from \(\cos x=-1\): \(x=\pi\)
✓ 1
Exam tip: After solving for sec x, convert to cos x = 1/sec x before using the unit circle; the exact values then come from standard angles.
166IBMA3H-3-20r40-043.8 Solving trigonometric equationsPaper 1Medium6 marks
No calculator
Consider the equation \(2\cos^2x+\sin x-1=0\) for \(0\le x\le2\pi\).
(a)
Show that the equation can be written as \(2\sin^2x-\sin x-1=0\).
(2)
(b)
Hence solve the equation, giving your answers in exact form.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses \(\cos^2x=1-\sin^2x\)
✓ 1
A1 \(2-2\sin^2x+\sin x-1=0\) and rearranges to \(2\sin^2x-\sin x-1=0\) (AG)
A1 \(x=\frac{7\pi}{6}\) and \(x=\frac{11\pi}{6}\) (≈ 5.76)
✓ 1
Common error: Dividing both sides by \(\sin x\) loses the solutions where \(\sin x=0\). Factorise instead.
167IBMA3H-3-20r40-073.9 Reciprocal and inverse trigonometric functionsPaper 1Medium9 marks
No calculator
The vibration of an engine mount in a Formula 1 car is modelled by \(g(x)=a\sin(bx)+c\) for \(x\ge0\), where \(a>0\), \(b>0\) and \(c\) are constants and \(x\) is in radians. The graph of \(y=g(x)\) has a maximum point at \(\left(\frac{\pi}{6},7\right)\) and the next minimum point after it is at \(\left(\frac{\pi}{2},-1\right)\).
(a)
Find the values of \(a\), \(b\) and \(c\).
(4)
(b)
Find the exact value of \(g\left(\frac{\pi}{9}\right)\).
(2)
(c)
Solve \(g(x)=5\) for \(0\le x\le\frac{\pi}{3}\), giving your answers in exact form.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(a=\frac{7-(-1)}{2}=4\)
✓ 1
A1 \(c=\frac{7+(-1)}{2}=3\)
✓ 1
M1 half the period = \(\frac{\pi}{2}-\frac{\pi}{6}=\frac{\pi}{3}\) so period = \(\frac{2\pi}{3}\)
✓ 1
A1 \(b=3\)
✓ 1
Part (b)
M1 \(4\sin\frac{\pi}{3}+3\)
✓ 1
A1 \(2\sqrt{3}+3\)
✓ 1
Part (c)
M1 \(\sin 3x=\frac{1}{2}\)
✓ 1
A1 \(3x=\frac{\pi}{6}\) so \(x=\frac{\pi}{18}\)
✓ 1
A1 \(3x=\frac{5\pi}{6}\) so \(x=\frac{5\pi}{18}\) (≈ 0.873)
✓ 1
Common error: The horizontal gap between a maximum and the next minimum is half a period, so double it to get the period.
168IBMA3H-3-20r40-103.9 Reciprocal and inverse trigonometric functionsPaper 1Medium6 marks
No calculator
This question is about reciprocal and inverse trigonometric functions.
(a)
Solve \(2\operatorname{cosec}x=4\) for \(0\le x\le2\pi\).
(3)
(b)
Find the exact value of \(\arccos\left(-\frac{\sqrt3}{2}\right)+2\arcsin\left(-\frac{1}{2}\right)\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\operatorname{cosec}x=2\) so \(\sin x=\frac{1}{2}\)
✓ 1
A1 \(x=\frac{\pi}{6}\)
✓ 1
A1 \(x=\frac{5\pi}{6}\)
✓ 1
Part (b)
M1 uses principal values: \(\arccos\) gives an angle in \([0,\pi]\) and \(\arcsin\) in \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\)
✓ 1
A1 \(\arccos\left(-\frac{\sqrt3}{2}\right)=\frac{5\pi}{6}\) and \(\arcsin\left(-\frac{1}{2}\right)=-\frac{\pi}{6}\)
✓ 1
A1 \(\frac{5\pi}{6}-\frac{\pi}{3}=\frac{\pi}{2}\)
✓ 1
Common error: The output of \(\arcsin\) is always between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\). A positive angle such as \(\frac{7\pi}{6}\) is never the value of \(\arcsin\left(-\frac{1}{2}\right)\).
169IBMA3H-3-20r40-113.9 Reciprocal and inverse trigonometric functionsPaper 1Medium6 marks
No calculator
This question is about the identity \(\operatorname{cosec}x-\sin x\equiv\cos x\cot x\).
(a)
Prove the identity.
(3)
(b)
Hence solve \(\operatorname{cosec}x-\sin x=0\) for \(0\le x\le2\pi\), where \(\sin x\neq0\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 writes the left-hand side as \(\frac{1}{\sin x}-\sin x\)
✓ 1
M1 common denominator: \(\frac{1-\sin^2x}{\sin x}=\frac{\cos^2x}{\sin x}\)
Exam tip: In a proof, work on one side only and finish on the exact expression you were given. Convert reciprocal functions to sine and cosine at the start.
170IBMA3H-3-20r41-043.4 Radians, arc length and sector areaPaper 1Medium7 marks
No calculator
The floor of a circular library hall includes a sector-shaped reading area with radius r metres and centre angle θ radians, where 0 < θ < π. The perimeter of the reading area is 40 m.
Diagram not to scale
(a)
Show that the area of the reading area is A = 20r − r2 m2.
(3)
(b)
The area of the reading area is 75 m2. Find the value of r and the value of θ.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 perimeter: 2r + rθ = 40
✓ 1
M1 rθ = 40 − 2r substituted into A = ½r2θ = ½r(rθ)
✓ 1
A1 A = ½r(40 − 2r) = 20r − r2
✓ 1
Part (b)
M1 20r − r2 = 75, so r2 − 20r + 75 = 0
✓ 1
A1 (r − 5)(r − 15) = 0, so r = 5 or r = 15
✓ 1
R1 r = 5 gives θ = 30/5 = 6 > π, which is rejected
✓ 1
A1 r = 15 and θ = (40 − 30)/15 = 2/3
✓ 1
Common error: Both roots of the quadratic must be tested against the condition 0 < θ < π; stopping at the first root gives an invalid sector.
171IBMA3H-3-20r41-073.9 Reciprocal and inverse trigonometric functionsPaper 1Medium5 marks
No calculator
Consider the equation sin 2x = sin(π/5) for 0 ≤ x ≤ 2π.
(a)
Write down an expression, in terms of θ, that is equal to sin(π − θ).
(1)
(b)
Hence find all the solutions of sin 2x = sin(π/5) in the given interval. Give your answers in terms of π.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 sin θ
✓ 1
Part (b)
M1 2x = π/5 or 2x = π − π/5 = 4π/5, and the range for 2x is 0 ≤ 2x ≤ 4π
✓ 1
A1 x = π/10 and x = 2π/5
✓ 1
M1 adding 2π to find further values: 2x = 11π/5 and 14π/5
✓ 1
A1 x = 11π/10 and x = 7π/5
✓ 1
Common error: Because the argument is 2x, the interval for x becomes an interval of length 4π for 2x; solving only within 0 to 2π misses half of the solutions.
172IBMA3H-3-20r41-113.17 Equations of a planePaper 1Medium7 marks
No calculator
Two flat panels on a wind-farm building lie in the planes Π1: x + 2y − 2z = 5 and Π2: 2x + 2y + z = 6. The point A(1, 2, 0) lies on both planes.
(a)
Show that the acute angle between Π1 and Π2 is arccos(4/9).
(3)
(b)
Find a vector parallel to the line of intersection of the two planes.
(2)
(c)
Find a vector equation of the line of intersection of the two planes.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 normal vectors (1, 2, −2) and (2, 2, 1)
✓ 1
M1 use of cos θ = |n1 · n2| / (|n1||n2|)
✓ 1
A1 n1 · n2 = 4 and |n1| = |n2| = 3, so cos θ = 4/9
✓ 1
Part (b)
M1 vector (cross) product n1 × n2
✓ 1
A1 (6, −5, −2) or any non-zero multiple
✓ 1
Part (c)
M1 line through A with the direction found in (b), in the form r = a + td
✓ 1
A1 r = (1, 2, 0) + t(6, −5, −2)
✓ 1
Exam tip: A line lying in both planes is perpendicular to both normal vectors, so its direction is their vector product.
173IBMA3H-3-18r42-143.8 Solving trigonometric equationsPaper 1Medium6 marks
No calculator
The angle \(x\), in radians, of a solar tracker satisfies \(2\sin^{2}x+\cos x-1=0\) for \(0\le x\le2\pi\). Solve the equation, giving exact answers.
(a)
Find all solutions for \(x\).
(6)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses \(\sin^{2}x=1-\cos^{2}x\)
✓ 1
A1 \(2\cos^{2}x-\cos x-1=0\)
✓ 1
M1 factorises: \((2\cos x+1)(\cos x-1)=0\)
✓ 1
A1 \(\cos x=1\) or \(\cos x=-\tfrac12\)
✓ 1
A1 \(x=0\) and \(x=2\pi\)
✓ 1
A1 \(x=\dfrac{2\pi}{3}\) and \(x=\dfrac{4\pi}{3}\)
✓ 1
Exam tip: Convert to a quadratic in a single trigonometric function using sin²x + cos²x = 1, then factorise before looking for angles.
174IBMA3H-3-19r43-053.6 Pythagorean and double angle identitiesPaper 1Medium8 marks
No calculator
Consider the identity \(\dfrac{1-\cos2x}{\sin2x}\equiv\tan x\), valid wherever both sides are defined.
(a)
Prove the identity.
(3)
(b)
Hence solve \(\dfrac{1-\cos2x}{\sin2x}=\sqrt3\) for \(0<x<\pi\).
(2)
(c)
Use the identity with a suitable value of \(x\) to show that \(\tan\frac{\pi}{8}=\sqrt2-1\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(1-\cos2x=1-(1-2\sin^2x)=2\sin^2x\) (use of \(\cos2x=1-2\sin^2x\))
Exam tip: In a 'show that' question, finish with the exact result you were asked to reach and write each line so that the route from the given expression is clear.
175AAHL3-q00033.1 Three-dimensional geometry: distance, solids and anglesPaper 1Hard10 marks
A solid right circular cone has base radius \(6\) cm and height \(8\) cm. A sphere of radius \(r\) cm fits exactly inside the cone, touching the base and touching the curved surface all the way round. The diagram shows the cross-section through the axis of the cone: the triangle \(VPQ\), where \(O\) is the centre of the base, and the circle in which the plane meets the sphere.
Diagram not to scale
(a)
Write down the slant height of the cone.
(1)
(b)
Show that \(r=3\).
(3)
(c)
Find the fraction of the volume of the cone that is occupied by the sphere.
(3)
(d)
Show that the ratio of the total surface area of the cone to the surface area of the sphere is equal to the ratio of the volume of the cone to the volume of the sphere.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(10\) cm
A1
(b)
the centre \(C\) of the circle lies on \([VO]\) with \(CO=r\), \(VC=8-r\), and the radius to the point of contact \(X\) on \([VQ]\) is perpendicular to \([VQ]\)
M1
\(\sin O\hat{V}Q=\dfrac{6}{10}=\dfrac{CX}{VC}\), so \(\dfrac{r}{8-r}=\dfrac{3}{5}\)
A1
Or similar triangles \(VXC\) and \(VOQ\).
\(5r=24-3r\), so \(8r=24\)
A1
\(r=3\)
AG
(c)
volume of sphere \(=\dfrac{4}{3}\pi(3)^{3}=36\pi\)
A1
volume of cone \(=\dfrac{1}{3}\pi(6)^{2}(8)=96\pi\)
A1
\(\dfrac{36\pi}{96\pi}=\dfrac{3}{8}\)
A1
(d)
total surface area of cone \(=\pi(6)(10)+\pi(6)^{2}=96\pi\)
A1
surface area of sphere \(=4\pi(3)^{2}=36\pi\)
A1
both ratios equal \(\dfrac{96\pi}{36\pi}=\dfrac{8}{3}\)
A1
Must compare with the volume ratio \(\dfrac{96\pi}{36\pi}\) from (c).
the two ratios are equal
AG
Answers: (a) 10 cm · (c) \(\dfrac{3}{8}\)
176AAHL3-q00163.4 Radians, arc length and sector areaPaper 1Hard14 marks
The diagram shows an equilateral triangle \(ABC\) with sides of length \(a\) cm. Three circular arcs are drawn: arc \(BC\) with centre \(A\), arc \(CA\) with centre \(B\) and arc \(AB\) with centre \(C\), each of radius \(a\) cm. The shaded region enclosed by the three arcs is called a Reuleaux triangle.
Diagram not to scale
(a)
Write down the angle, in radians, subtended at \(A\) by the arc \(BC\).
(1)
(b)
Show that the perimeter of the Reuleaux triangle is \(\pi a\).
(2)
(c)
Find, in terms of \(a\), the exact area of the segment bounded by the arc \(BC\) and the chord \([BC]\).
(3)
(d)
Hence show that the area of the Reuleaux triangle is \(\dfrac{(\pi-\sqrt3)a^{2}}{2}\).
(3)
(e)
A circle has diameter \(a\) cm. Show that its circumference is equal to the perimeter of the Reuleaux triangle, and find, in its simplest form, the ratio of the area of the Reuleaux triangle to the area of the circle.
(3)
(f)
Given that \(\pi\lt3.2\), show that the area of the Reuleaux triangle is less than the area of the circle.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(\dfrac{\pi}{3}\)
A1
(b)
each arc has length \(a\times\dfrac{\pi}{3}\)
M1
\(3\times\dfrac{\pi a}{3}\)
A1
\(=\pi a\)
AG
(c)
sector area \(=\dfrac{1}{2}a^{2}\left(\dfrac{\pi}{3}\right)=\dfrac{\pi a^{2}}{6}\)
(M1)
triangle area \(=\dfrac{1}{2}a^{2}\sin\dfrac{\pi}{3}=\dfrac{\sqrt3a^{2}}{4}\)
180AAHL3-q00393.9 Reciprocal and inverse trigonometric functionsPaper 1Hard14 marks
Let \(f(x)=\sec x+\tan x\), for \(-\dfrac{\pi}{2}\lt x\lt\dfrac{\pi}{2}\).
(a)
Show that \(f(x)=\dfrac{1+\sin x}{\cos x}\).
(1)
(b)
Show that \(f(x)=\dfrac{\cos x}{1-\sin x}\).
(2)
(c)
By writing \(\sin x\) and \(\cos x\) in terms of \(\sin\dfrac{x}{2}\) and \(\cos\dfrac{x}{2}\), show that \(f(x)=\tan\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)\).
\(1+\sin x=\left(\cos\dfrac{x}{2}+\sin\dfrac{x}{2}\right)^{2}\) and \(\cos x=\left(\cos\dfrac{x}{2}-\sin\dfrac{x}{2}\right)\left(\cos\dfrac{x}{2}+\sin\dfrac{x}{2}\right)\)
Multiplying numerator and denominator by \(1-t^{2}\).
\(=\dfrac{3t-t^{3}}{1-3t^{2}}\)
AG
(b)
\(\tan\dfrac{3\pi}{12}=\tan\dfrac{\pi}{4}=1\), so \(\dfrac{3t-t^{3}}{1-3t^{2}}=1\)
M1
\(3t-t^{3}=1-3t^{2}\)
A1
\(t^{3}-3t^{2}-3t+1=0\)
AG
(c)
\(t=-1\) is a root, so \((t+1)\) is a factor
M1
\((t+1)(t^{2}-4t+1)=0\)
A1
\(t=2\pm\sqrt3\)
A1
\(0\lt\tan\dfrac{\pi}{12}\lt\tan\dfrac{\pi}{4}=1\), so \(\tan\dfrac{\pi}{12}=2-\sqrt3\)
R1
Must reject both \(-1\) and \(2+\sqrt3\) with a valid reason.
Answers: (c) \(\tan\frac{\pi}{12}=2-\sqrt3\)
182AAHL3-q00493.10 Compound and double angle identitiesPaper 1Hard8 marks
The angles of a triangle are \(A\), \(B\) and \(C\). None of the angles is a right angle.
(a)
Show that \(\tan A+\tan B+\tan C=\tan A\tan B\tan C\).
(3)
(b)
Given that \(\tan A=\dfrac12\) and \(\tan B=3\), find \(\tan C\).
(2)
(c)
The values of \(\tan A\), \(\tan B\) and \(\tan C\) are three consecutive positive integers. Find these integers.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(\tan C=\tan\left(\pi-(A+B)\right)=-\tan(A+B)\)
M1
\(\tan C=-\dfrac{\tan A+\tan B}{1-\tan A\tan B}\)
A1
Valid since \(\tan A\tan B\neq1\) (\(C\neq\frac{\pi}{2}\)).
\(\tan C-\tan A\tan B\tan C=-\tan A-\tan B\)
M1
\(\tan A+\tan B+\tan C=\tan A\tan B\tan C\)
AG
(b)
\(\dfrac12+3+\tan C=\dfrac32\tan C\)
M1
Or direct use of the expression for tan C in (a).
\(\tan C=7\)
A1
(c)
\(n+(n+1)+(n+2)=n(n+1)(n+2)\)
M1
\(3(n+1)=n(n+1)(n+2)\Rightarrow n(n+2)=3\)
A1
Dividing by \(n+1\gt0\).
\(n=1\): the integers are \(1,\ 2,\ 3\)
A1
Award A0 if \(n=-3\) is not rejected.
Answers: (b) \(\tan C=7\) · (c) 1, 2, 3
183AAHL3-q00543.11 Symmetry properties of trigonometric graphsPaper 1Hard15 marks
Let \(f(x)=\sin x+\sin3x\), for \(0\le x\le2\pi\).
(a)
Show that \(f(\pi-x)=f(x)\).
(2)
(b)
Show that \(f(x+\pi)=-f(x)\).
(2)
(c)
By writing \(x=2x-x\) and \(3x=2x+x\), use the compound angle identities to show that \(f(x)=4\sin x\cos^{2}x\).
(3)
(d)
The greatest value of \(f\) occurs at \(x=a\), where \(0\lt a\lt\dfrac{\pi}{2}\). By considering \(f'(x)\), show that \(\sin a=\dfrac{1}{\sqrt3}\).
(3)
(e)
Find the exact greatest value of \(f\).
(2)
(f)
Using parts (a) and (b), write down, in terms of \(a\), the other value of \(x\) at which \(f\) attains its greatest value, and the two values of \(x\) at which \(f\) attains its least value.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(f(\pi-x)=\sin(\pi-x)+\sin(3\pi-3x)\)
M1
\(\sin(\pi-x)=\sin x\) and \(\sin(3\pi-3x)=\sin(\pi-3x)=\sin3x\)
A1
\(f(\pi-x)=f(x)\)
AG
(b)
\(f(x+\pi)=\sin(x+\pi)+\sin(3x+3\pi)\)
M1
\(\sin(x+\pi)=-\sin x\) and \(\sin(3x+3\pi)=\sin(3x+\pi)=-\sin3x\)
Answers: (e) \(\frac{8\sqrt3}{9}\) · (f) greatest at \(\pi-a\); least at \(\pi+a\) and \(2\pi-a\)
184AAHL3-q00623.12 Vectors: concepts and algebraPaper 1Hard14 marks
OABC is a parallelogram with \(\overrightarrow{OA}=\boldsymbol{a}\) and \(\overrightarrow{OC}=\boldsymbol{c}\). The point M is the midpoint of [BC]. The lines (OM) and (AC) meet at P, as shown.
Diagram not to scale
(a)(i)
Find, in terms of \(\boldsymbol{a}\) and \(\boldsymbol{c}\),
\(\overrightarrow{OB}\);
(1)
(a)(ii)
\(\overrightarrow{OM}\);
(1)
(a)(iii)
\(\overrightarrow{AC}\).
(1)
(b)
Given that \(\overrightarrow{OP}=\lambda\overrightarrow{OM}\) and \(\overrightarrow{OP}=\overrightarrow{OA}+\mu\overrightarrow{AC}\), find the value of \(\lambda\) and the value of \(\mu\).
(4)
(c)
Hence write down the ratio \(OP:PM\) and the ratio \(AP:PC\).
(2)
(d)
Now let O be the origin, A\((6,0,3)\) and C\((2,4,-1)\). Find the coordinates of B.
185AAHL3-q00773.15 Intersecting, parallel and skew linesPaper 1Hard10 marks
The lines \(L_{1}\) and \(L_{2}\) have equations \(\boldsymbol{r}=\begin{pmatrix}1\\0\\2\end{pmatrix}+\lambda\begin{pmatrix}2\\1\\k\end{pmatrix}\) and \(\boldsymbol{r}=\begin{pmatrix}0\\2\\1\end{pmatrix}+\mu\begin{pmatrix}1\\k\\1\end{pmatrix}\) respectively, where \(k\in\mathbb{R}\).
(a)
Show that \(L_{1}\) and \(L_{2}\) are not parallel for any value of \(k\).
(2)
(b)
Find the values of \(k\) for which \(L_{1}\) and \(L_{2}\) intersect.
(5)
(c)
For the positive value of \(k\) found in part (b), find the coordinates of the point of intersection.
(2)
(d)
State the geometrical relationship between \(L_{1}\) and \(L_{2}\) for all other values of \(k\).
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(\begin{pmatrix}2\\1\\k\end{pmatrix}=t\begin{pmatrix}1\\k\\1\end{pmatrix}\) gives \(t=2\) from the \(x\)-components
M1
then \(k=t=2\) from the \(z\)-components but \(1=2k\Rightarrow k=\tfrac12\) from the \(y\)-components: contradiction
186AAHL3-q00803.16 Vector productPaper 1Hard8 marks
Let \(\boldsymbol{a}\) and \(\boldsymbol{b}\) be non-zero, non-parallel vectors.
(a)
Show that \((\boldsymbol{a}+\boldsymbol{b})\times(\boldsymbol{a}-\boldsymbol{b})=2(\boldsymbol{b}\times\boldsymbol{a})\).
(3)
(b)
Given that \(|\boldsymbol{a}|=3\), \(|\boldsymbol{b}|=4\) and \(\boldsymbol{a}\cdot\boldsymbol{b}=6\), find the exact value of \(|\boldsymbol{a}\times\boldsymbol{b}|\).
(3)
(c)
Hence find the exact area of the parallelogram that has \(\boldsymbol{a}+\boldsymbol{b}\) and \(\boldsymbol{a}-\boldsymbol{b}\) as adjacent sides.
area \(=\left|(\boldsymbol{a}+\boldsymbol{b})\times(\boldsymbol{a}-\boldsymbol{b})\right|=2|\boldsymbol{b}\times\boldsymbol{a}|=2|\boldsymbol{a}\times\boldsymbol{b}|\)
M1
\(12\sqrt3\)
A1
FT from (b).
Answers: (b) \(6\sqrt3\) · (c) \(12\sqrt3\)
187AAHL3-q00893.18 Intersections and angles of lines and planesPaper 1Hard9 marks
Consider the planes
\(\Pi_{1}:\ x+y+z=2,\qquad \Pi_{2}:\ 2x-y+3z=1,\qquad \Pi_{3}:\ x+ay+2z=b\), where \(a,b\in\mathbb{R}\).
(a)
Show that the three planes meet at a single point unless \(a=-2\).
(3)
(b)
Given that \(a=-2\), find the value of \(b\) for which the three planes meet in a line, and find a vector equation of this line.
(4)
(c)
Given that \(a=-2\) and \(b\) does not take the value found in part (b), describe the geometrical arrangement of the three planes, justifying your answer.
\(0y=b+1\neq0\) has no solution, so the planes have no common point
R1
no two normals are parallel, so each pair of planes meets in a line; the three lines are parallel and the planes form a triangular prism
R1
Answers: (b) \(b=-1\), \(\boldsymbol{r}=(5,0,-3)+\lambda(-4,1,3)\) · (c) triangular prism (no common point)
188AAHL3-q00913.18 Intersections and angles of lines and planesPaper 1Hard16 marks
The plane \(\Pi\) has equation \(x-2y+2z=8\) and the point P has coordinates \((4,-3,8)\).
(a)
Write down a vector equation of the line through P that is perpendicular to \(\Pi\).
(1)
(b)
Find the coordinates of F, the foot of the perpendicular from P to \(\Pi\).
(3)
(c)
Hence find the shortest distance from P to \(\Pi\).
(2)
(d)
Find the coordinates of P′, the reflection of P in \(\Pi\).
(2)
(e)
The line \(L\) passes through P and has direction vector \(\begin{pmatrix}0\\1\\-2\end{pmatrix}\). Find the coordinates of R, the point where \(L\) meets \(\Pi\).
(3)
(f)
Find the exact value of the sine of the acute angle between \(L\) and \(\Pi\).
(3)
(g)
Find a vector equation of the line \(L'\), the reflection of \(L\) in \(\Pi\).
189IBMA2H-049-05-113.15 Intersecting, parallel and skew linesPaper 1Hard13 marks
No calculator
In a chemical plant, two straight pipes are modelled by \(L_1:\ \boldsymbol{r}=\begin{pmatrix}1\\0\\2\end{pmatrix}+\lambda\begin{pmatrix}1\\2\\-1\end{pmatrix}\) and \(L_2:\ \boldsymbol{r}=\begin{pmatrix}3\\1\\k\end{pmatrix}+\mu\begin{pmatrix}2\\k\\-2\end{pmatrix}\), where \(k\in\mathbb{R}\) can be changed by moving an adjustable mounting.
(a)
Find the value of \(k\) for which the pipes are parallel, and show that they are then distinct lines.
(3)
(b)
Show that the pipes intersect only when \(k=0\), and find the point of intersection in this case.
(6)
(c)
Write down the set of values of \(k\) for which the pipes are skew.
(1)
(d)
For \(k=0\), find the acute angle between the pipes.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 require \((2,k,-2)\) to be a multiple of \((1,2,-1)\)
✓ 1
A1 \(k=4\)
✓ 1
R1 \((3,1,4)\) is not on \(L_1\), since \(x=3\) gives \(\lambda=2\) but then \(y=4\neq1\), so the lines are distinct
Exam tip: When a parameter appears in the equations, look for a combination that removes both λ and μ. The condition on k then follows at once, and the parallel case can be treated separately.
190IBMA2H-049-06-113.18 Intersections and angles of lines and planesPaper 1Hard13 marks
No calculator
A CNC cutting machine uses three planar cutting surfaces, modelled by Π1: x + y + z = 3 Π2: 2x − y + z = 2 Π3: 4x + y + az = b where a, b ∈ ℝ.
(a)
Find the value of a for which the three planes do not meet at a unique point.
(4)
(b)
For this value of a, find the value of b for which the three planes meet in a line.
(3)
(c)
Find a vector equation of this line.
(4)
(d)
For a = 3 and b = 5, describe the geometrical arrangement of the three planes, justifying your answer.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt determinant of coefficient matrix (or row reduction)
✓ 1
A1 1(−a − 1) − 1(2a − 4) + 1(2 + 4)
✓ 1
M1 set −3a + 9 = 0
✓ 1
A1 a = 3
✓ 1
Part (b)
M1 attempt to write Π3 as a combination of Π1 and Π2 (or row reduce augmented matrix)
✓ 1
A1 2 × Π1 + Π2 gives 4x + y + 3z = 8
✓ 1
A1 b = 8
✓ 1
Part (c)
M1 eliminate a variable or find vector product of normals of Π1 and Π2
✓ 1
A1 direction 2i + j − 3k
✓ 1
A1 a point on the line, e.g. (−1, 0, 4)
✓ 1
A1 r = −i + 4k + λ(2i + j − 3k)
✓ 1
Part (d)
R1 the system is inconsistent (2Π1 + Π2 gives 4x + y + 3z = 8 ≠ 5), so there is no common point
✓ 1
A1 no two normals are parallel, so the planes meet pairwise in three parallel lines forming a triangular prism
✓ 1
Exam tip: Check your line by substituting its point into all three planes; with b = 8, (−1, 0, 4) gives −4 + 0 + 12 = 8 in Π₃.
191IBMA2H-049-00-113.6 Pythagorean and double angle identitiesPaper 1Hard14 marks
No calculator
Let \(t=\tan x\), where \(x\ne\pm\frac{\pi}{4}\) and all tangents are defined.
(a)
Show that \(\tan\left(\frac{\pi}{4}+x\right)-\tan\left(\frac{\pi}{4}-x\right)=2\tan2x\).
(4)
(b)
Hence solve \(\tan\left(\frac{\pi}{4}+x\right)-\tan\left(\frac{\pi}{4}-x\right)=2\sqrt3\) for \(-\frac{\pi}{2}<x<\frac{\pi}{2}\).
(4)
(c)
Find the exact value of \(\tan\frac{3\pi}{8}-\tan\frac{\pi}{8}\).
(2)
(d)
Given that \(\tan\frac{3\pi}{8}\tan\frac{\pi}{8}=1\), find the exact value of \(\tan\frac{3\pi}{8}\).
M1 lets \(u=\tan\frac{3\pi}{8}\), so \(u-\frac{1}{u}=2\)
✓ 1
A1 \(u^2-2u-1=0\)
✓ 1
M1 solves: \(u=1\pm\sqrt2\)
✓ 1
R1 \(u>0\) because \(\frac{3\pi}{8}\) is acute, so \(u=1+\sqrt2\)
✓ 1
Exam tip: In a 'hence' part, choose \(x\) so the bracketed angles match the ones you need. Here \(\frac{\pi}{4}\pm\frac{\pi}{8}\) gives \(\frac{3\pi}{8}\) and \(\frac{\pi}{8}\) at once.
192IBMA2H-049-01-103.9 Reciprocal and inverse trigonometric functionsPaper 1Hard12 marks
No calculator
Let \(\theta = \arctan x\), where \(x \in \mathbb{R}\).
(a)
Show that \(\cos2\theta = \frac{1-x^2}{1+x^2}\).
(3)
(b)
Show that \(\sin2\theta = \frac{2x}{1+x^2}\).
(2)
(c)
Hence solve \(2\arctan x = \arccos\frac35\), justifying your answer.
(4)
(d)
Find the exact value of \(\sin\left(4\arctan\frac13\right)\).
Common error: Squaring or taking cosines hides the sign of x. Use the range of arctan: 2 arctan x has the same sign as x, so it must match the positive value of arccos(3/5).
193IBMA2H-049-02-113.16 Vector productPaper 1Hard14 marks
No calculator
Two cables attached to a module on a space station exert forces represented by the vectors \(\boldsymbol{a}\) and \(\boldsymbol{b}\), where \(|\boldsymbol{a}|=3\), \(|\boldsymbol{b}|=4\) and the angle between \(\boldsymbol{a}\) and \(\boldsymbol{b}\) is \(\frac{\pi}{3}\).
(a)
Find \(\boldsymbol{a}\cdot\boldsymbol{b}\).
(2)
(b)
Find the exact magnitude of the resultant force \(\boldsymbol{a}+\boldsymbol{b}\).
(3)
(c)
Find the value of \(t\) for which \(\boldsymbol{a}+t\boldsymbol{b}\) is perpendicular to \(\boldsymbol{a}\).
(3)
(d)
Show that \((\boldsymbol{a}+\boldsymbol{b})\times(\boldsymbol{a}-\boldsymbol{b})=-2(\boldsymbol{a}\times\boldsymbol{b})\).
(3)
(e)
Hence find the exact area of the parallelogram with adjacent sides \(\boldsymbol{a}+\boldsymbol{b}\) and \(\boldsymbol{a}-\boldsymbol{b}\).
Common error: The vector product is not commutative: \(\boldsymbol{b}\times\boldsymbol{a}=-\boldsymbol{a}\times\boldsymbol{b}\). Treating the two middle terms as cancelling gives the false result \(\boldsymbol{0}\).
194IBMA2H-049-02-123.7 Circular functions and their transformationsPaper 1Hard15 marks
No calculator
A cyclist pedals at a steady cadence of 90 revolutions per minute. The crank has length 0.17 m and the pedal axle is 0.30 m above the ground. At \(t=0\) the right pedal is at its lowest point. The height, in metres, of the right pedal above the ground \(t\) seconds later is modelled by \(h(t)=a\cos(bt)+c\), \(b>0\).
(a)
Find the values of \(a\), \(b\) and \(c\).
(4)
(b)
Find the first time at which the right pedal is 0.385 m above the ground.
(3)
(c)
Find the total time in the first 2 seconds during which the right pedal is higher than 0.385 m.
(3)
(d)
The left pedal is always diametrically opposite the right pedal. Write down an expression for its height and hence find all times in \(0\le t\le 1\) at which both pedals are at the same height.
(3)
(e)
On a hill the cyclist changes cadence, and the model becomes \(h(t)=0.30-0.17\cos(kt)\), with the right pedal first reaching its highest point at \(t=0.25\) s. Find the new cadence in revolutions per minute.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(c=0.30\)
✓ 1
A1 \(a=-0.17\) (lowest point at \(t=0\))
✓ 1
M1 1.5 revolutions per second, so period \(=\frac{2}{3}\) s
✓ 1
A1 \(b=3\pi\)
✓ 1
Part (b)
M1 \(0.30-0.17\cos(3\pi t)=0.385\)
✓ 1
A1 \(\cos(3\pi t)=-\frac{1}{2}\) so \(3\pi t=\frac{2\pi}{3}\)
✓ 1
A1 \(t=\frac{2}{9}\) s
✓ 1
Part (c)
M1 \(\frac{2\pi}{3}<3\pi t<\frac{4\pi}{3}\) in each cycle, i.e. \(\frac{2}{9}<t<\frac{4}{9}\), duration \(\frac{2}{9}\) s per cycle
✓ 1
M1 2 s contains exactly 3 cycles
✓ 1
A1 total \(\frac{2}{3}\) s
✓ 1
Part (d)
M1 left pedal height \(0.30+0.17\cos(3\pi t)\)
✓ 1
M1 equal heights \(\Rightarrow\cos(3\pi t)=0\)
✓ 1
A1 \(t=\frac{1}{6},\frac{1}{2},\frac{5}{6}\) s
✓ 1
Part (e)
M1 \(0.25k=\pi\), so \(k=4\pi\)
✓ 1
A1 period 0.5 s, cadence 120 revolutions per minute
✓ 1
Exam tip: Convert the cadence to revolutions per second before finding \(b\): \(b=2\pi\times\)(revolutions per second). The highest point is reached after half a revolution, not a full one.
195IBMA2H-049-03-113.1 Three-dimensional geometry: distance, solids and anglesPaper 1Hard15 marks
No calculator
A triangular glass panel on a sports stadium roof has vertices A(1, 0, 2), B(3, 1, 0) and C(0, 2, 1), with distances in metres. The panel lies in the plane Π. The z-axis is vertically upwards, and the ground is the plane z = 0. A support cable is attached to the top of a mast at D(4, 4, 5).
(a)
Show that \(\cos B\hat{A}C = \frac{\sqrt{6}}{9}\).
(4)
(b)
Hence find the exact area of the panel.
(3)
(c)
Find a Cartesian equation of Π.
(3)
(d)
Find the angle between Π and the ground.
(2)
(e)
Find the exact volume of the tetrahedron ABCD.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 AB = (2, 1, −2) and AC = (−1, 2, −1)
✓ 1
M1 AB·AC = −2 + 2 + 2 = 2
✓ 1
M1 |AB| = 3, |AC| = √6 and uses cos BÂC = 2/(3√6)
✓ 1
A1 rationalises to √6/9
✓ 1
Part (b)
M1 uses sin²BÂC = 1 − 6/81 = 75/81
✓ 1
A1 sin BÂC = 5√3/9
✓ 1
A1 area = ½ × 3 × √6 × 5√3/9 = 5√2/2 m²
✓ 1
Part (c)
M1 attempts AB × AC
✓ 1
A1 normal (3, 4, 5)
✓ 1
A1 3x + 4y + 5z = 13
✓ 1
Part (d)
M1 uses normals (3, 4, 5) and (0, 0, 1): cos θ = 5/√50
✓ 1
A1 θ = π/4
✓ 1
Part (e)
M1 perpendicular distance from D to Π: |12 + 16 + 25 − 13| / √50
✓ 1
A1 = 40/(5√2) = 4√2
✓ 1
A1 volume = ⅓ × 5√2/2 × 4√2 = 20/3 m³
✓ 1
Exam tip: When an angle's cosine comes from a scalar product, the identity sin²θ + cos²θ = 1 gives the exact sine for the area formula. Check it against ½|AB × AC| once you have the normal vector.
196IBMA2H-049-03-123.6 Pythagorean and double angle identitiesPaper 1Hard13 marks
No calculator
This question uses the identity for cos 3x to solve a cubic equation exactly.
(a)
Show that \(\cos 3x = 4\cos^3 x - 3\cos x\).
(4)
(b)
Hence solve \(8\cos^3 x - 6\cos x + 1 = 0\) for 0 ≤ x ≤ π.
(4)
(c)
By considering the polynomial \(8c^3 - 6c + 1 = 0\), find the exact value of \(\cos\frac{2\pi}{9} + \cos\frac{4\pi}{9} + \cos\frac{8\pi}{9}\).
(3)
(d)
Hence find the exact value of \(\sec\frac{2\pi}{9}\sec\frac{4\pi}{9}\sec\frac{8\pi}{9}\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 writes cos 3x = cos(2x + x) = cos 2x cos x − sin 2x sin x
✓ 1
M1 substitutes cos 2x = 2cos²x − 1 and sin 2x = 2 sin x cos x
✓ 1
A1 replaces sin²x by 1 − cos²x: 2cos³x − cos x − 2cos x(1 − cos²x)
✓ 1
A1 simplifies to 4cos³x − 3cos x
✓ 1
Part (b)
M1 recognises the equation as 2cos 3x + 1 = 0
✓ 1
A1 cos 3x = −1/2
✓ 1
M1 uses 0 ≤ 3x ≤ 3π: 3x = 2π/3, 4π/3, 8π/3
✓ 1
A1 x = 2π/9, 4π/9, 8π/9
✓ 1
Part (c)
R1 cos is strictly decreasing on [0, π], so the three values of cos x are distinct and are therefore the three roots of the cubic in c
✓ 1
M1 sum of roots = −(coefficient of c²)/8
✓ 1
A1 0
✓ 1
Part (d)
M1 product of roots = −1/8
✓ 1
A1 product of secants = 1 ÷ (−1/8) = −8
✓ 1
Common error: Linking the roots of a cubic to cosine values needs a reason: the three x-values must give three different cosines. Without this justification the sum and product cannot be read off safely.
197IBMA2H-049-04-123.9 Reciprocal and inverse trigonometric functionsPaper 1Hard14 marks
No calculator
A designer is building a logo from a regular pentagon and needs exact trigonometric values. You are given that \(\cos\frac{\pi}{5} = \frac{1 + \sqrt5}{4}\).
(a)
Using symmetry properties of trigonometric graphs, find the exact values of \(\cos\frac{4\pi}{5}\), \(\cos\frac{6\pi}{5}\), \(\sin\frac{3\pi}{10}\) and \(\sin\frac{13\pi}{10}\).
(4)
(b)
Find the exact value of \(\cos\frac{2\pi}{5}\).
(3)
(c)
Hence find the exact value of \(\cos\frac{3\pi}{5}\).
(2)
(d)
Hence solve \(4\cos^2 x - 2\cos x - 1 = 0\) for \(0 \le x \le 2\pi\).
M1 recognises the roots as \(\cos\frac{\pi}{5}\) and \(\cos\frac{3\pi}{5}\) from the earlier parts
✓ 1
A1 \(x = \frac{\pi}{5}\) and \(\frac{9\pi}{5}\)
✓ 1
A1 \(x = \frac{3\pi}{5}\) and \(\frac{7\pi}{5}\)
✓ 1
Common error: Each value of cos x in (0, 1) or (−1, 0) gives two solutions in [0, 2π], because cos(2π − θ) = cos θ. Stopping at the first solution from each root loses half the answers.
198IBMA3H-3-19r5-073.10 Compound and double angle identitiesPaper 1Hard7 marks
No calculator
Two support cables on a radio mast have direction vectors \(\mathbf{a}=\begin{pmatrix}1\\2\\k\end{pmatrix}\) and \(\mathbf{b}=\begin{pmatrix}2\\-1\\2\end{pmatrix}\). The angle between the cables is \(60^\circ\). Find the exact value of \(k\).
(a)
Find the exact value of \(k\).
(7)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\mathbf{a}\cdot\mathbf{b}=2k\)
✓ 1
A1 \(|\mathbf{b}|=3\) and \(|\mathbf{a}|=\sqrt{5+k^2}\)
Common error: When the equation is in terms of \(3x\), the interval for \(3x\) is three times as long, so extra solutions appear; widen the range before solving.
201IBMA3H-3-18r11-133.9 Reciprocal and inverse trigonometric functionsPaper 1Hard12 marks
No calculator
The shadow-flicker intensity near a wind farm is modelled by \(f(x)=3\csc(2x)+1\).
(a)
Find the equations of the vertical asymptotes of the graph of \(y=f(x)\) for \(0<x<2\pi\).
(2)
(b)
Write down the range of \(f\).
(2)
(c)
Solve \(f(x)=7\) for \(0<x<\pi\).
(3)
(d)
The function \(f\) is restricted to the domain \(\frac{\pi}{4}\leq x<\frac{\pi}{2}\). Find \(f^{-1}(x)\) and state its domain.
(4)
(e)
Find \(f^{-1}(7)\) and comment on its relation to your answer to (c).
A1 \(f^{-1}(7)=\frac12\left(\pi-\frac{\pi}{6}\right)=\frac{5\pi}{12}\), the solution of (c) lying in the restricted domain
✓ 1
Common error: \(\arcsin\) only returns angles in \([-\frac{\pi}{2},\frac{\pi}{2}]\); when the original angle lies in the second quadrant, use \(\pi-\arcsin(\cdot)\).
202IBMA3H-3-18r11-143.5 The unit circle, exact values and the ambiguous casePaper 1Hard10 marks
No calculator
Three wind turbines \(P\), \(Q\) and \(R\) stand at the corners of a triangle with \(PQ=9\) km, \(QR=7\) km and \(PR=12\) km. Let \(\theta\) be angle \(PQR\). Give all answers in exact form.
Diagram not to scale
(a)
Find the exact value of \(\cos\theta\).
(2)
(b)
Show that \(\sin\theta=\frac{4\sqrt5}{9}\).
(2)
(c)
Find the exact value of \(\sin2\theta\).
(2)
(d)
Find the exact area of triangle \(PQR\).
(2)
(e)
The turbines lie on a circle. Find the exact radius of this circle.
A1 \(r=\frac{27}{2\sqrt5}=\frac{27\sqrt5}{10}\) km
✓ 1
Exam tip: With three sides known, find one angle by the cosine rule; the exact sine then follows from \(\sin^2\theta+\cos^2\theta=1\), and the sine rule gives the circumradius via \(2r=\frac{a}{\sin A}\).
203IBMA3H-3-18r15-053.10 Compound and double angle identitiesPaper 1Hard8 marks
No calculator
In a Formula 1 simulation, two steering angles \(A\) and \(B\) are both acute, with \(\tan A=\frac12\) and \(\tan B=\frac13\).
Exam tip: When asked to prove an identity, work on one side only and finish with the exact expression stated in the question.
206IBMA3H-3-18r22-073.18 Intersections and angles of lines and planesPaper 1Hard10 marks
No calculator
The supports of a bridge model lie in the planes \(\Pi_1: x+2y+z=4\) and \(\Pi_2: 2x+y-z=5\). A third plane is \(\Pi_3: 3x+3y+kz=c\), where \(k\) and \(c\) are constants.
(a)
Show that \(\Pi_1\) and \(\Pi_2\) intersect in a line, and find a vector equation of this line.
(4)
(b)
Show that the point with parameter \(t\) on this line lies on \(\Pi_3\) when \(9+kt=c\).
(2)
(c)
Hence find the conditions on \(k\) and \(c\) for which the three planes meet (i) at exactly one point, (ii) in a line, (iii) at no point.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
R1 the normals \((1,2,1)\) and \((2,1,-1)\) are not parallel, so the planes meet in a line
✓ 1
M1 direction \(=(1,2,1)\times(2,1,-1)=(-3,3,-3)\), so \((1,-1,1)\)
✓ 1
A1 a point on both planes, for example \((2,1,0)\)
✓ 1
A1 \(\mathbf r=(2,1,0)+t(1,-1,1)\)
✓ 1
Part (b)
M1 substitute \(x=2+t\), \(y=1-t\), \(z=t\) into \(\Pi_3\): \(3(2+t)+3(1-t)+kt\)
✓ 1
A1 \(9+kt=c\)
✓ 1
Part (c)
M1 consider the equation \(kt=c-9\)
✓ 1
A1 (i) \(k\ne0\), any \(c\) (unique \(t=\frac{c-9}{k}\))
✓ 1
A1 (ii) \(k=0\) and \(c=9\) (every \(t\) works)
✓ 1
A1 (iii) \(k=0\) and \(c\ne9\) (no \(t\) works)
✓ 1
Exam tip: Reduce three planes to one equation in the line's parameter; the number of solutions of that single equation tells you the geometry.
207IBMA3H-3-18r23-123.11 Symmetry properties of trigonometric graphsPaper 1Hard8 marks
No calculator
Consider the identity \(\sin3x\equiv3\sin x-4\sin^3x\).
A1 \(\sin x=0\) rejected for \(0<x<\pi\); \(\sin x=-\frac{1}{\sqrt2}\) rejected as \(\sin x>0\) here, so \(\sin x=\frac{1}{\sqrt2}\)
✓ 1
A1 \(x=\frac{\pi}{4}\) and \(x=\frac{3\pi}{4}\)
✓ 1
Common error: Do not divide both sides by \(\sin x\) when solving; factorise instead so the root \(\sin x=0\) is considered (and rejected only because of the stated interval).
208IBMA3H-3-18r32-133.18 Intersections and angles of lines and planesPaper 1Hard11 marks
No calculator
A cable follows the line \(L:\ \mathbf{r}=\begin{pmatrix}1\\2\\0\end{pmatrix}+t\begin{pmatrix}2\\2\\1\end{pmatrix}\) (distances in metres). A monitoring unit is at \(P(6,4,4)\).
(a)
Let \(F\) be the foot of the perpendicular from \(P\) to \(L\). Find the coordinates of \(F\).
(4)
(b)
Hence find the shortest distance from \(P\) to \(L\).
(2)
(c)
There are two points \(R\) on \(L\) such that \(PR=5\) m. Find the coordinates of both points.
(5)
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Part (a)
M1 general point \(Q(1+2t,\ 2+2t,\ t)\) and \(\overrightarrow{PQ}=\begin{pmatrix}2t-5\\2t-2\\t-4\end{pmatrix}\)
Common error: When the angle is \(2x\), solve for \(2x\) over the doubled interval first, and only then halve. Halving the range or the answers too early loses solutions.
210IBMA3H-3-20r41-133.17 Equations of a planePaper 1Hard10 marks
No calculator
A lamp is at the point P(3, −1, 2) in front of a flat mirrored wall in a library. The wall lies in the plane Π with equation 2x − 3y + 6z = −14. Coordinates are in metres.
(a)
Find a vector equation of the line l through P that is perpendicular to Π.
(2)
(b)
Find the coordinates of the point F where l meets Π.
(4)
(c)
Find the distance PF.
(2)
(d)
The image of the lamp in the mirror is the point P′, where F is the midpoint of PP′. Find the coordinates of P′.
(2)
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Part (a)
M1 direction of l is the normal to the plane, (2, −3, 6)
✓ 1
A1 r = (3, −1, 2) + λ(2, −3, 6)
✓ 1
Part (b)
M1 substituting (3 + 2λ, −1 − 3λ, 2 + 6λ) into the equation of Π
M1 P′ is on l with λ = 2 × (−5/7) = −10/7 (or P′ = 2F − P)
✓ 1
A1 P′ = (1/7, 23/7, −46/7)
✓ 1
Common error: The normal vector of a plane gives the direction of the perpendicular line, but the parameter λ is not the distance unless the direction vector has length 1.
211IBMA3H-3-20r41-143.17 Equations of a planePaper 1Hard12 marks
No calculator
A laser beam from a nacelle on a wind turbine starts at A(1, −2, 3) and travels along the line L: r = (1, −2, 3) + λ(2, −1, k), where k is a constant and coordinates are in metres. The beam hits a flat panel that lies in the plane Π: x + 2y + 2z = 15. The angle φ between L and Π satisfies sin φ = 4/9.
(a)
Show that k2 = 4.
(5)
(b)
Given that k > 0, find the coordinates of the point where L meets Π.
(5)
(c)
Find the distance the beam travels from A to the panel.
(2)
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Part (a)
M1 normal to Π is (1, 2, 2), with magnitude 3
✓ 1
A1 d · n = 2 − 2 + 2k = 2k
✓ 1
M1 sin φ = |2k| / (3√(5 + k2)) = 4/9
✓ 1
M1 squaring and rearranging: 9k2 = 4(5 + k2)
✓ 1
A1 5k2 = 20, so k2 = 4
✓ 1
Part (b)
A1 k = 2
✓ 1
M1 substituting (1 + 2λ, −2 − λ, 3 + 2λ) into the equation of Π
✓ 1
A1 (1 + 2λ) + 2(−2 − λ) + 2(3 + 2λ) = 3 + 4λ = 15
✓ 1
A1 λ = 3
✓ 1
A1 point (7, −5, 9)
✓ 1
Part (c)
M1 |d| = √(4 + 1 + 4) = 3 (distance = λ × |d|)
✓ 1
A1 distance = 3 × 3 = 9 m
✓ 1
Exam tip: If an unknown appears in the direction vector, leave it as a letter until the angle condition has been used; squaring both sides removes the modulus and the square root.
212IBMA3H-3-19r42-143.6 Pythagorean and double angle identitiesPaper 1Hard12 marks
No calculator
Give all values in exact form.
(a)
Use \(\frac{\pi}{12}=\frac{\pi}{4}-\frac{\pi}{6}\) to show that \(\sin\frac{\pi}{12}=\frac{\sqrt6-\sqrt2}{4}\).
(3)
(b)
Find the exact value of \(\cos\frac{\pi}{12}\).
(2)
(c)
Hence show that \(\tan\frac{\pi}{12}=2-\sqrt3\).
(3)
(d)
Solve \(2\sin^2x+\cos x-1=0\) for \(0\le x\le2\pi\).
(4)
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Part (a)
M1 \(\frac{\pi}{12}=\frac{\pi}{4}-\frac{\pi}{6}\) used in \(\sin(A-B)\)
Common error: In a quadratic in \(\cos x\), divide-by-\(\cos x\) shortcuts lose the root \(\cos x=1\); factorise instead and list every solution in the interval.
213IBMA3H-3-19r43-123.11 Symmetry properties of trigonometric graphsPaper 1Hard12 marks
No calculator
Let \(f(x)=\dfrac{6}{5-\cos x+\sqrt3\sin x}\), for \(0\le x\le2\pi\).
(a)
Write \(\cos x-\sqrt3\sin x\) in the form \(R\cos(x+\alpha)\), where \(R>0\) and \(0<\alpha<\frac{\pi}{2}\).
(4)
(b)
Hence solve \(\cos x-\sqrt3\sin x=\sqrt2\), giving exact answers in the interval \(0\le x\le2\pi\).
(5)
(c)
Find the greatest value of \(f(x)\).
(3)
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Part (a)
M1 expand \(R\cos(x+\alpha)=R\cos\alpha\cos x-R\sin\alpha\sin x\) and compare coefficients
✓ 1
A1 \(R\cos\alpha=1\) and \(R\sin\alpha=\sqrt3\)
✓ 1
M1 \(R=\sqrt{1^2+(\sqrt3)^2}\) or \(\tan\alpha=\sqrt3\)
✓ 1
A1 \(R=2\) and \(\alpha=\frac{\pi}{3}\)
✓ 1
Part (b)
M1 write the equation as \(2\cos\left(x+\frac{\pi}{3}\right)=\sqrt2\)
M1 consider the range \(\frac{\pi}{3}\le x+\frac{\pi}{3}\le\frac{7\pi}{3}\)
✓ 1
A1 \(x+\frac{\pi}{3}=\frac{7\pi}{4}\) or \(\frac{9\pi}{4}\), so \(x=\frac{17\pi}{12}\)
✓ 1
A1 \(x=\frac{23\pi}{12}\)
✓ 1
Part (c)
M1 \(f(x)=\dfrac{6}{5-2\cos\left(x+\frac{\pi}{3}\right)}\) is greatest when the (positive) denominator is least
✓ 1
A1 least denominator is \(5-2=3\) since \(\cos\left(x+\frac{\pi}{3}\right)\le1\)
✓ 1
A1 greatest value of \(f(x)\) is \(2\)
✓ 1
Common error: When rewriting \(a\cos x+b\sin x\) in the form \(R\cos(x+\alpha)\), take care with the signs: compare the coefficients of \(\cos x\) and \(\sin x\) separately before choosing \(\alpha\).
214IBMA3H-3-19r43-143.11 Symmetry properties of trigonometric graphsPaper 1Hard13 marks
No calculator
The vertical displacement \(f(x)\), in centimetres, of a cyclist's helmet over a bumpy track section is modelled by \(f(x)=\cos2x+3\sin x\), for \(0\le x\le2\pi\).
(a)
Show that \(f(x)=-2\sin^2x+3\sin x+1\).
(2)
(b)
Find the greatest value of \(f(x)\).
(3)
(c)
Solve \(f(x)=2\) for \(0\le x\le2\pi\), giving exact answers.
(4)
(d)
Find all values of \(k\) for which the equation \(f(x)=k\) has exactly two solutions in \(0\le x\le2\pi\).
A1 \((2s-1)(s-1)=0\), so \(\sin x=\frac12\) or \(\sin x=1\)
✓ 1
A1 \(x=\frac{\pi}{6}\) and \(x=\frac{5\pi}{6}\)
✓ 1
A1 \(x=\frac{\pi}{2}\)
✓ 1
Part (d)
M1 consider the number of solutions of \(\sin x=s\) for each value of \(s\) that satisfies \(-2s^2+3s+1=k\) (two solutions for \(-1<s<1\), one for \(s=\pm1\))
✓ 1
A1 \(-4<k<2\) (one value of \(s\) in \((-1,\frac34)\), giving two values of \(x\))
✓ 1
A1 \(k=\frac{17}{8}\) (only \(s=\frac34\), giving two values of \(x\))
✓ 1
R1 \(k=2\) is excluded since it gives \(\sin x=\frac12\) and \(\sin x=1\), which is three solutions; \(k=-4\) gives only one solution
✓ 1
Exam tip: For 'number of solutions' questions, substitute \(s=\sin x\), solve for \(s\), then count how many angles each admissible value of \(s\) gives; the values \(s=\pm1\) give only one angle each.
215AAHL3-q00063.2 Right-angled and non-right-angled trianglesPaper 2Easy7 marks
The diagram shows a kite \(ABCD\) with \(AB=AD=5\) cm, \(CB=CD=9\) cm and \(B\hat{A}D=110^{\circ}\).
217IBMA2H-049-05-023.4 Radians, arc length and sector areaPaper 2Easy5 marks
Calculator
On a training track, a cyclist rides round a bend that is an arc of a circle of radius 25.0 m. The cyclist travels 61.0 m along the arc.
Diagram not to scale
(a)
Find the angle, in radians, subtended at the centre of the circle by this arc.
(2)
(b)
Convert this angle to degrees, giving your answer to one decimal place.
(1)
(c)
Find the area of the sector enclosed by the arc and the two radii to its ends.
(2)
Show mark scheme
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Notes
Part (a)
M1 \(61.0=25.0\,\theta\)
✓ 1
A1 \(\theta=2.44\) (radians)
✓ 1
Part (b)
A1 \(139.8°\) (accept \(140°\))
✓ 1
Part (c)
M1 \(\frac{1}{2}\times25.0^2\times2.44\) or equivalent, e.g. \(\frac{1}{2}\times25.0\times61.0\)
✓ 1
A1 \(762.5\) m2 (accept \(763\) m2)
✓ 1
Exam tip: The formulas s = rθ and A = ½r²θ only work with θ in radians. Use the radian value from part (a), not the degree value from part (b). Alternatively, A = ½rs avoids the angle altogether.
218IBMA2H-049-06-033.18 Intersections and angles of lines and planesPaper 2Easy5 marks
Calculator
In a laser light show, beam 1 passes through A(1, 0, 2) and B(2, 2, 4). Beam 2 passes through C(−1, 2, 0) and D(5, 2, 8). The two beams intersect at B.
(a)
Find \(\vec{AB}\) and \(\vec{CD}\).
(2)
(b)
Find the acute angle between the two beams.
(3)
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Part (a)
A1 \(\vec{AB}\) = i + 2j + 2k
✓ 1
A1 \(\vec{CD}\) = 6i + 8k
✓ 1
Part (b)
M1 use of scalar product with the direction vectors: \(\vec{AB}\)·\(\vec{CD}\) = 6 + 0 + 16 = 22
✓ 1
A1 |\(\vec{AB}\)| = 3, |\(\vec{CD}\)| = 10, so cos θ = 22/30 = 11/15
✓ 1
A1 θ = 42.8°
✓ 1
Exam tip: The angle between two lines depends only on their direction vectors. Any non-zero multiple of a direction vector gives the same angle, so (3, 0, 4) works just as well as (6, 0, 8).
219IBMA2H-049-00-023.2 Right-angled and non-right-angled trianglesPaper 2Easy4 marks
Calculator
The inside of a shipping container is a cuboid of length 12.0 m, width 2.35 m and height 2.39 m. A straight steel rod is placed inside the container, running from one bottom corner to the diagonally opposite top corner.
Rod in the container, from a bottom corner A to the opposite top corner G. Diagram not to scale
(a)
Find the length of the rod.
(2)
(b)
Find the angle between the rod and the floor of the container.
(2)
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Notes
Part (a)
M1 \(\sqrt{12.0^2+2.35^2+2.39^2}\)
✓ 1
A1 12.5 m (12.459…)
✓ 1
Part (b)
M1 \(\sin\theta=\frac{2.39}{12.459\ldots}\), or equivalently \(\tan\theta=\frac{2.39}{12.227\ldots}\) using the floor diagonal \(\sqrt{12.0^2+2.35^2}=12.227\ldots\)
✓ 1
A1 \(\theta=11.1^\circ\)
✓ 1
Common error: The angle with the floor lies between the rod and its projection onto the floor, which is the floor diagonal. The 12.0 m edge is not that projection. Use unrounded values from part 1, because the rounded 12.5 gives 11.0°.
220IBMA2H-049-01-023.3 Applications of trigonometry: elevation, depression and bearingsPaper 2Easy6 marks
Calculator
A survey drone takes off from a launch point L. It flies 1.8 km on a bearing of 065° to a point P, then 2.4 km on a bearing of 160° to a point Q. Assume the flight is horizontal.
Exam tip: Find the angle at P using the back-bearing (065° + 180° = 245°) before applying the cosine rule. Keep the unrounded LQ in your calculator for part (b).
221IBMA2H-049-01-123.7 Circular functions and their transformationsPaper 2Easy5 marks
Calculator
A security camera is fixed to a vertical wall at a height of 4.2 m above a horizontal car-park floor. It is aimed at a point F on the floor, with an angle of depression of 32°.
Side view. Diagram not to scale
(a)
Find the horizontal distance from the foot of the wall to F.
(2)
(b)
The camera is re-aimed at a point G on the floor. G lies on the straight line from the foot of the wall to F, and is 3 m closer to the wall than F. Find the new angle of depression.
(3)
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Part (a)
M1 \(\tan32° = \frac{4.2}{d}\)
✓ 1
A1 d = 6.72 m (6.7214…)
✓ 1
Part (b)
M1 new horizontal distance = 6.7214 − 3 = 3.7214 m
✓ 1
M1 \(\tan\alpha = \frac{4.2}{3.7214}\)
✓ 1
A1 α = 48.5°
✓ 1
Common error: The angle of depression is measured down from the horizontal, so it equals the angle of elevation from the floor point. Do not use the angle measured from the wall. Keep the unrounded distance from part (a) in your calculator for part (b).
222IBMA2H-049-07-023.2 Right-angled and non-right-angled trianglesPaper 2Easy5 marks
Calculator
A survey drone flies in a straight line from point A to point B, a distance of 350 m over flat farmland. A signal beacon is at point C. The drone's camera measures angle CAB = 48° and angle ABC = 67°.
Diagram not to scale
(a)
Find the distance AC.
(3)
(b)
Find the area of the triangular region ABC.
(2)
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Notes
Part (a)
A1 angle ACB = 180° − 48° − 67° = 65° ; M1 sine rule \(\frac{AC}{\sin 67^\circ}=\frac{350}{\sin 65^\circ}\) ; A1 AC = 355 m (355.483…)
✓ 3
Part (b)
M1 area = ½ × 350 × 355.48… × sin 48° ; A1 46 200 m² (46 230.7…)
✓ 2
Exam tip: Keep the unrounded value of AC in your calculator memory for part (b). Using 355 instead of 355.483… still gives 46 200 here, but premature rounding often shifts the third significant figure.
223IBMA2H-049-03-023.14 Vector equation of a linePaper 2Easy5 marks
Calculator
A delivery drone flies two straight legs. Relative to a fixed set of axes, with distances in metres, the displacement of the first leg is u = (120, 50, 30) and the displacement of the second leg is v = (−40, 90, 20).
(a)
Find u·v.
(2)
(b)
Hence find the angle between the directions of the two legs, giving your answer in degrees.
(3)
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Notes
Part (a)
M1 (120)(−40) + (50)(90) + (30)(20)
✓ 1
A1 300
✓ 1
Part (b)
M1 |u| = √17800 ≈ 133.4 and |v| = √10100 ≈ 100.5
✓ 1
M1 cos θ = 300 / (√17800 × √10100) ≈ 0.0224
✓ 1
A1 θ = 88.7°
✓ 1
Exam tip: A scalar product that is small compared with the product of the magnitudes signals an angle close to 90°. Use this as a quick sense-check of your final answer.
224IBMA2H-049-03-043.7 Circular functions and their transformationsPaper 2Easy5 marks
Calculator
During a medical test, the blood pressure of a patient, P mmHg, is modelled by P(t) = 100 + 20 sin(7t), where t is the time in seconds after the start of the recording and the argument of sine is in radians.
(a)
Write down the maximum blood pressure predicted by the model.
(1)
(b)
Find the first two values of t, t ≥ 0, for which the blood pressure is 112 mmHg.
(4)
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Part (a)
A1 120 mmHg
✓ 1
Part (b)
M1 sets 100 + 20 sin(7t) = 112
✓ 1
A1 sin(7t) = 0.6
✓ 1
A1 7t = 0.6435…, so t = 0.0919 s
✓ 1
A1 7t = π − 0.6435… = 2.498…, so t = 0.357 s
✓ 1
Exam tip: Make sure your calculator is in radian mode, and find the second solution from π − (first angle) before dividing by 7.
225IBMA2H-049-04-043.18 Intersections and angles of lines and planesPaper 2Easy4 marks
Calculator
A section of ski slope is modelled by the plane \(x + 2y - 2z = 4\). A straight ski-lift cable runs in the direction of the vector 2i + 3j + 6k.
(a)
Find the acute angle between the cable and the slope, in degrees.
(4)
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Notes
Part (a)
M1 normal n = i + 2j − 2k identified
✓ 1
M1 use of \(\sin\phi = \frac{|\boldsymbol{n}\cdot\boldsymbol{d}|}{|\boldsymbol{n}||\boldsymbol{d}|}\)
Common error: Using cosine with the normal gives the angle between the cable and the normal (79.0°). The angle with the plane is its complement, so use sine.
226IBMA3H-3-19r5-023.4 Radians, arc length and sector areaPaper 2Easy6 marks
Calculator
A rotating sprinkler waters a sector of a circular field. Its water reaches a distance of 18 m and it sweeps through an angle of 1.4 radians. Give answers to 3 significant figures.
Diagram not to scale
(a)
Find the length of the curved edge of the watered region.
(2)
(b)
Find the area of the watered region.
(2)
(c)
A fence is to be placed all the way round the watered region. Find the length of fence needed.
Exam tip: Check that the calculator is in radian mode, or avoid angle functions altogether when the angle is already in radians.
227IBMA3H-3-19r5-093.5 The unit circle, exact values and the ambiguous casePaper 2Easy5 marks
Calculator
A fan-shaped sound-absorbing panel in a music studio is a sector of a circle. Its area is 90 cm2 and the length of its curved edge is 15 cm.
Diagram not to scale
(a)
Find the radius of the sector.
(2)
(b)
Find the angle of the sector in radians.
(1)
(c)
Convert this angle to degrees, correct to 3 significant figures.
(2)
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Notes
Part (a)
M1 \(A=\tfrac12r^2\theta\) and \(l=r\theta\) combined to give \(A=\tfrac12rl\)
✓ 1
A1 \(r=\dfrac{2\times90}{15}=12\) cm
✓ 1
Part (b)
A1 \(\theta=\dfrac{15}{12}=1.25\)
✓ 1
Part (c)
M1 \(1.25\times\dfrac{180}{\pi}\)
✓ 1
A1 \(71.6^\circ\)
✓ 1
Exam tip: If area and arc length are both given, \(A=\tfrac12 r l\) reaches the radius in one step.
228IBMA3H-3-18r8-033.17 Equations of a planePaper 2Easy4 marks
Calculator
Two guy ropes from a mountain tent have direction vectors \(\mathbf{a}=\begin{pmatrix}2\\-1\\3\end{pmatrix}\) and \(\mathbf{b}=\begin{pmatrix}4\\5\\-2\end{pmatrix}\).
(a)
Find \(\mathbf{a}\cdot\mathbf{b}\).
(2)
(b)
Find the angle between the two ropes, to the nearest tenth of a degree.
(2)
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Notes
Part (a)
M1 \(2(4)+(-1)(5)+3(-2)\)
✓ 1
A1 \(-3\)
✓ 1
Part (b)
M1 use of \(\cos\theta=\frac{\mathbf a\cdot\mathbf b}{|\mathbf a||\mathbf b|}\) with \(|\mathbf a|=\sqrt{14}\), \(|\mathbf b|=\sqrt{45}\)
✓ 1
A1 \(\theta=96.9^\circ\)
✓ 1
Exam tip: A negative scalar product means the angle between the vectors is obtuse.
229IBMA3H-3-18r8-043.17 Equations of a planePaper 2Easy4 marks
Calculator
A survey drone's position (in km) at time \(t\) minutes is given by \(\mathbf r=\begin{pmatrix}2\\1\\5\end{pmatrix}+t\begin{pmatrix}3\\-2\\1\end{pmatrix}\). A telecoms relay mast is at the point \((14,-7,9)\).
(a)
Show that the drone passes through the mast position.
(2)
(b)
Find the speed of the drone, in km per minute, to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 equate components: \(2+3t=14\) gives \(t=4\)
✓ 1
A1 check \(1-2(4)=-7\) and \(5+4=9\), so the point lies on the line
✓ 1
Part (b)
M1 magnitude of direction vector \(\sqrt{3^2+(-2)^2+1^2}\)
✓ 1
A1 \(\sqrt{14}=3.74\) km per minute
✓ 1
Exam tip: The speed along \(\mathbf r=\mathbf a+t\mathbf d\) is \(|\mathbf d|\) when \(t\) is time; the position vector \(\mathbf a\) does not affect it.
230IBMA3H-3-19r10-023.1 Three-dimensional geometry: distance, solids and anglesPaper 2Easy6 marks
Calculator
A cylindrical coffee-roasting drum lies with its axis horizontal. It has diameter 0.60 m and length 0.90 m. Give answers to three significant figures.
The horizontal cylindrical drum, with a rod touching the rim of one end and the opposite rim of the other end. Diagram not to scale
(a)
A straight metal rod is placed inside the drum, touching the rim of one circular end and the opposite rim of the other end. Find the greatest possible length of such a rod.
(3)
(b)
The drum is filled to 40% of its volume with green beans. Find the volume of beans, in m\(^3\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 recognise the longest rod is a diagonal of the rectangle formed by the diameter and the length
✓ 1
M1 \(\sqrt{0.60^2+0.90^2}\)
✓ 1
A1 \(1.08\) m
✓ 1
Part (b)
M1 \(V=\pi\times0.30^2\times0.90\) (\(=0.2545\))
✓ 1
M1 \(0.40\times V\)
✓ 1
A1 \(0.102\) m\(^3\)
✓ 1
Exam tip: In a cylinder the longest straight line lies in the axial rectangle, so treat it as a right-angled triangle using the diameter, not the radius.
231IBMA3H-3-18r11-023.1 Three-dimensional geometry: distance, solids and anglesPaper 2Easy6 marks
Calculator
In an open-water swim, a swimmer starts at \(S\), swims 45 m to buoy \(P\), turns, and swims 62 m to buoy \(Q\). Angle \(SPQ=118^\circ\).
Diagram not to scale
(a)
Find the straight-line distance \(SQ\), to 3 significant figures.
M1 sine rule: \(\frac{\sin PSQ}{62}=\frac{\sin118^\circ}{SQ}\)
✓ 1
A1 \(\sin PSQ=0.594\) (approximately)
✓ 1
A1 \(PSQ=36.5^\circ\) (acute, since \(PQ<SQ\) and so it is not the largest angle)
✓ 1
Common error: When using the sine rule to find an angle, check whether the angle could be obtuse; here the angle faces the shorter side so it must be acute.
232IBMA3H-3-18r12-013.1 Three-dimensional geometry: distance, solids and anglesPaper 2Easy7 marks
Calculator
A straight section of railway track rises from point A to point B at a constant angle of 2.4° above the horizontal. The length of track AB is 850 m.
Diagram not to scale
(a)
Calculate the height of B above A. Give your answer to 3 significant figures.
(2)
(b)
Calculate the horizontal distance between A and B. Give your answer to 3 significant figures.
(2)
(c)
The gradient of the track is '1 in \(n\)', meaning a rise of 1 for every \(n\) units along the horizontal. Find \(n\) to 3 significant figures, and hence state whether the track is steeper than 1 in 30.
(3)
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Mark
Notes
Part (a)
M1 \(850\sin 2.4°\)
✓ 1
A1 35.6 m
✓ 1
Part (b)
M1 \(850\cos 2.4°\) (or Pythagoras with the height from (a))
✓ 1
A1 849 m
✓ 1
Part (c)
M1 \(\tan 2.4° = \frac{1}{n}\)
✓ 1
A1 \(n = 23.9\)
✓ 1
R1 \(23.9 < 30\), so the track is steeper than 1 in 30
✓ 1
Common error: Set your calculator to degrees when the angle is given in degrees; a result of about 850 for the height shows radian mode.
233IBMA3H-3-18r12-093.14 Vector equation of a linePaper 2Easy5 marks
Calculator
The velocity of a shuttlecock just before a racket hits it is \(\mathbf{u} = \begin{pmatrix}3\\-2\\6\end{pmatrix}\) m s-1 and just after is \(\mathbf{v} = \begin{pmatrix}4\\-4\\2\end{pmatrix}\) m s-1.
(a)
Find \(\mathbf{u}\cdot\mathbf{v}\).
(1)
(b)
Find \(|\mathbf{u}|\) and \(|\mathbf{v}|\).
(2)
(c)
Find the angle between \(\mathbf{u}\) and \(\mathbf{v}\), to the nearest 0.1°.
(2)
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Notes
Part (a)
A1 \(12 + 8 + 12 = 32\)
✓ 1
Part (b)
A1 \(|\mathbf{u}| = \sqrt{9+4+36} = 7\)
✓ 1
A1 \(|\mathbf{v}| = \sqrt{16+16+4} = 6\)
✓ 1
Part (c)
M1 \(\cos\theta = \frac{32}{7\times6}\) (ECF from (a) and (b))
✓ 1
A1 40.4°
✓ 1
Exam tip: Compute the scalar product and both magnitudes separately, then substitute into \(\cos\theta=\frac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{u}||\mathbf{v}|}\).
234IBMA3H-3-19r12-013.7 Circular functions and their transformationsPaper 2Easy6 marks
Calculator
The number of new cases of a seasonal infection reported per week in a town is modelled by \(N(t)=150+70\cos\left(\frac{\pi t}{26}\right)\), where \(t\) is the number of weeks after 1 January and \(0\le t\le 52\). The argument of the cosine is in radians.
(a)
State the maximum and minimum values of \(N\).
(2)
(b)
Find the first time at which \(N=200\). Give your answer in weeks to 3 significant figures.
(2)
(c)
Find the total number of weeks in the interval \(0\le t\le 52\) for which \(N>200\). Use the unrounded answer to (b).
A1 \(t=\frac{26}{\pi}\arccos\frac{5}{7}=6.4155\), so \(t=6.42\)
✓ 1
Part (c)
M1 recognises two equal intervals at the start and end of the year, by symmetry of the cosine curve, total \(2t_1\)
✓ 1
A1 \(2\times6.4155=12.83\), so 12.8 weeks
✓ 1
Exam tip: Keep your calculator in radian mode whenever the argument of a trigonometric function contains \(\pi\) without a degree sign.
235IBMA3H-3-19r12-033.15 Intersecting, parallel and skew linesPaper 2Easy7 marks
Calculator
Two straight support cables for a wind turbine are modelled by the lines \(L_1:\ \mathbf r=\begin{pmatrix}2\\-1\\3\end{pmatrix}+\lambda\begin{pmatrix}1\\2\\-1\end{pmatrix}\) and \(L_2:\ \mathbf r=\begin{pmatrix}7\\4\\0\end{pmatrix}+\mu\begin{pmatrix}-3\\-1\\1\end{pmatrix}\), where coordinates are in metres.
(a)
Show that the two cables intersect, and find the coordinates of the point of intersection.
(4)
(b)
Find the acute angle between the two cables, in degrees, to 3 significant figures.
A1 solving two of them gives \(\lambda=2\), \(\mu=1\)
✓ 1
R1 checks the remaining equation: \(2+2=4=7-3\) is satisfied, so the lines intersect
✓ 1
A1 point \((4,3,1)\)
✓ 1
Part (b)
M1 scalar product of direction vectors \((1,2,-1)\cdot(-3,-1,1)=-6\)
✓ 1
M1 \(\cos\theta=\frac{6}{\sqrt6\sqrt{11}}=\frac{6}{\sqrt{66}}=0.7385\) using the modulus to give an acute angle
✓ 1
A1 \(\theta=42.4^\circ\), so 42.4°
✓ 1
Exam tip: When the scalar product of the direction vectors is negative, take its modulus if the question asks for the acute angle between lines.
236IBMA3H-3-19r16-013.14 Vector equation of a linePaper 2Easy4 marks
Calculator
Two space probes are tracked relative to a ground station. Relative to the station, probe A is at the point with position vector \(\mathbf{a}=(2,-1,4)\) km and probe B is at \(\mathbf{b}=(8,3,-8)\) km.
(a)
Find the vector \(\overrightarrow{AB}\).
(2)
(b)
Find the distance between the two probes.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to subtract position vectors, \(\mathbf{b}-\mathbf{a}\)
✓ 1
A1 \(\overrightarrow{AB}=(6,4,-12)\) km
✓ 1
Part (b)
M1 \(\sqrt{6^2+4^2+(-12)^2}\)
✓ 1
A1 14 km
✓ 1
Exam tip: Distance is the magnitude of the displacement vector; do not add the components.
237IBMA3H-3-19r16-023.13 Scalar productPaper 2Easy5 marks
Calculator
The direction vectors of two support cables meeting at a joint on a footbridge are \(\mathbf{u}=2\mathbf{i}+\mathbf{j}-2\mathbf{k}\) and \(\mathbf{v}=\mathbf{i}-4\mathbf{j}+8\mathbf{k}\).
(a)
Find \(\mathbf{u}\cdot\mathbf{v}\).
(2)
(b)
Find the angle between the two cables, giving your answer in degrees to one decimal place.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(2\times1+1\times(-4)+(-2)\times8\)
✓ 1
A1 \(-18\)
✓ 1
Part (b)
M1 use of \(\mathbf{u}\cdot\mathbf{v}=|\mathbf{u}||\mathbf{v}|\cos\theta\)
✓ 1
A1 \(|\mathbf{u}|=3\) and \(|\mathbf{v}|=9\) (so \(\cos\theta=-\frac{2}{3}\))
✓ 1
A1 \(\theta=131.8^\circ\)
✓ 1
Common error: A negative scalar product means the angle between the vectors is obtuse; do not take the absolute value unless the question asks for the acute angle.
238IBMA3H-3-19r16-063.2 Right-angled and non-right-angled trianglesPaper 2Easy6 marks
Calculator
Wind causes a footbridge deck to sway. At time \(t\) seconds the horizontal displacement of the deck centre from its rest position is \(d=4.5\cos(0.8\pi t)\) cm.
(a)
Write down the amplitude of the sway and find its period.
(2)
(b)
Find the displacement when \(t=0.5\). Give your answer to three significant figures.
(2)
(c)
Find the first time after \(t=0\) at which \(d=-2.0\) cm. Give your answer to three significant figures.
(2)
Show mark scheme
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Mark
Notes
Part (a)
A1 amplitude 4.5 cm
✓ 1
A1 period \(\dfrac{2\pi}{0.8\pi}=2.5\) s
✓ 1
Part (b)
M1 \(4.5\cos(0.4\pi)\) (calculator in radian mode)
✓ 1
A1 1.39 cm
✓ 1
Part (c)
M1 \(\cos(0.8\pi t)=-\frac{2}{4.5}\) and \(0.8\pi t=\cos^{-1}\left(-\frac{2}{4.5}\right)\)
✓ 1
A1 \(t=0.808\) s
✓ 1
Common error: Set the calculator to radians when the argument of a trigonometric function contains \(\pi\) or no degree sign.
239IBMA3H-3-19r16-113.12 Vectors: concepts and algebraPaper 2Easy4 marks
Calculator
The two tower tops of a suspension bridge are at the points \(A(1,5,-2)\) and \(B(7,-1,4)\), with coordinates in metres.
(a)
Find the coordinates of the midpoint \(M\) of \(AB\).
(2)
(b)
The point \(P\) lies on \(AB\) such that \(AP:PB=1:2\). Find the coordinates of \(P\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 average of the position vectors, \(\frac12(\mathbf{a}+\mathbf{b})\)
✓ 1
A1 \(M(4,2,1)\)
✓ 1
Part (b)
M1 \(\overrightarrow{OP}=\mathbf{a}+\frac13\overrightarrow{AB}\) with \(\overrightarrow{AB}=(6,-6,6)\)
✓ 1
A1 \(P(3,3,0)\)
✓ 1
Exam tip: For a ratio \(m:n\) along a segment, move the fraction \(\frac{m}{m+n}\) of the way from the first point.
240IBMA3H-3-19r16-123.4 Radians, arc length and sector areaPaper 2Easy6 marks
Calculator
A sector of a circular roasting tray has radius 14 cm and an arc of length 29.4 cm.
Diagram not to scale
(a)
Find the angle of the sector in radians.
(2)
(b)
Find the area of the sector.
(2)
(c)
Convert the angle to degrees, giving your answer to one decimal place.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(s=r\theta\) so \(\theta=\dfrac{29.4}{14}\)
✓ 1
A1 2.1
✓ 1
Part (b)
M1 \(\frac12 r^2\theta\)
✓ 1
A1 \(205.8\) cm2
✓ 1
Part (c)
M1 \(2.1\times\dfrac{180}{\pi}\)
✓ 1
A1 \(120.3^\circ\)
✓ 1
Common error: The formulas \(s=r\theta\) and \(A=\frac12 r^2\theta\) are only valid when \(\theta\) is in radians.
241IBMA3H-3-18r17-013.3 Applications of trigonometry: elevation, depression and bearingsPaper 2Easy5 marks
Calculator
A glider is flying at a height of 640 m above level ground. The pilot sees the end of a runway, T, on the ground ahead at an angle of depression of 7.5°.
Diagram not to scale
(a)
Write down the angle of elevation of the glider from T.
(1)
(b)
Find the horizontal distance from the glider to T. Give your answer to the nearest 10 m.
(2)
(c)
Find the straight-line distance from the glider to T. Give your answer to the nearest 10 m.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 7.5° (alternate angles)
✓ 1
Part (b)
M1 tan 7.5° = 640 / d, so d = 640 / tan 7.5°
✓ 1
A1 4860 m
✓ 1
Part (c)
M1 sin 7.5° = 640 / s (or uses Pythagoras with the unrounded horizontal distance)
✓ 1
A1 4900 m
✓ 1
Common error: The angle of depression is measured from the horizontal at the observer; it equals the angle of elevation from the object, not the angle to the vertical.
242IBMA3H-3-18r17-023.3 Applications of trigonometry: elevation, depression and bearingsPaper 2Easy6 marks
Calculator
On a farm, a barn B is 450 m from the farmhouse F on a bearing of 062°. A silo S is 300 m from F on a bearing of 140°.
Diagram not to scale
(a)
Write down the size of angle BFS.
(1)
(b)
Find the distance BS. Give your answer to 3 significant figures.
(3)
(c)
Find the area of triangle FBS in hectares (1 hectare = 10 000 m2), to 3 significant figures.
Exam tip: Use both bearings measured from north at the same point to find the angle between the two directions: subtract 062° from 140°.
243IBMA3H-3-18r17-033.3 Applications of trigonometry: elevation, depression and bearingsPaper 2Easy5 marks
Calculator
A crop-spraying aircraft leaves an airstrip A and flies 12 km on a bearing of 040° to a point B. It then turns and flies 9 km on a bearing of 130° to a point C.
Diagram not to scale
(a)
Find the distance AC.
(2)
(b)
Find the bearing on which the aircraft must fly from C to return directly to A. Give your answer to the nearest 0.1°.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 recognises the turn is 130° − 40° = 90° and applies Pythagoras: AC2 = 122 + 92
✓ 1
A1 AC = 15 km
✓ 1
Part (b)
M1 angle BAC = arctan(9/12) = 36.87…°
✓ 1
M1 bearing of C from A = 040° + 36.87…° = 76.87…°, so the return bearing is 180° added to this
✓ 1
A1 256.9° (accept 257°)
✓ 1
Exam tip: For a return bearing, add or subtract 180° to the outward bearing, and keep the answer between 000° and 360°.
244IBMA3H-3-19r18-023.10 Compound and double angle identitiesPaper 2Easy6 marks
Calculator
A shipping container is a cuboid of internal length 12.0 m, width 2.4 m and height 2.5 m. Give lengths and angles to 3 significant figures unless stated otherwise.
The container with its longest rod, which makes an angle θ with the floor. Diagram not to scale
(a)
Calculate the length of the longest straight rod that can be placed inside the container.
(2)
(b)
Calculate the angle between this longest rod and the floor of the container.
(2)
(c)
The container is filled with identical cuboid crates measuring 0.8 m by 0.6 m by 0.5 m, all packed with edges parallel to the container's edges. Find the greatest number of crates that fit.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of 3D Pythagoras, \(\sqrt{12.0^2+2.4^2+2.5^2}\)
✓ 1
A1 12.5 m (12.490…)
✓ 1
Part (b)
M1 floor diagonal \(\sqrt{12.0^2+2.4^2}=12.24\) and \(\tan\theta=\frac{2.5}{12.24}\)
✓ 1
A1 \(\theta=11.5^\circ\) (11.546…)
✓ 1
Part (c)
M1 \(\frac{12.0}{0.8}=15\), \(\frac{2.4}{0.6}=4\), \(\frac{2.5}{0.5}=5\) along the three edges
✓ 1
A1 \(15\times4\times5=300\) crates
✓ 1
Exam tip: For a space diagonal, form a right-angled triangle using the floor diagonal and the vertical edge; keep unrounded values until the final line.
245IBMA3H-3-18r22-023.18 Intersections and angles of lines and planesPaper 2Easy6 marks
Calculator
A straight bridge cable lies along the line \(L\) with equation \(\mathbf r=\begin{pmatrix}1\\0\\4\end{pmatrix}+\lambda\begin{pmatrix}2\\3\\6\end{pmatrix}\). The flat bridge deck lies in the plane \(\Pi\) with equation \(2x-y+2z=23\). Distances are in metres.
(a)
Find the coordinates of the point where the cable meets the deck.
(3)
(b)
Find the acute angle between the cable and the deck. Give your answer in degrees to 3 significant figures.
(3)
Show mark scheme
Marking point
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Notes
Part (a)
M1 substitute \(x=1+2\lambda\), \(y=3\lambda\), \(z=4+6\lambda\) into the plane equation: \(10+13\lambda=23\)
Common error: The angle between a line and a plane uses sine with the normal vector; using cosine gives the angle between the line and the normal instead.
246IBMA3H-3-18r22-093.4 Radians, arc length and sector areaPaper 2Easy4 marks
Calculator
A sector-shaped scoop for green coffee beans has radius 8 cm and arc length 14 cm.
Exam tip: Arc length and sector area formulae only work with the angle in radians.
247IBMA3H-3-18r23-013.10 Compound and double angle identitiesPaper 2Easy7 marks
Calculator
A library reading room is a cuboid \(ABCDEFGH\) with floor $ABCD$, where \(AB=14\) m, \(BC=9\) m and the height \(CG=4.5\) m. A sensor cable runs in a straight line from corner \(A\) on the floor to corner \(G\) on the ceiling. Give lengths to 3 significant figures and angles to the nearest 0.1°, using unrounded values from earlier parts in later calculations.
Diagram not to scale
(a)
Find the length of the cable \(AG\).
(3)
(b)
Find the angle between \(AG\) and the floor.
(2)
(c)
Find the angle \(GAB\) between the cable and the edge \(AB\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(AC^2=14^2+9^2\)
✓ 1
M1 \(AG^2=AC^2+4.5^2=297.25\)
✓ 1
A1 \(AG=17.2\) m
✓ 1
Part (b)
M1 \(\tan\theta=\frac{4.5}{AC}\) with \(AC=\sqrt{277}\)
✓ 1
A1 \(15.1^\circ\)
✓ 1
Part (c)
M1 triangle \(ABG\) is right-angled at \(B\), so \(\cos\theta=\frac{14}{AG}\) (or use the cosine rule)
✓ 1
A1 \(35.7^\circ\)
✓ 1
Exam tip: Split a 3D problem into right-angled triangles and name each one; the angle between a line and a plane uses the line's projection onto that plane.
248IBMA3H-3-18r23-063.4 Radians, arc length and sector areaPaper 2Easy6 marks
Calculator
A swimming pool is 25 m long and 10 m wide. Its floor slopes uniformly, so that the water depth increases from 1 m at the shallow end to 3 m at the deep end. Give lengths to 3 significant figures and angles to 3 significant figures.
Side view of the pool, not to scale
(a)
Find the length of the sloping floor, measured along the slope.
(2)
(b)
Find the angle the floor makes with the horizontal.
(2)
(c)
Find the volume of water when the pool is full to the top.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\sqrt{25^2+(3-1)^2}\)
✓ 1
A1 \(25.1\) m
✓ 1
Part (b)
M1 \(\tan\theta=\frac{2}{25}\)
✓ 1
A1 \(4.57^\circ\)
✓ 1
Part (c)
M1 cross-section area (trapezium) \(\frac{1+3}{2}\times25=50\) m2, multiplied by width 10
✓ 1
A1 \(500\) m3
✓ 1
Exam tip: A pool with a sloping floor is a prism with a trapezium cross-section: find the cross-section area first, then multiply by the width.
249IBMA3H-3-19r25-033.14 Vector equation of a linePaper 2Easy6 marks
Calculator
During a practice lap, a Formula 1 car travels along a straight section of track. Relative to a marshal post at the origin \(O\), its position at time \(t\) seconds is given by \(\mathbf{r}=\begin{pmatrix}120\\-40\end{pmatrix}+t\begin{pmatrix}55\\30\end{pmatrix}\), where distances are in metres and \(t\ge0\).
Not to scale. Distances in metres
(a)
Find the position vector of the car when \(t=4\).
(2)
(b)
Find the speed of the car, in m s\(^{-1}\), correct to 3 significant figures.
(2)
(c)
The car passes a timing line at which the \(x\)-coordinate is 450. Find the value of \(t\) when the car reaches this line.
Exam tip: The speed is the magnitude of the direction vector, not of the position vector; the starting point plays no part.
250IBMA3H-3-19r25-043.7 Circular functions and their transformationsPaper 2Easy7 marks
Calculator
A hospital monitor models a patient's arterial pressure \(P\) millimetres of mercury (mmHg) at time \(t\) seconds by \(P(t)=100+20\sin(2.5\pi t)\), where the angle is in radians.
(a)
Write down the maximum and minimum values of \(P\).
(2)
(b)
Find the period of \(P\), in seconds.
(2)
(c)
Find the first time, \(t>0\), at which \(P=110\). Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 maximum \(=120\) mmHg
✓ 1
A1 minimum \(=80\) mmHg
✓ 1
Part (b)
M1 period \(=\frac{2\pi}{2.5\pi}\)
✓ 1
A1 \(0.8\) s
✓ 1
Part (c)
M1 \(20\sin(2.5\pi t)=10\) so \(\sin(2.5\pi t)=0.5\)
✓ 1
A1 first solution \(2.5\pi t=\frac{\pi}{6}\)
✓ 1
A1 \(t=\frac{1}{15}=0.0667\) s
✓ 1
Common error: Leaving the calculator in degree mode: the angle 2.5πt is in radians, so degree mode gives a completely different answer.
251IBMA3H-3-18r32-013.7 Circular functions and their transformationsPaper 2Easy6 marks
Calculator
The number of new cases per day of a seasonal infection in a town is modelled by \(C(t)=80+30\cos\!\left(\frac{\pi t}{6}\right)\), where \(t\) is the time in months since 1 January and \(0\le t\le12\). Angles are in radians.
(a)
Write down the maximum and the minimum value of \(C\).
(2)
(b)
Find the period of the model.
(2)
(c)
Find the number of new cases per day when \(t=1\), correct to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 maximum 110
✓ 1
A1 minimum 50
✓ 1
Part (b)
M1 \(\frac{2\pi}{\pi/6}\)
✓ 1
A1 12 months
✓ 1
Part (c)
M1 \(80+30\cos\!\left(\frac{\pi}{6}\right)\)
✓ 1
A1 106
✓ 1
Common error: Make sure the calculator is in radian mode when the argument of the cosine contains \(\pi\); degree mode gives a very different value.
252IBMA3H-3-18r32-033.14 Vector equation of a linePaper 2Easy6 marks
Calculator
A drone inspecting a wind turbine moves in a straight line. Its position, in metres relative to a base station, at time \(t\) seconds is given by \(\mathbf{r}=\begin{pmatrix}50\\120\\30\end{pmatrix}+t\begin{pmatrix}4\\-4\\2\end{pmatrix}\).
(a)
Find the position of the drone when \(t=10\).
(2)
(b)
Determine whether the drone passes through the point \((130,\ 40,\ 70)\).
(2)
(c)
Find the speed of the drone in m s−1.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 substitute \(t=10\)
✓ 1
A1 \((90,\ 80,\ 50)\)
✓ 1
Part (b)
M1 solve one component for \(t\), e.g. \(50+4t=130\) so \(t=20\)
✓ 1
A1 other components give \(t=20\) also, so yes
✓ 1
Part (c)
M1 \(\sqrt{4^{2}+(-4)^{2}+2^{2}}\)
✓ 1
A1 6 m s−1
✓ 1
Exam tip: The speed along \(\mathbf{r}=\mathbf{a}+t\mathbf{d}\) is \(|\mathbf{d}|\) when \(t\) is time; the direction vector is not itself the speed.
253IBMA3H-3-19r35-013.3 Applications of trigonometry: elevation, depression and bearingsPaper 2Easy7 marks
Calculator
A wind farm has a control hut H on level ground. Turbine T1 is 4.5 km from H on a bearing of 070°. Turbine T2 is 3.8 km from H on a bearing of 160°.
Diagram not to scale
(a)
Write down the size of angle T1HT2.
(1)
(b)
Find the distance T1T2, in km, to 3 significant figures.
(2)
(c)
Find the bearing of T2 from T1, to the nearest 0.1°.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 160° − 70° = 90°
✓ 1
Part (b)
M1 use of Pythagoras (or the cosine rule with 90°): T1T22 = 4.52 + 3.82
✓ 1
A1 T1T2 = 5.89 km
✓ 1
Part (c)
M1 bearing of H from T1 is 070° + 180° = 250°
✓ 1
M1 attempt at angle HT1T2 using trigonometry, e.g. tan−1(3.8 ÷ 4.5) (accept any valid method, e.g. components)
✓ 1
A1 angle HT1T2 = 40.2°
✓ 1
A1 bearing = 250° − 40.2° = 209.8°
✓ 1
Common error: A bearing is measured clockwise from north at the point you are travelling from; draw a north line at T₁ before adding or subtracting angles.
254IBMA3H-3-19r35-063.3 Applications of trigonometry: elevation, depression and bearingsPaper 2Easy5 marks
Calculator
On a wind farm, turbine B is due east of turbine A. Turbine C is 2.6 km from A on a bearing of 052°, and C is due north of B.
Diagram not to scale
(a)
Find the distance AB, in km, to 3 significant figures.
(2)
(b)
Find the distance BC, in km, to 3 significant figures.
(2)
(c)
Write down the bearing of A from C.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 AB = 2.6 sin 52°
✓ 1
A1 AB = 2.05 km
✓ 1
Part (b)
M1 BC = 2.6 cos 52° (accept Pythagoras or tan)
✓ 1
A1 BC = 1.60 km
✓ 1
Part (c)
A1 052° + 180° = 232°
✓ 1
Exam tip: Sketch the right-angled triangle first: the bearing from north gives the angle at A between AC and the north line, so the opposite side is the east distance.
255IBMA3H-3-20r40-013.7 Circular functions and their transformationsPaper 2Easy5 marks
Calculator
The number of bees at the entrance of a hive is modelled by \(N(t)=60+35\cos\left(\frac{\pi}{12}(t-13)\right)\), where \(t\) is the time in hours after midnight and \(0\le t\le24\). The argument of the cosine is in radians.
(a)
Write down the maximum and minimum numbers of bees at the entrance according to the model.
(2)
(b)
Write down the period of the function \(N\).
(1)
(c)
Find the number of bees at the entrance at 08:00. Give your answer to the nearest whole number.
Common error: Make sure the calculator is in radian mode when the argument contains \(\pi\). Degree mode gives a value close to 95 here and the error is easy to miss.
256IBMA3H-3-20r41-013.4 Radians, arc length and sector areaPaper 2Easy6 marks
Calculator
A circular pool has radius 8.4 m. A sector of the pool, with centre angle 0.75 radians at the centre of the pool, is roped off for a swimming class.
Diagram not to scale
(a)
Find the length of the curved rope that forms the arc of the sector.
(2)
(b)
Find the area of the sector.
(2)
(c)
Two straight ropes run along the radii bounding the sector. Find the total length of rope needed for the whole boundary of the sector.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of s = rθ
✓ 1
A1 arc length = 6.3 m
✓ 1
Part (b)
M1 use of A = ½r2θ
✓ 1
A1 area = 26.46 m2
✓ 1
Part (c)
M1 2r + arc length = 2 × 8.4 + 6.3
✓ 1
A1 total length = 23.1 m
✓ 1
Common error: The formulas s = rθ and A = ½r²θ only work when θ is in radians; check the calculator is in radian mode.
257IBMA3H-3-20r41-033.7 Circular functions and their transformationsPaper 2Easy6 marks
Calculator
A wind turbine blade has length 45 m and turns at a constant rate of 15 revolutions per minute. Its tip moves along a circular path.
(a)
Find the angle, in radians, that the blade turns through in 12 seconds. Give your answer in terms of π.
(2)
(b)
Find the distance travelled by the tip of the blade in 12 seconds. Give your answer to 3 significant figures.
(2)
(c)
Find the speed of the tip of the blade, in m s−1, giving your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 15 revolutions per minute = 0.25 revolutions per second, so 12 s is 3 revolutions
✓ 1
A1 angle = 3 × 2π = 6π
✓ 1
Part (b)
M1 use of s = rθ = 45 × 6π (ECF from (a))
✓ 1
A1 distance = 848 m
✓ 1
Part (c)
M1 distance ÷ time (use of the unrounded distance from (b))
✓ 1
A1 speed = 70.7 m s−1
✓ 1
Exam tip: Convert revolutions to radians first (1 revolution = 2π), then use the arc length formula; keep π unrounded until the final answer.
258IBMA3H-3-18r42-063.17 Equations of a planePaper 2Easy6 marks
Calculator
In a retail warehouse, a robotic picker moves in a straight line from shelf \(A(2,-1,4)\) to shelf \(B(8,2,10)\), where coordinates are in metres and the third coordinate is height.
(a)
Find the distance \(AB\).
(2)
(b)
Find a vector equation of the line \(AB\).
(2)
(c)
The picker continues along the same line beyond \(B\) until it is at a height of 13 m. Find its coordinates at this point.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\overrightarrow{AB}=\begin{pmatrix}6\\3\\6\end{pmatrix}\) and \(|AB|=\sqrt{36+9+36}\)
✓ 1
A1 \(9\) m
✓ 1
Part (b)
M1 uses a point on the line and a direction parallel to \(\overrightarrow{AB}\)
Find the exact shortest distance from C to the line AB.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\overrightarrow{AB}=\begin{pmatrix}2\\-1\\2\end{pmatrix}\), \(\overrightarrow{AC}=\begin{pmatrix}1\\2\\3\end{pmatrix}\) and an attempt at the vector product
✓ 1
A1 \(\begin{pmatrix}-7\\-4\\5\end{pmatrix}\)
✓ 1
Part (b)
M1 area \(=\frac12\left|\overrightarrow{AB}\times\overrightarrow{AC}\right|=\frac12\sqrt{49+16+25}\)
✓ 1
A1 \(\frac{3\sqrt{10}}{2}\) m2 (ECF from (a))
✓ 1
Part (c)
M1 \(d=\frac{\left|\overrightarrow{AB}\times\overrightarrow{AC}\right|}{\left|\overrightarrow{AB}\right|}=\frac{3\sqrt{10}}{3}\), or area \(=\frac12|\text{AB}|d\) with \(|\text{AB}|=3\)
✓ 1
A1 \(\sqrt{10}\) m
✓ 1
Exam tip: \(|\mathbf{a}\times\mathbf{b}|\) is the area of the parallelogram; halve it for the triangle, then reuse it to get a height.
261IBMA3H-3-19r43-033.12 Vectors: concepts and algebraPaper 2Easy6 marks
Calculator
A maintenance drone flies along a straight line inside a data centre hall. Its position, in metres relative to one corner of the hall, at time \(t\) seconds (\(t\ge0\)) is given by \(\mathbf{r}=\begin{pmatrix}2\\-1\\4\end{pmatrix}+t\begin{pmatrix}2\\3\\6\end{pmatrix}\).
(a)
Find the position of the drone when \(t=3\).
(2)
(b)
Find the speed of the drone.
(2)
(c)
Show that the drone does not pass through the point \(P(6,\,5,\,15)\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 substitute \(t=3\) into the vector equation
✓ 1
A1 \((8,\,8,\,22)\)
✓ 1
Part (b)
M1 attempt to find the magnitude of the direction vector, \(\sqrt{2^2+3^2+6^2}\)
✓ 1
A1 \(7\) m s-1
✓ 1
Part (c)
M1 equate \(x\) and \(y\) components: \(2+2t=6\) and \(-1+3t=5\) both give \(t=2\)
✓ 1
A1 at \(t=2\) the \(z\)-component is \(4+12=16\neq15\), so \(P\) is not on the path
✓ 1
Common error: The speed is the magnitude of the direction vector only when the parameter is time in seconds; do not include the position vector in the magnitude.
262IBMA3H-3-19r43-043.14 Vector equation of a linePaper 2Easy7 marks
Calculator
Two survey lines in an archaeological trench start at a datum point and have direction vectors \(\mathbf{a}=2\mathbf{i}-\mathbf{j}+3\mathbf{k}\) and \(\mathbf{b}=\mathbf{i}+4\mathbf{j}-2\mathbf{k}\).
(a)
Find \(\mathbf{a}\cdot\mathbf{b}\).
(2)
(b)
Find the angle between \(\mathbf{a}\) and \(\mathbf{b}\), giving your answer in degrees to the nearest degree.
(3)
(c)
The vector \(\mathbf{c}=p\mathbf{i}+2\mathbf{j}+\mathbf{k}\) is perpendicular to \(\mathbf{a}\), where \(p\) is a constant. Find the value of \(p\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to multiply corresponding components and add: \(2\times1+(-1)\times4+3\times(-2)\)
✓ 1
A1 \(\mathbf{a}\cdot\mathbf{b}=-8\)
✓ 1
Part (b)
M1 use of \(\cos\theta=\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}\)
✓ 1
A1 \(|\mathbf{a}|=\sqrt{14}\) and \(|\mathbf{b}|=\sqrt{21}\)
✓ 1
A1 \(\theta=117.8\ldots^\circ=118^\circ\)
✓ 1
Part (c)
M1 set \(\mathbf{a}\cdot\mathbf{c}=0\): \(2p-2+3=0\)
✓ 1
A1 \(p=-\frac12\)
✓ 1
Exam tip: Two non-zero vectors are perpendicular exactly when their scalar product is zero; set up the equation before expanding.
263AAHL3-q00093.3 Applications of trigonometry: elevation, depression and bearingsPaper 2Medium10 marks
Points \(A\) and \(B\) lie on horizontal ground, \(250\) m apart. A vertical tower \([FT]\) stands with its base \(F\) on the ground, where \(F\hat{A}B=30^{\circ}\) and \(F\hat{B}A=50^{\circ}\). The angle of elevation of the top \(T\) of the tower from \(A\) is \(25^{\circ}\).
Diagram not to scale
(a)
Show that \(AF=194\) m, correct to three significant figures.
(2)
(b)
Find the height of the tower.
(2)
(c)
Find the angle of elevation of \(T\) from \(B\).
(3)
(d)
Find \(A\hat{T}B\).
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(A\hat{F}B=100^{\circ}\) and \(\dfrac{AF}{\sin50^{\circ}}=\dfrac{250}{\sin100^{\circ}}\)
264AAHL3-q00103.3 Applications of trigonometry: elevation, depression and bearingsPaper 2Medium11 marks
At noon, two ships leave port \(P\) at the same time. Ship \(A\) sails on a bearing of \(040^{\circ}\) at a constant speed of \(18\) km h−1. Ship \(B\) sails on a bearing of \(115^{\circ}\) at a constant speed of \(24\) km h−1.
Plan view, not to scale
(a)
Write down the size of the angle \(A\hat{P}B\).
(1)
(b)
Find the distance between the ships at 14:00.
(3)
(c)
Find the bearing of ship \(B\) from ship \(A\) at 14:00.
(4)
(d)
Explain why the distance between the ships is directly proportional to the time since noon. Hence find, to the nearest minute, the time at which the ships are \(50\) km apart.
63.046…° The obtuse value is impossible since \(P\hat{A}B+P\hat{B}A=105^{\circ}\).
bearing of \(P\) from \(A\) is \(220^{\circ}\), and \(B\) is \(63.0^{\circ}\) anticlockwise from it
(M1)
\(157^{\circ}\)
A1
156.953…°
(d)
after \(t\) hours \(PA=18t\), \(PB=24t\) and \(A\hat{P}B=75^{\circ}\) always, so all the triangles are similar (enlargements) and \(AB=t\sqrt{18^{2}+24^{2}-2(18)(24)\cos75^{\circ}}=26.0t\)
265AAHL3-q00143.4 Radians, arc length and sector areaPaper 2Medium10 marks
A tank is a closed cylinder of radius \(1.2\) m and length \(3.5\) m, lying with its axis horizontal. The diagram shows its circular cross-section, with centre \(O\). The tank contains water to a depth of \(0.4\) m. The surface of the water meets the cross-section at \(P\) and \(Q\), and \(P\hat{O}Q=\theta\) radians.
Diagram not to scale
(a)
Show that \(\theta=1.68\), correct to three significant figures.
(2)
(b)
Find the area of the shaded segment.
(3)
(c)
Find the volume of water in the tank.
(1)
(d)
More water is added until the tank is exactly one quarter full. Find the new depth of the water.
\(P\hat{Q}R=72.2^{\circ}\) or \(23.8^{\circ}\), and area \(=\dfrac{1}{2}(15)(11)\sin P\hat{Q}R=78.5\) or \(33.4\)
(M1)
area \(\gt70\) requires \(P\hat{Q}R=72.2^{\circ}\) \((P\hat{R}Q=65.8^{\circ})\)
(A1)
\(PR=\dfrac{11\sin72.15^{\circ}}{\sin42^{\circ}}=15.6\) cm
A1
15.6482…
Answers: (b) 65.8° or 114° · (c) 15.6 cm
267AAHL3-q00253.7 Circular functions and their transformationsPaper 2Medium12 marks
A lighthouse lamp \(L\) is \(400\) m from a long straight sea wall. The point \(O\) on the wall is the point closest to \(L\). The lamp rotates at a constant rate, completing one revolution every \(12\) seconds, and casts a spot of light on the wall. At time \(t=0\) seconds the beam points at \(O\). While the beam is hitting the wall, the displacement of the spot from \(O\), \(d\) metres, is modelled by \(d(t)=400\tan\left(\dfrac{\pi t}{6}\right)\), for \(-3\lt t\lt3\).
Diagram not to scale
(a)
Explain why the angle between the beam and \([LO]\) at time \(t\) is \(\dfrac{\pi t}{6}\) radians, and hence why \(d(t)=400\tan\left(\dfrac{\pi t}{6}\right)\).
(2)
(b)
Write down the equations of the vertical asymptotes of the graph of \(d\) and interpret them in context.
(2)
(c)
Find the value of \(d\) when \(t=2\).
(2)
(d)
A walker stands on the wall \(250\) m from \(O\), on the side where \(d\gt0\). Find the length of time, during one revolution, for which the spot of light is within \(100\) m of the walker.
(3)
(e)
The rate of rotation is changed so that, in each revolution, the spot of light is within \(400\) m of \(O\) for exactly \(2\) seconds. Find the new time for one revolution.
(3)
Show mark scheme
Working
Mark
Notes
(a)
angular speed \(=\dfrac{2\pi}{12}=\dfrac{\pi}{6}\) radians per second, so the angle after \(t\) seconds is \(\dfrac{\pi t}{6}\)
R1
triangle \(LOS\) (spot \(S\)) is right-angled at \(O\), so \(\tan\dfrac{\pi t}{6}=\dfrac{d}{400}\)
R1
(b)
\(t=-3\) and \(t=3\)
A1
as \(t\to\pm3\) the beam becomes parallel to the wall, so \(|d|\) increases without bound and the beam stops hitting the wall
R1
(c)
\(400\tan\dfrac{\pi}{3}\)
(M1)
\(=400\sqrt3\)
\(693\) m
A1
692.820…
(d)
\(150\le d\le350\), e.g. \(400\tan\left(\dfrac{\pi t}{6}\right)=150\) and \(=350\), or graph
the beam turns through \(\dfrac{\pi}{2}\) in \(2\) seconds, so angular speed \(=\dfrac{\pi}{4}\) radians per second
(M1)
\(8\) seconds
A1
Answers: (b) \(t=\pm3\) · (c) 693 m · (d) 0.688 s · (e) 8 s
268AAHL3-q00303.8 Solving trigonometric equationsPaper 2Medium8 marks
A triangle has sides of length \(5\) cm and \(8\) cm, with included angle \(\theta\), where \(0\lt\theta\lt\pi\). The numerical value of the area of the triangle, in cm², is equal to the numerical value of the length of the third side, in cm.
Diagram not to scale
(a)
Write down expressions, in terms of \(\theta\), for the area of the triangle and for the length of the third side.
(2)
(b)
Show that \(400\cos^{2}\theta-80\cos\theta-311=0\).
(2)
(c)
Find the possible values of \(\theta\).
(3)
(d)
Explain why neither of your answers to part (c) needs to be rejected.
for \(0\lt\theta\lt\pi\), \(20\sin\theta\gt0\) and the square root is positive, so squaring introduced no extra roots; both values give area \(=\) side (3.16 and 12.3)
R1
Answers: (c) \(\theta=0.159\) or \(\theta=2.48\)
269AAHL3-q00383.9 Reciprocal and inverse trigonometric functionsPaper 2Medium6 marks
Let \(f(x)=\arccos x\), for \(-1\le x\le1\), and \(g(x)=\arctan2x\), for \(x\in\mathbb{R}\). The diagram shows the graphs of \(f\) and of \(g\) for \(-1\le x\le1\). The graphs intersect at the point P. The shaded region \(R\) is enclosed by the graph of \(f\), the graph of \(g\) and the \(y\)-axis.
Diagram not to scale
(a)
Write down the range of \(f\).
(1)
(b)
Write down the range of \(g\).
(1)
(c)
Find the coordinates of P.
(2)
(d)
Find the area of \(R\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(0\le y\le\pi\)
A1
(b)
\(-\dfrac{\pi}{2}\lt y\lt\dfrac{\pi}{2}\)
A1
Award A0 for the range of the part shown (\(\pm\arctan 2\)), or for \(\le\).
(c)
solving \(\arccos x=\arctan2x\) with GDC (graph or solver)
\(\dfrac{-1-\sqrt5}{2}\lt-1\) has no solutions, and \(\sin x=\dfrac{\sqrt5-1}{2}\in\,]0,1[\) has exactly two solutions in \([0,2\pi]\) (one in each of the first two quadrants)
272AAHL3-q00613.12 Vectors: concepts and algebraPaper 2Medium8 marks
A drone takes off from a point O. Relative to O, the unit vectors \(\boldsymbol{i}\), \(\boldsymbol{j}\) and \(\boldsymbol{k}\) point east, north and vertically upwards respectively, and distances are in metres. The drone makes three straight-line flights with displacements
Find the position vector of the drone after the second flight.
(1)
(b)
Find the distance of the drone from O after the second flight.
(1)
(c)
Find the distance of the drone from O after the third flight.
(3)
(d)
The drone then flies in a straight line directly back to O at a constant speed of 12 m s−1. Find a unit vector in the direction of this flight and the time it takes.
Exact: \(-\dfrac{1}{\sqrt{86}}\begin{pmatrix}2\\9\\1\end{pmatrix}\). Award A0 for the positive direction.
\(\dfrac{278.2\ldots}{12}\)
(M1)
\(23.2\) s
A1
\(23.1840\ldots\)
Answers: (a) \((120,360,65)\) · (b) 385 m · (c) 278 m · (d) \((-0.216,-0.970,-0.108)\), 23.2 s
273AAHL3-q00723.14 Vector equation of a linePaper 2Medium7 marks
A weather balloon rises in a straight line at constant velocity. Its position vector, in metres, relative to a tracking station O is given by \(\boldsymbol{r}=\begin{pmatrix}300\\-200\\40\end{pmatrix}+t\begin{pmatrix}120\\-90\\80\end{pmatrix}\), where \(t\) is the time in minutes after tracking begins. The unit vectors \(\boldsymbol{i}\), \(\boldsymbol{j}\) and \(\boldsymbol{k}\) point east, north and vertically upwards, and the ground is the horizontal plane \(z=0\).
(a)
Find the speed of the balloon.
(2)
(b)
Find the time at which the balloon is 1000 m above the ground, and its position vector at this time.
(3)
(c)
Find the angle between the path of the balloon and the ground, giving your answer in degrees.
\(5.63067\ldots,\ -4.83067\ldots\) Both values required.
(c)
minimising \(5k^{2}-4k+89\) (or the area) as a function of \(k\)
M1
Eg vertex, derivative or GDC.
\(k=0.4\), giving \(5k^{2}-4k+89=88.2\)
(A1)
minimum area \(=\dfrac12\sqrt{88.2}=4.70\)
A1
\(4.69574\ldots\)
Answers: (b) \(k=5.63\) or \(k=-4.83\) · (c) 4.70
275AAHL3-q00903.18 Intersections and angles of lines and planesPaper 2Medium8 marks
A hillside is modelled by the plane \(\Pi:\ x+2y-20z=0\), where distances are in metres and the \(z\)-axis is vertical. A vertical mast stands on the hillside at M\((10,5,1)\) and its top is T\((10,5,25)\). A straight cable runs from T in the direction of \(\begin{pmatrix}2\\-1\\-4\end{pmatrix}\) to an anchor point A on the hillside.
(a)
Find the coordinates of A.
(3)
(b)
Find the length of the cable.
(2)
(c)
Find the angle between the cable and the hillside, giving your answer in degrees.
Answers: (a) A\((22,-1,1)\) · (b) 27.5 m · (c) \(60.2^{\circ}\)
276IBMA2H-049-05-043.4 Radians, arc length and sector areaPaper 2Medium7 marks
Calculator
In a model of a partially blocked artery, the cross-section of the artery is a circle of radius 2.0 mm. A plaque deposit fills a minor segment of this circle. The flat surface of the plaque is a chord of length 3.2 mm. Let \(\theta\) be the angle, in radians, subtended by this chord at the centre of the circle.
Cross-section of the artery. Diagram not to scale
(a)
Show that \(\theta=1.85\), correct to 3 significant figures.
(2)
(b)
Find the cross-sectional area of the plaque.
(3)
(c)
Find the percentage of the artery's cross-sectional area that is blocked.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\sin\frac{\theta}{2}=\frac{1.6}{2.0}\) (or the cosine rule with sides 2.0, 2.0, 3.2)
✓ 1
A1 \(\theta=2\arcsin0.8=1.8546\ldots\approx1.85\)
✓ 1
Part (b)
M1 segment area \(=\frac{1}{2}r^2(\theta-\sin\theta)\)
✓ 1
A1 \(\frac{1}{2}(2.0)^2(1.8546-\sin1.8546)\), where \(\sin\theta=0.96\)
✓ 1
A1 \(1.79\) mm2
✓ 1
Part (c)
M1 \(\frac{1.789}{\pi(2.0)^2}\times100\)
✓ 1
A1 \(14.2\%\)
✓ 1
Common error: Keep the calculator in radian mode when you evaluate sin θ in the segment formula. Rounding θ to 1.85 before substituting can also change the third significant figure, so use the stored value.
277IBMA2H-049-05-073.14 Vector equation of a linePaper 2Medium10 marks
Calculator
Relative to a control tower, with distances in kilometres, the straight flight paths of two aircraft are modelled by \(A:\ \boldsymbol{r}=\begin{pmatrix}0\\0\\3\end{pmatrix}+t\begin{pmatrix}4\\3\\0\end{pmatrix}\) and \(B:\ \boldsymbol{r}=\begin{pmatrix}10\\0\\1\end{pmatrix}+s\begin{pmatrix}-2\\4\\1\end{pmatrix}\).
Exam tip: The vector product is perpendicular to both lines. Projecting any connecting vector onto it gives the shortest distance directly, and the absolute value removes any negative sign.
278IBMA2H-049-06-073.18 Intersections and angles of lines and planesPaper 2Medium7 marks
Calculator
A hillside is modelled by the plane Π: 2x − y + 2z = 5. A zip-line cable is anchored at A(1, 1, 2) and runs along the line r = i + j + 2k + λ(i + 2j + 4k). Distances are in metres.
(a)
Show that A lies on Π.
(1)
(b)
Find the acute angle between the cable and the hillside.
(4)
(c)
A support tower is attached to the cable at the point B where λ = 2. Find the perpendicular distance from B to the hillside.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 substitutes A: 2×1 − 1 + 2×2 = 5, so A lies on Π (AG)
✓ 1
Part (b)
M1 uses normal n = 2i − j + 2k with direction d = i + 2j + 4k
✓ 1
A1 n·d = 8, |n| = 3, |d| = √21
✓ 1
M1 sin θ = 8/(3√21), or finds the angle with the normal and subtracts it from 90°
✓ 1
A1 θ = 35.6°
✓ 1
Part (c)
M1 finds B(3, 5, 10) and uses distance = |2×3 − 5 + 2×10 − 5|/3, or AB sin θ = 2√21 × 8/(3√21)
✓ 1
A1 16/3 = 5.33 m
✓ 1
Common error: Using cos θ with the plane's normal gives the angle between the line and the normal, not the plane. The angle to the plane is its complement, so use sin θ.
279IBMA2H-049-06-093.14 Vector equation of a linePaper 2Medium6 marks
Calculator
Two delivery drones fly in straight lines at constant velocity. At time t minutes their positions, in kilometres, are rA = 10i + 5k + t(−2i + j) rB = 4j + k + t(i + k) Let D be the distance between the drones at time t.
(a)
Show that D2 = 11t2 − 76t + 132.
(3)
(b)
Find the time at which the drones are closest together and the minimum distance between them.
Exam tip: When the side opposite the given angle is the shorter of the two given sides, test \(180^\circ-C\). The shorter third side comes from the obtuse value of \(C\).
281IBMA2H-049-00-063.2 Right-angled and non-right-angled trianglesPaper 2Medium10 marks
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A football pitch is the rectangle ABCD with AB = 105 m and BC = 68 m. A camera drone hovers at point X, vertically above the centre of the pitch. The angle of elevation of X from corner A is 32°.
Plan view: the drone X hovers vertically above the centre of the pitch. Diagram not to scale
(a)
Find the length of the diagonal AC.
(2)
(b)
Find the height of the drone above the pitch.
(2)
(c)
Find the angle AXB.
(4)
(d)
Find the area of triangle AXB.
(2)
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Part (a)
M1 \(\sqrt{105^2+68^2}\)
✓ 1
A1 125 m (125.096…)
✓ 1
Part (b)
M1 \(h=62.548\ldots\times\tan32^\circ\), using half the diagonal
✓ 1
A1 \(h=39.1\) m
✓ 1
Part (c)
M1 \(XA=\frac{62.548}{\cos32^\circ}\) or \(\sqrt{62.548^2+39.08^2}\)
✓ 1
A1 \(XA=XB=73.76\) m
✓ 1
M1 cosine rule \(\cos A\hat{X}B=\frac{2(73.76)^2-105^2}{2(73.76)^2}\), or \(\sin\frac{1}{2}A\hat{X}B=\frac{52.5}{73.76}\)
✓ 1
A1 \(A\hat{X}B=90.8^\circ\)
✓ 1
Part (d)
M1 \(\frac{1}{2}(73.76)^2\sin90.77^\circ\), or \(\frac{1}{2}\times105\times\sqrt{h^2+34^2}\)
✓ 1
A1 2720 m²
✓ 1
Exam tip: The drone is above the centre, so the horizontal distance to any corner is half the diagonal. Draw the vertical right-angled triangle first, then work in the sloping triangle.
282IBMA2H-049-00-083.2 Right-angled and non-right-angled trianglesPaper 2Medium6 marks
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A car windscreen wiper arm is 52 cm long and pivots about one end. The rubber blade covers only the outer 40 cm of the arm. The arm sweeps through an angle of 2.2 radians.
Blade shown in red. Diagram not to scale
(a)
Find the area of windscreen cleared by the blade.
(3)
(b)
Find the straight-line distance between the two extreme positions of the tip of the arm.
(3)
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Part (a)
M1 uses \(\frac{1}{2}r^2\theta\)
✓ 1
M1 \(\frac{1}{2}(52^2-12^2)(2.2)\)
✓ 1
A1 2820 cm² (2816)
✓ 1
Part (b)
M1 cosine rule, or \(2r\sin\frac{\theta}{2}\)
✓ 1
A1 \(\sqrt{2(52^2)(1-\cos2.2)}\) or \(104\sin1.1\)
✓ 1
A1 92.7 cm
✓ 1
Common error: The blade starts 12 cm from the pivot, so the cleared region is an annular sector. Using \(\frac{1}{2}(40)^2\theta\) treats it as a full sector of radius 40 cm, which is wrong.
283IBMA2H-049-00-123.4 Radians, arc length and sector areaPaper 2Medium7 marks
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A medicine capsule is a cylinder with a hemisphere at each end, all of radius \(r\) mm. The total length of the capsule is 18.0 mm and its volume is 500 mm³.
Side view of the capsule. Diagram not to scale
(a)
Show that \(18\pi r^2-\frac{2}{3}\pi r^3=500\).
(2)
(b)
Find the value of \(r\), giving your answer to 3 significant figures.
(2)
(c)
Show that the total surface area of the capsule is \(36\pi r\), and hence find its value.
(3)
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Part (a)
M1 \(\pi r^2(18-2r)+\frac{4}{3}\pi r^3\), with cylinder length \(18-2r\)
✓ 1
A1 simplifies to \(18\pi r^2-\frac{2}{3}\pi r^3=500\)
✓ 1
Part (b)
M1 solves the equation with a GDC, rejecting roots that do not satisfy \(0<r\le9\)
✓ 1
A1 \(r=3.16\) mm (3.1648…)
✓ 1
Part (c)
M1 \(2\pi r(18-2r)+4\pi r^2\)
✓ 1
A1 \(36\pi r-4\pi r^2+4\pi r^2=36\pi r\)
✓ 1
A1 358 mm² using the unrounded \(r\); accept 357 mm² from \(r=3.16\)
✓ 1
Common error: The 18.0 mm includes both hemispherical ends, so the cylinder is only \(18-2r\) long. Using 18 for the cylinder length overstates both the volume and the surface area. Keep the full calculator value of \(r\) for later parts.
284IBMA2H-049-01-043.3 Applications of trigonometry: elevation, depression and bearingsPaper 2Medium8 marks
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A weather balloon B is held at a fixed height h metres vertically above a point T on level horizontal ground. An observer at point A on the ground sees the balloon due north at an angle of elevation of 35°. A second observer at point C on the ground, 600 m due east of A, sees the balloon at an angle of elevation of 24°.
Plan view (the balloon is vertically above T). Diagram not to scale
(a)
Write down expressions for AT and CT in terms of h.
(2)
(b)
Find h.
(3)
(c)
Find the bearing of the balloon from C.
(3)
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Part (a)
A1 \(AT = \frac{h}{\tan35°}\)
✓ 1
A1 \(CT = \frac{h}{\tan24°}\)
✓ 1
Part (b)
M1 triangle ATC is right-angled at A, so \(CT^2 = AT^2 + 600^2\)
M1 correct angle at C: angle between CA (due west) and CT from \(\tan\alpha = \frac{494.3}{600}\), giving α = 39.5°, or angle west of north from \(\tan\beta = \frac{600}{494.3}\), giving β = 50.5°
Exam tip: Draw the horizontal ground triangle ATC on its own. T is due north of A and C is due east of A, so the right angle is at A. Pythagoras then links the two horizontal distances. Keep unrounded values in your calculator until the final answer.
285IBMA2H-049-01-093.3 Applications of trigonometry: elevation, depression and bearingsPaper 2Medium8 marks
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A search-and-rescue helicopter H hovers 250 m above the sea. The pilot sees a lifeboat A on a bearing of 040° at an angle of depression of 18°. The pilot also sees a yacht Y on a bearing of 115° at an angle of depression of 11°. The lifeboat and the yacht are on the sea surface.
Plan view; H′ is the point on the sea directly below H. Diagram not to scale
(a)
Find the horizontal distance from the point directly below H to each of A and Y.
(2)
(b)
Find the distance between the lifeboat and the yacht.
(3)
(c)
Find the bearing of the yacht from the lifeboat.
(3)
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Part (a)
A1 \(\frac{250}{\tan18°} = 769\) m to A
✓ 1
A1 \(\frac{250}{\tan11°} = 1290\) m (1286 m) to Y
✓ 1
Part (b)
A1 angle between the horizontal directions = 115° − 40° = 75°
M1 sine rule with O the point below H: \(\sin(O\hat{A}Y) = \frac{1286.1\sin75°}{1316.8}\)
✓ 1
A1 angle OAY = 70.6° (acute, since OY < AY)
✓ 1
A1 bearing = 220° − 70.6° = 149°
✓ 1
Exam tip: Angles of depression give horizontal distances only. Work in the horizontal sea-level triangle for the cosine and sine rules. Then use the back-bearing 220° from A to the point below the helicopter.
286IBMA2H-049-01-113.10 Compound and double angle identitiesPaper 2Medium8 marks
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A large decorative fountain in a city park fires a jet of water from ground level with speed v m s−1, at an angle θ above the horizontal. Ignoring air resistance, the horizontal range is \(R = \frac{v^2\sin2\theta}{g}\) and the maximum height is \(H = \frac{v^2\sin^2\theta}{2g}\). Use g = 9.8 m s−2 and v = 14 m s−1.
Path of the water jet, landing 15 m from the nozzle. Diagram not to scale
(a)
The jet must land 15 m from the nozzle. Find the two possible values of θ, where 0° < θ < 90°.
(4)
(b)
Show that \(H = \frac{v^2(1-\cos2\theta)}{4g}\).
(2)
(c)
Find the maximum height of the jet for the larger value of θ.
A1 substitutes into \(H = \frac{v^2\sin^2\theta}{2g}\) to obtain \(H = \frac{v^2(1-\cos2\theta)}{4g}\)
✓ 1
Part (c)
M1 either uses \(\cos2\theta = -\sqrt{1-0.75^2} = -0.661\) (2θ is obtuse) in the result of the previous part, or substitutes θ = 65.7° into \(H = \frac{v^2\sin^2\theta}{2g}\)
✓ 1
A1 H = 8.31 m
✓ 1
Exam tip: The equation sin 2θ = 0.75 has two solutions for 2θ between 0° and 180°. Halve both to get the two launch angles, which always add up to 90°. Keep full calculator accuracy for θ before working out H.
287IBMA2H-049-02-053.18 Intersections and angles of lines and planesPaper 2Medium9 marks
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The triangular mainsail of a racing yacht has corners A, B and C. Relative to the foot of the mast, with distances in metres, the coordinates are A(0, 0, 0), B(4, 1, 6) and C(3, 5, 2).
(a)
Find \(\vec{AB}\) and \(\vec{AC}\).
(2)
(b)
Find \(\vec{AB}\times\vec{AC}\).
(2)
(c)
Hence find the area of the sail.
(2)
(d)
Use the scalar product to find the angle BAC, in degrees.
Exam tip: The magnitude of the cross product gives the area of the parallelogram; the triangle is half of it. You can check your angle: \(\sin\theta=\frac{|\vec{AB}\times\vec{AC}|}{|\vec{AB}||\vec{AC}|}\) should agree with your answer.
288IBMA2H-049-02-063.7 Circular functions and their transformationsPaper 2Medium8 marks
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In a car engine, the height of the top of a piston above a fixed reference point is modelled by \(h(t)=a\cos(bt)+c\), \(a,b,c>0\), where \(h\) is in cm and \(t\) is the time in seconds. At \(t=0\) the piston is at its highest point, 13 cm. Its lowest point is 5 cm. The crankshaft completes 50 revolutions per second, and each revolution is one full cycle of the piston.
(a)
Find the values of \(a\) and \(c\).
(2)
(b)
Find the value of \(b\).
(2)
(c)
Find the total time during one cycle for which the piston is more than 10 cm above the reference point.
(4)
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Part (a)
A1 \(a=\frac{13-5}{2}=4\)
✓ 1
A1 \(c=\frac{13+5}{2}=9\)
✓ 1
Part (b)
M1 period \(=\frac{1}{50}=0.02\) s, so \(b=\frac{2\pi}{0.02}\)
M1 uses symmetry: time above \(=\frac{2\times1.318\ldots}{100\pi}\) (or \(0.02-\) time below)
✓ 1
A1 \(8.39\times10^{-3}\) s
✓ 1
Common error: Finding one solution of \(\cos(100\pi t)=0.25\) is not enough. The piston is above 10 cm both just after \(t=0\) and just before the end of the cycle, so the time above is double the first solution.
289IBMA2H-049-02-093.17 Equations of a planePaper 2Medium8 marks
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A rescue team pulls a supply sled across an ice sheet. The rope exerts a constant force \(\boldsymbol{F}=\begin{pmatrix}30\\10\\15\end{pmatrix}\) N and the sled undergoes a displacement \(\boldsymbol{d}=\begin{pmatrix}40\\-5\\0\end{pmatrix}\) m. The work done by the force is \(W=\boldsymbol{F}\cdot\boldsymbol{d}\).
(a)
Find the work done by the rope.
(2)
(b)
Find the angle between the force and the displacement.
(3)
(c)
Find the magnitude of the component of \(\boldsymbol{F}\) perpendicular to the direction of motion.
(3)
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Part (a)
M1 \(W=30\times40+10\times(-5)+15\times0\)
✓ 1
A1 \(W=1150\) J
✓ 1
Part (b)
M1 \(|\boldsymbol{F}|=35\) and \(|\boldsymbol{d}|=\sqrt{1625}\approx40.3\)
M1 a valid method: uses \(|\boldsymbol{F}|\sin\theta\), or \(\frac{|\boldsymbol{F}\times\boldsymbol{d}|}{|\boldsymbol{d}|}\), or \(\sqrt{|\boldsymbol{F}|^2-\left(\frac{\boldsymbol{F}\cdot\boldsymbol{d}}{|\boldsymbol{d}|}\right)^2}\)
✓ 1
M1 correct substitution, e.g. \(\boldsymbol{F}\times\boldsymbol{d}=\begin{pmatrix}75\\600\\-550\end{pmatrix}\), or \(\sin\theta\approx0.579\), or parallel component \(\frac{1150}{\sqrt{1625}}\approx28.5\) N
✓ 1
A1 20.3 N
✓ 1
Exam tip: Keep full calculator accuracy for \(\theta\) before finding \(\sin\theta\). The cross-product and Pythagoras routes avoid rounding the angle altogether.
290IBMA2H-049-02-103.11 Symmetry properties of trigonometric graphsPaper 2Medium10 marks
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The apparent brightness of a variable star is modelled by \(B(t)=p\cos(q(t-r))+s\), where \(t\) is time in days, \(p,q>0\) and \(0<r<5\). The first maximum brightness, 4.2 units, occurs at \(t=1.5\). The next minimum brightness, 2.6 units, occurs at \(t=4.2\).
(a)
Find the values of \(p\) and \(s\).
(2)
(b)
Find the exact value of \(q\).
(2)
(c)
Write down the value of \(r\).
(1)
(d)
Using your exact value of \(q\), find the brightness predicted at \(t=10\). Give your answer to 3 significant figures.
(2)
(e)
Find the length of time during each cycle for which the brightness exceeds 3.8 units.
(3)
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Part (a)
A1 \(p=\frac{4.2-2.6}{2}=0.8\)
✓ 1
A1 \(s=\frac{4.2+2.6}{2}=3.4\)
✓ 1
Part (b)
M1 half period \(=4.2-1.5=2.7\), so period \(=5.4\) days
M1 \(-\frac{\pi}{3}<q(t-r)<\frac{\pi}{3}\), so interval length \(=\frac{2\pi/3}{q}\) (or solves \(B(t)=3.8\) to get \(t=0.6\) and \(t=2.4\))
✓ 1
A1 1.8 days
✓ 1
Common error: The time between a maximum and the next minimum is half a period, not a full period. Using 2.7 days as the period doubles \(q\) and makes every later answer wrong. Keep \(q\) exact, or store it in your calculator, so that rounding does not change the later answers.
291IBMA2H-049-07-053.3 Applications of trigonometry: elevation, depression and bearingsPaper 2Medium7 marks
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A lighthouse is at point A and a coastguard station is at point B, where AB = 12 km. A ship at point C is known to be 9 km from B, and angle BAC = 40°. There are two possible positions for the ship.
The two possible positions of the ship. Diagram not to scale
(a)
Find the two possible values of angle ACB.
(3)
(b)
By forming a quadratic equation in x = AC, find the distance between the two possible positions of the ship.
(4)
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Part (a)
M1 sine rule \(\frac{\sin C}{12}=\frac{\sin 40^\circ}{9}\), so \(\sin C = \frac{12\sin 40^\circ}{9} = 0.857…\) ; A1 C = 59.0° ; R1 C = 180° − 59.0° = 121° is also possible because 40° + 121° < 180°
✓ 3
Part (b)
M1 cosine rule 9² = 12² + x² − 2(12)x cos 40° ; A1 x² − 18.385x + 63 = 0 ; A1 x = 13.83 or x = 4.56 ; A1 both positions lie on the same ray from A, so distance = 13.83 − 4.56 = 9.27 km
✓ 4
Exam tip: Both positions lie on the line AC, so the distance between them is the difference of the two roots. For x² − px + q = 0 the difference of the roots is √(p² − 4q), which gives a quick check.
292IBMA2H-049-07-073.17 Equations of a planePaper 2Medium8 marks
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A flat solar panel is modelled as the plane Π through the points P(0, 0, 2), Q(4, 0, 3) and R(0, 5, 1), with coordinates in metres. At a certain time, the Sun's rays travel in the direction d = i + 2j − 3k.
Find the acute angle between the Sun's rays and the panel.
(4)
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Part (a)
M1 \(\overrightarrow{PQ}\) = (4, 0, 1), \(\overrightarrow{PR}\) = (0, 5, −1) and a correct method for the product ; A1 −5i + 4j + 20k
✓ 2
Part (b)
M1 −5x + 4y + 20z = d and substitute P (or Q or R) ; A1 −5x + 4y + 20z = 40
✓ 2
Part (c)
M1 n·d = −5 + 8 − 60 = −57 ; A1 |n| = 21 and |d| = √14 ; M1 \(\sin\theta=\frac{57}{21\sqrt{14}}\) (using sine because the angle is with a plane) ; A1 θ = 46.5°
✓ 4
Common error: The scalar product of the ray with the normal gives the angle to the normal. The angle with the plane is its complement, so use sin θ rather than cos θ, or subtract from 90°.
293IBMA2H-049-07-083.1 Three-dimensional geometry: distance, solids and anglesPaper 2Medium7 marks
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A camping tent is modelled as a right pyramid with a square base of side 2.4 m. A vertical pole of height 1.8 m runs from the centre of the base to the apex V. Four straight sloping edges run from V to the corners of the base.
Tent modelled as a right pyramid with a vertical pole from the centre of the base to V. Diagram not to scale
(a)
Find the length of one sloping edge.
(3)
(b)
Find the angle between a sloping edge and the base.
(2)
(c)
Find the angle at V between two adjacent sloping edges.
(2)
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Part (a)
M1 half-diagonal of base = ½√(2.4² + 2.4²) = 1.697… m ; M1 edge² = 1.8² + 1.697…² = 6.12 ; A1 edge = 2.47 m
Exam tip: Work with edge² = 6.12 exactly in part (c). This avoids rounding the edge length and makes the cosine rule substitution neat.
294IBMA2H-049-07-103.13 Scalar productPaper 2Medium8 marks
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Two adjacent edges of a parallelogram-shaped carbon-fibre plate for a drone are represented by the vectors a = pi + j + 2k and b = i − j + 3k, where p ∈ ℝ. The area of the plate is \(3\sqrt{10}\) units².
(a)
Show that a × b = 5i + (2 − 3p)j − (p + 1)k.
(2)
(b)
Find the possible values of p.
(4)
(c)
For p = 3, find the angle between a and b.
(2)
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Part (a)
M1 correct determinant expansion seen for at least two components. For example, the i component is 1×3 − 2×(−1) and the j component is 2×1 − 3p ; A1 all three components correct: 5, 2 − 3p and −p − 1, leading to the given answer
✓ 2
Part (b)
M1 uses area = |a × b|, with |a × b|² = 25 + (2 − 3p)² + (p + 1)² ; A1 10p² − 10p + 30 ; M1 sets this equal to 90 and obtains p² − p − 6 = 0 ; A1 p = 3 or p = −2
✓ 4
Part (c)
M1 \(\cos\theta=\frac{3-1+6}{\sqrt{14}\sqrt{11}}=\frac{8}{\sqrt{154}}\), or \(\sin\theta=\frac{\sqrt{90}}{\sqrt{154}}\) with θ taken as acute because a·b > 0 ; A1 θ = 49.9°
✓ 2
Common error: The area of a parallelogram is |a × b|, not half of it (half is the triangle's area). Square both sides to get |a × b|² = 90 before solving, so you work with a quadratic rather than a square root.
295IBMA2H-049-03-053.18 Intersections and angles of lines and planesPaper 2Medium7 marks
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A laser beam from a satellite travels along the line L: r = (1, 0, 3) + λ(3, 4, 1), λ ∈ ℝ. It strikes a flat solar panel lying in the plane Π: 2x − y + 2z = 7.
(a)
Find the coordinates of the point where the beam meets the panel.
(3)
(b)
Find the acute angle between the beam and the panel.
M1 uses direction (3, 4, 1) and normal (2, −1, 2): scalar product = 4
✓ 1
M1 magnitudes √26 and 3
✓ 1
M1 uses sin θ = 4 / (3√26) (or finds angle with normal and subtracts from 90°)
✓ 1
A1 θ = 15.2°
✓ 1
Common error: The scalar product of the direction and the normal gives the angle between the line and the normal. The angle with the plane is its complement, so use sine rather than cosine.
296IBMA2H-049-03-073.12 Vectors: concepts and algebraPaper 2Medium7 marks
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A winch applies a constant force F = (30, −10, 20) newtons to a crate on a construction site. The crate moves in a straight line from A(1, 2, 0) to B(6, 4, 3), where distances are in metres. The work done by the force is F · \(\overrightarrow{AB}\).
(a)
Find the work done by the force.
(2)
(b)
Find the angle between the force and the direction of motion.
(3)
(c)
Find the component of the force in the direction of motion.
(2)
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Part (a)
M1 finds \(\overrightarrow{AB}\) = (5, 2, 3) and forms the scalar product 150 − 20 + 60
M1 uses F · \(\overrightarrow{AB}\) / |\(\overrightarrow{AB}\)| = 190/√38, or equivalently |F| cos θ = √1400 × cos 34.5°
✓ 1
A1 30.8 N
✓ 1
Exam tip: The component of F along a direction is the scalar product divided by the magnitude of that direction vector only. Do not also divide by |F|.
297IBMA2H-049-03-093.15 Intersecting, parallel and skew linesPaper 2Medium13 marks
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Two drone flight corridors are modelled by straight lines, with distances in kilometres. L1: r = (1, −2, 3) + s(2, 1, −2) L2: r = (4, 0, −1) + t(1, −1, 1), where s, t ∈ ℝ.
(a)
Find the acute angle between the two corridors.
(4)
(b)
Show that the corridors are skew.
(4)
(c)
Find a vector perpendicular to both corridors.
(2)
(d)
Find the shortest distance between the two corridors, giving your answer in metres.
Exam tip: To show lines are skew you need two facts: they are not parallel and they do not intersect. State both explicitly to secure the reasoning mark.
298IBMA2H-049-03-103.7 Circular functions and their transformationsPaper 2Medium12 marks
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The height above the ground, h metres, of the tip of one blade of a wind turbine is modelled by h(t) = 72 − 38 cos(1.2t), where t is the time in seconds after the tip passes its lowest point.
(a)
Write down the minimum and maximum heights of the blade tip.
(2)
(b)
Find the time taken for one complete revolution of the blade.
(2)
(c)
Find the first two times, t > 0, at which the tip is 90 m above the ground.
(4)
(d)
Find the length of time, during each revolution, for which the tip is more than 90 m above the ground.
(2)
(e)
Hence find the percentage of each revolution for which the tip is more than 90 m above the ground.
(2)
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Part (a)
A1 minimum 34 m
✓ 1
A1 maximum 110 m
✓ 1
Part (b)
M1 period = 2π / 1.2
✓ 1
A1 5.24 s
✓ 1
Part (c)
M1 sets 72 − 38 cos(1.2t) = 90
✓ 1
A1 cos(1.2t) = −9/19
✓ 1
A1 1.2t = 2.064…, t = 1.72 s
✓ 1
A1 1.2t = 2π − 2.064… = 4.219…, t = 3.52 s
✓ 1
Part (d)
M1 3.5157… − 1.7202…
✓ 1
A1 1.80 s
✓ 1
Part (e)
M1 1.7955… / 5.2359… × 100
✓ 1
A1 34.3%
✓ 1
Common error: Rounding the times in part (c) before subtracting can shift the answer to part (d). Keep full calculator values until the final step.
299IBMA2H-049-04-083.17 Equations of a planePaper 2Medium10 marks
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In a radiotherapy plan, a beam leaves a source S at (1, 4, −2) and travels along the line r = (i + 4j − 2k) + t(2i + j + 3k). The boundary of a tissue layer is modelled by the plane \(3x + 2y - z = 20\). All distances are in centimetres.
(a)
Find the coordinates of the point P where the beam meets the boundary.
(3)
(b)
Find the acute angle between the beam and the boundary.
(3)
(c)
Find the distance SP.
(2)
(d)
Find the shortest distance from S to the boundary plane.
(2)
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Part (a)
M1 substitute (1 + 2t, 4 + t, −2 + 3t) into the plane
✓ 1
A1 \(13 + 5t = 20\), so \(t = 1.4\)
✓ 1
A1 P(3.8, 5.4, 2.2)
✓ 1
Part (b)
M1 \(\sin\phi = \frac{|\boldsymbol{n}\cdot\boldsymbol{d}|}{|\boldsymbol{n}||\boldsymbol{d}|}\) with n = (3, 2, −1)
M1 SP sin φ, or \(\frac{|3 + 8 + 2 - 20|}{\sqrt{14}}\)
✓ 1
A1 \(\frac{7}{\sqrt{14}}\) = 1.87 cm
✓ 1
Common error: Using the parameter value t = 1.4 as the distance SP ignores the length of the direction vector. Multiply by |d| unless the direction is a unit vector.
300IBMA3H-3-19r5-043.16 Vector productPaper 2Medium8 marks
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A sloping orchard is surveyed with a drone. Relative to a marker O, the corners of one triangular section are at points A(0, 0, 0), B(60, 20, 3) and C(10, 50, 5), where all coordinates are in metres and the third coordinate is height. Give answers to 3 significant figures.
Find the angle the plane ABC makes with the horizontal.
(3)
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Part (a)
M1 attempt at the vector product of \((60,20,3)\) and \((10,50,5)\)
✓ 1
A1 first two components \(-50\) and \(-270\)
✓ 1
A1 third component \(2800\), giving \((-50,-270,2800)\)
✓ 1
Part (b)
M1 area \(=\tfrac12|\overrightarrow{AB}\times\overrightarrow{AC}|=\tfrac12\sqrt{(-50)^2+(-270)^2+2800^2}\)
✓ 1
A1 1410 m2 (1406.7 unrounded)
✓ 1
Part (c)
M1 angle between normal \((-50,-270,2800)\) and \((0,0,1)\) equals angle between planes
✓ 1
A1 \(\cos\varphi=\dfrac{2800}{\sqrt{7915400}}\)
✓ 1
A1 \(\varphi=5.60^\circ\)
✓ 1
Common error: The vector product gives a vector perpendicular to the plane; the angle between the plane and the horizontal equals the angle between their normals, not between the plane and a vertical line.
301IBMA3H-3-19r5-053.16 Vector productPaper 2Medium7 marks
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Two edges of a triangular winglet panel on an aircraft are represented by the vectors \(\mathbf{a}=\begin{pmatrix}2\\-1\\3\end{pmatrix}\) and \(\mathbf{b}=\begin{pmatrix}1\\4\\-2\end{pmatrix}\), in metres.
(a)
Find \(\mathbf{a}\times\mathbf{b}\).
(2)
(b)
Show that \(\mathbf{a}\times\mathbf{b}\) is perpendicular to both \(\mathbf{a}\) and \(\mathbf{b}\).
(2)
(c)
Find a unit vector perpendicular to the panel.
(1)
(d)
Find the area of the panel, correct to 3 significant figures.
(2)
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Part (a)
M1 attempt at the vector product
✓ 1
A1 \((-10,7,9)\)
✓ 1
Part (b)
R1 \(-10\times2+7\times(-1)+9\times3=0\) so perpendicular to \(\mathbf{a}\)
✓ 1
R1 \(-10\times1+7\times4+9\times(-2)=0\) so perpendicular to \(\mathbf{b}\)
✓ 1
Part (c)
A1 \(\dfrac{1}{\sqrt{230}}(-10,7,9)\) or equivalent
✓ 1
Part (d)
M1 area \(=\tfrac12|\mathbf{a}\times\mathbf{b}|=\tfrac12\sqrt{230}\)
✓ 1
A1 7.58 m2
✓ 1
Exam tip: A quick scalar-product check against both original vectors catches arithmetic slips in a vector product.
302IBMA3H-3-19r5-063.14 Vector equation of a linePaper 2Medium7 marks
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An aircraft flies with velocity \(\mathbf{v}=\begin{pmatrix}180\\-40\\12\end{pmatrix}\) km h−1 relative to the ground while the wind has velocity \(\mathbf{w}=\begin{pmatrix}-30\\25\\0\end{pmatrix}\) km h−1. Give answers to 3 significant figures unless stated otherwise.
(a)
Find the angle between the directions of \(\mathbf{v}\) and \(\mathbf{w}\), in degrees to 1 decimal place.
(4)
(b)
Find the component of the wind velocity in the direction of \(\mathbf{v}\), to 3 significant figures.
R1 negative: this part of the wind blows against the direction of flight (a headwind component)
✓ 1
Common error: The scalar projection divides by the magnitude of the vector you project onto only, not by both magnitudes.
303IBMA3H-3-19r5-083.4 Radians, arc length and sector areaPaper 2Medium8 marks
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A farmer fences a segment-shaped part of a circular field. The circle has centre O and radius 25 m. The straight boundary AB is a chord of length 40 m. Use unrounded values in later parts and give final answers to 3 significant figures.
Diagram not to scale
(a)
Find the size of angle AOB in radians.
(3)
(b)
Find the length of the arc AB.
(2)
(c)
Find the area of the segment between the chord AB and the arc AB.
M1 subtract triangle area \(\tfrac12r^2\sin\theta\)
✓ 1
A1 280 m2 (279.56 unrounded)
✓ 1
Exam tip: Segment area = sector area minus triangle area: keep the full calculator value of the angle until the end.
304IBMA3H-3-19r5-103.14 Vector equation of a linePaper 2Medium8 marks
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A technician tightens a bolt on an aircraft propeller hub using a spanner. Relative to the axis of the bolt, the force is applied at position \(\mathbf{r}=\begin{pmatrix}0.30\\0.10\\0\end{pmatrix}\) m and the force is \(\mathbf{F}=\begin{pmatrix}0\\20\\15\end{pmatrix}\) N. The torque is \(\boldsymbol{\tau}=\mathbf{r}\times\mathbf{F}\).
(a)
Find \(\boldsymbol{\tau}\).
(3)
(b)
Find the magnitude of \(\boldsymbol{\tau}\), correct to 3 significant figures.
(1)
(c)
Use the scalar product to find the angle between \(\mathbf{r}\) and \(\mathbf{F}\), correct to 3 significant figures.
(4)
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Part (a)
M1 attempt at \(\mathbf{r}\times\mathbf{F}\)
✓ 1
A1 two components correct
✓ 1
A1 \((1.5,-4.5,6.0)\) N m
✓ 1
Part (b)
A1 \(\sqrt{1.5^2+4.5^2+6.0^2}=7.65\) N m
✓ 1
Part (c)
M1 \(\mathbf{r}\cdot\mathbf{F}=2.0\)
✓ 1
A1 \(|\mathbf{r}|=\sqrt{0.10}\) and \(|\mathbf{F}|=25\)
✓ 1
M1 \(\cos\theta=\dfrac{2.0}{25\sqrt{0.10}}\)
✓ 1
A1 \(\theta=75.3^\circ\)
✓ 1
Exam tip: The scalar product gives an unambiguous angle; \(\sin\theta\) from the vector product cannot distinguish \(\theta\) from \(180^\circ-\theta\).
305IBMA3H-3-19r5-113.11 Symmetry properties of trigonometric graphsPaper 2Medium8 marks
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A sound-pressure model gives the equation \(2\cos2x+3\sin x=1\) for \(0\le x\le2\pi\).
(a)
Show that the equation can be written as \(4\sin^2x-3\sin x-1=0\).
(3)
(b)
Hence solve the equation, giving answers in radians to 3 significant figures except where exact.
(5)
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Part (a)
M1 uses \(\cos2x=1-2\sin^2x\)
✓ 1
M1 \(2(1-2\sin^2x)+3\sin x=1\)
✓ 1
A1 rearranges to \(4\sin^2x-3\sin x-1=0\) (AG)
✓ 1
Part (b)
M1 factorises \((4\sin x+1)(\sin x-1)=0\) or uses the quadratic formula
✓ 1
A1 \(\sin x=1\) or \(\sin x=-\dfrac14\)
✓ 1
A1 \(x=\dfrac{\pi}{2}\)
✓ 1
M1 attempts \(\sin x=-0.25\) with solutions in the third and fourth quadrants: \(\pi+0.2527\) and \(2\pi-0.2527\)
✓ 1
A1 \(x=3.39\) and \(x=6.03\) (6.0305 unrounded)
✓ 1
Common error: Solving \(\sin x=-\tfrac14\) with only the calculator value \(-0.253\) misses the required solutions in the interval \([0,2\pi]\).
306IBMA3H-3-19r5-143.4 Radians, arc length and sector areaPaper 2Medium11 marks
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Two circular irrigation zones overlap. Zone A has centre A and radius 25 m. Zone B has centre B and radius 20 m. The centres are 35 m apart. The boundaries cross at points P and Q. Use unrounded values in later parts and give final answers to 3 significant figures.
The region watered by both zones is shaded. Diagram not to scale
(a)
Find the size of angle PAB in radians.
(3)
(b)
Find the size of angle PBA in radians.
(2)
(c)
Find the area of the region watered by both zones.
(4)
(d)
Find the percentage of zone B that lies inside zone A.
(2)
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Part (a)
M1 cosine rule in triangle PAB: \(20^2=25^2+35^2-2(25)(35)\cos\alpha\)
✓ 1
A1 \(\cos\alpha=\dfrac{1450}{1750}\)
✓ 1
A1 \(\alpha=0.594\) rad (0.59425 unrounded)
✓ 1
Part (b)
M1 cosine rule at B: \(\cos\beta=\dfrac{20^2+35^2-25^2}{2(20)(35)}\)
✓ 1
A1 \(\beta=0.775\) rad (0.77519 unrounded)
✓ 1
Part (c)
M1 overlap is two circular segments, one from each circle, with angles \(2\alpha\) and \(2\beta\)
✓ 1
M1 segment area \(\tfrac12r^2(\theta-\sin\theta)\) used for each
✓ 1
A1 segment in A: \(\tfrac12(25^2)(1.18849-\sin1.18849)=81.5\) m2
✓ 1
A1 segment in B 110 m2 (110.1 unrounded); total 192 m2 (191.58 unrounded)
Common error: The overlap of two circles is the sum of two segments, one from each circle, each with its own radius and its own angle at its own centre.
307IBMA3H-3-18r8-053.7 Circular functions and their transformationsPaper 2Medium5 marks
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The temperature \(T\,{}^\circ\)C at a mountain camp is modelled by \(T=a\cos\!\big(b(t-14)\big)+c\), where \(t\) is the time in hours after midnight and \(b>0\). The highest temperature is \(-2\,{}^\circ\)C at 14:00 and the lowest is \(-12\,{}^\circ\)C at 02:00.
(a)
Find the values of \(a\) and \(c\).
(2)
(b)
Find the exact value of \(b\).
(1)
(c)
Find the temperature at 17:00, to 3 significant figures.
(2)
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Part (a)
M1 \(a=\frac{-2-(-12)}{2}\) or midline \(\frac{-2+(-12)}{2}\)
✓ 1
A1 \(a=5,\ c=-7\)
✓ 1
Part (b)
A1 period 24 so \(b=\frac{2\pi}{24}=\frac{\pi}{12}\)
✓ 1
Part (c)
M1 \(T=5\cos\!\left(\frac{\pi}{12}\times3\right)-7\) (calculator in radians)
✓ 1
A1 \(-3.46\,{}^\circ\)C
✓ 1
Common error: Make sure the calculator is in radian mode when \(b\) is written with \(\pi\); degrees mode would give a very different temperature.
308IBMA3H-3-18r8-063.7 Circular functions and their transformationsPaper 2Medium6 marks
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The temperature of a bread oven varies around its set point and is modelled by \(T=180+12\sin(0.5t)\) degrees Celsius, where \(t\) is the time in minutes after the oven starts to cycle.
(a)
State the maximum temperature and the exact period of the cycle.
(2)
(b)
During the first cycle (\(0\le t\le4\pi\)) the temperature is above \(186\,{}^\circ\)C for a single interval. Find the length of this interval, to 3 significant figures.
(4)
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Part (a)
A1 maximum \(192\,{}^\circ\)C
✓ 1
A1 period \(\frac{2\pi}{0.5}=4\pi\) minutes
✓ 1
Part (b)
M1 \(12\sin(0.5t)>6\), so \(\sin(0.5t)>0.5\)
✓ 1
A1 boundaries \(0.5t=\frac{\pi}{6}\) and \(\frac{5\pi}{6}\)
✓ 1
M1 \(t=\frac{\pi}{3}\) and \(t=\frac{5\pi}{3}\), subtract
✓ 1
A1 \(\frac{4\pi}{3}=4.19\) minutes
✓ 1
Common error: Dividing the angle range by 0.5 changes the times, so convert from \(0.5t\) back to \(t\) before finding the length of the interval.
309IBMA3H-3-18r8-083.18 Intersections and angles of lines and planesPaper 2Medium7 marks
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Two cables are modelled by the lines \(L_1:\ \mathbf r=\begin{pmatrix}1\\2\\0\end{pmatrix}+s\begin{pmatrix}2\\1\\1\end{pmatrix}\) and \(L_2:\ \mathbf r=\begin{pmatrix}8\\3\\2\end{pmatrix}+t\begin{pmatrix}-1\\2\\1\end{pmatrix}\), with coordinates in metres.
(a)
Show that the cables meet and find the coordinates of the meeting point.
(4)
(b)
Find the acute angle between the cables, to the nearest tenth of a degree.
(3)
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Part (a)
M1 equate components: \(1+2s=8-t\) and \(2+s=3+2t\)
✓ 1
A1 solve to get \(s=3,\ t=1\)
✓ 1
R1 verify third component \(0+3=2+1\)
✓ 1
A1 point \((7,5,3)\)
✓ 1
Part (b)
M1 \(\cos\theta=\frac{|\mathbf d_1\cdot\mathbf d_2|}{|\mathbf d_1||\mathbf d_2|}\) with \(\mathbf d_1\cdot\mathbf d_2=1\)
✓ 1
A1 \(|\mathbf d_1|=|\mathbf d_2|=\sqrt6\) so \(\cos\theta=\frac16\)
✓ 1
A1 \(80.4^\circ\)
✓ 1
Common error: Two equations determine \(s\) and \(t\); the third component must be checked, since non-parallel lines in 3D need not meet.
310IBMA3H-3-18r8-093.18 Intersections and angles of lines and planesPaper 2Medium6 marks
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A hiker follows a straight path with equation \(\mathbf r=\begin{pmatrix}1\\0\\2\end{pmatrix}+t\begin{pmatrix}2\\1\\-2\end{pmatrix}\) (coordinates in km, \(t\) in hours). A mountain hut is at \(P(4,4,-2)\). Let \(F\) be the point on the path closest to the hut.
(a)
Write down the vector from \(P\) to a general point on the path, and use a scalar product to find the value of \(t\) at \(F\).
(3)
(b)
Find the coordinates of \(F\) and the shortest distance from the hut to the path, to 3 significant figures.
Exam tip: The shortest distance from a point to a line occurs where the line from the point is perpendicular to the direction vector, so the scalar product is zero.
311IBMA3H-3-18r8-103.12 Vectors: concepts and algebraPaper 2Medium5 marks
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Three relay masts A, B and C have position vectors (in km) \(\overrightarrow{OA}=\begin{pmatrix}1\\2\\-1\end{pmatrix}\), \(\overrightarrow{OB}=\begin{pmatrix}4\\-1\\5\end{pmatrix}\) and \(\overrightarrow{OC}=\begin{pmatrix}5\\-2\\7\end{pmatrix}\).
(a)
Find \(\overrightarrow{AB}\) and \(\overrightarrow{BC}\).
(1)
(b)
Show that A, B and C lie on a straight line.
(2)
(c)
Find the exact distance AC, written as \(k\sqrt6\).
(2)
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Part (a)
A1 \(\overrightarrow{AB}=\begin{pmatrix}3\\-3\\6\end{pmatrix}\) and \(\overrightarrow{BC}=\begin{pmatrix}1\\-1\\2\end{pmatrix}\)
✓ 1
Part (b)
M1 \(\overrightarrow{AB}=3\overrightarrow{BC}\)
✓ 1
R1 parallel vectors with common point B
✓ 1
Part (c)
M1 \(\overrightarrow{AC}=\begin{pmatrix}4\\-4\\8\end{pmatrix}\) and \(|\overrightarrow{AC}|=\sqrt{16+16+64}\)
✓ 1
A1 \(4\sqrt6\) km (\(\approx9.798\))
✓ 1
Exam tip: To show three points are collinear, show two of the vectors between them are parallel and that they share a point.
312IBMA3H-3-18r8-113.17 Equations of a planePaper 2Medium5 marks
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Two antenna direction vectors are \(\mathbf a=\begin{pmatrix}k\\2\\-1\end{pmatrix}\) and \(\mathbf b=\begin{pmatrix}3\\k\\4\end{pmatrix}\), where \(k\) is a constant.
(a)
The antennas are perpendicular. Find the value of \(k\).
(2)
(b)
When \(k=2\), find the angle between the antennas, to the nearest tenth of a degree.
(3)
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Part (a)
M1 set \(\mathbf a\cdot\mathbf b=0\): \(3k+2k-4=0\)
✓ 1
A1 \(k=\frac45\)
✓ 1
Part (b)
A1 \(\mathbf a\cdot\mathbf b=6\), \(|\mathbf a|=3\) and \(|\mathbf b|=\sqrt{29}\)
✓ 1
M1 \(\cos\theta=\frac{6}{3\sqrt{29}}\)
✓ 1
A1 \(68.2^\circ\)
✓ 1
Common error: Perpendicular vectors have scalar product zero, not magnitude product zero; do not set \(|\mathbf a||\mathbf b|\cos\theta\) equal to 1 or to the lengths.
313IBMA3H-3-19r10-043.7 Circular functions and their transformationsPaper 2Medium6 marks
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The air temperature \(T\) °C in a coffee roaster, \(t\) minutes after a roast begins, is modelled by \(T=190+25\sin(0.4t)\) for \(0\le t\le10\), where the angle is in radians.
(a)
Find the two times in the interval at which \(T=205\). Give your answers to three significant figures.
(4)
(b)
Hence find the total time within the interval for which the temperature exceeds 205 °C. Use unrounded values from part (a) and give your answer to three significant figures.
(2)
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Part (a)
M1 \(\sin(0.4t)=0.6\)
✓ 1
A1 \(0.4t=0.6435\), so \(t=1.61\) minutes
✓ 1
M1 use the second solution \(0.4t=\pi-0.6435\ldots\)
Exam tip: On \(0\le t\le10\) the angle \(0.4t\) runs from 0 to 4 radians, so check how many sine solutions fit before you stop.
314IBMA3H-3-19r10-053.1 Three-dimensional geometry: distance, solids and anglesPaper 2Medium7 marks
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A glass canopy at the entrance of a visitor centre beside a bridge is in the shape of a right pyramid VABCD. The base ABCD is a horizontal square of side 8.0 m and the apex V is 6.0 m vertically above the centre M of the base. Give lengths to three significant figures and angles to three significant figures.
Right pyramid VABCD with square base ABCD and centre M. Diagram not to scale
(a)
Find the length of the sloping edge VA.
(2)
(b)
Find the angle between VA and the base.
(2)
(c)
Find the angle between the face VAB and the base.
(3)
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Part (a)
M1 \(AM=4\sqrt2\) (or \(AM^2=32\))
✓ 1
A1 \(VA=\sqrt{6^2+32}=8.25\) m
✓ 1
Part (b)
M1 \(\tan\hat{V}AM=\frac{6}{4\sqrt2}\)
✓ 1
A1 \(46.7^\circ\)
✓ 1
Part (c)
M1 N is the midpoint of AB and \(MN=4\) m
✓ 1
M1 \(\tan\hat{V}NM=\frac{6}{4}\)
✓ 1
A1 \(56.3^\circ\)
✓ 1
Exam tip: The angle between a sloping face and the base is measured along a line at right angles to their common edge, so use the midpoint of the base edge, not the corner.
315IBMA3H-3-19r10-123.17 Equations of a planePaper 2Medium7 marks
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A landing platform lies in the plane \(\Pi\) with equation \(2x-3y+6z=19\). A probe's approach beam follows the line \(L:\ \mathbf r=\begin{pmatrix}3\\1\\0\end{pmatrix}+\lambda\begin{pmatrix}1\\2\\2\end{pmatrix}\). Coordinates are in km.
(a)
Find the coordinates of the point where \(L\) meets \(\Pi\).
(3)
(b)
Find the acute angle between \(L\) and \(\Pi\). Give your answer to three significant figures.
(4)
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Part (a)
M1 substitute \((3+\lambda,\ 1+2\lambda,\ 2\lambda)\) into the equation of \(\Pi\)
✓ 1
A1 \(3+8\lambda=19\), so \(\lambda=2\)
✓ 1
A1 \((5,5,4)\)
✓ 1
Part (b)
M1 normal \(\mathbf n=(2,-3,6)\) and direction \(\mathbf d=(1,2,2)\)
Common error: The angle between a line and a plane uses \(\sin\theta\) with the normal vector; using \(\cos\theta\) gives the angle between the line and the normal instead.
316IBMA3H-3-18r11-043.7 Circular functions and their transformationsPaper 2Medium8 marks
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A wind-turbine blade of length 45 m rotates at 10 revolutions per minute. The hub is 80 m above the ground. At \(t=0\) the blade tip is level with the hub and moving upwards. The height of the tip is modelled by \(h(t)=a\sin(bt)+c\) metres, where \(t\) is in seconds and \(b>0\).
Side view at t = 0; not to scale
(a)
Find the values of \(a\), \(b\) and \(c\).
(3)
(b)
Find the two times in the first revolution when the tip is 100 m above the ground. Give your answers to 3 significant figures (angles in radians).
(3)
(c)
Hence find the time in one revolution for which the tip is above 100 m. Give your answer to 3 significant figures.
(2)
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Part (a)
A1 \(a=45\)
✓ 1
A1 \(b=\frac{2\pi}{6}=\frac{\pi}{3}\)
✓ 1
A1 \(c=80\)
✓ 1
Part (b)
M1 \(45\sin\left(\frac{\pi t}{3}\right)=20\) so \(\sin\left(\frac{\pi t}{3}\right)=\frac49\)
✓ 1
A1 \(t=0.440\) s
✓ 1
A1 \(t=2.56\) s (using \(\pi-\arcsin\frac49\))
✓ 1
Part (c)
M1 difference of the two times
✓ 1
A1 \(2.12\) s
✓ 1
Common error: Set the calculator to radians when the model uses \(\pi\) in the argument, and remember the second solution of \(\sin\alpha=k\) is \(\pi-\alpha\).
317IBMA3H-3-18r11-053.2 Right-angled and non-right-angled trianglesPaper 2Medium7 marks
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On an open-water swim course, buoys \(A\), \(B\) and \(C\) are placed so that \(AB=40\) m, angle \(BAC=35^\circ\) and \(BC=28\) m. Angles are to be given to 1 decimal place and lengths to 3 significant figures.
Diagram not to scale
(a)
Show that there are two possible sizes for angle \(ACB\) and find both.
(4)
(b)
Find the greater of the two possible lengths of \(AC\).
(3)
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Part (a)
M1 sine rule: \(\frac{\sin C}{40}=\frac{\sin35^\circ}{28}\)
✓ 1
A1 \(\sin C=0.819\) (approximately)
✓ 1
A1 \(C=55.0^\circ\)
✓ 1
A1 \(C=180^\circ-55.0^\circ=125.0^\circ\) (also valid since \(BC<AB\))
M1 sine rule: \(AC=\frac{28\sin B}{\sin35^\circ}\)
✓ 1
A1 \(AC=48.8\) m
✓ 1
Common error: Given two sides and a non-included angle (the ambiguous case), always test whether the obtuse angle \(180^\circ-C\) also gives a valid triangle.
318IBMA3H-3-18r11-063.6 Pythagorean and double angle identitiesPaper 2Medium6 marks
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A library lighting controller uses the signal \(f(x)=6\sin^2x+\cos x\) for \(0\leq x\leq 2\pi\) (radians). The controller switches on when \(f(x)=5\).
(a)
Show that the switching values satisfy \(6\cos^2x-\cos x-1=0\).
(2)
(b)
Solve the equation to find all values of \(x\) in the domain, to 3 significant figures where not exact.
(4)
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Part (a)
M1 use \(\sin^2x=1-\cos^2x\): \(6-6\cos^2x+\cos x=5\)
✓ 1
A1 \(6\cos^2x-\cos x-1=0\)
✓ 1
Part (b)
M1 factorise: \((3\cos x+1)(2\cos x-1)=0\)
✓ 1
A1 \(\cos x=\frac12\) or \(\cos x=-\frac13\)
✓ 1
A1 \(x=1.05\) and \(x=5.24\)
✓ 1
A1 \(x=1.91\) and \(x=4.37\)
✓ 1
Exam tip: An equation mixing \(\sin^2x\) and \(\cos x\) should be converted entirely to \(\cos x\) using \(\sin^2x+\cos^2x=1\) before solving.
319IBMA3H-3-18r12-033.3 Applications of trigonometry: elevation, depression and bearingsPaper 2Medium6 marks
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In a simplified model, the Sun S, Earth E and Mars M lie in a plane. The distance SE is 1.00 AU, the distance SM is 1.52 AU and angle ESM is 35°.
Diagram not to scale
(a)
Calculate the distance EM to 3 significant figures.
(3)
(b)
Find the size of angle SEM, giving your answer in degrees to the nearest 0.1°. Use the unrounded value of EM.
Exam tip: When the sine rule could give an acute or obtuse angle, use the cosine rule instead, or check which angle faces the longest side.
320IBMA3H-3-18r12-043.2 Right-angled and non-right-angled trianglesPaper 2Medium6 marks
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A railway surveyor models three signal posts as the vertices of triangle ABC. The distance BC is 8.0 km, the distance AC is 11.0 km and angle BAC is 40°.
Diagram not to scale
(a)
Find the two possible sizes of angle ABC, to the nearest 0.1°.
(3)
(b)
Hence find the two possible lengths of AB, to 3 significant figures.
(3)
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Part (a)
M1 sine rule: \(\frac{\sin B}{11.0} = \frac{\sin40°}{8.0}\)
✓ 1
A1 \(B = 62.1°\)
✓ 1
A1 \(B = 117.9°\) (the obtuse solution)
✓ 1
Part (b)
M1 find angle C for each case: \(77.9°\) and \(22.1°\) and use the sine rule \(AB = \frac{8.0\sin C}{\sin40°}\)
✓ 1
A1 \(AB = 12.2\) km
✓ 1
A1 \(AB = 4.68\) km
✓ 1
Common error: Because \(\sin\theta = \sin(180° - \theta)\), the sine rule gives two angles when the side opposite the known angle is the shorter of the two given sides; check both.
321IBMA3H-3-18r12-053.3 Applications of trigonometry: elevation, depression and bearingsPaper 2Medium7 marks
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An orienteering athlete runs 4.1 km from the start O on a bearing of 035° to checkpoint P. She then runs 3.3 km on a bearing of 110° to checkpoint Q, and finally runs in a straight line from Q back to O.
Diagram not to scale
(a)
Show that angle OPQ is 105°.
(2)
(b)
Calculate the length of QO to 3 significant figures.
(3)
(c)
Calculate the area enclosed by her three legs, to 3 significant figures.
(2)
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Part (a)
M1 the bearing of O from P is 215°, and the bearing of Q from P is 110°, so the angle between them is \(215° - 110°\) (or co-interior angles with the north lines)
Exam tip: For bearings questions, mark the north line at each point and use parallel lines to find the interior angle before using any triangle rule.
322IBMA3H-3-18r12-103.15 Intersecting, parallel and skew linesPaper 2Medium7 marks
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Two straight railway lines are modelled in three dimensions, with distances in kilometres, by \(L_1\): \(\mathbf{r} = \begin{pmatrix}1\\2\\0\end{pmatrix} + s\begin{pmatrix}2\\1\\1\end{pmatrix}\) and \(L_2\): \(\mathbf{r} = \begin{pmatrix}0\\4\\-2\end{pmatrix} + t\begin{pmatrix}3\\-1\\3\end{pmatrix}\).
(a)
Show that \(L_1\) and \(L_2\) intersect, and find the coordinates of the point of intersection.
(4)
(b)
Find the acute angle between \(L_1\) and \(L_2\), to the nearest 0.1°.
Common error: Solve for the parameters using two equations only, then check the third; if it fails the lines are skew, not intersecting.
323IBMA3H-3-18r12-113.17 Equations of a planePaper 2Medium6 marks
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A tunnel survey models a boring line by \(\mathbf{r} = \begin{pmatrix}1\\0\\-1\end{pmatrix} + \lambda\begin{pmatrix}1\\2\\1\end{pmatrix}\) and a layer of rock by the plane \(\Pi\): \(2x - y + 2z = 9\), with distances in hundreds of metres.
(a)
Find the coordinates of the point where the boring line meets \(\Pi\).
(3)
(b)
Find the acute angle between the boring line and \(\Pi\), to the nearest 0.1°.
(3)
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Part (a)
M1 substitute \((1+\lambda, 2\lambda, -1+\lambda)\) into the plane equation: \(2(1+\lambda) - 2\lambda + 2(-1+\lambda) = 9\)
✓ 1
A1 \(2\lambda = 9\), so \(\lambda = 4.5\)
✓ 1
A1 \((5.5, 9, 3.5)\)
✓ 1
Part (b)
M1 \(\sin\theta = \frac{|\mathbf{d}\cdot\mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}\) with \(\mathbf{n} = (2,-1,2)\) and \(\mathbf{d} = (1,2,1)\)
✓ 1
A1 \(\sin\theta = \frac{2}{\sqrt{6}\times3}\)
✓ 1
A1 15.8°
✓ 1
Common error: The angle between a line and a plane uses \(\sin\theta\) with the normal vector; using \(\cos\theta\) gives the angle between the line and the normal instead.
324IBMA3H-3-19r12-053.7 Circular functions and their transformationsPaper 2Medium7 marks
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The depth of water \(h\) metres in a tidal channel is modelled by \(h(t)=d+a\cos\big(b(t-3.5)\big)\), where \(t\) is the time in hours after midnight and the argument is in radians. A high tide of 9.4 m occurs at \(t=3.5\). The next low tide of 2.6 m occurs at \(t=9.7\).
(a)
Find the values of \(a\), \(d\) and \(b\). Give \(b\) to 3 significant figures.
(4)
(b)
A turbine generates power only when \(h\ge 7.5\). Find the number of hours, in the interval \(0\le t\le12.4\), for which the turbine generates power. Use the unrounded value of \(b\) and give your answer to 3 significant figures.
(3)
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Part (a)
A1 \(a=\frac{9.4-2.6}{2}=3.4\)
✓ 1
A1 \(d=\frac{9.4+2.6}{2}=6.0\)
✓ 1
M1 period \(=2(9.7-3.5)=12.4\) hours
✓ 1
A1 \(b=\frac{2\pi}{12.4}=0.5067\), so \(b=0.507\)
✓ 1
Part (b)
M1 \(6+3.4\cos(b(t-3.5))=7.5\), so \(\cos(b(t-3.5))=\frac{1.5}{3.4}=0.4412\)
✓ 1
M1 half-width \(\frac{\arccos(0.4412)}{b}=2.1983\) hours either side of \(t=3.5\) (so \(t\) from 1.30 to 5.70)
✓ 1
A1 duration \(4.397\), so 4.40 hours
✓ 1
Exam tip: Half the time between a maximum and the next minimum is half a period, so double it to find the period before calculating \(b=\frac{2\pi}{\text{period}}\).
325IBMA3H-3-19r12-093.14 Vector equation of a linePaper 2Medium11 marks
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Two drones inspect a wind farm. Relative to a fixed origin, with distances in metres and time \(t\) in seconds (\(t\ge0\)), the position vectors are \(\mathbf r_A=\begin{pmatrix}20\\10\\30\end{pmatrix}+t\begin{pmatrix}4\\3\\-2\end{pmatrix}\) for drone A and \(\mathbf r_B=\begin{pmatrix}105\\70\\25\end{pmatrix}+t\begin{pmatrix}-3\\-2\\-1\end{pmatrix}\) for drone B.
(a)
Show that the paths of the two drones intersect, and find the coordinates of the point of intersection.
(4)
(b)
Show that the drones do not collide at the point of intersection.
(2)
(c)
Show that the square of the distance between the drones at time \(t\) is \(75t^2-1800t+10850\).
(2)
(d)
Hence find the minimum distance between the drones for \(t\ge0\). Give your answer in metres to 3 significant figures.
(3)
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Part (a)
M1 equates components using separate parameters \(s\) and \(u\): \(20+4s=105-3u\), \(10+3s=70-2u\), \(30-2s=25-u\)
✓ 1
A1 solving two equations gives \(s=10\), \(u=15\)
✓ 1
R1 checks the third equation: \(30-20=10=25-15\) is satisfied, so the paths intersect
✓ 1
A1 point \((60,40,10)\)
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Part (b)
A1 drone A is at \((60,40,10)\) when \(t=10\) and drone B is there when \(t=15\)
✓ 1
R1 they are at the point at different times, so they are not there together
M1 completes the square, \(75(t-12)^2+50\), or sets the derivative \(150t-1800=0\)
✓ 1
A1 minimum at \(t=12\) s, with squared distance 50
✓ 1
A1 minimum distance \(\sqrt{50}=7.071\), so 7.07 m
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Exam tip: Intersecting paths are not the same as a collision: use separate parameters to find where paths cross, then compare the times.
326IBMA3H-3-19r12-103.7 Circular functions and their transformationsPaper 2Medium7 marks
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The number of customers entering a shop per hour is modelled by \(F(t)=40+30\sin^2\left(\frac{\pi t}{12}\right)\), where \(t\) is the number of hours after opening and \(0\le t\le12\). The argument is in radians.
(a)
Show that \(F(t)=55-15\cos\left(\frac{\pi t}{6}\right)\).
(3)
(b)
Write down the minimum value of \(F\).
(1)
(c)
Find the total number of hours, in \(0\le t\le12\), for which \(F>60\). Give your answer to 3 significant figures.
(3)
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Part (a)
M1 \(\cos2A=1-2\sin^2A\) with \(A=\frac{\pi t}{12}\), so \(\sin^2A=\frac{1-\cos2A}{2}\)
M1 \(55-15\cos\left(\frac{\pi t}{6}\right)=60\), so \(\cos\left(\frac{\pi t}{6}\right)=-\frac13\)
✓ 1
A1 boundaries \(t=3.649\) and \(t=8.351\) (3.65 and 8.35)
✓ 1
A1 duration \(4.702\), so 4.70 hours
✓ 1
Exam tip: To write a squared sine or cosine in terms of a double angle, rearrange \(\cos2A=1-2\sin^2A\) or \(\cos2A=2\cos^2A-1\).
327IBMA3H-3-19r12-113.11 Symmetry properties of trigonometric graphsPaper 2Medium6 marks
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In a model of two overlapping seasonal waves of infection, the combined weekly index is \(I(x)=\sin\left(x+\frac\pi6\right)+\cos\left(x+\frac\pi3\right)\), where \(x\) is in radians.
(a)
Show that \(I(x)=\cos x\).
(3)
(b)
Hence solve \(I(x)=0.4\) for \(-\pi\le x\le\pi\). Give your answers to 3 significant figures.
Common error: Do not write \(\sin(x+\frac\pi6)\) as \(\sin x+\sin\frac\pi6\); expand with the compound angle identity.
328IBMA3H-3-18r15-073.12 Vectors: concepts and algebraPaper 2Medium12 marks
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A drone filming a Formula 1 pit lane flies in a straight line from \(A(1,0,5)\) to \(B(9,-4,13)\). Coordinates are in metres relative to a fixed origin on the ground; the \(z\)-axis points vertically upwards.
(a)
Write down the vector \(\overrightarrow{AB}\).
(1)
(b)
Find the distance \(AB\).
(2)
(c)
The drone passes through point \(C\) where \(AC:CB=1:3\). Find the coordinates of \(C\).
(3)
(d)
Find the coordinates of the point \(P\) on line \(AB\) at which the drone is at a height of 10 m.
(3)
(e)
Find the acute angle between \(AB\) and the vertical, in degrees to 3 significant figures.
Exam tip: For a ratio \(AC:CB=1:3\), point \(C\) is one quarter of the way from \(A\) to \(B\), so use \(\frac{1}{1+3}\overrightarrow{AB}\).
329IBMA3H-3-18r15-103.18 Intersections and angles of lines and planesPaper 2Medium10 marks
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A laser beam in a hospital theatre travels along the line \(L:\ \mathbf r=\begin{pmatrix}1\\3\\-2\end{pmatrix}+t\begin{pmatrix}1\\1\\2\end{pmatrix}\) and meets a flat screen modelled by the plane \(\Pi:\ 2x-y+2z=5\). Distances are in metres.
(a)
Find the coordinates of the point where \(L\) meets \(\Pi\).
(3)
(b)
Find the acute angle between \(L\) and \(\Pi\), in degrees to 3 significant figures.
(4)
(c)
Find the perpendicular distance from the point \((1,3,-2)\) to \(\Pi\).
(3)
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Part (a)
M1 substitute \(x=1+t,\ y=3+t,\ z=-2+2t\) into plane
M1 \(\frac{|2\times1-3+2\times(-2)-5|}{\sqrt{2^{2}+(-1)^{2}+2^{2}}}\) or foot of perpendicular method
✓ 1
A1 \(\frac{10}{3}\)
✓ 1
A1 \(3.33\) m
✓ 1
Common error: The angle between a line and a plane uses \(\sin\theta\) with the normal vector; using \(\cos\theta\) gives the angle to the normal instead.
330IBMA3H-3-18r15-143.14 Vector equation of a linePaper 2Medium8 marks
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Two drones film an apiary. Relative to a fixed origin (in metres) their paths are the lines \(L_1:\ \mathbf r=\begin{pmatrix}1\\2\\3\end{pmatrix}+s\begin{pmatrix}2\\1\\-1\end{pmatrix}\) and \(L_2:\ \mathbf r=\begin{pmatrix}1\\5\\-2\end{pmatrix}+t\begin{pmatrix}1\\-1\\2\end{pmatrix}\).
(a)
Show that the two paths intersect.
(4)
(b)
Find the coordinates of the point of intersection.
(1)
(c)
Find the acute angle between the paths, in degrees to 3 significant figures.
Common error: If \(\mathbf d_1\cdot\mathbf d_2\) is negative, the angle found is obtuse; take the modulus to get the acute angle between the lines.
331IBMA3H-3-19r16-033.14 Vector equation of a linePaper 2Medium6 marks
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A space probe moves in a straight line. Its position, in km, relative to a tracking station at time \(t\) hours is given by \(\mathbf{r}=\begin{pmatrix}4\\-2\\5\end{pmatrix}+t\begin{pmatrix}1\\2\\-2\end{pmatrix}\).
(a)
Find the speed of the probe.
(2)
(b)
Show that the probe passes through the point \(Q(7,4,-1)\).
(2)
(c)
The probe later reaches the point where \(x=10\). Find the distance the probe travels between \(Q\) and this point.
(2)
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Part (a)
M1 magnitude of the direction vector \(\sqrt{1^2+2^2+(-2)^2}\)
✓ 1
A1 3 km h−1
✓ 1
Part (b)
M1 equate \(x\)-components: \(4+t=7\) so \(t=3\)
✓ 1
A1 substitute \(t=3\) to get \(y=-2+6=4\) and \(z=5-6=-1\), so the probe passes through \(Q\)
✓ 1
Part (c)
M1 \(t=6\), so 3 hours after \(Q\) (accept finding both points and using their separation)
✓ 1
A1 \(3\times3=9\) km
✓ 1
Exam tip: Check the last two components after finding the parameter from the first; one matching component is not enough to show a point lies on a line.
332IBMA3H-3-19r16-043.18 Intersections and angles of lines and planesPaper 2Medium7 marks
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Two straight cables on a model of a bridge are represented by the lines \(L_1\) and \(L_2\), where lengths are in metres. \(L_1:\ \mathbf{r}=\begin{pmatrix}1\\0\\2\end{pmatrix}+s\begin{pmatrix}1\\2\\-1\end{pmatrix}\) and \(L_2:\ \mathbf{r}=\begin{pmatrix}5\\-2\\1\end{pmatrix}+t\begin{pmatrix}-2\\6\\-1\end{pmatrix}\)
(a)
Show that the cables meet and find the coordinates of the point where they meet.
(4)
(b)
Find the acute angle between the two cables, in degrees to one decimal place.
A1 solve two of the equations, e.g. \(s=1+t\) and \(t=1\), giving \(s=2\)
✓ 1
R1 check the third equation: \(4=4\), so the lines meet
✓ 1
A1 \((3,4,0)\)
✓ 1
Part (b)
M1 scalar product of direction vectors \(1\times(-2)+2\times6+(-1)\times(-1)=11\)
✓ 1
A1 \(\cos\theta=\dfrac{11}{\sqrt{6}\sqrt{41}}\)
✓ 1
A1 \(45.5^\circ\)
✓ 1
Exam tip: When equating components of two lines use different parameters, and always test the unused equation.
333IBMA3H-3-19r16-083.7 Circular functions and their transformationsPaper 2Medium8 marks
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The temperature \(T\) °C of the drum of a coffee roaster oscillates because of its thermostat. The temperature is modelled by \(T=a\sin(bt)+c\), where \(t\) is time in seconds, \(t\ge 0\) and \(b>0\). The temperature varies between a minimum of 195 °C and a maximum of 215 °C, the period is 40 s, and \(T=205\) when \(t=0\) with the temperature rising.
(a)
Find the values of \(a\), \(b\) and \(c\). Give \(b\) in terms of \(\pi\).
(3)
(b)
Find the first time at which the temperature reaches 212 °C. Give your answer to three significant figures.
(3)
(c)
Find the length of time in each cycle for which the temperature is above 212 °C. Use the unrounded value of \(t\) from (b) and give your answer to three significant figures.
(2)
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Part (a)
A1 \(a=10\)
✓ 1
A1 \(c=205\)
✓ 1
A1 \(b=\dfrac{2\pi}{40}=\dfrac{\pi}{20}\)
✓ 1
Part (b)
M1 \(10\sin\left(\frac{\pi t}{20}\right)+205=212\) so \(\sin\left(\frac{\pi t}{20}\right)=0.7\)
✓ 1
M1 \(\dfrac{\pi t}{20}=\sin^{-1}0.7=0.7754\)
✓ 1
A1 \(t=4.94\) s
✓ 1
Part (c)
M1 by symmetry the temperature is above 212 °C between \(t_1\) and \(20-t_1\), so the duration is \(20-2t_1\)
✓ 1
A1 10.1 s
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Exam tip: Maximum and minimum values give the amplitude and the vertical shift; the period gives \(b\) through \(b=2\pi/\text{period}\).
334IBMA3H-3-19r16-103.14 Vector equation of a linePaper 2Medium6 marks
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Two forces, in kN, act at a joint on a bridge. They are represented by the vectors \(\mathbf{a}=\begin{pmatrix}k\\3\\-2\end{pmatrix}\) and \(\mathbf{b}=\begin{pmatrix}2\\k\\5\end{pmatrix}\), where \(k\) is a constant.
(a)
Find the value of \(k\) for which the two forces are perpendicular.
(3)
(b)
Find the angle between the forces when \(k=4\). Give your answer in degrees to one decimal place.
(3)
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Part (a)
M1 set \(\mathbf{a}\cdot\mathbf{b}=0\)
✓ 1
A1 \(2k+3k-10=0\)
✓ 1
A1 \(k=2\)
✓ 1
Part (b)
M1 \(\mathbf{a}\cdot\mathbf{b}=8+12-10=10\) and \(|\mathbf{a}|=\sqrt{29}\), \(|\mathbf{b}|=\sqrt{45}\)
✓ 1
M1 \(\cos\theta=\dfrac{10}{\sqrt{29}\sqrt{45}}\)
✓ 1
A1 \(73.9^\circ\)
✓ 1
Common error: Perpendicular vectors have a scalar product of zero; setting the vectors' components equal to zero is not the same condition.
335IBMA3H-3-18r17-043.3 Applications of trigonometry: elevation, depression and bearingsPaper 2Medium5 marks
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A stage lighting rig is attached to the top of a vertical tower of height h metres on level ground. A technician at point A, in line with the base of the tower, measures the angle of elevation of the top of the tower as 25°. She walks 18 m directly towards the tower to point B, where the angle of elevation is 32°. Let x metres be the distance from B to the base of the tower.
Diagram not to scale
(a)
Show that x(tan 32° − tan 25°) = 18 tan 25°.
(2)
(b)
Find the value of h. Use the unrounded value of x and give your answer to 3 significant figures.
(3)
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Part (a)
M1 writes h = x tan 32° and h = (x + 18) tan 25°
✓ 1
A1 equates and rearranges to the given result (AG)
✓ 1
Part (b)
M1 x = 18 tan 25° / (tan 32° − tan 25°) (= 52.9…)
✓ 1
M1 h = x tan 32°
✓ 1
A1 h = 33.1 m
✓ 1
Exam tip: With two observation points in line with the object, write the common height in two ways and equate them.
336IBMA3H-3-18r17-053.5 The unit circle, exact values and the ambiguous casePaper 2Medium6 marks
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A triangular field PQR has PQ = 85 m, QR = 70 m and angle QPR = 48°. The angle PRQ is not known to be acute or obtuse. In this question, use unrounded values in later working and give angles to 3 significant figures.
Diagram not to scale
(a)
Find the two possible sizes of angle PRQ.
(3)
(b)
Find the largest possible area of the field. Give your answer to 3 significant figures.
(3)
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Part (a)
M1 sine rule: sin R / 85 = sin 48° / 70
✓ 1
A1 64.5°
✓ 1
A1 115.5° (180° − 64.5°)
✓ 1
Part (b)
M1 finds angle PQR = 180° − 48° − 64.5° = 67.5° (the larger angle Q gives the larger area)
✓ 1
M1 area = ½ × 85 × 70 × sin(angle PQR)
✓ 1
A1 2750 m2
✓ 1
Common error: The sine rule can give two angles in (0°, 180°) with the same sine; check whether the triangle is still possible with the obtuse angle before discarding it.
337IBMA3H-3-18r17-063.3 Applications of trigonometry: elevation, depression and bearingsPaper 2Medium6 marks
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An air traffic radar is at a point O on the ground. Aircraft A is 12 km from O, on a bearing of 070°, at an angle of elevation of 15° from O. Aircraft B is 9 km from O, on a bearing of 160°, at an angle of elevation of 25° from O. Let A′ and B′ be the points on the ground directly below A and B.
(a)
Find the lengths OA′ and OB′, to 3 significant figures.
(2)
(b)
Find the distance A′B′, giving your answer to 3 significant figures.
(2)
(c)
Find the distance AB between the two aircraft. Use unrounded values and give your answer to 3 significant figures.
M1 heights 12 sin 15° = 3.11 and 9 sin 25° = 3.80, so height difference 0.698 km
✓ 1
A1 AB = √(A′B′2 + 0.6982) = 14.2 km
✓ 1
Exam tip: For 3D bearing problems, split each position into a horizontal distance (using cos of the elevation) and a height (using sin of the elevation).
338IBMA3H-3-18r17-113.14 Vector equation of a linePaper 2Medium7 marks
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Relative to a control tower, the flight paths of two aircraft are modelled by the lines L1: r = (1, −2, 3) + λ(2, 1, −1) and L2: r = (4, 1, −1) + μ(1, −1, 2), where coordinates are in kilometres, the third coordinate is altitude, and λ, μ are real parameters.
(a)
Show that the two lines intersect, and find the position vector of the point of intersection.
(4)
(b)
Find the acute angle between the two flight paths. Give your answer to the nearest 0.1°.
M1 solves two of the equations simultaneously, e.g. λ = 2 and μ = 1
✓ 1
M1 checks the third equation is satisfied (3 − 2 = 1 = −1 + 2)
✓ 1
A1 position vector (5, 0, 1)
✓ 1
Part (b)
M1 attempts the scalar product of the direction vectors: (2, 1, −1)·(1, −1, 2) = −1
✓ 1
A1 magnitudes √6 and √6, so cos θ = 1/6 (using the absolute value for the acute angle)
✓ 1
A1 80.4°
✓ 1
Common error: If the scalar product is negative, the angle found is obtuse: take the absolute value of the cosine to get the acute angle between the lines.
339IBMA3H-3-18r17-143.10 Compound and double angle identitiesPaper 2Medium7 marks
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At an open-air concert, a vertical video screen has its bottom edge 2 m above the eye level of a spectator and its top edge 8 m above eye level. The spectator stands at a horizontal distance d metres from the screen. The angle subtended at her eye by the screen is θ, so θ = arctan(8/d) − arctan(2/d).
Diagram not to scale
(a)
Show that tan θ = 6d / (d2 + 16).
(3)
(b)
Find the value of θ, in radians, when d = 10. Give your answer to 3 significant figures.
(1)
(c)
The spectator moves until θ = π/6. Find the greater of the two possible values of d, to 3 significant figures.
(3)
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Part (a)
M1 applies tan(A − B) = (tan A − tan B)/(1 + tan A tan B) with tan A = 8/d, tan B = 2/d
✓ 1
M1 obtains (6/d) / (1 + 16/d2)
✓ 1
A1 multiplies numerator and denominator by d2 to reach the given result (AG)
Exam tip: Whenever a question gives an angle as a difference of two arctangents, take the tangent of both sides and use the compound angle formula for tan(A − B).
340IBMA3H-3-19r18-073.18 Intersections and angles of lines and planesPaper 2Medium9 marks
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Two taut cables pass through a theatre fly tower, with coordinates in metres. Cable \(L_1\) has equation \(\boldsymbol{r}=\begin{pmatrix}1\\0\\2\end{pmatrix}+s\begin{pmatrix}1\\2\\-1\end{pmatrix}\) and cable \(L_2\) has equation \(\boldsymbol{r}=\begin{pmatrix}1\\5\\-1\end{pmatrix}+u\begin{pmatrix}2\\-1\\1\end{pmatrix}\).
(a)
Show that the cables intersect and find the coordinates of the point of intersection.
(5)
(b)
Find the acute angle between the cables, correct to 3 significant figures.
Common error: Use the modulus of the scalar product (or subtract from \(180^\circ\)) when the question asks for the acute angle between lines.
341IBMA3H-3-19r18-083.1 Three-dimensional geometry: distance, solids and anglesPaper 2Medium8 marks
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A stage prop is a solid square-based pyramid $ABCDV$. The base \(ABCD\) is a horizontal square of side 2.4 m and the apex \(V\) is 1.8 m vertically above the centre of the base. Give answers to 3 significant figures, using unrounded intermediate values.
Diagram not to scale
(a)
Calculate the length of the sloping edge \(VA\).
(2)
(b)
Calculate the angle between \(VA\) and the base.
(2)
(c)
Calculate the angle between the sloping face \(VAB\) and the base.
(2)
(d)
Calculate the volume of the prop.
(2)
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Part (a)
M1 half-diagonal \(=\frac{2.4\sqrt2}{2}=1.697\), then \(VA=\sqrt{1.697^2+1.8^2}\)
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A1 \(VA=2.47\) m
✓ 1
Part (b)
M1 \(\tan\theta=\frac{1.8}{1.697}\)
✓ 1
A1 \(\theta=46.7^\circ\)
✓ 1
Part (c)
M1 \(\tan\phi=\frac{1.8}{1.2}\) (using the foot of the perpendicular to the midpoint of \(AB\))
✓ 1
A1 \(\phi=56.3^\circ\)
✓ 1
Part (d)
M1 \(V=\frac13\times2.4^2\times1.8\)
✓ 1
A1 \(3.46\) m3 (3.456)
✓ 1
Exam tip: The angle between a face and the base is measured along a line perpendicular to the common edge, so use the midpoint of the base edge, not the corner.
342IBMA3H-3-19r18-093.14 Vector equation of a linePaper 2Medium11 marks
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Two seismic ray paths from a station are modelled by the vectors \(\boldsymbol{u}=\begin{pmatrix}k\\2\\-1\end{pmatrix}\) and \(\boldsymbol{v}=\begin{pmatrix}2\\1\\4\end{pmatrix}\).
(a)
Find the value of \(k\) for which the paths are perpendicular.
(3)
(b)
For \(k=2\), find the angle between the paths, correct to 3 significant figures.
(4)
(c)
For \(k=2\), the two vectors form two sides of a triangle. Find the area of the triangle, correct to 3 significant figures.
M1 area \(=\frac12|\boldsymbol{u}\times\boldsymbol{v}|=\frac12\sqrt{81+100+4}\)
✓ 1
A1 \(6.80\) (6.8007…)
✓ 1
Exam tip: The area of a triangle with sides given by two vectors is half the magnitude of their vector product; do not forget the factor \(\frac12\).
343IBMA3H-3-19r18-123.3 Applications of trigonometry: elevation, depression and bearingsPaper 2Medium9 marks
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A volcano's summit \(S\) is vertically above the point \(T\) on level ground. Two observation posts \(A\) and \(B\) are on the same level ground, with \(\angle ATB=90^\circ\) and \(AB=3.2\) km. The angle of elevation of \(S\) from \(A\) is \(28^\circ\) and from \(B\) is \(19^\circ\). Let the height of the summit above the ground be \(h\) km.
Sketch of the situation, not to scale
(a)
Write \(AT\) and \(BT\) in terms of \(h\).
(2)
(b)
Show that \(h^2\left(\frac{1}{\tan^228^\circ}+\frac{1}{\tan^219^\circ}\right)=3.2^2\).
(2)
(c)
Find \(h\), correct to 3 significant figures.
(3)
(d)
\(M\) is the midpoint of \(AB\). Find the angle of elevation of \(S\) from \(M\), correct to 3 significant figures. Use unrounded values.
(2)
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Part (a)
A1 \(AT=\frac{h}{\tan28^\circ}\)
✓ 1
A1 \(BT=\frac{h}{\tan19^\circ}\)
✓ 1
Part (b)
M1 Pythagoras in triangle \(ATB\): \(AT^2+BT^2=AB^2\)
✓ 1
A1 substitution of the expressions from part (a) (AG)
✓ 1
Part (c)
M1 \(h^2=\frac{3.2^2}{3.5371+8.4345}\)
✓ 1
A1 \(h^2=0.8554\)
✓ 1
A1 \(h=0.925\) km
✓ 1
Part (d)
M1 \(TM=\frac12AB=1.6\) km (median to the hypotenuse of a right-angled triangle) and \(\tan\alpha=\frac{h}{1.6}\)
✓ 1
A1 \(\alpha=30.0^\circ\)
✓ 1
Common error: Calculators must be in degree mode for this question; check by confirming \(\tan 45^\circ=1\).
344IBMA3H-3-18r22-013.14 Vector equation of a linePaper 2Medium7 marks
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Two space probes move in straight lines, with positions in kilometres relative to a station. Probe A has position \(\mathbf r=\begin{pmatrix}10\\-4\\6\end{pmatrix}+t\begin{pmatrix}1\\2\\2\end{pmatrix}\) and probe B has position \(\mathbf r=\begin{pmatrix}0\\5\\3\end{pmatrix}+s\begin{pmatrix}2\\3\\6\end{pmatrix}\), where \(t\) and \(s\) are parameters.
(a)
Find the acute angle between the paths of the two probes. Give your answer in degrees to 3 significant figures.
A1 solving the first two equations gives \(s=29\), \(t=48\)
✓ 1
M1 substitute into the third equation: left side 102, right side 177
✓ 1
R1 the values are not equal, so the paths do not intersect (they are skew)
✓ 1
Exam tip: If the angle between lines is asked for, use the direction vectors only; the position vectors do not matter.
345IBMA3H-3-18r22-043.18 Intersections and angles of lines and planesPaper 2Medium8 marks
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Three flat heating plates in a coffee roaster lie in the planes \(\Pi_1: x+y+z=6\), \(\Pi_2: 2x-y+z=3\) and \(\Pi_3: 3x+2z=9\).
(a)
Show that the three planes meet in a line, and not in a single point.
(2)
(b)
Find a vector equation of the line of intersection.
(3)
(c)
A fourth plate lies in the plane \(\Pi_4: x-y+2z=10\). Find the coordinates of the point where the line in (b) meets \(\Pi_4\).
(3)
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Part (a)
M1 adding the equations of \(\Pi_1\) and \(\Pi_2\) gives the equation of \(\Pi_3\)
✓ 1
R1 \(\Pi_1\) and \(\Pi_2\) are not parallel (normals \((1,1,1)\) and \((2,-1,1)\) are not multiples), so they meet in a line that also lies in \(\Pi_3\)
✓ 1
Part (b)
M1 set \(z=3s\) (or similar) and solve \(\Pi_3\) for \(x\), then \(\Pi_1\) for \(y\)
✓ 1
A1 \(x=3-2s\), \(y=3-s\), \(z=3s\)
✓ 1
A1 \(\mathbf r=(3,3,0)+s(-2,-1,3)\) or equivalent
✓ 1
Part (c)
M1 substitute \((3-2s)-(3-s)+2(3s)=10\)
✓ 1
A1 \(5s=10\), so \(s=2\)
✓ 1
A1 \((-1,1,6)\)
✓ 1
Common error: Showing that two equations add to give a third proves the planes share a line only if the first two planes are not parallel; state this.
346IBMA3H-3-18r22-063.18 Intersections and angles of lines and planesPaper 2Medium7 marks
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In a colour sorter for roasted coffee beans, two laser beams travel along the lines \(l_1: \mathbf r=\begin{pmatrix}1\\2\\3\end{pmatrix}+s\begin{pmatrix}1\\-1\\2\end{pmatrix}\) and \(l_2: \mathbf r=\begin{pmatrix}1\\-1\\8\end{pmatrix}+t\begin{pmatrix}2\\1\\-1\end{pmatrix}\), with distances in centimetres.
(a)
Show that the beams cross and find the coordinates of the crossing point.
(4)
(b)
Find the acute angle between the beams. Give your answer in degrees to 3 significant figures.
A1 \(\mathbf d_1\cdot\mathbf d_2=-1\) and \(|\mathbf d_1|=|\mathbf d_2|=\sqrt6\), giving \(\cos\theta=\frac16\)
✓ 1
A1 \(\theta=80.4^\circ\)
✓ 1
Common error: A negative dot product gives an obtuse angle; take the modulus of the dot product if the acute angle is required.
347IBMA3H-3-18r22-113.2 Right-angled and non-right-angled trianglesPaper 2Medium7 marks
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Tracking stations are at \(A\) and \(B\), and a space probe is at \(C\). Distances are in thousands of kilometres. In triangle \(ABC\), \(AC=9.0\), \(BC=6.5\) and angle \(CAB=40^\circ\). Use unrounded values in later working and give final answers to 3 significant figures.
Diagram not to scale. Distances in thousands of km
(a)
Find the two possible sizes of angle \(ABC\), in degrees to 3 significant figures.
(4)
(b)
Find the smaller possible distance \(AB\).
(3)
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Part (a)
M1 sine rule: \(\frac{\sin B}{9.0}=\frac{\sin40^\circ}{6.5}\)
✓ 1
A1 \(\sin B=0.890\) (0.89002…)
✓ 1
A1 \(B=62.9^\circ\)
✓ 1
A1 \(B=117^\circ\) (180° − 62.9°, i.e. 117.1°)
✓ 1
Part (b)
M1 the smaller \(AB\) comes from the obtuse \(B\): angle \(ACB=180^\circ-40^\circ-117.1^\circ=22.9^\circ\)
Common error: When the given angle is opposite the shorter given side, two triangles exist; check whether 180° minus the first answer also fits.
348IBMA3H-3-18r22-133.11 Symmetry properties of trigonometric graphsPaper 2Medium7 marks
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The angle \(x\), in radians, of a probe's antenna satisfies the equation \(2\cos2x+3\sin x=1\), for \(0\le x\le2\pi\).
(a)
Show that the equation can be written as \(4\sin^2x-3\sin x-1=0\).
(2)
(b)
Hence find all solutions of the equation. Give non-exact answers to 3 significant figures.
(5)
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Part (a)
M1 use \(\cos2x=1-2\sin^2x\): \(2-4\sin^2x+3\sin x=1\)
✓ 1
A1 rearrange to \(4\sin^2x-3\sin x-1=0\)
✓ 1
Part (b)
M1 factorise \((4\sin x+1)(\sin x-1)=0\) or use the quadratic formula
✓ 1
A1 \(\sin x=1\) or \(\sin x=-\frac14\)
✓ 1
A1 \(x=\frac\pi2\) (1.57)
✓ 1
A1 \(x=3.39\)
✓ 1
A1 \(x=6.03\)
✓ 1
Common error: After factorising, solve each factor separately and check every solution lies in the stated interval; sin x = −1/4 gives two values.
349IBMA3H-3-18r23-043.1 Three-dimensional geometry: distance, solids and anglesPaper 2Medium7 marks
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A library skylight is a square-based pyramid $VABCD$. The base \(ABCD\) is horizontal with side 6 m, and the apex \(V\) is 4 m vertically above the centre \(O\) of the base. Give lengths to 3 significant figures and angles to the nearest 0.1°.
Diagram not to scale
(a)
Find the length of the edge \(VA\).
(2)
(b)
Find the angle between the edge \(VA\) and the base.
(2)
(c)
Find the angle between the face \(VAB\) and the base.
(3)
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Part (a)
M1 \(OA=3\sqrt2\) (half the diagonal), so \(VA^2=(3\sqrt2)^2+4^2=34\)
M1 let \(M\) be the midpoint of \(AB\); the required angle is \(VMO\)
✓ 1
A1 \(OM=3\)
✓ 1
A1 \(\tan(VMO)=\frac43\), so \(53.1^\circ\)
✓ 1
Common error: The angle between a face and the base is measured along the line perpendicular to their common edge (to the midpoint \(M\)), not along an edge such as \(VA\).
350IBMA3H-3-18r23-113.10 Compound and double angle identitiesPaper 2Medium6 marks
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The identity \(\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}\) may be used.
(a)
Show that \(\tan\left(x+\frac{\pi}{4}\right)=\frac{1+\tan x}{1-\tan x}\).
(3)
(b)
Hence solve \(\tan\left(x+\frac{\pi}{4}\right)=3\) for \(0\le x\le\pi\). Give your answer to 3 significant figures.
A1 \(x=0.464\) (the only solution in the interval)
✓ 1
Exam tip: Using the part-(a) result turns a compound-angle equation into a linear equation in \(\tan x\).
351IBMA3H-3-19r25-053.2 Right-angled and non-right-angled trianglesPaper 2Medium7 marks
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A hospital \(H\), a clinic \(C\) and an ambulance station \(S\) are the vertices of a triangle. \(HC=9.0\) km, \(CS=7.0\) km and angle \(CHS=40^\circ\).
Diagram not to scale
(a)
Find the two possible sizes of angle \(HSC\), to the nearest 0.1°.
(3)
(b)
Explain why both of these values give a possible triangle.
(1)
(c)
Find the shorter of the two possible distances \(HS\), in km, to 3 significant figures. Use unrounded angles from (a).
(3)
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Part (a)
M1 sine rule: \(\frac{\sin HSC}{9.0}=\frac{\sin 40^\circ}{7.0}\), so \(\sin HSC=0.8264\ldots\)
✓ 1
A1 \(HSC=55.7^\circ\)
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A1 \(HSC=124.3^\circ\)
✓ 1
Part (b)
R1 for \(124.3^\circ\): \(40^\circ+124.3^\circ=164.3^\circ<180^\circ\), so the angles can still sum to \(180^\circ\) (accept the same check for \(55.7^\circ\))
✓ 1
Part (c)
M1 angle \(HCS=180^\circ-40^\circ-124.27^\circ=15.73^\circ\) (shorter distance comes from the obtuse \(HSC\))
A1 \(HS=2.95\) km (the longer distance is \(10.8\) km)
✓ 1
Exam tip: When the given angle is not between the two given sides, always check whether the sine rule also gives an obtuse angle 180° − θ.
352IBMA3H-3-19r25-083.14 Vector equation of a linePaper 2Medium7 marks
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Two bees fly along straight paths. In suitable units, the path of bee A is \(L_1:\ \mathbf{r}=\begin{pmatrix}1\\2\\3\end{pmatrix}+s\begin{pmatrix}2\\-1\\1\end{pmatrix}\) and the path of bee B is \(L_2:\ \mathbf{r}=\begin{pmatrix}1\\-1\\8\end{pmatrix}+t\begin{pmatrix}1\\1\\-2\end{pmatrix}\), where \(s\) and \(t\) are scalar parameters.
(a)
Show that the two paths intersect and find the coordinates of the point of intersection.
(4)
(b)
Find the acute angle between the two paths, to the nearest 0.1°.
M1 check the third equation: \(3+1=4\) and \(8-4=4\), so it is satisfied
✓ 1
A1 point of intersection \((3,\ 1,\ 4)\)
✓ 1
Part (b)
M1 scalar product of direction vectors: \(2\times1+(-1)\times1+1\times(-2)=-1\) with magnitudes \(\sqrt6\) and \(\sqrt6\)
✓ 1
A1 \(\cos\theta=\frac{|-1|}{6}=\frac16\)
✓ 1
A1 \(\theta=80.4^\circ\)
✓ 1
Common error: Giving 99.6° (the obtuse angle) because the scalar product is negative; the question asks for the acute angle, so use the modulus of the scalar product.
353IBMA3H-3-19r25-093.11 Symmetry properties of trigonometric graphsPaper 2Medium6 marks
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Solve the equation \(2\cos2x=3\sin x\) for \(0\le x\le2\pi\). Give your answers correct to 3 significant figures.
(a)
Solve the equation for \(0\le x\le2\pi\).
(6)
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Part (a)
M1 use \(\cos2x=1-2\sin^2x\)
✓ 1
A1 \(4\sin^2x+3\sin x-2=0\)
✓ 1
M1 solve the quadratic in \(\sin x\)
✓ 1
A1 \(\sin x=\frac{-3+\sqrt{41}}{8}=0.4254\ldots\)
✓ 1
R1 reject \(\sin x=\frac{-3-\sqrt{41}}{8}=-1.175\ldots\) since it is less than \(-1\)
✓ 1
A1 \(x=0.439\) and \(x=2.70\)
✓ 1
Exam tip: Choose the form of cos 2x that leaves the equation in a single trigonometric function, here sin x, so that it becomes a quadratic.
354IBMA3H-3-19r25-103.7 Circular functions and their transformationsPaper 2Medium11 marks
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The temperature \(T\,{}^\circ\)C inside a beehive, \(t\) hours after midnight, is modelled for \(0\le t\le24\) by \(T(t)=c+a\cos\big(b(t-15)\big)\), where \(a>0\), \(b>0\) and \(b\) is in radians per hour. The highest temperature is \(36\,{}^\circ\)C, reached at 15:00, the lowest temperature is \(30\,{}^\circ\)C, and the pattern repeats every 24 hours.
(a)
Find the values of \(a\) and \(c\).
(2)
(b)
Show that \(b=\frac{\pi}{12}\).
(2)
(c)
Find the times, to the nearest 0.1 hour, between which the temperature is above \(34\,{}^\circ\)C.
(4)
(d)
Find the total time in the 24 hours from \(t=0\) to \(t=24\) for which the temperature is below \(31\,{}^\circ\)C. Give your answer to 3 significant figures.
(3)
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Part (a)
A1 \(a=\frac{36-30}{2}=3\)
✓ 1
A1 \(c=\frac{36+30}{2}=33\)
✓ 1
Part (b)
M1 period \(=\frac{2\pi}{b}=24\)
✓ 1
A1 \(b=\frac{2\pi}{24}=\frac{\pi}{12}\) (AG)
✓ 1
Part (c)
M1 \(33+3\cos\left(\frac{\pi}{12}(t-15)\right)>34\) so \(\cos\left(\frac{\pi}{12}(t-15)\right)>\frac13\)
M1 use symmetry about the minimum at \(t=3\): the temperature is below \(31\,{}^\circ\)C for \(|t-3|<\frac{12}{\pi}\left(\pi-\cos^{-1}(-\tfrac23)\right)=3.21\) (the interval wraps round midnight, so the total is \(24-2\times\frac{12}{\pi}\cos^{-1}(-\tfrac23)\))
✓ 1
A1 \(6.43\) hours
✓ 1
Common error: Treating the interval near the minimum as a single block from t = 0 to a single upper limit; the cooler period wraps around midnight, so use symmetry about the minimum instead.
355IBMA3H-3-19r25-143.7 Circular functions and their transformationsPaper 2Medium7 marks
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At a hospital charity fair, the height \(h\) metres above the ground of a seat on a Ferris wheel, \(t\) seconds after the seat passes its lowest point, is modelled by \(h(t)=14-12\cos\left(\frac{\pi t}{45}\right)\), where the angle is in radians.
(a)
Write down the minimum height of the seat and the time taken for one complete revolution.
(2)
(b)
Find the two times in the first revolution, \(0\le t\le90\), at which the seat is 20 m above the ground.
(3)
(c)
Find the height of the seat when \(t=20\), to 3 significant figures.
(2)
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Part (a)
A1 minimum height \(=14-12=2\) m
✓ 1
A1 one revolution takes \(\frac{2\pi}{\pi/45}=90\) s
✓ 1
Part (b)
M1 \(14-12\cos\left(\frac{\pi t}{45}\right)=20\) so \(\cos\left(\frac{\pi t}{45}\right)=-\frac12\)
✓ 1
A1 \(\frac{\pi t}{45}=\frac{2\pi}{3}\) so \(t=30\) s
✓ 1
A1 \(\frac{\pi t}{45}=\frac{4\pi}{3}\) so \(t=60\) s
Common error: Using degree mode: the argument πt/45 is in radians, so cos(πt/45) at t = 20 must be evaluated in radian mode.
356IBMA3H-3-18r32-053.7 Circular functions and their transformationsPaper 2Medium8 marks
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The height of the tide at a tidal-energy site is modelled by \(h(t)=5.2+3.5\sin(0.51t)\) metres, where \(t\) is the time in hours after midnight and the argument is in radians. A turbine generates power only when \(h\ge7.0\).
(a)
Find the period of \(h\), correct to 3 significant figures.
(2)
(b)
Find the first time after midnight at which \(h=7.0\), correct to 3 significant figures.
(3)
(c)
Find the length of time, in the first cycle, for which the turbine can generate power. Give your answer to 3 significant figures.
M1 second solution of \(\sin(0.51t)=\frac{1.8}{3.5}\) is \(0.51t=\pi-0.540\ldots\)
✓ 1
M1 time is the difference between the two solutions, \(\frac{\pi-2\times0.5401\ldots}{0.51}\)
✓ 1
A1 4.04 hours
✓ 1
Exam tip: A sine equation has a second solution in each cycle, \(\pi-\theta\), and the time between the two is the duration the condition holds.
357IBMA3H-3-18r32-063.1 Three-dimensional geometry: distance, solids and anglesPaper 2Medium7 marks
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A shop display is a solid right pyramid with a square base of side 60 cm. The vertex is 45 cm vertically above the centre of the base.
Diagram not to scale
(a)
Find the slant height of a triangular face, measured from the vertex to the midpoint of a base edge. Give your answer to 3 significant figures.
(2)
(b)
Find the angle that a sloping edge makes with the base. Give your answer to 3 significant figures.
(3)
(c)
The four triangular faces are to be covered in fabric. Use the unrounded slant height to find the area of fabric needed, correct to 3 significant figures.
(2)
Show mark scheme
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Part (a)
M1 \(\sqrt{45^{2}+30^{2}}\)
✓ 1
A1 54.1 cm
✓ 1
Part (b)
M1 half the diagonal of the base \(=30\sqrt2=42.4\ldots\) cm
✓ 1
M1 \(\tan\theta=\frac{45}{30\sqrt2}\)
✓ 1
A1 46.7°
✓ 1
Part (c)
M1 \(4\times\frac12\times60\times\sqrt{2925}\)
✓ 1
A1 6490 cm2
✓ 1
Common error: The angle between an edge and the base is measured from the half-diagonal of the base, not from half a side.
358IBMA3H-3-18r32-073.1 Three-dimensional geometry: distance, solids and anglesPaper 2Medium8 marks
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A warehouse is a cuboid with horizontal rectangular floor \(ABCD\) and roof $EFGH$, where \(E\), \(F\), \(G\), \(H\) are vertically above \(A\), \(B\), \(C\), \(D\) respectively. \(AB=24\) m, \(AD=18\) m and \(AE=40\) m. A support cable runs in a straight line from corner \(A\) of the floor to corner \(G\) of the roof.
Diagram not to scale
(a)
Find the length of \(AC\).
(2)
(b)
Find the length of the cable \(AG\).
(2)
(c)
Find the angle between the cable and the floor, correct to 3 significant figures.
(2)
(d)
Find the angle between the cable \(AG\) and the edge \(AB\), correct to 3 significant figures.
(2)
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Part (a)
M1 \(\sqrt{24^{2}+18^{2}}\)
✓ 1
A1 30 m
✓ 1
Part (b)
M1 \(\sqrt{30^{2}+40^{2}}\)
✓ 1
A1 50 m
✓ 1
Part (c)
M1 \(\tan\theta=\frac{40}{30}\)
✓ 1
A1 53.1°
✓ 1
Part (d)
M1 \(\cos\theta=\frac{24}{50}\) (triangle \(ABG\) has a right angle at \(B\))
✓ 1
A1 61.3°
✓ 1
Exam tip: Before using a trigonometric ratio in a 3D problem, identify the right-angled triangle and state which angle in it you are finding.
359IBMA3H-3-18r32-083.18 Intersections and angles of lines and planesPaper 2Medium9 marks
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Two underground power cables follow straight lines, with coordinates in metres: \(L_1:\ \mathbf{r}=\begin{pmatrix}1\\2\\3\end{pmatrix}+s\begin{pmatrix}2\\1\\-1\end{pmatrix}\) and \(L_2:\ \mathbf{r}=\begin{pmatrix}3\\8\\-5\end{pmatrix}+t\begin{pmatrix}1\\-2\\3\end{pmatrix}\).
(a)
Show that the cables meet and find the coordinates of the point where they meet.
(5)
(b)
Find the acute angle between the two cables, correct to 3 significant figures.
Common error: If \(\mathbf{d}_1\cdot\mathbf{d}_2\) is negative, the formula gives an obtuse angle; the acute angle between the lines uses the absolute value.
360IBMA3H-3-18r32-093.16 Vector productPaper 2Medium8 marks
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Three wind turbines are at \(A(1,2,3)\), \(B(4,-1,5)\) and \(C(2,3,-1)\), where coordinates are in hundreds of metres relative to a control point.
(a)
Find the vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\).
(2)
(b)
Find angle \(BAC\), correct to 3 significant figures.
(4)
(c)
Find the area of triangle \(ABC\), in square hundreds of metres, correct to 3 significant figures. Use unrounded values.
M1 \(\cos A=\frac{\overrightarrow{AB}\cdot\overrightarrow{AC}}{|\overrightarrow{AB}||\overrightarrow{AC}|}\) with \(|\overrightarrow{AB}|=\sqrt{22}\), \(|\overrightarrow{AC}|=\sqrt{18}\)
✓ 1
A1 \(\cos A=\frac{-8}{\sqrt{396}}\)
✓ 1
A1 114° (113.7…° before rounding)
✓ 1
Part (c)
M1 \(\frac12\sqrt{22}\sqrt{18}\sin A\) or \(\frac12|\overrightarrow{AB}\times\overrightarrow{AC}|\)
✓ 1
A1 9.11
✓ 1
Exam tip: When an angle comes out obtuse, a negative scalar product is the check: it must give a negative cosine.
361IBMA3H-3-18r32-113.18 Intersections and angles of lines and planesPaper 2Medium6 marks
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A delivery drone flies along the line \(L:\ \mathbf{r}=\begin{pmatrix}1\\0\\2\end{pmatrix}+t\begin{pmatrix}2\\1\\-2\end{pmatrix}\), with distances in metres. A parcel locker is at the point \(P(4,5,3)\). Let \(Q\) be the point on \(L\) that is closest to \(P\).
(a)
Find \(\overrightarrow{PQ}\) in terms of \(t\) for a general point \(Q\) on \(L\), and hence find the value of \(t\) at the closest point.
(3)
(b)
Write down the coordinates of \(Q\).
(1)
(c)
Find the shortest distance from the locker to the line, correct to 3 significant figures.
Exam tip: The shortest distance from a point to a line is along the perpendicular, so set the scalar product of \(\overrightarrow{PQ}\) and the direction vector to zero.
362IBMA3H-3-19r35-033.1 Three-dimensional geometry: distance, solids and anglesPaper 2Medium6 marks
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A lifeguard sits on a chair with her eyes 2.4 m above the water surface of a swimming pool. Two swimmers, A and B, are in a straight line directly in front of her, with A nearer. Treat each swimmer as a point on the water surface. The angle of depression of A from her eyes is 15° and the angle of depression of B is 9°. Give all answers to 3 significant figures and use unrounded values in later parts.
Diagram not to scale
(a)
Find the horizontal distance from the lifeguard to swimmer A.
(2)
(b)
Find the horizontal distance from the lifeguard to swimmer B.
(2)
(c)
Find the distance between the two swimmers.
(2)
Show mark scheme
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Part (a)
M1 \(\tan 15^\circ = \dfrac{2.4}{x}\) (angle of depression equals angle of elevation from the swimmer)
✓ 1
A1 x = 8.96 m
✓ 1
Part (b)
M1 \(\tan 9^\circ = \dfrac{2.4}{y}\)
✓ 1
A1 y = 15.15 m
✓ 1
Part (c)
M1 subtract the two horizontal distances
✓ 1
A1 6.20 m
✓ 1
Common error: The angle of depression is measured from the horizontal at the observer, not from the vertical; draw the right-angled triangle with the angle at the swimmer.
363IBMA3H-3-19r35-043.3 Applications of trigonometry: elevation, depression and bearingsPaper 2Medium7 marks
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A maintenance boat leaves a harbour H and sails 12 km on a bearing of 042° to reach a turbine platform A. It then sails 9 km on a bearing of 125° to reach a second platform B.
Diagram not to scale
(a)
Show that angle HAB = 97°.
(1)
(b)
Find the distance HB, in km, to 3 significant figures.
(3)
(c)
Find the bearing of B from H, to the nearest 0.1°.
(3)
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Notes
Part (a)
A1 bearing of H from A is 042° + 180° = 222°, so HAB = 222° − 125° = 97°
✓ 1
Part (b)
M1 cosine rule: HB2 = 122 + 92 − 2(12)(9)cos 97°
✓ 1
A1 HB2 = 251.3 (accept 251)
✓ 1
A1 HB = 15.9 km
✓ 1
Part (c)
M1 sine rule: \(\dfrac{\sin \angle AHB}{9}=\dfrac{\sin 97^\circ}{HB}\)
✓ 1
A1 angle AHB = 34.3°
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A1 bearing of B from H = 042° + 34.3° = 76.3°
✓ 1
Exam tip: When the sine rule gives an angle, check that it is the acute one that fits the diagram before adding it to or subtracting it from a bearing.
364IBMA3H-3-19r35-083.14 Vector equation of a linePaper 2Medium7 marks
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A radio mast stands at M, with its top 40 m above level ground. Take the origin at the foot of the mast, with \(\mathbf{i}\) pointing east, \(\mathbf{j}\) north and \(\mathbf{k}\) vertically upwards, units in metres, so M = (0, 0, 40). Two straight support cables run from M to anchor points A(30, 0, 0) and B(−10, 24, 0) on the ground.
(a)
Write down \(\overrightarrow{MA}\) and \(\overrightarrow{MB}\).
(2)
(b)
Find the angle AMB between the cables, to the nearest 0.1°.
(3)
(c)
Find the area of triangle MAB, in m2, to the nearest m2, using the unrounded angle from (b).
Exam tip: Use the scalar product (a·b = |a||b|cos θ) for angles and the vector product only when the question needs an area or a normal vector; here a cross-product check also gives 1000.
365IBMA3H-3-19r35-113.18 Intersections and angles of lines and planesPaper 2Medium7 marks
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A hillside on a wind farm site is modelled by the plane \(2x-y+2z=9\), where \(z\) is the height above sea level, in metres. A survey drone at P(2, −1, 5) sends a laser beam in the direction \(\mathbf{d}=\mathbf{i}+2\mathbf{j}-2\mathbf{k}\).
(a)
Find the coordinates of the point where the beam meets the hillside.
(4)
(b)
Find the acute angle between the beam and the hillside, to 3 significant figures.
(3)
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Notes
Part (a)
M1 line \(\mathbf{r}=(2,-1,5)+\lambda(1,2,-2)\) substituted into the plane
Common error: The angle between a line and a plane uses sin of the angle to the normal's dot product; using cos gives the angle between the line and the normal instead.
366IBMA3H-3-20r40-023.7 Circular functions and their transformationsPaper 2Medium7 marks
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A hospital monitor models a patient's arterial blood pressure, \(P\) mmHg, by \(P(t)=95+25\sin(kt)\), where \(t\) is the time in seconds after the start of a heartbeat and \(k>0\). The heart beats once every 0.8 seconds, so the function has period 0.8. Angles are in radians.
(a)
Show that \(k=\frac{5\pi}{2}\).
(2)
(b)
Find the two times during the first heartbeat, \(0\le t<0.8\), at which \(P=105\). Give your answers to 3 significant figures.
(3)
(c)
Find the length of time in each heartbeat for which the pressure exceeds 105 mmHg. Use the unrounded values of \(t\) and give your answer to 3 significant figures.
(2)
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Part (a)
M1 period = \(\frac{2\pi}{k}\) so \(\frac{2\pi}{k}=0.8\)
✓ 1
A1 \(k=\frac{2\pi}{0.8}=\frac{5\pi}{2}\) (AG)
✓ 1
Part (b)
M1 \(\sin(kt)=0.4\)
✓ 1
M1 \(kt=\arcsin 0.4\) and \(kt=\pi-\arcsin 0.4\)
✓ 1
A1 \(t=0.0524\) s and \(t=0.348\) s
✓ 1
Part (c)
M1 difference of the two times found in (b)
✓ 1
A1 0.295 seconds
✓ 1
Exam tip: A sine equation in one period has two solutions when the value is between −1 and 1. Find the first with arcsin, then use symmetry about the peak to find the second.
367IBMA3H-3-20r40-053.14 Vector equation of a linePaper 2Medium8 marks
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A Formula 1 car travels at a constant speed of 60 m s−1. Its wheels have radius 0.33 m and roll without slipping, so each wheel turns through \(\omega=\frac{v}{r}\) radians per second, where \(v\) is the speed of the car and \(r\) is the radius. The air valve on a wheel is at the lowest point of the wheel, at ground level, when \(t=0\). The height of the valve above the ground at time \(t\) seconds is \(h=0.33(1-\cos\omega t)\) metres.
(a)
Calculate \(\omega\). Give your answer to 3 significant figures.
(2)
(b)
Find the height of the valve when \(t=0.01\) s. Use the unrounded value of \(\omega\) and give your answer in metres to 3 significant figures.
(2)
(c)
During one complete turn of the wheel, find the fraction of the turn for which the valve is higher than 0.5 m above the ground. Give your answer to 3 significant figures.
(4)
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Part (a)
M1 \(\omega=\frac{60}{0.33}\)
✓ 1
A1 \(\omega=182\) rad s−1 (181.8…)
✓ 1
Part (b)
M1 \(0.33\left(1-\cos(\omega\times0.01)\right)\) with the argument in radians
Exam tip: Convert a time interval in a repeating motion into a fraction of the period by dividing the angle swept by \(2\pi\). The speed value then drops out of the final fraction.
368IBMA3H-3-20r40-123.17 Equations of a planePaper 2Medium6 marks
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A drone delivers medicine to a sloping hospital roof. Relative to a fixed origin, with distances in metres, the drone is launched from the point (0, −1, 0) and flies along the line \(L\) with equation \(\boldsymbol{r}=\begin{pmatrix}0\\-1\\0\end{pmatrix}+t\begin{pmatrix}2\\3\\6\end{pmatrix}\). The roof lies in the plane \(\Pi\) with equation \(2x+y+2z=18\).
(a)
Find the coordinates of the point where the line \(L\) meets the plane \(\Pi\).
(3)
(b)
Find the acute angle between \(L\) and \(\Pi\). Give your answer in degrees to 3 significant figures.
(3)
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Part (a)
M1 substitutes \(x=2t\), \(y=-1+3t\), \(z=6t\) into \(2x+y+2z=18\)
✓ 1
A1 \(19t-1=18\) so \(t=1\)
✓ 1
A1 (2, 2, 6)
✓ 1
Part (b)
M1 uses \(\sin\theta=\frac{|\boldsymbol{d}\cdot\boldsymbol{n}|}{|\boldsymbol{d}||\boldsymbol{n}|}\) with \(\boldsymbol{d}=(2,3,6)\) and \(\boldsymbol{n}=(2,1,2)\)
Common error: The angle between a line and a plane uses the sine of the angle between the direction vector and the normal. Using cosine gives the angle between the line and the normal instead (25.2° here).
369IBMA3H-3-20r41-053.4 Radians, arc length and sector areaPaper 2Medium8 marks
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A circular paddling pool has radius 5 m. A straight rope of length 8 m is stretched across the pool as a chord, so that it joins two points on the edge of the pool. The chord subtends an angle θ radians at the centre of the pool.
Diagram not to scale. The rope is a chord of the circle
(a)
Find θ, giving your answer to 3 significant figures.
(3)
(b)
Find the length of the shorter arc joining the two points where the rope meets the edge of the pool. Use the unrounded value of θ and give your answer to 3 significant figures.
(2)
(c)
Find the area of the smaller region of the pool that lies between the rope and the edge of the pool. Use the unrounded value of θ and give your answer to 3 significant figures.
(3)
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Part (a)
M1 attempt to use the cosine rule: 82 = 52 + 52 − 2 × 5 × 5 × cos θ (or a right-angled triangle with sin(θ/2) = 4/5)
✓ 1
A1 cos θ = −0.28 (or sin(θ/2) = 0.8)
✓ 1
A1 θ = 1.85
✓ 1
Part (b)
M1 use of s = rθ = 5θ
✓ 1
A1 arc length = 9.27 m
✓ 1
Part (c)
M1 area of sector = ½ × 25 × θ
✓ 1
M1 area of triangle = ½ × 25 × sin θ = 12 (sin θ = 0.96)
✓ 1
A1 area of segment = 11.2 m2
✓ 1
Exam tip: The area between a chord and its arc is the sector area minus the triangle area, ½r²(θ − sin θ), with θ in radians.
370IBMA3H-3-20r41-063.7 Circular functions and their transformationsPaper 2Medium11 marks
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In a wave pool, the height of the water above the floor at a fixed point is modelled by h(t) = 1.2 + 0.35 sin(0.5t), where h is in metres and t is the time in seconds, with the argument of sine in radians.
(a)
Find the period of the wave. Give your answer in terms of π.
(2)
(b)
Write down the maximum and minimum heights of the water.
(2)
(c)
Find the two times, in the interval 0 ≤ t ≤ 12, at which the height of the water is 1.4 m. Give each answer to 3 significant figures.
(5)
(d)
Hence find the length of time, in one cycle, for which the water is higher than 1.4 m. Use the unrounded times from (c) and give your answer to 3 significant figures.
(2)
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Part (a)
M1 period = 2π ÷ 0.5
✓ 1
A1 4π seconds
✓ 1
Part (b)
A1 maximum = 1.55 m
✓ 1
A1 minimum = 0.85 m
✓ 1
Part (c)
M1 sin(0.5t) = 0.2/0.35 = 4/7
✓ 1
M1 0.5t = 0.6082... (principal value)
✓ 1
A1 t = 1.22
✓ 1
M1 use of symmetry of the sine curve: 0.5t = π − 0.6082... = 2.5333...
✓ 1
A1 t = 5.07
✓ 1
Part (d)
M1 difference of the two times (the water is above 1.4 m between them)
✓ 1
A1 duration = 3.85 s
✓ 1
Common error: The calculator returns only the principal value of sin⁻¹; the second solution in the cycle comes from the symmetry sin θ = sin(π − θ), not from adding π.
371IBMA3H-3-20r41-093.15 Intersecting, parallel and skew linesPaper 2Medium8 marks
Calculator
Two cables on a wind farm follow the lines L1: r = (1, 2, 0) + λ(2, 1, −1) and L2: r = (2, 10, −5) + μ(1, −2, 1).
(a)
Show that L1 and L2 intersect, and find the coordinates of the point of intersection.
(4)
(b)
Find the acute angle between L1 and L2, giving your answer in degrees to 3 significant figures.
M1 solving two equations, e.g. from the third μ = 5 − λ, then λ = 2 and μ = 3
✓ 1
R1 substituting into the remaining equation: 2 + 2 = 4 and 10 − 6 = 4, so the lines intersect
✓ 1
A1 point of intersection (5, 4, −2)
✓ 1
Part (b)
M1 use of cos θ = |d1 · d2| / (|d1||d2|)
✓ 1
A1 d1 · d2 = 2 − 2 − 1 = −1 and |d1| = |d2| = √6
✓ 1
A1 cos θ = 1/6
✓ 1
A1 θ = 80.4°
✓ 1
Common error: The angle between two lines is found from the direction vectors; a negative scalar product gives the obtuse angle, so use the modulus to obtain the acute angle.
372IBMA3H-3-20r41-103.18 Intersections and angles of lines and planesPaper 2Medium8 marks
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In a library, a sensor at A(2, 4, 1) sends a beam in the direction of the vector (1, 2, 2) towards a sloping glass roof. All coordinates are in metres. The roof lies in the plane Π with equation 2x + y + 2z = 30.
(a)
Find the coordinates of the point where the beam meets the roof.
(4)
(b)
Find the angle between the beam and the roof, giving your answer in degrees to 3 significant figures.
(4)
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Part (a)
M1 beam: r = (2, 4, 1) + λ(1, 2, 2); substitute (2 + λ, 4 + 2λ, 1 + 2λ) into the equation of Π
M1 normal to the plane n = (2, 1, 2), with |n| = 3 and |d| = 3
✓ 1
M1 sin φ = |d · n| / (|d||n|)
✓ 1
A1 sin φ = 8/9
✓ 1
A1 φ = 62.7°
✓ 1
Common error: The angle between a line and a plane uses the sine formula with the normal vector; using cosine gives the angle between the line and the normal instead.
373IBMA3H-3-18r42-023.18 Intersections and angles of lines and planesPaper 2Medium7 marks
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Two buried cables on a solar farm are modelled by the lines \(l_1:\ \mathbf r=\begin{pmatrix}1\\2\\0\end{pmatrix}+s\begin{pmatrix}2\\1\\-1\end{pmatrix}\) and \(l_2:\ \mathbf r=\begin{pmatrix}-1\\0\\4\end{pmatrix}+t\begin{pmatrix}3\\2\\-3\end{pmatrix}\). Coordinates are in metres, with the third coordinate measuring height relative to ground level.
(a)
Show that the cables meet and find the coordinates of the point where they meet.
M1 solves two equations: \(s=2t-2\) gives \(1+4t-4=-1+3t\), so \(t=2\), \(s=2\)
✓ 1
A1 checks the third equation: \(-2=4-6\), so the lines intersect
✓ 1
A1 point \((5,4,-2)\)
✓ 1
Part (b)
M1 scalar product of direction vectors \(=6+2+3=11\)
✓ 1
M1 \(\cos\theta=\dfrac{11}{\sqrt6\,\sqrt{22}}\)
✓ 1
A1 \(\theta=16.8^{\circ}\) (16.78...)
✓ 1
Common error: The angle between two lines uses the direction vectors only; use the modulus of the scalar product to obtain the acute angle.
374IBMA3H-3-18r42-033.18 Intersections and angles of lines and planesPaper 2Medium5 marks
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A tethered inspection balloon at a wind farm moves along the line \(l:\ \mathbf r=\begin{pmatrix}2\\1\\3\end{pmatrix}+\lambda\begin{pmatrix}1\\2\\2\end{pmatrix}\), where coordinates are in kilometres. A radar station is at \(P(6,5,7)\). Find the shortest distance from \(P\) to \(l\), giving your answer to 3 significant figures.
(a)
Find the shortest distance from \(P\) to \(l\).
(5)
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Part (a)
M1 a general point on \(l\) is \(F(2+\lambda,\,1+2\lambda,\,3+2\lambda)\), so \(\overrightarrow{PF}=\begin{pmatrix}\lambda-4\\2\lambda-4\\2\lambda-4\end{pmatrix}\)
A1 \(9\lambda-20=0\), so \(\lambda=\dfrac{20}{9}\)
✓ 1
M1 \(\overrightarrow{PF}=\dfrac19\begin{pmatrix}-16\\4\\4\end{pmatrix}\), so \(|PF|=\dfrac{\sqrt{288}}{9}\)
✓ 1
A1 \(1.89\) km (1.8856...)
✓ 1
Exam tip: The shortest distance from a point to a line is along the perpendicular, so set the scalar product of PF and the direction vector to zero.
375IBMA3H-3-18r42-083.7 Circular functions and their transformationsPaper 2Medium10 marks
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The weekly number of new cases of a seasonal infection in a town is modelled by \(C(t)=120+85\cos\!\left(\dfrac{\pi t}{26}\right)\), where \(t\) is the number of weeks after the peak of the season, for \(0\le t\le52\).
(a)
State the maximum weekly number of new cases predicted by the model.
(1)
(b)
Find the number of new cases in week \(t=10\), giving your answer to the nearest whole number.
(2)
(c)
Find the two values of \(t\) for which \(C(t)=100\). Give your answers to 3 significant figures.
(4)
(d)
Find the total number of weeks, in the 52 weeks, for which the model gives more than 150 new cases per week. Give your answer to 3 significant figures.
M1 \(120+85\cos\!\left(\dfrac{\pi t}{26}\right)>150\), so \(\cos\!\left(\dfrac{\pi t}{26}\right)>\dfrac{30}{85}\)
✓ 1
M1 boundary at \(\dfrac{\pi t}{26}=\cos^{-1}\dfrac{30}{85}=1.2094\ldots\), so \(t=10.01\ldots\); the cases exceed 150 for \(0\le t<10.01\) and \(41.99<t\le52\)
✓ 1
A1 total \(=2\times10.01\ldots=20.0\) weeks
✓ 1
Exam tip: A cosine equation in a full period has a second solution from the symmetry of the graph about the minimum or maximum; look for it every time.
376IBMA3H-3-18r42-103.4 Radians, arc length and sector areaPaper 2Medium6 marks
Calculator
A shop's logo is a circular segment bounded by the chord \(PQ\) and the minor arc \(PQ\) of a circle with centre \(O\) and radius 15 cm. Angle \(POQ=1.8\) radians.
Diagram not to scale. The segment lies between the chord PQ and the minor arc
(a)
Find the area of the segment. Give your answer to 3 significant figures.
(3)
(b)
Find the perimeter of the segment. Give your answer to 3 significant figures.
(3)
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Part (a)
M1 sector area \(=\dfrac12\times15^{2}\times1.8=202.5\) cm2
✓ 1
M1 triangle area \(=\dfrac12\times15^{2}\sin1.8=109.56\ldots\) cm2
✓ 1
A1 segment area \(=92.9\) cm2 (92.94...)
✓ 1
Part (b)
M1 arc length \(=15\times1.8=27\) cm
✓ 1
M1 chord \(PQ=2\times15\sin0.9=23.50\ldots\) cm (or by the cosine rule)
✓ 1
A1 perimeter \(=50.5\) cm (50.4998...)
✓ 1
Exam tip: Use the half-angle in the right-angled triangle to find the chord, or use the cosine rule with the full angle.
377IBMA3H-3-18r42-113.1 Three-dimensional geometry: distance, solids and anglesPaper 2Medium6 marks
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A retailer's storage unit is a cuboid $ABCDEFGH$. The base \(ABCD\) is horizontal, with \(AB=8\) m, \(BC=6\) m and vertical edge \(AE=5\) m, where \(E\) is directly above \(A\) and \(G\) is directly above \(C\).
Diagram not to scale
(a)
Find the length of the diagonal \(AG\). Give your answer to 3 significant figures.
(3)
(b)
Find the angle between \(AG\) and the base $ABCD$. Give your answer to 3 significant figures.
(3)
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Part (a)
M1 \(AC^{2}=8^{2}+6^{2}=100\)
✓ 1
M1 \(AG^{2}=AC^{2}+CG^{2}=100+25=125\)
✓ 1
A1 \(AG=11.2\) m (11.18...)
✓ 1
Part (b)
M1 identifies the angle as \(\angle GAC\) in the right-angled triangle \(GCA\)
✓ 1
M1 \(\tan\angle GAC=\dfrac{5}{10}\)
✓ 1
A1 \(26.6^{\circ}\) (26.565...)
✓ 1
Common error: The angle between a line and a plane is measured to its projection onto the plane, here AC, not to an edge of the cuboid.
378IBMA3H-3-19r42-043.2 Right-angled and non-right-angled trianglesPaper 2Medium7 marks
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Two tracking stations S and T observe a space probe P. The distance SP is 10.5 million km, the distance PT is 7.2 million km and angle PST is 35°. Give angles to the nearest 0.1° and other answers to 3 significant figures. Use unrounded values in later calculations.
Diagram not to scale. Distances in millions of km
(a)
Find the two possible values of angle PTS.
(3)
(b)
Find the two possible values of the distance ST.
(2)
(c)
A third station shows that angle PTS is obtuse. Find the area of triangle PST in this case.
(2)
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Part (a)
M1 sine rule: \(\frac{\sin\text{PTS}}{10.5}=\frac{\sin35^\circ}{7.2}\)
✓ 1
A1 acute value 56.8°
✓ 1
A1 obtuse value \(180^\circ-56.8^\circ\) = 123.2°
✓ 1
Part (b)
M1 finds angle SPT = \(180^\circ-35^\circ-\text{PTS}\) (88.2° or 21.8°) and applies the sine rule (or cosine rule) to find ST
✓ 1
A1 12.5 million km and 4.66 million km
✓ 1
Part (c)
M1 \(\frac12\times\text{SP}\times\text{ST}\times\sin35^\circ\) with ST = 4.66
✓ 1
A1 14.0 million km2
✓ 1
Exam tip: When the given angle is not between the two given sides, check whether the side opposite it is shorter than the other given side; if so, a second, obtuse triangle exists.
379IBMA3H-3-19r42-053.4 Radians, arc length and sector areaPaper 2Medium10 marks
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The arch of a footbridge is an arc AB of a circle with centre O and radius \(r\) metres. The straight line AB is horizontal with length 36 m. The highest point of the arch is 4.5 m above the midpoint of AB. Give answers to 3 significant figures and use unrounded values in later parts.
Diagram not to scale
(a)
Show that \(r=38.25\).
(3)
(b)
Find angle AOB in radians.
(2)
(c)
Find the length of the arc AB.
(2)
(d)
Find the area enclosed between the arc AB and the line AB.
(3)
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Part (a)
M1 right-angled triangle with hypotenuse \(r\) and sides 18 and \(r-4.5\): \(r^2=18^2+(r-4.5)^2\)
Common error: The triangle height is \(r-4.5\), not \(r\) and not 4.5; draw the right-angled triangle from the centre to the midpoint of the chord.
380IBMA3H-3-19r42-083.15 Intersecting, parallel and skew linesPaper 2Medium7 marks
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Two straight bridge cables are modelled, in metres, by the lines \(l_1:\ \mathbf{r}=\begin{pmatrix}1\\2\\-1\end{pmatrix}+\lambda\begin{pmatrix}2\\-1\\1\end{pmatrix}\) and \(l_2:\ \mathbf{r}=\begin{pmatrix}3\\6\\-3\end{pmatrix}+\mu\begin{pmatrix}1\\-3\\2\end{pmatrix}\).
(a)
Show that the cables meet, and find the coordinates of the point where they meet.
(4)
(b)
Find the acute angle between the cables, to the nearest 0.1°.
(3)
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Part (a)
M1 equates the components: \(1+2\lambda=3+\mu\), \(2-\lambda=6-3\mu\), \(-1+\lambda=-3+2\mu\)
✓ 1
A1 solves two equations to get \(\lambda=2\), \(\mu=2\)
✓ 1
R1 checks the third equation: \(1=1\), so the lines intersect
Common error: If the scalar product is negative the formula gives an obtuse angle; take the modulus of the numerator to get the acute angle between the lines.
381IBMA3H-3-19r42-093.15 Intersecting, parallel and skew linesPaper 2Medium6 marks
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The flight paths of two space probes are modelled by the lines \(l_1:\ \mathbf{r}=\begin{pmatrix}1\\0\\2\end{pmatrix}+\lambda\begin{pmatrix}1\\2\\-1\end{pmatrix}\) and \(l_2:\ \mathbf{r}=\begin{pmatrix}3\\1\\0\end{pmatrix}+\mu\begin{pmatrix}2\\-1\\1\end{pmatrix}\), with distances in thousands of km.
(a)
Show that the two flight paths are skew lines.
(4)
(b)
Find the acute angle between the two flight paths, to the nearest 0.1°.
(2)
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Part (a)
M1 the direction vectors are not parallel (neither is a multiple of the other)
Exam tip: Two lines in three dimensions can fail to meet without being parallel; show both facts before concluding the lines are skew.
382IBMA3H-3-19r42-103.15 Intersecting, parallel and skew linesPaper 2Medium12 marks
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Two steam pipes in a coffee roasting plant are modelled, in metres, by the lines \(l_1:\ \mathbf{r}=\begin{pmatrix}0\\1\\2\end{pmatrix}+\lambda\begin{pmatrix}1\\2\\-1\end{pmatrix}\) and \(l_2:\ \mathbf{r}=\begin{pmatrix}-2\\2\\-6\end{pmatrix}+\mu\begin{pmatrix}2\\-1\\1\end{pmatrix}\).
(a)
Show that the pipes do not meet.
(3)
(b)
Find a vector perpendicular to both pipes.
(2)
(c)
Find the shortest distance between the pipes, to 3 significant figures.
(3)
(d)
Find the coordinates of the point on each pipe where the pipes are closest.
A1 \(\begin{pmatrix}1\\-3\\-5\end{pmatrix}\) (or any non-zero multiple)
✓ 1
Part (c)
M1 vector between a point on each line, e.g. \(\begin{pmatrix}-2\\1\\-8\end{pmatrix}\)
✓ 1
M1 component along the common perpendicular: \(\frac{|(-2)\times1+1\times(-3)+(-8)\times(-5)|}{\sqrt{35}}\)
✓ 1
A1 \(\frac{35}{\sqrt{35}}=\sqrt{35}=5.92\) m
✓ 1
Part (d)
M1 writes the vector joining a general point of \(l_1\) to a general point of \(l_2\)
✓ 1
M1 sets its scalar product with each direction vector equal to zero: \(8-\mu-6\lambda=0\) and \(-13+6\mu+\lambda=0\)
✓ 1
A1 \(\lambda=1\), \(\mu=2\)
✓ 1
A1 \((1,3,1)\) on \(l_1\) and \((2,0,-4)\) on \(l_2\)
✓ 1
Exam tip: The shortest distance between skew lines is the length of the projection of any joining vector onto the common perpendicular direction.
383IBMA3H-3-19r42-123.15 Intersecting, parallel and skew linesPaper 2Medium6 marks
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Two support cables on a bridge are modelled, in metres, by the lines \(l_1:\ \mathbf{r}=\begin{pmatrix}1\\2\\3\end{pmatrix}+\lambda\begin{pmatrix}2\\1\\-2\end{pmatrix}\) and \(l_2:\ \mathbf{r}=\begin{pmatrix}4\\-2\\5\end{pmatrix}+\mu\begin{pmatrix}-4\\-2\\4\end{pmatrix}\).
(a)
Show that the cables are parallel and do not lie along the same line.
(2)
(b)
Find the shortest distance between the cables, to 3 significant figures.
(4)
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Notes
Part (a)
R1 \(\begin{pmatrix}-4\\-2\\4\end{pmatrix}=-2\begin{pmatrix}2\\1\\-2\end{pmatrix}\), so the directions are parallel
✓ 1
R1 \((4,-2,5)\) is not on \(l_1\): \(4=1+2\lambda\) gives \(\lambda=1.5\), but then \(y=3.5\ne-2\)
✓ 1
Part (b)
M1 vector between the two given points, \(\begin{pmatrix}3\\-4\\2\end{pmatrix}\)
✓ 1
M1 vector product of this with a direction vector: \(\begin{pmatrix}6\\10\\11\end{pmatrix}\)
A1 5.34 m (any valid method, e.g. foot of the perpendicular, earns full marks)
✓ 1
Exam tip: For parallel lines, any point on one line works as a starting point; the cross product with the shared direction, divided by its length, gives the perpendicular distance.
384IBMA3H-3-19r42-133.1 Three-dimensional geometry: distance, solids and anglesPaper 2Medium10 marks
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A conical hopper for roasted coffee beans is made by bending a flat sector of sheet metal of radius 0.60 m and angle \(\theta\) radians until the two straight edges meet. The sector's radius becomes the slant height of the cone, and the sector's curved edge becomes the circular rim of radius 0.25 m. The hopper has its apex at the bottom and the rim at the top. Give answers to 3 significant figures and use unrounded values in later parts.
Flat sector before bending; not to scale
(a)
Find the circumference of the rim.
(1)
(b)
Show that \(\theta=\frac{5\pi}{6}\).
(2)
(c)
Find the area of sheet metal in the sector.
(2)
(d)
Find the volume of the hopper.
(3)
(e)
Roasted beans have bulk density 380 kg m−3. Find the mass of beans when the hopper is full.
(2)
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Mark
Notes
Part (a)
A1 \(2\pi\times0.25=1.57\) m
✓ 1
Part (b)
M1 curved edge of sector \(=0.6\theta=2\pi(0.25)\)
M1 height \(h=\sqrt{0.6^2-0.25^2}=0.5454\ldots\) m
✓ 1
M1 \(V=\frac13\pi r^2h\)
✓ 1
A1 0.0357 m3
✓ 1
Part (e)
M1 mass = density × volume \(=380\times0.035698\ldots\)
✓ 1
A1 13.6 kg
✓ 1
Common error: The slant height of the cone is the radius of the sector, not the height of the cone; use Pythagoras to find the height before using the volume formula.
385IBMA3H-3-19r43-063.8 Solving trigonometric equationsPaper 2Medium6 marks
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The imbalance \(D\) between the pushes of a cyclist's two legs depends on the crank angle \(\theta\) (in radians) according to \(D(\theta)=3\sin2\theta-2\cos\theta\), for \(0\le\theta\le2\pi\).
(a)
Show that \(D(\theta)=2\cos\theta\,(3\sin\theta-1)\).
(2)
(b)
Hence solve \(D(\theta)=0\) for \(0\le\theta\le2\pi\), giving your answers to three significant figures.
(4)
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Part (a)
M1 use of \(\sin2\theta=2\sin\theta\cos\theta\) to obtain \(6\sin\theta\cos\theta-2\cos\theta\)
✓ 1
A1 factorise \(2\cos\theta\) to obtain \(2\cos\theta(3\sin\theta-1)\) (AG)
✓ 1
Part (b)
M1 set each factor to zero: \(\cos\theta=0\) or \(\sin\theta=\frac13\)
✓ 1
A1 \(\theta=1.57\) and \(\theta=4.71\) (from \(\frac{\pi}{2}\) and \(\frac{3\pi}{2}\))
✓ 1
A1 \(\theta=0.340\)
✓ 1
A1 \(\theta=2.80\) (from \(\pi-0.3398\ldots\))
✓ 1
Common error: Do not divide both sides by \(\cos\theta\) when solving; that loses the solutions where \(\cos\theta=0\). Factorise instead.
386IBMA3H-3-19r43-073.8 Solving trigonometric equationsPaper 2Medium6 marks
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A laser scanner records the angle \(x\), in radians with \(0\le x\le\pi\), at which its beam meets a wall. The angle satisfies \(6\sin^2x+\cos x=4\).
(a)
Show that \(6\cos^2x-\cos x-2=0\).
(2)
(b)
Hence find the possible values of \(x\), giving your answers to three significant figures.
(4)
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Part (a)
M1 use of \(\sin^2x=1-\cos^2x\) to obtain \(6-6\cos^2x+\cos x=4\)
✓ 1
A1 rearrange to \(6\cos^2x-\cos x-2=0\) (AG)
✓ 1
Part (b)
M1 attempt to solve the quadratic in \(\cos x\), for example \((3\cos x-2)(2\cos x+1)=0\)
✓ 1
A1 \(\cos x=\frac23\) and \(\cos x=-\frac12\)
✓ 1
A1 \(x=0.841\)
✓ 1
A1 \(x=2.09\)
✓ 1
Exam tip: The Pythagorean identity lets you convert an equation containing both \(\sin^2x\) and \(\cos x\) into a quadratic in one ratio; choose the substitution that leaves a single ratio.
387IBMA3H-3-19r43-093.18 Intersections and angles of lines and planesPaper 2Medium9 marks
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Two straight cable trays in a data centre lie along the lines \(L_1\) and \(L_2\), with positions in metres: \(L_1:\ \mathbf{r}=\begin{pmatrix}1\\2\\-1\end{pmatrix}+s\begin{pmatrix}2\\-1\\3\end{pmatrix}\) and \(L_2:\ \mathbf{r}=\begin{pmatrix}3\\-4\\7\end{pmatrix}+t\begin{pmatrix}1\\2\\-1\end{pmatrix}\).
(a)
Show that the two trays cross, and find the coordinates of the crossing point.
(5)
(b)
Find the acute angle between the two trays, giving your answer in degrees to three significant figures.
(4)
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Part (a)
M1 equate \(x\) and \(y\) components: \(1+2s=3+t\) and \(2-s=-4+2t\)
✓ 1
M1 solve the two equations simultaneously
✓ 1
A1 \(s=2\) and \(t=2\)
✓ 1
R1 check the \(z\)-components: \(-1+3(2)=5\) and \(7-2=5\), so the lines intersect
✓ 1
A1 crossing point \((5,\,0,\,5)\)
✓ 1
Part (b)
M1 scalar product of direction vectors: \(2\times1+(-1)\times2+3\times(-1)=-3\)
✓ 1
A1 magnitudes \(\sqrt{14}\) and \(\sqrt6\)
✓ 1
M1 use of \(\cos\theta=\frac{|\mathbf{d}_1\cdot\mathbf{d}_2|}{|\mathbf{d}_1||\mathbf{d}_2|}=\frac{3}{\sqrt{84}}\)
✓ 1
A1 \(\theta=70.9^\circ\)
✓ 1
Exam tip: The angle between two lines uses their direction vectors only; take the modulus of the scalar product (or subtract an obtuse result from 180°) to report the acute angle.
388IBMA3H-3-19r43-103.16 Vector productPaper 2Medium8 marks
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An archaeological team records three post holes in a buried structure, with coordinates in metres relative to a datum: \(A(5,\,1,\,5)\), \(B(2,\,1,\,1)\) and \(C(0,\,5,\,5)\).
(a)
Find the vectors \(\overrightarrow{BA}\) and \(\overrightarrow{BC}\).
(2)
(b)
Find angle \(ABC\), giving your answer in degrees to one decimal place.
(3)
(c)
Find the area of triangle \(ABC\), giving your answer in m2 to three significant figures. Use an unrounded value of angle \(ABC\).
A1 \(|\overrightarrow{BA}|=5\) and \(|\overrightarrow{BC}|=6\), so \(\cos B=\frac{10}{30}=\frac13\)
✓ 1
A1 \(B=70.5^\circ\)
✓ 1
Part (c)
M1 use of area \(=\frac12|\overrightarrow{BA}||\overrightarrow{BC}|\sin B\)
✓ 1
A1 \(\sin B=\sqrt{1-\frac19}=\frac{2\sqrt2}{3}\) (or use \(70.5\ldots^\circ\))
✓ 1
A1 area \(=\frac12\times5\times6\times\frac{2\sqrt2}{3}=14.1\) m2
✓ 1
Exam tip: To find an area from two vectors, first find the included angle with the scalar product, then use \(\frac12ab\sin C\); keep exact values until the end.
389IBMA3H-3-19r43-113.18 Intersections and angles of lines and planesPaper 2Medium8 marks
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A cooling pipe in a data centre runs along the line \(L:\ \mathbf{r}=\begin{pmatrix}1\\0\\2\end{pmatrix}+t\begin{pmatrix}1\\2\\2\end{pmatrix}\), with distances in metres. A temperature sensor is fixed at the point \(S(5,\,5,\,4)\).
(a)
Show that the vector from the point on \(L\) with parameter \(t\) to \(S\) is \(\begin{pmatrix}4-t\\5-2t\\2-2t\end{pmatrix}\).
(2)
(b)
The point \(F\) on \(L\) is closest to \(S\). Find the value of \(t\) at \(F\).
(3)
(c)
Write down the coordinates of \(F\) and find the shortest distance from the sensor to the pipe.
(3)
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Part (a)
M1 subtract the position vector of the point on \(L\) from \(\overrightarrow{OS}\): \((5,5,4)-(1+t,\,2t,\,2+2t)\)
✓ 1
A1 \((4-t,\,5-2t,\,2-2t)\) (AG)
✓ 1
Part (b)
M1 the vector in (a) is perpendicular to the direction vector, so its scalar product with \((1,2,2)\) is zero
✓ 1
A1 \((4-t)+2(5-2t)+2(2-2t)=18-9t=0\)
✓ 1
A1 \(t=2\)
✓ 1
Part (c)
A1 \(F=(3,\,4,\,6)\)
✓ 1
M1 magnitude of \((2,\,1,\,-2)\): \(\sqrt{2^2+1^2+(-2)^2}\)
✓ 1
A1 shortest distance \(=3\) m
✓ 1
Common error: The shortest distance to a line occurs where the connecting vector is perpendicular to the direction vector; choosing a convenient point on the line without checking perpendicularity gives a distance that is too large.
390AAHL3-q00043.1 Three-dimensional geometry: distance, solids and anglesPaper 2Hard16 marks
The diagram shows the roof of a building. The base \(ABCD\) is a horizontal rectangle with \(AB=12\) m and \(BC=8\) m. The ridge \([EF]\) is horizontal, has length \(6\) m, and is \(3\) m above the base. Seen from above, \([EF]\) lies on the line of symmetry of \(ABCD\) that is parallel to \([AB]\), with its midpoint directly above the centre of \(ABCD\). The faces \(ABFE\) and \(DCFE\) are congruent trapezia and the faces \(ADE\) and \(BCF\) are congruent isosceles triangles.
Diagram not to scale
(a)
Find the length of the edge \([AE]\).
(2)
(b)
Find the angle between \([AE]\) and the base \(ABCD\).
(2)
(c)
Find the angle between the face \(ADE\) and the base \(ABCD\).
(2)
(d)
Find the total area of the four faces of the roof.
(4)
(e)
Find the volume of the space enclosed between the roof and the base \(ABCD\).
(3)
(f)
Find the size of the angle \(A\hat{E}B\).
(3)
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(a)
the point \(E'\) vertically below \(E\) is \(3\) m from \([AD]\) and \(4\) m from \([AB]\), so \(AE'=5\); \(AE=\sqrt{5^{2}+3^{2}}\)
(M1)
Or \(\sqrt{3^{2}+4^{2}+3^{2}}\).
\(\sqrt{34}=5.83\) m
A1
5.83095…
(b)
\(\tan\theta=\dfrac{3}{5}\) or \(\sin\theta=\dfrac{3}{\sqrt{34}}\)
(M1)
\(31.0^{\circ}\)
A1
30.9637…° (0.540 radians).
(c)
line from \(E\) perpendicular to \([AD]\) meets \([AD]\) at its midpoint, horizontal distance \(3\) and rise \(3\): \(\tan\phi=\dfrac{3}{3}\)
(M1)
\(45^{\circ}\)
A1
Accept \(\dfrac{\pi}{4}\).
(d)
slant height of a trapezium \(=\sqrt{4^{2}+3^{2}}=5\)
(A1)
area of one trapezium \(=\dfrac{1}{2}(12+6)(5)=45\)
(A1)
slant height of a triangle \(=\sqrt{3^{2}+3^{2}}=3\sqrt2\); area \(=\dfrac{1}{2}(8)(3\sqrt2)=16.97\)
(A1)
total \(=2(45)+2(16.97)=124\text{ m}^{2}\)
A1
123.941… FT from their slant heights.
(e)
splitting the roof space into a triangular prism (between the planes through \(E\) and \(F\) perpendicular to \([AB]\)) and two pyramids, or equivalent
391AAHL3-q00113.3 Applications of trigonometry: elevation, depression and bearingsPaper 2Hard10 marks
A hot-air balloon \(H\) is vertically above the point \(P\) on horizontal ground, at a height of \(h\) metres. From a point \(A\) on the ground the bearing of \(P\) is \(030^{\circ}\) and the angle of elevation of \(H\) is \(40^{\circ}\). The point \(C\) is on the ground \(800\) m due east of \(A\). From \(C\) the angle of elevation of \(H\) is \(35^{\circ}\).
Plan view (H is vertically above P). Diagram not to scale
(a)
Write down expressions, in terms of \(h\), for \(AP\) and \(CP\).
(2)
(b)
Show that \(h^{2}\left(\dfrac{1}{\tan^{2}35^{\circ}}-\dfrac{1}{\tan^{2}40^{\circ}}\right)+\dfrac{800h}{\tan40^{\circ}}-640\,000=0\).
(3)
(c)
Hence find the value of \(h\).
(2)
(d)
Find the bearing of \(P\) from \(C\).
(3)
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Notes
(a)
\(AP=\dfrac{h}{\tan40^{\circ}}\)
A1
\(CP=\dfrac{h}{\tan35^{\circ}}\)
A1
(b)
\(P\hat{A}C=90^{\circ}-30^{\circ}=60^{\circ}\)
(A1)
May be seen on a diagram.
cosine rule in triangle \(APC\): \(CP^{2}=AP^{2}+800^{2}-2(800)(AP)\cos60^{\circ}\)
attempt to solve \(0.6193h^{2}+953.4h-640\,000=0\) (e.g. GDC)
(M1)
Roots 505.37… and −2044.8…
\(h=505\) m
A1
Award A1 only if the negative root is rejected or not given.
(d)
\(AP=602.3\), \(CP=721.7\) and \(\dfrac{\sin A\hat{C}P}{602.3}=\dfrac{\sin60^{\circ}}{721.7}\), or coordinates \(P(301.1,\ 521.6)\) relative to \(A\)
(M1)
\(A\hat{C}P=46.3^{\circ}\)
(A1)
46.2759…°
bearing \(=270^{\circ}+46.3^{\circ}=316^{\circ}\)
A1
316.276…°
Answers: (c) \(h=505\) m · (d) 316°
392AAHL3-q00153.4 Radians, arc length and sector areaPaper 2Hard10 marks
Two circular pulleys, with centres \(O_{1}\) and \(O_{2}\) and radii \(10\) cm and \(4\) cm, lie in the same plane with \(O_{1}O_{2}=18\) cm. A taut belt passes around both pulleys without crossing itself. The straight section \([AB]\) of the belt touches the larger pulley at \(A\) and the smaller pulley at \(B\). Let \(\alpha\) be the acute angle between \([AB]\) and \([O_{1}O_{2}]\).
Diagram not to scale
(a)
Show that \(AB=12\sqrt2\) cm.
(2)
(b)
Show that \(\sin\alpha=\dfrac{1}{3}\) and that the belt is in contact with the larger pulley along an arc that subtends an angle \(\pi+2\alpha\) at \(O_{1}\).
(3)
(c)
Find the total length of the belt.
(3)
(d)
The larger pulley turns at \(120\) revolutions per minute and the belt does not slip. Find the number of revolutions per minute of the smaller pulley.
(2)
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(a)
\([O_{1}A]\) and \([O_{2}B]\) are both perpendicular to \([AB]\); the line through \(O_{2}\) parallel to \([AB]\) meets \([O_{1}A]\) at \(N\), with \(O_{1}N=10-4=6\) and \(O_{2}N=AB\)
M1
\(AB^{2}=18^{2}-6^{2}=288\)
A1
\(AB=\sqrt{288}=12\sqrt2\)
AG
(b)
\(N\hat{O_{2}}O_{1}=\alpha\) and \(\sin\alpha=\dfrac{O_{1}N}{O_{1}O_{2}}=\dfrac{6}{18}=\dfrac{1}{3}\)
M1
\(A\hat{O_{1}}O_{2}=\dfrac{\pi}{2}-\alpha\) (angle sum in triangle \(O_{1}NO_{2}\)), and by symmetry the same on the other side of \((O_{1}O_{2})\)
R1
angle of contact \(=2\pi-2\left(\dfrac{\pi}{2}-\alpha\right)\)
A1
\(=\pi+2\alpha\)
AG
(c)
contact angle on the smaller pulley \(=\pi-2\alpha\), with \(\alpha=0.3398\)
393AAHL3-q00263.7 Circular functions and their transformationsPaper 2Hard15 marks
The temperature inside a greenhouse, \(T\) °C, \(t\) hours after midnight, is modelled by \(T(t)=a\cos\big(b(t-c)\big)+d\), for \(0\le t\le24\), where \(a,b\gt0\) and \(0\le c\lt24\). The model has a period of \(24\) hours, a maximum temperature of \(31\) °C at 14:00 and a minimum temperature of \(9\) °C at 02:00.
(a)
Find the values of \(a\), \(b\), \(c\) and \(d\).
(4)
(b)
Find the temperature at 09:30.
(2)
(c)
The plants are damaged if the temperature is above \(28\) °C. Find the total length of time during the day for which the temperature is above \(28\) °C.
(3)
(d)
A heater is switched on automatically whenever the temperature falls below \(12\) °C, and switched off when it rises above \(12\) °C. Find, to the nearest minute, the times at which the heater switches off and switches on during the day.
(3)
(e)
The temperature in a second greenhouse is modelled by \(U(t)=8\cos\left(\dfrac{\pi}{12}(t-15)\right)+22\), for \(0\le t\le24\). Find, to the nearest minute, the times during the day at which the two greenhouses are at the same temperature.
\(T(t)=28\), or \(\cos\left(\dfrac{\pi}{12}(t-14)\right)=\dfrac{8}{11}\)
(M1)
\(t=11.11\) and \(t=16.89\)
(A1)
11:07 and 16:53
\(5.78\) hours
A1
5.77890… (5 h 47 min).
(d)
\(T(t)=12\), or \(\cos\left(\dfrac{\pi}{12}(t-14)\right)=-\dfrac{8}{11}\)
(M1)
switches off at 04:53 \((t=4.889)\)
A1
switches on at 23:07 \((t=23.11)\)
A1
(e)
\(T(t)=U(t)\), e.g. intersection of graphs on GDC
(M1)
07:55 \((t=7.918)\)
A1
15:46 \((t=15.77)\)
A1
Answers: (a) \(a=11,\ b=\dfrac{\pi}{12},\ c=14,\ d=20\) · (b) 24.2 °C · (c) 5.78 h · (d) off 04:53, on 23:07 · (e) 07:55 and 15:46
394AAHL3-q00503.10 Compound and double angle identitiesPaper 2Hard14 marks
A painting hangs on a vertical wall. Its bottom edge B is 2 m above the eye level of an observer and its top edge T is 5 m above eye level. The observer's eye E is at a horizontal distance \(x\) metres from the point H on the wall at eye level, as shown. The angle \(T\hat{E}B=\theta\), where \(0\lt\theta\lt\dfrac{\pi}{2}\).
Diagram not to scale
(a)
Show that \(\tan\theta=\dfrac{3x}{x^{2}+10}\).
(3)
(b)
Find \(\theta\) when \(x=4\).
(2)
(c)
By considering \(\dfrac{\mathrm{d}}{\mathrm{d}x}(\tan\theta)\), show that \(\theta\) is greatest when \(x=\sqrt{10}\).
(4)
(d)
Find the greatest value of \(\theta\), giving your answer in degrees.
(2)
(e)
The observer finds the painting comfortable to view when \(\theta\ge0.4\). Find the set of values of \(x\) for which this is true.
(3)
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(a)
\(\theta=T\hat{E}H-B\hat{E}H\), where \(\tan T\hat{E}H=\dfrac{5}{x}\) and \(\tan B\hat{E}H=\dfrac{2}{x}\)
395AAHL3-q00733.14 Vector equation of a linePaper 2Hard16 marks
Relative to a control tower at O, the unit vectors \(\boldsymbol{i}\), \(\boldsymbol{j}\) and \(\boldsymbol{k}\) point east, north and vertically upwards, and distances are in kilometres. Two light aircraft, A and B, fly in straight lines with constant velocities. Their position vectors \(t\) minutes after 10:00 are
\(\boldsymbol{r}_{A}=\begin{pmatrix}0\\0\\1.2\end{pmatrix}+t\begin{pmatrix}4\\3\\0.1\end{pmatrix}\) and \(\boldsymbol{r}_{B}=\begin{pmatrix}26\\-9\\2.3\end{pmatrix}+t\begin{pmatrix}-1\\4\\-0.1\end{pmatrix}\).
(a)
Find the speed of aircraft A in km h−1.
(2)
(b)
Show that the paths of the two aircraft intersect, and find the coordinates of the point X where they intersect.
(4)
(c)
Explain why the aircraft do not collide at X.
(1)
(d)
Find the acute angle between the two paths, in degrees.
(3)
(e)(i)
Let \(d\) km be the distance between the aircraft at time \(t\).
Show that \(d^{2}=26.04t^{2}-278.44t+758.21\).
(3)
(e)(ii)
Find the minimum distance between the aircraft and the time, to the nearest second after 10:00, at which it occurs.
(3)
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Mark
Notes
(a)
\(\sqrt{4^{2}+3^{2}+0.1^{2}}=\sqrt{25.01}\) km min−1, multiplied by 60
(M1)
\(300\) km h−1
A1
\(300.059\ldots\)
(b)
equating with different parameters: \(4s=26-u\), \(3s=-9+4u\), \(1.2+0.1s=2.3-0.1u\)
M1
Award M0 if the same parameter is used for both lines.
solving two equations: \(s=5,\ u=6\)
A1
checking the third: \(1.2+0.5=1.7=2.3-0.6\), so the paths intersect
R1
X\((20,15,1.7)\)
A1
(c)
A is at X when \(t=5\) but B is at X when \(t=6\); they are not there at the same time
minimising \(d^{2}\) (or \(d\)), eg \(t=\dfrac{278.44}{2\times26.04}\) or GDC
(M1)
\(t=5.3464\ldots\) min, ie at 10:05:21
A1
Accept 10:05:20 to 10:05:21.
minimum distance \(3.73\) km
A1
\(3.72633\ldots\)
Answers: (a) 300 km h−1 · (b) X\((20,15,1.7)\) · (d) \(67.2^{\circ}\) · (e)(ii) 3.73 km at 10:05:21
396AAHL3-q00923.18 Intersections and angles of lines and planesPaper 2Hard15 marks
The diagram shows a pyramid VABCD with rectangular base ABCD in the plane \(z=0\). The vertices have coordinates A\((0,0,0)\), B\((8,0,0)\), C\((8,6,0)\), D\((0,6,0)\) and V\((3,2,7)\). Distances are in centimetres.
normal \(\begin{pmatrix}0\\-7\\2\end{pmatrix}\) (or \(\begin{pmatrix}0\\7\\-2\end{pmatrix}\)) and the plane passes through A\((0,0,0)\), so \(-7y+2z=0\)
M1
checking V: \(7(2)-2(7)=0\) (or B: \(0=0\))
A1
Either a check or a correct use of a point with the normal.
397IBMA2H-049-05-093.1 Three-dimensional geometry: distance, solids and anglesPaper 2Hard13 marks
Calculator
Model the Earth as a sphere with centre O and radius \(R=6371\) km. A satellite S is at height \(h\) km above the surface. P is the point on the surface directly below S, and H is a point on the horizon as seen from S, so that SH is a tangent to the Earth. Let \(\alpha=\angle\text{SOH}\), measured in radians.
Not to scale (h is exaggerated)
(a)
Show that \(\cos\alpha=\frac{R}{R+h}\).
(2)
(b)
A space station orbits at \(h=420\) km. (i) Find the straight-line distance SH. (ii) Find \(\alpha\), and hence the distance along the Earth's surface from P to H.
(5)
(c)
Identical satellites at a height of 420 km are placed equally spaced in a single circular orbit above the equator. Find the least number of satellites needed so that every point on the equator is on or inside the horizon of at least one satellite. Justify your answer.
(3)
(d)
Find the height at which a satellite must orbit so that the distance along the surface from P to H is 2000 km.
(3)
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Part (a)
R1 a tangent is perpendicular to the radius, so \(\angle\text{OHS}=\frac{\pi}{2}\)
✓ 1
A1 in right-angled triangle OHS, \(\cos\alpha=\frac{\text{OH}}{\text{OS}}=\frac{R}{R+h}\)
✓ 1
Part (b)
(i) M1 \(\text{SH}=\sqrt{6791^2-6371^2}\)
✓ 1
A1 \(2350\) km (2351.2…)
✓ 1
(ii) A1 \(\alpha=\arccos\frac{6371}{6791}=0.354\) (0.35356…)
✓ 1
M1 arc length \(=R\alpha=6371\times0.35356\ldots\), using the unrounded value of α
✓ 1
A1 \(2250\) km (2252.5…)
✓ 1
Part (c)
M1 each satellite covers an angle of \(2\alpha=0.7071\) rad of the equator
✓ 1
A1 \(\frac{2\pi}{0.7071}=8.89\)
✓ 1
R1 8 satellites cover only \(8\times0.7071<2\pi\), so the number must be a whole number at least 8.89: 9 satellites
✓ 1
Part (d)
M1 \(\alpha=\frac{2000}{6371}=0.3139\)
✓ 1
M1 \(R+h=\frac{6371}{\cos0.3139}\) (\(=6698\))
✓ 1
A1 \(h=327\) km
✓ 1
Exam tip: Keep α in radians and store its unrounded value in your calculator. You need it in s = Rα and again when you divide 2π by the angle covered. Rounding α early can change your final distance.
398IBMA2H-049-06-123.15 Intersecting, parallel and skew linesPaper 2Hard12 marks
Calculator
In a railway tunnel, two straight cable runs are modelled by L1: r = i + 2k + s(i + 2j − k) L2: r = 4i + j + t(2i − j + k) Distances are in metres. For safety, the cables must be at least 1.5 m apart.
(a)
Show that L1 and L2 are skew.
(4)
(b)
Find a vector perpendicular to both lines.
(2)
(c)
Find the Cartesian equation of the plane Π that contains L1 and is parallel to L2.
(3)
(d)
Find the shortest distance between the cables and determine whether they meet the safety requirement.
(3)
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Marking point
Mark
Notes
Part (a)
R1 direction vectors are not scalar multiples, so the lines are not parallel
✓ 1
M1 equate components: 1 + s = 4 + 2t, 2s = 1 − t, 2 − s = t
✓ 1
A1 solve two equations, e.g. s = −1, t = 3
✓ 1
R1 remaining equation fails (0 ≠ 10), so no intersection; not parallel and not intersecting, hence skew (AG)
A1 1.69 m, which is at least 1.5 m so the requirement is met
✓ 1
Common error: The distance between two arbitrary points on skew lines is not the shortest distance; you must project the joining vector onto the common perpendicular.
399IBMA2H-049-00-103.10 Compound and double angle identitiesPaper 2Hard12 marks
Calculator
A glass museum entrance is a right pyramid with a square base of side 28 m and vertical height 21 m. The apex is V and the centre of the base is O. Let \(\alpha\) be the angle between a triangular face and the base, and let \(\beta\) be the angle between a sloping edge and the base. M1 and M2 are the midpoints of two opposite base edges.
Cross-section through V, M₁ and M₂. Diagram not to scale
(a)
Show that \(\tan\alpha=\frac{3}{2}\).
(1)
(b)
Find \(\beta\).
(2)
(c)
By using a double angle identity, show that \(\cos M_1\hat{V}M_2=\frac{5}{13}\).
(4)
(d)
Place O at the origin with base vertices \((\pm14,\pm14,0)\) and V at \((0,0,21)\). Find the acute angle between the planes containing two adjacent triangular faces.
Common error: Do not confuse the face angle \(\alpha\), measured to the midpoint of a base edge, with the edge angle \(\beta\), measured to a vertex. They use different horizontal distances: 14 and \(14\sqrt2\).
400IBMA2H-049-01-073.10 Compound and double angle identitiesPaper 2Hard11 marks
Calculator
In a rugby match, the goal posts P and Q stand on the goal line, 5.6 m apart. A kick is to be taken from a point K on a line that is perpendicular to the goal line. This line meets the goal line at O, where O, P and Q are collinear, OP = 10 m and OQ = 15.6 m. The kicker may choose the distance OK = x metres, x > 0. Let θ be the angle PKQ.
Plan view. OP = 10 m, PQ = 5.6 m. Diagram not to scale
(a)
Show that \(\theta = \arctan\left(\frac{15.6}{x}\right) - \arctan\left(\frac{10}{x}\right)\).
(2)
(b)
Find \(\frac{d\theta}{dx}\), simplifying your answer.
(3)
(c)
Find the exact value of x that maximises θ, justifying that it gives a maximum.
(4)
(d)
Find the maximum value of θ, in degrees.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 Uses the right angle at O to get angle OKQ = \(\arctan\frac{15.6}{x}\) and angle OKP = \(\arctan\frac{10}{x}\)
✓ 1
A1 States that P lies between O and Q, so θ = angle OKQ − angle OKP, giving the result
✓ 1
Part (b)
M1 Uses the chain rule with \(\frac{d}{du}\arctan u = \frac{1}{1+u^2}\) and \(u = \frac{a}{x}\)
R1 Valid justification: \(\frac{d\theta}{dx} = \frac{873.6-5.6x^2}{(x^2+100)(x^2+243.36)}\) has a positive denominator. Its numerator is positive for \(0 < x < 2\sqrt{39}\) and negative for \(x > 2\sqrt{39}\), so this is a maximum. Sign tests at values such as x = 10 and x = 15 also earn this mark
✓ 1
Part (d)
M1 Substitutes \(x = \sqrt{156}\) into θ, or uses \(\tan\theta = \frac{5.6}{2\sqrt{156}}\)
✓ 1
A1 θ = 12.6°
✓ 1
Exam tip: Write the derivative of arctan(a/x) as −a/(x² + a²) straight away. The cross-multiplied equation is then linear in x², and the optimum works out as the geometric mean √(10 × 15.6).
401IBMA2H-049-07-113.17 Equations of a planePaper 2Hard15 marks
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On a space station, a triangular solar array has corners A(2, 1, 0), B(5, 1, 4) and C(2, 7, 0). A sensor is fixed at D(5, 4, −6). All coordinates are in metres. The array lies in the plane Π.
Exam tip: Because the normal (4, 0, −3) has magnitude 5, the distance DF is simply 5|t|. Spotting this saves calculating F to D explicitly, and it provides a check on the plane-distance formula.
402IBMA2H-049-07-123.10 Compound and double angle identitiesPaper 2Hard13 marks
Calculator
A robotic arm in a pharmacy has an upper segment of length 40 cm attached to a fixed shoulder S, and a lower segment of length 25 cm ending at a gripper G. The angle between the two segments at the elbow is θ radians, where 0 < θ < π. Let d cm be the distance SG.
Not drawn to scale
(a)
Find d when θ = \(\frac{2\pi}{3}\).
(2)
(b)
Find θ when d = 50.
(3)
(c)
For d = 50, find the angle at S between the upper segment and the line SG.
(3)
(d)
In normal use, θ varies between \(\frac{\pi}{3}\) and \(\frac{5\pi}{6}\). Find the range of possible values of d.
(3)
(e)
Find the value of θ for which the triangle SEG, where E is the elbow, is isosceles with SG = SE.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 d² = 40² + 25² − 2(40)(25)cos(\(\frac{2\pi}{3}\)) = 3225 ; A1 d = 56.8 cm
✓ 2
Part (b)
M1 2500 = 2225 − 2000 cos θ ; A1 cos θ = −0.1375 ; A1 θ = 1.71
✓ 3
Part (c)
M1 sine rule \(\frac{\sin\alpha}{25}=\frac{\sin\theta}{50}\) with θ = 1.708… ; A1 sin α = 0.495… ; A1 α = 0.518 rad (acute, as it is opposite the shorter segment)
✓ 3
Part (d)
R1 d² = 2225 − 2000 cos θ and cos θ is decreasing on [0, π], so d is increasing ; A1 θ = π/3 gives d = 35 ; A1 θ = 5π/6 gives d = 62.9, so 35 ≤ d ≤ 62.9
✓ 3
Part (e)
M1 1600 = 2225 − 2000 cos θ, giving cos θ = 0.3125 ; A1 θ = 1.25 (which lies within the range in (d))
✓ 2
Common error: Having the calculator in degree mode while θ is in radians ruins every part. Set radian mode, and in part (d) justify the range by monotonicity rather than just evaluating the endpoints.
403IBMA2H-049-04-113.18 Intersections and angles of lines and planesPaper 2Hard15 marks
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Relative to the centre of the Earth O, the orbits of two satellites lie in the planes Π1: \(x - 2y + 2z = 0\) Π2: \(2x + y - 2z = 0\). A space probe travels along the line r = (10i + 5k) + t(−i + j). Units are thousands of kilometres.
(a)
Find the acute angle between the two orbital planes.
(3)
(b)
Find a vector equation of the line of intersection of the two planes.
(3)
(c)
The probe crosses Π1 at P and Π2 at Q. Find the coordinates of P and Q.
(4)
(d)
Find the acute angle between the probe's path and Π1.
(3)
(e)
Find the distance PQ, giving your answer in kilometres.
(2)
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Marking point
Mark
Notes
Part (a)
M1 dot product of normals (1, −2, 2) and (2, 1, −2)
M1 \(\sin\phi = \frac{|\boldsymbol{n}_1\cdot\boldsymbol{d}|}{|\boldsymbol{n}_1||\boldsymbol{d}|}\), or the angle with the normal then subtracted from 90°
M1 PQ = \(\left(10 - \frac{20}{3}\right)\sqrt2\) or \(\sqrt{\left(\frac{10}{3}\right)^2 + \left(\frac{10}{3}\right)^2}\)
✓ 1
A1 4710 km; accept answers from 4710 km to 4714 km
✓ 1
Exam tip: Check a line-of-intersection direction by confirming that its dot product with each normal is zero. It takes seconds and catches sign errors in the cross product.
404IBMA3H-3-19r5-133.16 Vector productPaper 2Hard13 marks
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A drone surveys a triangular plot of land. Relative to a base point O, the corners are at A(0, 0, 0), B(120, 50, 10) and C(40, 130, −5), with all coordinates in metres and the third coordinate the height. Use unrounded values in later parts and give answers to 3 significant figures unless stated otherwise.
Exam tip: When a vector product is used for a plane, any point on the plane gives the constant; with the origin on the plane the constant is zero.
405IBMA3H-3-18r8-123.7 Circular functions and their transformationsPaper 2Hard8 marks
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The height \(h\) cm of dough in a proving container is modelled by \(h=p-q\cos(rt)\), where \(q>0\), \(r>0\) and \(t\) is the time in minutes. At \(t=0\) the height is a minimum of 6 cm. The height reaches its first maximum of 14 cm at \(t=40\).
(a)
Find \(p\) and \(q\).
(2)
(b)
Find the exact value of \(r\).
(2)
(c)
Find the first time at which the height is 12.5 cm, to 3 significant figures.
(2)
(d)
During the first 80 minutes, the height is above 12.5 cm for one interval. Find its length, to 3 significant figures. Use unrounded values.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 midline \(\frac{14+6}{2}\) or amplitude \(\frac{14-6}{2}\)
✓ 1
A1 \(p=10,\ q=4\)
✓ 1
Part (b)
M1 half a period is 40 so period is 80
✓ 1
A1 \(r=\frac{2\pi}{80}=\frac{\pi}{40}\)
✓ 1
Part (c)
M1 \(10-4\cos\!\left(\frac{\pi t}{40}\right)=12.5\) so \(\cos\!\left(\frac{\pi t}{40}\right)=-0.625\)
✓ 1
A1 \(t=28.6\) minutes
✓ 1
Part (d)
M1 symmetry of the cosine graph about \(t=40\): second time is \(80-28.596\ldots\)
✓ 1
A1 \(22.8\) minutes
✓ 1
Exam tip: Cosine is symmetric about its maximum and minimum; use symmetry to get the second crossing time instead of solving again.
406IBMA3H-3-18r8-133.14 Vector equation of a linePaper 2Hard8 marks
Calculator
Two relay drones fly in straight lines. With \(t\) in minutes and distances in km, the positions are \(\mathbf r_A=\begin{pmatrix}1\\2\\3\end{pmatrix}+t\begin{pmatrix}2\\0\\-1\end{pmatrix}\) and \(\mathbf r_B=\begin{pmatrix}9\\-2\\5\end{pmatrix}+t\begin{pmatrix}-1\\2\\-1\end{pmatrix}\).
(a)
Find the vector \(\overrightarrow{AB}\) from drone A to drone B at time \(t\).
(2)
(b)
Show that \(AB^2=13t^2-64t+84\).
(2)
(c)
Find the time at which the drones are closest, to 3 significant figures.
(2)
(d)
Find the minimum distance between the drones, to 3 significant figures.
(2)
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Marking point
Mark
Notes
Part (a)
M1 \(\mathbf r_B-\mathbf r_A\)
✓ 1
A1 \(\begin{pmatrix}8-3t\\2t-4\\2\end{pmatrix}\)
✓ 1
Part (b)
M1 sum of squares \((8-3t)^2+(2t-4)^2+4\)
✓ 1
A1 expand to \(13t^2-64t+84\) (AG)
✓ 1
Part (c)
M1 minimum of the quadratic: \(t=\frac{64}{26}\) (or differentiate)
✓ 1
A1 \(t=2.46\) minutes
✓ 1
Part (d)
M1 substitute \(t=\frac{32}{13}\) into \(AB^2\) to get \(5.23\ldots\)
✓ 1
A1 \(\sqrt{5.2308}=2.29\) km
✓ 1
Common error: Minimise \(AB^2\) rather than \(AB\); the quadratic is easier to handle, and the minimum occurs at the same time. Remember to take the square root at the end.
407IBMA3H-3-18r8-143.16 Vector productPaper 2Hard8 marks
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In a model of a fat crystal, three atoms are at A\((1,2,2)\), B\((-1,0,3)\) and C\((4,-2,1)\), with coordinates in ångström units (1 Å \(=10^{-10}\) m).
(a)
Find \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\).
(2)
(b)
Find angle BAC, to the nearest tenth of a degree.
(3)
(c)
Find the area of triangle ABC, to 3 significant figures. Use the unrounded angle.
M1 \(\overrightarrow{AB}\cdot\overrightarrow{AC}=-6+8-1=1\) with \(|\overrightarrow{AB}|=3\) and \(|\overrightarrow{AC}|=\sqrt{26}\)
✓ 1
M1 \(\cos A=\frac{1}{3\sqrt{26}}\)
✓ 1
A1 \(86.3^\circ\)
✓ 1
Part (c)
M1 area \(=\frac12|\overrightarrow{AB}||\overrightarrow{AC}|\sin A\)
✓ 1
A1 \(\frac12(3)\sqrt{26}\sin(86.25\ldots^\circ)\)
✓ 1
A1 \(7.63\) Å\(^2\)
✓ 1
Exam tip: After finding the angle with the scalar product, use \(\frac12ab\sin C\) for the area, and keep the unrounded angle in your calculator.
408IBMA3H-3-19r10-063.1 Three-dimensional geometry: distance, solids and anglesPaper 2Hard11 marks
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The concrete anchor block of a suspension bridge is a cuboid ABCDEFGH. The base ABCD is horizontal, E, F, G, H are vertically above A, B, C, D respectively, AB = 5.0 m, BC = 4.0 m and BF = 3.0 m. Give answers to three significant figures and use unrounded values in later parts.
The cuboid ABCDEFGH with the diagonal AG and the triangle AFC. Diagram not to scale
(a)
Find the length of the diagonal AG.
(2)
(b)
Find the size of angle \(F\hat{A}C\).
(4)
(c)
Find the area of triangle AFC.
(2)
(d)
The tetrahedron BAFC has triangle ABC as one face. Find the perpendicular distance from B to the plane AFC.
A1 13.9 m\(^2\) (exact value \(\frac{\sqrt{769}}{2}\))
✓ 1
Part (d)
M1 volume of tetrahedron BAFC \(=\frac13\times\left(\frac12\times5\times4\right)\times3=10\) m\(^3\)
✓ 1
M1 volume \(=\frac13\times\text{area }AFC\times h\), so \(h=\frac{30}{\text{area }AFC}\)
✓ 1
A1 \(h=30\div13.865=2.16\) m
✓ 1
Exam tip: A volume found two different ways gives a quick route to a perpendicular height when the direct geometry is awkward.
409IBMA3H-3-19r10-133.18 Intersections and angles of lines and planesPaper 2Hard10 marks
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A flat mirror panel on a space probe lies in the plane \(\Pi\) with equation \(x+2y-2z=17\). A sensor is at the point P\((2,1,-2)\). Distances are in metres.
(a)
Find a vector equation of the line through P perpendicular to \(\Pi\).
(2)
(b)
Find the coordinates of the point F where this line meets \(\Pi\).
(3)
(c)
The image of P in the mirror is P\('\). Find the coordinates of P\('\).
(2)
(d)
Find the acute angle between \(\Pi\) and the horizontal plane \(z=0\). Give your answer to three significant figures.
M1 \(\lambda=2\), or \(\overrightarrow{OP'}=2\overrightarrow{OF}-\overrightarrow{OP}\)
✓ 1
A1 \(P'=(4,5,-6)\)
✓ 1
Part (d)
M1 normals \((1,2,-2)\) and \((0,0,1)\)
✓ 1
M1 \(\cos\theta=\frac{|{-2}|}{3\times1}=\frac23\)
✓ 1
A1 \(48.2^\circ\)
✓ 1
Exam tip: The image of a point in a plane lies on the same perpendicular, the same distance beyond the plane, so it corresponds to doubling the parameter at the foot.
410IBMA3H-3-18r11-073.7 Circular functions and their transformationsPaper 2Hard10 marks
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The depth of water at a point in a swimming wave pool is modelled by \(d(t)=A\cos\big(B(t-1.5)\big)+D\) metres, where \(t\) is in seconds, \(A>0\) and \(B>0\). The maximum depth is 3.2 m, first reached at \(t=1.5\), and the minimum depth is 1.4 m, first reached at \(t=4.5\). Use radians.
(a)
Find the values of \(A\) and \(D\).
(2)
(b)
Find the value of \(B\).
(2)
(c)
Find the depth at \(t=8\), to 3 significant figures.
(2)
(d)
Find the length of time in the first cycle during which the depth is at least 2.9 m. Give your answer to 3 significant figures.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(A=\frac{3.2-1.4}{2}=0.9\)
✓ 1
A1 \(D=\frac{3.2+1.4}{2}=2.3\)
✓ 1
Part (b)
M1 half a period is \(4.5-1.5=3\) so the period is 6
M1 \(0.9\cos u=0.6\) with \(u=\frac{\pi}{3}(t-1.5)\), so \(\cos u=\frac23\)
✓ 1
M1 \(u=\pm0.841\) (from \(\arccos\frac23\))
✓ 1
M1 \(t=1.5\pm\frac{3}{\pi}(0.841)\), i.e. \(t=0.697\) and \(t=2.30\)
✓ 1
A1 length of time \(=1.61\) s
✓ 1
Exam tip: For \(A\cos(B(t-C))+D\): \(A\) is half the difference between maximum and minimum, \(D\) their mean, and \(C\) the time of a maximum.
411IBMA3H-3-18r12-063.1 Three-dimensional geometry: distance, solids and anglesPaper 2Hard10 marks
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A climbing structure is modelled as a solid right pyramid VABCD. The base ABCD is a horizontal rectangle with AB = 8 m and BC = 6 m. The vertex V is 6 m vertically above the centre O of the base.
Diagram not to scale
(a)
Show that \(AO = 5\) m and hence find the length VA, to 3 significant figures.
(3)
(b)
Find the angle between edge VA and the base ABCD, to the nearest 0.1°.
(2)
(c)
Let M be the midpoint of AB. Find the angle between the face VAB and the base, to the nearest 0.1°.
(3)
(d)
Find the area of triangle VAB, to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 diagonal \(AC = \sqrt{8^2+6^2} = 10\)
✓ 1
A1 \(AO = 5\) m (AG)
✓ 1
A1 \(VA = \sqrt{5^2+6^2} = \sqrt{61} = 7.81\) m
✓ 1
Part (b)
M1 \(\tan\theta = \frac{6}{5}\) (angle VAO)
✓ 1
A1 50.2°
✓ 1
Part (c)
R1 OM is perpendicular to AB with \(OM = 3\) m (half of BC), and VM is perpendicular to AB
✓ 1
M1 \(\tan\phi = \frac{6}{3}\) (angle VMO)
✓ 1
A1 63.4°
✓ 1
Part (d)
M1 \(VM = \sqrt{6^2+3^2} = \sqrt{45}\) and area \(= \frac{1}{2}\times8\times VM\)
✓ 1
A1 26.8 m2
✓ 1
Exam tip: In pyramid problems, draw the right-angled triangle you will use (apex, centre of base, and the foot of the perpendicular to the edge) as a separate 2D sketch.
412IBMA3H-3-18r12-123.18 Intersections and angles of lines and planesPaper 2Hard8 marks
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Two sloping faces of a rail cutting are modelled by the planes \(\Pi_1\): \(x + 2y - z = 3\) and \(\Pi_2\): \(2x - y + z = 4\). The planes meet in a straight line \(L\).
(a)
Use a vector product of the normal vectors to find a direction vector of \(L\).
(2)
(b)
Show that the point \((1, 4, 6)\) lies on both planes, and write down a vector equation of \(L\).
(3)
(c)
Find the acute angle between the two planes, to the nearest 0.1°.
Common error: The angle between planes comes from their normals; if the scalar product is negative, take the absolute value to give the acute angle.
413IBMA3H-3-18r12-143.17 Equations of a planePaper 2Hard9 marks
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A climbing panel is flat and passes through three anchor points \(A(1,0,0)\), \(B(0,2,0)\) and \(C(0,0,3)\), with coordinates in metres. The \(z\)-axis is vertical.
(a)
Find \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\).
A1 \(|\mathbf{n}| = 7\), so \(\sin\theta = \frac{2}{7}\)
✓ 1
A1 16.6°
✓ 1
Common error: Be careful to distinguish the angle between a line and a plane (use \(\sin\)) from the angle between a line and the plane's normal (use \(\cos\)).
414IBMA3H-3-19r12-123.11 Symmetry properties of trigonometric graphsPaper 2Hard12 marks
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The daily energy yield \(E\) (in kWh) of a solar panel at tilt angle \(\theta\) radians to the horizontal, for \(0\le\theta\le\frac\pi2\), is modelled by \(E(\theta)=4\sin\theta\cos\theta+3\cos^2\theta\).
(a)
Show that \(E(\theta)=1.5+2\sin2\theta+1.5\cos2\theta\).
(3)
(b)
Write \(2\sin2\theta+1.5\cos2\theta\) in the form \(R\sin(2\theta+\alpha)\), where \(R>0\) and \(0<\alpha<\frac\pi2\). Give \(\alpha\) to 3 significant figures.
(3)
(c)
Hence find the maximum value of \(E\) and the value of \(\theta\) at which it occurs. Use the unrounded value of \(\alpha\) and give \(\theta\) to 3 significant figures.
(3)
(d)
Find the range of values of \(\theta\), for \(0\le\theta\le\frac\pi2\), for which \(E\ge3.5\). Use the unrounded value of \(\alpha\) and give your answers to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\sin\theta\cos\theta=\frac12\sin2\theta\), so \(4\sin\theta\cos\theta=2\sin2\theta\)
✓ 1
M1 \(\cos^2\theta=\frac12(1+\cos2\theta)\), so \(3\cos^2\theta=1.5+1.5\cos2\theta\)
✓ 1
A1 adds the two to reach the given form (AG)
✓ 1
Part (b)
M1 expands \(R\sin2\theta\cos\alpha+R\cos2\theta\sin\alpha\) and equates: \(R\cos\alpha=2\), \(R\sin\alpha=1.5\)
✓ 1
A1 \(R=\sqrt{2^2+1.5^2}=2.5\)
✓ 1
A1 \(\tan\alpha=\frac{1.5}{2}=0.75\), so \(\alpha=0.6435\), i.e. \(\alpha=0.644\)
✓ 1
Part (c)
A1 maximum \(E=1.5+2.5=4.0\)
✓ 1
M1 \(2\theta+\alpha=\frac\pi2\)
✓ 1
A1 \(\theta=\frac{\frac\pi2-0.6435}{2}=0.4636\), so \(\theta=0.464\)
✓ 1
Part (d)
M1 \(2.5\sin(2\theta+\alpha)\ge2\), so \(\sin(2\theta+\alpha)\ge0.8\)
✓ 1
A1 lower end \(\theta=\frac{\arcsin0.8-\alpha}{2}=0.1419\), so 0.142
✓ 1
A1 upper end \(\theta=\frac{\pi-\arcsin0.8-\alpha}{2}=0.7854\) (\(=\frac\pi4\)), so \(0.142\le\theta\le0.785\)
✓ 1
Exam tip: In the form \(R\sin(2\theta+\alpha)\), find \(R\) and \(\alpha\) from the two simultaneous equations, then use the largest value of the sine (\(1\)) to find the maximum.
415IBMA3H-3-19r12-143.8 Solving trigonometric equationsPaper 2Hard13 marks
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The monthly sales index of a seasonal product is modelled by \(S(x)=3\cos2x+4\sin x\), where \(x\) is in radians, \(0\le x\le2\pi\), and \(x=0\) is the start of the year.
(a)
Show that \(S(x)=3+4\sin x-6\sin^2x\).
(2)
(b)
Find the maximum value of \(S\) and the values of \(x\), in \(0\le x\le2\pi\), at which it occurs. Give \(x\) to 3 significant figures.
(4)
(c)
Solve \(S(x)=1\) for \(0\le x\le2\pi\). Give your answers to 3 significant figures where they are not exact.
A1 \((3\sin x+1)(\sin x-1)=0\), so \(\sin x=1\) or \(\sin x=-\frac13\)
✓ 1
A1 \(x=\frac\pi2\)
✓ 1
A1 \(x=\pi+0.3398=3.48\)
✓ 1
A1 \(x=2\pi-0.3398=5.94\)
✓ 1
Part (d)
M1 considers \(-1\le\sin x\le1\), with end values \(S=1\) at \(\sin x=1\) and \(S=-7\) at \(\sin x=-1\), and the maximum \(\frac{11}{3}\) inside the interval
✓ 1
A1 \(-7\le S\le\frac{11}{3}\)
✓ 1
Common error: After rewriting in terms of \(\sin x\), the quadratic in \(\sin x\) is only valid for \(-1\le\sin x\le1\), so its vertex may not be the minimum of the function.
416IBMA3H-3-19r16-053.17 Equations of a planePaper 2Hard10 marks
Calculator
A relay antenna is at the point \(P(8,1,0)\). A space probe travels along the straight line \(L\) with equation \(\mathbf{r}=\begin{pmatrix}2\\1\\-3\end{pmatrix}+\lambda\begin{pmatrix}2\\1\\2\end{pmatrix}\), where distances are in km. The point \(X\) on \(L\) has parameter \(\lambda\).
(a)
Write down \(\overrightarrow{PX}\) in terms of \(\lambda\).
(2)
(b)
The probe is closest to the antenna when \(\overrightarrow{PX}\) is perpendicular to the direction of \(L\). Find the value of \(\lambda\) at this point.
(3)
(c)
Find the coordinates of the point \(P'\) which is the reflection of \(P\) in the line \(L\).
(3)
(d)
Find the shortest distance from the antenna to the path of the probe.
(2)
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Part (a)
M1 subtract: position of \(X\) minus \(\overrightarrow{OP}\)
M1 magnitude of \(\overrightarrow{PX}=(-2,2,1)\) (ECF from (b))
✓ 1
A1 3 km
✓ 1
Common error: Perpendicular distance needs the scalar product with the direction vector to be zero; the distance to the line's given point is not the shortest distance.
417IBMA3H-3-19r16-143.14 Vector equation of a linePaper 2Hard14 marks
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Two space probes move with constant velocities. Relative to a base, after \(t\) hours the position of probe A is \(\mathbf{r}_A=\begin{pmatrix}1\\6\\3\end{pmatrix}+t\begin{pmatrix}1\\2\\0\end{pmatrix}\) and the position of probe B is \(\mathbf{r}_B=\begin{pmatrix}-1\\2\\8\end{pmatrix}+t\begin{pmatrix}3\\3\\-2\end{pmatrix}\), with distances in km.
(a)
Find \(\overrightarrow{AB}\) in terms of \(t\).
(2)
(b)
Show that \(|\overrightarrow{AB}|^2=9t^2-36t+45\).
(3)
(c)
Find the minimum distance between the probes and the time at which it occurs.
(4)
(d)
Show that when \(t=2\) the vector \(\overrightarrow{AB}\) is perpendicular to the velocity of B relative to A.
(2)
(e)
Find the angle between the directions of travel of the two probes. Give your answer in degrees to one decimal place.
M1 C lies to the right of AB (the turn at B is clockwise), so adds: 072° + 46.1°
✓ 1
A1 118.1°
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Common error: Draw a north line at every point and use co-interior angles to turn each bearing into an angle inside the triangle; do not read bearings straight into the cosine rule.
419IBMA3H-3-18r17-123.15 Intersecting, parallel and skew linesPaper 2Hard10 marks
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Two straight support wires in a greenhouse frame are modelled, with lengths in metres, by the lines L1: r = (2, 0, 4) + λ(1, 2, −1) and L2: r = (0, 3, 1) + μ(2, −1, 1).
(a)
Show that L1 and L2 are skew.
(4)
(b)
Find a vector perpendicular to both lines.
(2)
(c)
Find the shortest distance between the two wires. Give your answer to 3 significant figures.
M1 vector between a point on each line, e.g. (0, 3, 1) − (2, 0, 4) = (−2, 3, −3)
✓ 1
M1 scalar product with the common perpendicular: |(−2, 3, −3)·(1, −3, −5)| = 4
✓ 1
M1 divides by the magnitude √35
✓ 1
A1 0.676 m
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Exam tip: The shortest distance between skew lines is the length of the projection of the vector joining any two points of the lines onto the common perpendicular direction.
420IBMA3H-3-19r18-103.14 Vector equation of a linePaper 2Hard9 marks
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Two ships move at constant velocity. At time \(t\) hours after noon, with positions in kilometres relative to a lighthouse, ship \(A\) has position \(\begin{pmatrix}12t\\5t\end{pmatrix}\) and ship \(B\) has position \(\begin{pmatrix}30-8t\\-10+6t\end{pmatrix}\).
Not to scale. Paths shown for 0 ≤ t ≤ 3; distances in km
(a)
Show that the position of \(B\) relative to \(A\) is \(\begin{pmatrix}30-20t\\t-10\end{pmatrix}\).
(1)
(b)
Show that the square of the distance between the ships is \(d^2=401t^2-1220t+1000\).
(3)
(c)
Find the time at which the ships are closest, and the minimum distance between them. Use the unrounded time to find the distance, and give each answer to 3 significant figures.
(4)
(d)
A ship sounds a fog signal that can be heard within 9.0 km. State, with a reason, whether the crew of ship \(A\) hear a signal sounded by ship \(B\) at the time of closest approach.
M1 minimise \(d^2\): \(802t-1220=0\) (or vertex of quadratic)
✓ 1
A1 \(t=1.52\) hours (1.5212…)
✓ 1
M1 substitute into \(d^2\), giving \(72.07\)
✓ 1
A1 \(d_{min}=8.49\) km
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Part (d)
R1 yes, because \(8.49<9.0\) km
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Exam tip: Minimising \(d^2\) instead of \(d\) gives the same time and avoids differentiating a square root.
421IBMA3H-3-19r18-133.18 Intersections and angles of lines and planesPaper 2Hard14 marks
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A follow-spot at \(A(0,0,9)\) shines a narrow beam through the point \(B(4,2,5)\). Coordinates are in metres, with the stage floor at \(z=0\). A hanging prop is at \(P(5,5,3)\).
(a)
Find a vector equation of the line along which the beam travels.
(2)
(b)
Find the coordinates of the point where the beam hits the stage floor.
(3)
(c)
Find the shortest distance from the prop \(P\) to the beam.
(6)
(d)
Find the angle the beam makes with the stage floor, correct to 3 significant figures.
M1 \(\overrightarrow{PF}=\begin{pmatrix}1\\-2\\0\end{pmatrix}\) and its magnitude
✓ 1
A1 \(\sqrt5=2.24\) m
✓ 1
Part (d)
M1 floor is perpendicular to \(\begin{pmatrix}0\\0\\1\end{pmatrix}\); \(\sin\phi=\frac{|{-4}|}{\sqrt{16+4+16}}\)
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A1 \(\sin\phi=\frac23\)
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A1 \(\phi=41.8^\circ\)
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Exam tip: For the angle between a line and a plane, find the angle with the normal and take its complement, or use sine with the normal direction.
422IBMA3H-3-18r22-053.18 Intersections and angles of lines and planesPaper 2Hard13 marks
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A flat docking plane of a space station lies in the plane \(\Pi: 2x-y+2z=19\), with distances in kilometres. A probe travels along the line \(L_1: \mathbf r=\begin{pmatrix}0\\1\\2\end{pmatrix}+t\begin{pmatrix}1\\2\\2\end{pmatrix}\).
(a)
Find the coordinates of the point where \(L_1\) meets \(\Pi\).
(3)
(b)
Find the acute angle \(\theta\) between \(L_1\) and \(\Pi\). Give your answer in degrees to 3 significant figures.
(3)
(c)
A second probe travels along \(L_2: \mathbf r=\begin{pmatrix}0\\1\\2\end{pmatrix}+t\begin{pmatrix}1\\2\\k\end{pmatrix}\), where \(k\ne2\). The line \(L_2\) also makes angle \(\theta\) with \(\Pi\). Find the value of \(k\).
(4)
(d)
Find the distance between the point where \(L_1\) meets \(\Pi\) and the point where \(L_2\) meets \(\Pi\). Give your answer to 3 significant figures.
Exam tip: When an angle is given as an exact value, equate sin θ expressions exactly (here 4/9) rather than using a rounded angle.
423IBMA3H-3-18r22-143.4 Radians, arc length and sector areaPaper 2Hard10 marks
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The deck of a footbridge is supported by a circular arc. The arc has a horizontal chord (span) of 40 m and a maximum height (rise) of 8 m above the chord. The arc is part of a circle with centre \(O\) and radius \(R\) metres, and subtends angle \(\theta\) radians at \(O\).
Diagram not to scale
(a)
Show that \(R=29\).
(3)
(b)
Find \(\theta\), in radians to 3 significant figures.
(2)
(c)
Find the length of the arc. Give your answer to 3 significant figures.
(2)
(d)
Find the area of the segment between the arc and the chord. Give your answer to 3 significant figures.
(3)
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Part (a)
M1 the centre is \(R-8\) below the chord, with half-chord 20: \((R-8)^2+20^2=R^2\)
✓ 1
A1 \(-16R+64+400=0\)
✓ 1
A1 \(R=29\)
✓ 1
Part (b)
M1 \(\sin\frac\theta2=\frac{20}{29}\)
✓ 1
A1 \(\theta=1.52\) radians (1.5220…)
✓ 1
Part (c)
M1 \(s=R\theta=29\times1.5220…\) (unrounded)
✓ 1
A1 44.1 m
✓ 1
Part (d)
M1 segment area \(=\frac12R^2(\theta-\sin\theta)\) (sector minus triangle)
✓ 1
M1 substitute \(R=29\) and unrounded \(\theta\)
✓ 1
A1 220 m2
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Exam tip: A segment area is sector minus triangle: ½R²θ − ½R² sin θ; keep θ unrounded until the end.
424IBMA3H-3-18r23-053.3 Applications of trigonometry: elevation, depression and bearingsPaper 2Hard11 marks
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Two wind turbines stand on level ground. The bases of the towers are at points \(A\) and \(B\), and an observer stands at point \(O\) on the same level ground, with \(OA=240\) m, \(OB=330\) m and angle \(AOB=47^\circ\). The tower at \(A\) is 90 m high and the tower at \(B\) is 120 m high. Give lengths to 3 significant figures and angles to the nearest 0.1°, using unrounded values from earlier parts in later calculations.
Plan view of the ground, not to scale. A and B are the bases of the towers
(a)
Find the distance \(AB\).
(3)
(b)
Find the size of angle \(OBA\).
(3)
(c)
Find the angle of elevation of the top of the tower at \(A\) from the point \(B\).
(2)
(d)
A cable is stretched in a straight line between the tops of the two towers. Find its length.
(3)
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Part (a)
M1 cosine rule in triangle \(OAB\)
✓ 1
A1 \(AB^2=240^2+330^2-2(240)(330)\cos47^\circ\)
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A1 \(AB=242\) m
✓ 1
Part (b)
M1 sine rule: \(\frac{\sin OBA}{240}=\frac{\sin47^\circ}{AB}\)
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A1 \(\sin OBA=0.7259\ldots\)
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A1 \(46.5^\circ\) (acute, since \(OA\) is not the longest side)
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Part (c)
M1 \(\tan\theta=\frac{90}{AB}\)
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A1 \(20.4^\circ\)
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Part (d)
M1 vertical difference \(120-90=30\) m
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M1 Pythagoras: \(\sqrt{AB^2+30^2}\)
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A1 \(244\) m
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Exam tip: Draw the plan view (triangle \(OAB\) on the ground) separately from the vertical right-angled triangles; the ground distance \(AB\) links the two.
425IBMA3H-3-18r23-143.17 Equations of a planePaper 2Hard10 marks
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A power cable in a wind farm runs along the line \(L\) with equation \(\mathbf{r}=\begin{pmatrix}2\\1\\0\end{pmatrix}+\lambda\begin{pmatrix}3\\-1\\2\end{pmatrix}\), where coordinates are in km relative to a control hut and \(A(2,1,0)\) is a point on \(L\). A turbine is at \(P(5,4,1)\). Give distances to 3 significant figures.
(a)
Write down the vector \(\overrightarrow{AP}\) and find \(\overrightarrow{AP}\times\begin{pmatrix}3\\-1\\2\end{pmatrix}\).
(3)
(b)
Find the coordinates of the point \(Q\) on \(L\) that is closest to \(P\).
M1 \(\overrightarrow{PQ}\cdot\begin{pmatrix}3\\-1\\2\end{pmatrix}=0\) with \(\overrightarrow{PQ}=\begin{pmatrix}2+3\lambda-5\\1-\lambda-4\\2\lambda-1\end{pmatrix}\)
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A1 \(14\lambda-8=0\)
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A1 \(\lambda=\frac47\)
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A1 \(Q\left(\frac{26}{7},\frac37,\frac87\right)\)
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Part (c)
M1 distance \(=\frac{\left|\overrightarrow{AP}\times\mathbf{d}\right|}{|\mathbf{d}|}\) (or \(|\overrightarrow{PQ}|\) from (b))
Exam tip: The distance from a point to a line equals \(\frac{|\overrightarrow{AP}\times\mathbf{d}|}{|\mathbf{d}|}\): it is \(|AP|\sin\theta\) with \(\theta\) the angle between \(AP\) and the line.
426IBMA3H-3-19r25-113.8 Solving trigonometric equationsPaper 2Hard11 marks
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The identity for \(\cos3\theta\) is used to solve a cubic equation.
(a)
Show that \(\cos3\theta=4\cos^3\theta-3\cos\theta\).
(4)
(b)
Let \(x=\cos\theta\). Hence show that the equation \(8x^3-6x-1=0\) can be written as \(\cos3\theta=\frac12\).
(2)
(c)
Solve \(\cos3\theta=\frac12\) for \(0\le\theta\le\pi\), giving exact values of \(\theta\).
(3)
(d)
Hence write down the three roots of \(8x^3-6x-1=0\) and state the largest root correct to 3 significant figures.
A1 roots \(\cos\frac{\pi}{9},\ \cos\frac{5\pi}{9},\ \cos\frac{7\pi}{9}\) (ECF from (c))
✓ 1
A1 largest root \(=\cos\frac{\pi}{9}=0.940\)
✓ 1
Exam tip: When you substitute x = cos θ, list every θ in the matching interval for 3θ (here up to 3π), since each value gives a different root.
427IBMA3H-3-19r25-123.14 Vector equation of a linePaper 2Hard13 marks
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A camera drone at a Formula 1 circuit flies along a straight path with position \(\mathbf{r}=\begin{pmatrix}2\\-1\\4\end{pmatrix}+t\begin{pmatrix}1\\2\\-2\end{pmatrix}\), where \(t\ge0\) is the time in seconds and distances are in metres. A camera operator stands at the fixed point \(P(5,\ 3,\ 5)\).
(a)
Show that \(P\) does not lie on the path of the drone.
(2)
(b)
Find the value of \(t\) at which the drone is closest to \(P\).
(4)
(c)
Find the coordinates of the drone at this time and its minimum distance from \(P\), to 3 significant figures.
(3)
(d)
The camera operator's radio link works only while the drone is within 5 m of \(P\). Find, to 3 significant figures, the length of time for which the link works.
(4)
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Part (a)
M1 equate the first components: \(2+t=5\) so \(t=3\)
✓ 1
R1 then \(y=-1+2(3)=5\ne3\), so \(P\) is not on the path
✓ 1
Part (b)
A1 vector from drone to \(P\): \(\begin{pmatrix}3-t\\4-2t\\1+2t\end{pmatrix}\)
✓ 1
M1 set the scalar product with the direction \(\begin{pmatrix}1\\2\\-2\end{pmatrix}\) equal to zero
A1 \(t=1\pm\frac{2\sqrt2}{3}\) (\(0.0572\) and \(1.94\))
✓ 1
A1 duration \(=\frac{4\sqrt2}{3}=1.89\) s
✓ 1
Exam tip: The shortest distance from a point to a line always occurs where the vector joining them is perpendicular to the direction of the line, so set a scalar product equal to zero.
428IBMA3H-3-19r25-133.2 Right-angled and non-right-angled trianglesPaper 2Hard12 marks
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A hospital garden designer marks out a triangular flower bed \(ABC\) with \(AB=8\) cm on the plan, angle \(BAC=30^\circ\) and \(BC=k\) cm.
Diagram not to scale. One possible position of C is shown
(a)
Find the value of \(k\) for which exactly one triangle exists and angle \(ACB\) is a right angle.
(2)
(b)
Find the range of values of \(k\) for which two different triangles can be drawn.
(2)
(c)
For \(k=6\), find the two possible sizes of angle \(ACB\), to the nearest 0.1°.
(3)
(d)
For \(k=6\), find the difference between the areas of the two possible triangles, in cm\(^2\), to 3 significant figures. Use unrounded angles from (c).
(5)
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Part (a)
M1 \(k=8\sin30^\circ\) (shortest distance from \(B\) to line \(AC\))
✓ 1
A1 \(k=4\)
✓ 1
Part (b)
A1 \(k>4\) (so that \(\sin ACB=\frac{4}{k}<1\))
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A1 \(k<8\) (so that \(BC<AB\) and angle \(ACB\) can be obtuse)
✓ 1
Part (c)
M1 sine rule: \(\frac{\sin ACB}{8}=\frac{\sin30^\circ}{6}\) so \(\sin ACB=\frac23\)
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A1 \(ACB=41.8^\circ\)
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A1 \(ACB=138.2^\circ\)
✓ 1
Part (d)
M1 angle \(ABC=180^\circ-30^\circ-ACB\): \(108.19^\circ\) or \(11.81^\circ\)
✓ 1
M1 area \(=\frac12\times8\times6\times\sin ABC\)
✓ 1
A1 areas \(22.8\) cm\(^2\) and \(4.91\) cm\(^2\)
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M1 subtract the two areas
✓ 1
A1 difference \(=17.9\) cm\(^2\) (exact value \(8\sqrt5\))
✓ 1
Exam tip: Test the ambiguous case by comparing the side opposite the given angle with both the other given side and the perpendicular height 8 sin 30°.
429IBMA3H-3-18r32-123.8 Solving trigonometric equationsPaper 2Hard11 marks
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The hub of a wind turbine is 80 m above the ground. Blade 1 has length 45 m and completes one rotation every 4 seconds. The height above the ground of its tip at time \(t\) seconds is \(h_1(t)=80+45\cos\!\left(\frac{\pi t}{2}\right)\). Blade 2 follows one third of a rotation behind blade 1, so its tip height is \(h_2(t)=80+45\cos\!\left(\frac{\pi t}{2}-\frac{2\pi}{3}\right)\). Angles are in radians and \(0\le t\le4\).
(a)
Find the first time at which the tip of blade 1 is 100 m above the ground. Give your answer to 3 significant figures.
(3)
(b)
Let \(A=\frac{\pi t}{2}\). Show that, when the two tips are at the same height, \(A=\frac{\pi}{3}+k\pi\) for some integer \(k\).
(3)
(c)
Find the first time at which the two tips are at the same height, and find that height.
(3)
(d)
Find the second time in the interval \(0\le t\le4\) at which the two tips are at the same height, and find that height.
M1 \(A-\frac{2\pi}{3}=-A+2k\pi\) (the other branch \(A-\frac{2\pi}{3}=A+2k\pi\) is impossible)
✓ 1
A1 \(2A=\frac{2\pi}{3}+2k\pi\) so \(A=\frac{\pi}{3}+k\pi\) (AG)
✓ 1
Part (c)
A1 \(\frac{\pi t}{2}=\frac{\pi}{3}\) so \(t=\frac23\) s
✓ 1
M1 \(80+45\cos\!\left(\frac{\pi}{3}\right)\)
✓ 1
A1 102.5 m
✓ 1
Part (d)
A1 \(A=\frac{4\pi}{3}\) so \(t=\frac83\) s
✓ 1
A1 \(80+45\cos\!\left(\frac{4\pi}{3}\right)=57.5\) m
✓ 1
Exam tip: For \(\cos P=\cos Q\) use \(P=\pm Q+2k\pi\), and check which branch can actually occur before solving.
430IBMA3H-3-18r32-143.18 Intersections and angles of lines and planesPaper 2Hard12 marks
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A gift box is a right pyramid. In a coordinate system with units of centimetres, the square base has vertices \(A(12,0,0)\), \(B(0,12,0)\), \(C(-12,0,0)\) and \(D(0,-12,0)\), and the vertex is \(V(0,0,16)\). The point \(M\) is the midpoint of \(AB\) and \(O\) is the origin.
The gift box (a right pyramid), drawn as a perspective view. Diagram not to scale
(a)
Write down the vectors \(\overrightarrow{VA}\) and \(\overrightarrow{VB}\).
(2)
(b)
Find angle \(AVB\), correct to 3 significant figures.
(4)
(c)
Find the angle between the line \(MV\) and the base of the box, correct to 3 significant figures.
A1 \(M=(6,\ 6,\ 0)\) so \(\overrightarrow{MV}=\begin{pmatrix}-6\\-6\\16\end{pmatrix}\) and \(\overrightarrow{MO}=\begin{pmatrix}-6\\-6\\0\end{pmatrix}\)
M1 \(\frac13\times\left(\frac{24\times24}{2}\right)\times16\) (base area 288 cm2 from the diagonals)
✓ 1
A1 1536 cm3
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Common error: The angle between a line and a plane is found with the line's projection onto the plane (here \(MO\)), not with a line in the plane chosen at random.
431IBMA3H-3-19r35-093.14 Vector equation of a linePaper 2Hard10 marks
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Two laser guides in a library stacks hall are modelled by the lines \(L_1:\ \mathbf{r}=\begin{pmatrix}2\\-1\\3\end{pmatrix}+s\begin{pmatrix}1\\2\\-1\end{pmatrix}\) and \(L_2:\ \mathbf{r}=\begin{pmatrix}0\\a\\7\end{pmatrix}+t\begin{pmatrix}2\\1\\-3\end{pmatrix}\), where \(a\) is a constant and lengths are in metres. The two lines intersect at a point I.
(a)
Find the value of \(a\).
(4)
(b)
Write down the coordinates of I.
(1)
(c)
Find the acute angle between \(L_1\) and \(L_2\), to 3 significant figures.
(3)
(d)
The point Q = (6, 7, −1) lies on \(L_1\). Find the exact distance IQ.
(2)
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Part (a)
M1 equate components of \(L_1\) and \(L_2\)
✓ 1
A1 from x and z: 2 + s = 2t and 3 − s = 7 − 3t, giving t = 2 and s = 2
✓ 1
M1 substitute into y: −1 + 2(2) = a + 2
✓ 1
A1 a = 1
✓ 1
Part (b)
A1 I = (4, 3, 1)
✓ 1
Part (c)
M1 scalar product of direction vectors attempted: 1×2 + 2×1 + (−1)×(−3) = 7
✓ 1
A1 \(\cos\theta=\dfrac{7}{\sqrt6\sqrt{14}}\)
✓ 1
A1 \(\theta\) = 40.2°
✓ 1
Part (d)
M1 \(\overrightarrow{IQ}=(2,4,-2)\) (or use s = 4 and s = 2)
✓ 1
A1 IQ = \(\sqrt{4+16+4}=2\sqrt6\) m
✓ 1
Common error: When two components give you s and t, you must check the third component; it is that check that fixes the unknown constant.
432IBMA3H-3-19r35-123.17 Equations of a planePaper 2Hard10 marks
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A vertical glass partition in a library is modelled by a plane. It contains the line \(L:\ \mathbf{r}=\begin{pmatrix}1\\2\\0\end{pmatrix}+t\begin{pmatrix}1\\-1\\2\end{pmatrix}\) and the point P(2, 1, 3), with coordinates in metres. A scanner is at Q(6, 5, 4).
(a)
Find a vector normal to the partition.
(3)
(b)
Find the Cartesian equation of the partition.
(1)
(c)
Find the coordinates of the foot F of the perpendicular from Q to the partition.
(3)
(d)
Hence find the exact distance from Q to the partition.
(2)
(e)
Find the position vector of the image Q′ of Q in the partition.
(1)
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Part (a)
M1 vector in the plane from a point on L to P, e.g. (1, −1, 3)
✓ 1
M1 vector product of (1, −1, 2) and (1, −1, 3) attempted
✓ 1
A1 normal = (−1, −1, 0) (any non-zero multiple, e.g. (1, 1, 0))
✓ 1
Part (b)
A1 x + y = 3
✓ 1
Part (c)
M1 line through Q with direction (1, 1, 0): (6 + μ, 5 + μ, 4)
Exam tip: For distance from a point to a plane, the perpendicular line through the point always has the plane's normal vector as its direction.
433IBMA3H-3-20r40-063.7 Circular functions and their transformationsPaper 2Hard12 marks
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The temperature inside a beehive, \(T\) °C, is modelled by \(T(t)=a\cos\left(b(t-15)\right)+d\), where \(t\) is the time in hours after midnight, \(0\le t\le24\), and \(a\), \(b\) and \(d\) are positive constants with \(b\) in radians per hour. The maximum temperature of 36 °C occurs at 15:00 and the minimum temperature of 30 °C occurs at 03:00.
(a)
Find the values of \(a\), \(b\) and \(d\).
(3)
(b)
The beekeeper reports that the temperature is at least 34.5 °C for exactly 8 hours each day. Verify this claim using the model.
(4)
(c)
A second hive has temperature \(T_2(t)=33+3\cos\left(\frac{\pi}{12}(t-12)\right)\) for \(0\le t\le24\). Find the times at which the two hives are at the same temperature.
(3)
(d)
Find the common temperature at \(t=13.5\). Give your answer to 3 significant figures.
(2)
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Part (a)
A1 \(a=3\) (half the difference between maximum and minimum)
✓ 1
A1 \(d=33\)
✓ 1
A1 \(b=\frac{\pi}{12}\) (period 24 hours)
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Part (b)
M1 sets \(3\cos\left(\frac{\pi}{12}(t-15)\right)+33\ge34.5\), so \(\cos\left(\frac{\pi}{12}(t-15)\right)\ge\frac{1}{2}\)
Exam tip: To show a stated value is correct you must show the working that leads to it; reaching the interval 11 to 19 and then stating its length is the whole argument.
434IBMA3H-3-20r40-133.17 Equations of a planePaper 2Hard14 marks
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Two cables carrying cameras above a Formula 1 pit lane follow the lines \(L_1\) and \(L_2\), where \(L_1:\ \boldsymbol{r}=\begin{pmatrix}1\\0\\2\end{pmatrix}+s\begin{pmatrix}2\\1\\-1\end{pmatrix}\) and \(L_2:\ \boldsymbol{r}=\begin{pmatrix}2\\2\\-1\end{pmatrix}+t\begin{pmatrix}1\\-1\\2\end{pmatrix}\). Distances are in metres, the origin is a point on a gantry and the third coordinate is measured vertically upwards from the gantry level.
(a)
Show that \(L_1\) and \(L_2\) intersect, and find the coordinates of the point of intersection.
(4)
(b)
Find the acute angle between the two cables. Give your answer in degrees to 3 significant figures.
(3)
(c)
Find a Cartesian equation of the plane containing both cables.
(4)
(d)
Find the acute angle between this plane and the horizontal. Give your answer in degrees to 3 significant figures.
(3)
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Part (a)
M1 equates two components, e.g. \(1+2s=2+t\) and \(s=2-t\)
✓ 1
A1 solves to obtain \(s=1\) and \(t=1\)
✓ 1
M1 checks the third component: \(2-s=-1+2t\) gives \(1=1\)
✓ 1
A1 point of intersection (3, 1, 1)
✓ 1
Part (b)
M1 scalar product of the direction vectors \(=2-1-2=-1\)
A1 normal vector \(\begin{pmatrix}1\\-5\\-3\end{pmatrix}\) (or a multiple)
✓ 1
M1 substitutes a point on either line, e.g. (3, 1, 1)
✓ 1
A1 \(x-5y-3z=-5\)
✓ 1
Part (d)
M1 uses the normals \((1,-5,-3)\) and \((0,0,1)\)
✓ 1
M1 \(\cos\phi=\frac{3}{\sqrt{35}}\)
✓ 1
A1 \(\phi=59.5^\circ\)
✓ 1
Exam tip: The angle between two planes is the angle between their normal vectors; for the horizontal plane the normal is vertical, which makes the scalar product easy.
435IBMA3H-3-20r41-123.14 Vector equation of a linePaper 2Hard8 marks
Calculator
Two drones fly over a wind farm. Relative to a fixed origin, in metres, the position of drone A at time t seconds is rA = (10, −20, 30) + t(4, 3, −1) and the position of drone B is rB = (−30, 40, 20) + t(7, −1, 1), for t ≥ 0. The distance between the drones at time t is D metres.
(a)
Show that D2 = 29t2 − 760t + 5300.
(3)
(b)
Find the time at which the drones are closest together, giving your answer to 3 significant figures.
(2)
(c)
Find the minimum distance between the drones. Use the unrounded value of t from (b) and give your answer to 3 significant figures.
(2)
(d)
The safety rule requires the drones to stay at least 20 m apart. State, with a reason, whether the rule is broken.
R1 the rule is broken because the minimum distance is less than 20 m (ECF from (c))
✓ 1
Exam tip: Minimising the distance is the same as minimising its square, which avoids differentiating a square root.
436IBMA3H-3-18r42-053.18 Intersections and angles of lines and planesPaper 2Hard10 marks
Calculator
Two support cables for a wind-turbine test rig are modelled by the lines \(l_1:\ \mathbf r=\begin{pmatrix}1\\0\\2\end{pmatrix}+s\begin{pmatrix}1\\2\\-1\end{pmatrix}\) and \(l_2:\ \mathbf r=\begin{pmatrix}0\\3\\1\end{pmatrix}+t\begin{pmatrix}2\\-1\\1\end{pmatrix}\). Coordinates are in metres.
(a)
Show that \(l_1\) and \(l_2\) do not intersect.
(4)
(b)
Find the acute angle between the cables, giving your answer to 3 significant figures.
(3)
(c)
Find the shortest distance between the two cables, giving your answer to 3 significant figures.
M1 \(\mathbf n=\begin{pmatrix}1\\2\\-1\end{pmatrix}\times\begin{pmatrix}2\\-1\\1\end{pmatrix}=\begin{pmatrix}1\\-3\\-5\end{pmatrix}\), with \(|\mathbf n|=\sqrt{35}\)
✓ 1
M1 vector between points on the lines, e.g. \(\begin{pmatrix}-1\\3\\-1\end{pmatrix}\), projected onto \(\mathbf n\): \(\dfrac{|-1-9+5|}{\sqrt{35}}\)
✓ 1
A1 \(0.845\) m (0.8452...)
✓ 1
Exam tip: For skew lines the shortest distance is the length of the projection of a connecting vector onto the common perpendicular, which is parallel to the cross product of the directions.
437IBMA3H-3-18r42-073.14 Vector equation of a linePaper 2Hard10 marks
Calculator
Two inspection drones fly around a wind turbine at the same height. Relative to the base of the turbine, with distances in metres and time \(t\) in seconds, the position vectors of drone A and drone B are \(\mathbf r_A=\begin{pmatrix}3t\\4t\end{pmatrix}\) and \(\mathbf r_B=\begin{pmatrix}60-2t\\-20+6t\end{pmatrix}\) for \(t\ge0\).
Not to scale. Paths shown for 0 ≤ t ≤ 20; distances in metres
(a)
Find an expression for \(\overrightarrow{AB}=\mathbf r_B-\mathbf r_A\) in terms of \(t\).
(2)
(b)
Show that the square of the distance between the drones is \(29t^{2}-680t+4000\).
(3)
(c)
Find the minimum distance between the drones. Give your answer to 3 significant figures.
(3)
(d)
The drones must always stay at least 5 m apart. State whether this rule is broken, giving a reason.
(2)
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Marking point
Mark
Notes
Part (a)
M1 subtracts the components of \(\mathbf r_A\) from those of \(\mathbf r_B\)
M1 minimum of the quadratic occurs at \(t=\dfrac{680}{58}=11.72\ldots\) (by vertex, differentiation or graphing)
✓ 1
M1 minimum value of \(|AB|^{2}\) is \(4000-\dfrac{680^{2}}{4\times29}=13.79\ldots\)
✓ 1
A1 minimum distance \(3.71\) m (3.7139...)
✓ 1
Part (d)
M1 compares the minimum distance of 3.71 m with 5 m
✓ 1
R1 3.71 m is less than 5 m, so the rule is broken (near \(t=11.7\) s)
✓ 1
Common error: Minimise the squared distance, not the distance itself, to avoid a square root until the final step; both have their minimum at the same time.
438IBMA3H-3-18r42-123.3 Applications of trigonometry: elevation, depression and bearingsPaper 2Hard7 marks
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A vertical wind-turbine tower \(TP\) of height 30 m stands on level ground with its base at \(P\). Points \(A\) and \(B\) are on the ground. The angle of elevation of \(T\) from \(A\) is \(35^{\circ}\) and from \(B\) is \(42^{\circ}\), and \(\angle APB=110^{\circ}\). Use unrounded values in later parts and give lengths and areas to 3 significant figures.
(a)
Find the distances \(PA\) and \(PB\).
(3)
(b)
Find the distance \(AB\).
(2)
(c)
Find the area of triangle \(APB\).
(2)
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Notes
Part (a)
M1 \(PA=\dfrac{30}{\tan35^{\circ}}\)
✓ 1
M1 \(PB=\dfrac{30}{\tan42^{\circ}}\)
✓ 1
A1 \(PA=42.8\) m and \(PB=33.3\) m (42.84... and 33.32...)
Exam tip: Keep unrounded values from earlier parts in your calculator so the final answer is accurate to the required number of significant figures.
439IBMA3H-3-19r42-063.4 Radians, arc length and sector areaPaper 2Hard12 marks
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Two circular zones on a bridge deck are lit by floodlights. They are modelled by circles with centres O1 and O2 and radii 5.0 m and 3.0 m, where O1O2 = 6.0 m. The circles meet at points A and B. The common chord AB divides the overlap of the two zones into two circular segments, one in each circle. Let \(\alpha\) = angle AO1O2 and \(\beta\) = angle AO2O1, both in radians. Give answers to 3 significant figures and use unrounded values in later parts.
Diagram not to scale
(a)
Find the value of \(\alpha\).
(3)
(b)
Find the value of \(\beta\).
(2)
(c)
Find the area of the segment of the 5.0 m circle that lies inside the overlap.
(2)
(d)
Find the area of the segment of the 3.0 m circle that lies inside the overlap.
(2)
(e)
Hence find the percentage of the smaller zone that is also lit by the larger floodlight.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 cosine rule in triangle AO1O2: \(3^2=5^2+6^2-2(5)(6)\cos\alpha\)
✓ 1
A1 \(\cos\alpha=\frac{13}{15}\)
✓ 1
A1 \(\alpha=0.522\)
✓ 1
Part (b)
M1 \(5^2=3^2+6^2-2(3)(6)\cos\beta\), so \(\cos\beta=\frac59\)
✓ 1
A1 \(\beta=0.982\)
✓ 1
Part (c)
M1 sector minus triangle, \(25(\alpha-\sin\alpha\cos\alpha)\) (equivalently \(\frac12(25)(2\alpha)-\frac12(25)\sin2\alpha\))
✓ 1
A1 2.25 m2
✓ 1
Part (d)
M1 \(9(\beta-\sin\beta\cos\beta)\)
✓ 1
A1 4.68 m2
✓ 1
Part (e)
A1 overlap area \(=2.2486+4.6785=6.93\) m2 (ECF from (c) and (d))
✓ 1
M1 divides by the area of the smaller zone, \(9\pi\)
✓ 1
A1 24.5%
✓ 1
Exam tip: Find each segment as sector minus triangle using the angle at its own centre; the overlap is the sum of two segments, not a single sector.
440IBMA3H-3-19r42-113.17 Equations of a planePaper 2Hard14 marks
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A truss on a space probe is a tetrahedron ABCD with A(1, 1, 0), B(3, 2, 1), C(2, 4, 2) and D(3, −1, 2), where coordinates are in metres.
Hence find the Cartesian equation of the plane \(\Pi\) containing A, B and C.
(2)
(c)
Find the exact area of triangle ABC.
(2)
(d)
The line through D perpendicular to \(\Pi\) meets \(\Pi\) at E. Find the coordinates of E.
(4)
(e)
Hence find the volume of the tetrahedron ABCD.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\overrightarrow{AB}=\begin{pmatrix}2\\1\\1\end{pmatrix}\), \(\overrightarrow{AC}=\begin{pmatrix}1\\3\\2\end{pmatrix}\) and an attempt at the vector product
✓ 1
A1 \(\begin{pmatrix}-1\\-3\\5\end{pmatrix}\)
✓ 1
Part (b)
M1 uses \(\mathbf{n}\cdot\mathbf{r}=\mathbf{n}\cdot\mathbf{a}\) with \(\mathbf{n}\) from (a)
✓ 1
A1 \(x+3y-5z=4\) (or equivalent, e.g. \(-x-3y+5z=-4\))
Common error: The base of the tetrahedron is the triangle, so the volume formula is one third of base area times perpendicular height; the extra factor \(\frac12\) is already inside the triangle area.
441IBMA3H-3-19r43-133.14 Vector equation of a linePaper 2Hard14 marks
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Two cyclists, A and B, ride along straight, flat roads. Relative to a junction \(O\), distances are in km and \(t\) is the time in hours after 10:00. Their position vectors are \(\mathbf{r}_A=t\begin{pmatrix}16\\2\end{pmatrix}\) and \(\mathbf{r}_B=\begin{pmatrix}10\\5\end{pmatrix}+t\begin{pmatrix}12\\-1\end{pmatrix}\).
Not to scale. Paths shown for 0 ≤ t ≤ 1.5; distances in km
(a)
Find the speed of cyclist A, giving your answer in km h-1 to three significant figures.
(2)
(b)
Show that \(\overrightarrow{AB}=\begin{pmatrix}10-4t\\5-3t\end{pmatrix}\).
(2)
(c)
Show that the square of the distance between the cyclists is \(25t^2-110t+125\).
(2)
(d)
Find the time at which the cyclists are closest together, and the distance between them at that time.
(4)
(e)
The cyclists can see each other whenever they are at most 5 km apart. Find the length of time, in hours to three significant figures, for which they can see each other. Use unrounded values in your working.
A1 square of distance is \(4\), so the minimum distance is \(2\) km
✓ 1
Part (e)
M1 set \(25t^2-110t+125\le25\), giving \(5t^2-22t+20\le0\)
✓ 1
M1 solve the quadratic equation \(5t^2-22t+20=0\)
✓ 1
A1 \(t=1.28\ldots\) and \(t=3.12\ldots\) (that is, \(\frac{22\mp\sqrt{84}}{10}\))
✓ 1
A1 duration \(=\frac{\sqrt{84}}{5}=1.83\) hours
✓ 1
Exam tip: For closest-approach problems, work with the square of the distance: it is a quadratic in \(t\), so you can use the vertex or the discriminant without differentiating a square root.
442AAHL3-q00683.13 Scalar productPaper 3Hard28 marks
This question investigates how the direction of a ray of light changes when it is reflected by plane mirrors.
A ray travelling in the direction of a non-zero vector \(\boldsymbol{d}\) strikes a plane mirror with normal vector \(\boldsymbol{n}\). The vector \(\boldsymbol{d}\) can be written uniquely as \(\boldsymbol{d}=\boldsymbol{p}+\boldsymbol{q}\), where \(\boldsymbol{p}\) is parallel to \(\boldsymbol{n}\) and \(\boldsymbol{q}\) is perpendicular to \(\boldsymbol{n}\). The law of reflection states that the reflected ray travels in the direction \(\boldsymbol{d}'=\boldsymbol{q}-\boldsymbol{p}\).
The diagram is used in part (f).
Diagram not to scale
(a)(i)
A mirror lies in the plane \(z=0\) and an incoming ray has direction \(\boldsymbol{d}=2\boldsymbol{i}-\boldsymbol{j}-3\boldsymbol{k}\).
Write down \(\boldsymbol{p}\) and \(\boldsymbol{q}\).
(2)
(a)(ii)
Write down \(\boldsymbol{d}'\).
(1)
(b)(i)
Let \(\boldsymbol{p}=t\boldsymbol{n}\), where \(t\in\mathbb{R}\).
Show that \(t=\dfrac{\boldsymbol{d}\cdot\boldsymbol{n}}{\boldsymbol{n}\cdot\boldsymbol{n}}\).
(3)
(b)(ii)
Hence show that \(\boldsymbol{d}'=\boldsymbol{d}-2\left(\dfrac{\boldsymbol{d}\cdot\boldsymbol{n}}{\boldsymbol{n}\cdot\boldsymbol{n}}\right)\boldsymbol{n}\).
(1)
(c)(i)
Use the result in part (b)(ii) to answer the following.
Show that \(\boldsymbol{d}'\cdot\boldsymbol{n}=-\boldsymbol{d}\cdot\boldsymbol{n}\).
(2)
(c)(ii)
Show that \(|\boldsymbol{d}'|=|\boldsymbol{d}|\).
(2)
(c)(iii)
Hence explain why the incoming ray and the reflected ray make equal angles with the mirror.
(1)
(d)(i)
A mirror lies in the plane \(x-y+z=0\) and an incoming ray has direction \(\boldsymbol{d}=\boldsymbol{i}+2\boldsymbol{j}\).
Find \(\boldsymbol{d}'\).
(2)
(d)(ii)
Find the acute angle between the incoming ray and the mirror.
(2)
(e)
A corner reflector consists of three mirrors in the planes \(x=0\), \(y=0\) and \(z=0\). A ray with direction \(\boldsymbol{d}=d_{1}\boldsymbol{i}+d_{2}\boldsymbol{j}+d_{3}\boldsymbol{k}\) is reflected once by each mirror. Show that the final direction of the ray is \(-\boldsymbol{d}\), whatever the order in which the reflections occur.
(3)
(f)(i)
Two vertical mirrors meet along the \(z\)-axis. Mirror \(M_{1}\) lies in the plane \(y=0\) and mirror \(M_{2}\) has unit normal \(\boldsymbol{n}_{2}=-\sin\varphi\,\boldsymbol{i}+\cos\varphi\,\boldsymbol{j}\), where \(0\lt\varphi\lt\dfrac{\pi}{2}\), so that the angle between the mirrors is \(\varphi\). The diagram above shows the view from above of a horizontal ray that is reflected first by \(M_{1}\) and then by \(M_{2}\).
Show that a horizontal ray with direction \(\cos\beta\,\boldsymbol{i}+\sin\beta\,\boldsymbol{j}\) is reflected by \(M_{2}\) into the direction \(\cos(2\varphi-\beta)\,\boldsymbol{i}+\sin(2\varphi-\beta)\,\boldsymbol{j}\).
(5)
(f)(ii)
Write down the direction of a horizontal ray with direction \(\cos\alpha\,\boldsymbol{i}+\sin\alpha\,\boldsymbol{j}\) after it is reflected by \(M_{1}\), in the form \(\cos\theta\,\boldsymbol{i}+\sin\theta\,\boldsymbol{j}\).
(1)
(f)(iii)
Hence show that a horizontal ray reflected first by \(M_{1}\) and then by \(M_{2}\) is turned through an angle \(2\varphi\), whatever its initial direction.
(2)
(f)(iv)
Find the value of \(\varphi\) for which such a ray is turned through \(\dfrac{3\pi}{4}\).
Or \(|\boldsymbol{d}'|^{2}=|\boldsymbol{q}|^{2}+|\boldsymbol{p}|^{2}=|\boldsymbol{d}|^{2}\) since \(\boldsymbol{p}\perp\boldsymbol{q}\).
the last two terms cancel, so \(|\boldsymbol{d}'|^{2}=|\boldsymbol{d}|^{2}\)
A1
\(|\boldsymbol{d}'|=|\boldsymbol{d}|\)
AG
(c)(iii)
the angle \(\theta\) between a ray and the mirror satisfies \(\sin\theta=\dfrac{|\boldsymbol{d}\cdot\boldsymbol{n}|}{|\boldsymbol{d}||\boldsymbol{n}|}\); by (c)(i) and (c)(ii) this has the same value for \(\boldsymbol{d}'\)
reflection in \(x=0\) (\(\boldsymbol{n}=\boldsymbol{i}\)): \(\boldsymbol{d}'=\boldsymbol{d}-2d_{1}\boldsymbol{i}=-d_{1}\boldsymbol{i}+d_{2}\boldsymbol{j}+d_{3}\boldsymbol{k}\)
M1
similarly the mirrors \(y=0\) and \(z=0\) reverse only the \(\boldsymbol{j}\)- and \(\boldsymbol{k}\)-components, so after all three the direction is \(-d_{1}\boldsymbol{i}-d_{2}\boldsymbol{j}-d_{3}\boldsymbol{k}=-\boldsymbol{d}\)
A1
each reflection changes the sign of one component only, independently of the other components, so the order does not affect the result
R1
(f)(i)
\(\boldsymbol{d}\cdot\boldsymbol{n}_{2}=-\cos\beta\sin\varphi+\sin\beta\cos\varphi=\sin(\beta-\varphi)\), with \(\boldsymbol{n}_{2}\cdot\boldsymbol{n}_{2}=1\)
final direction \(\cos(2\varphi+\alpha)\,\boldsymbol{i}+\sin(2\varphi+\alpha)\,\boldsymbol{j}\): the direction angle has increased from \(\alpha\) to \(\alpha+2\varphi\)
A1
turned through \(2\varphi\), independent of \(\alpha\)
443AAHL3-q00823.16 Vector productPaper 3Hard28 marks
This question investigates a three-dimensional version of Pythagoras’ theorem.
The points A\((a,0,0)\), B\((0,b,0)\) and C\((0,0,c)\), where \(a,b,c\gt 0\), and the origin O are the vertices of a tetrahedron OABC, as shown in the diagram. The three faces that meet at O are mutually perpendicular. Let \(X\), \(Y\) and \(Z\) denote the areas of the faces OBC, OCA and OAB respectively, and let \(W\) denote the area of the face ABC.
Diagram not to scale
(a)(i)
Consider the case \(a=1,\ b=2,\ c=2\).
Find the values of \(X\), \(Y\) and \(Z\).
(2)
(a)(ii)
Find \(\overrightarrow{AB}\times\overrightarrow{AC}\) and hence find the exact value of \(W\).
(4)
(a)(iii)
Verify that \(W^{2}=X^{2}+Y^{2}+Z^{2}\).
(1)
(b)
Consider the case \(a=2,\ b=3,\ c=6\). Show that \(W^{2}=X^{2}+Y^{2}+Z^{2}\) also holds.
(3)
(c)(i)
Now consider the general case.
Show that \(\overrightarrow{AB}\times\overrightarrow{AC}=bc\,\boldsymbol{i}+ca\,\boldsymbol{j}+ab\,\boldsymbol{k}\).
(3)
(c)(ii)
Hence prove that \(W^{2}=X^{2}+Y^{2}+Z^{2}\).
(3)
(d)(i)
Let \(h\) be the shortest distance from O to the plane ABC.
Show that the plane ABC has Cartesian equation \(bcx+cay+abz=abc\).
(2)
(d)(ii)
By considering the line through O perpendicular to the plane ABC, show that \(h=\dfrac{abc}{\sqrt{b^{2}c^{2}+c^{2}a^{2}+a^{2}b^{2}}}\).
(3)
(d)(iii)
Hence show that \(\dfrac{1}{h^{2}}=\dfrac{1}{a^{2}}+\dfrac{1}{b^{2}}+\dfrac{1}{c^{2}}\).
(2)
(e)
Let \(\gamma\) be the acute angle between the plane ABC and the plane OAB. Show that \(\cos\gamma=\dfrac{Z}{W}\).
(3)
(f)
Let \(\alpha\) and \(\beta\) be the acute angles between the plane ABC and the planes OBC and OCA respectively. Prove that \(\cos^{2}\alpha+\cos^{2}\beta+\cos^{2}\gamma=1\).
\(\overrightarrow{AB}\times\overrightarrow{AC}=\begin{pmatrix}-2\\3\\0\end{pmatrix}\times\begin{pmatrix}-2\\0\\6\end{pmatrix}=\begin{pmatrix}18\\12\\6\end{pmatrix}\), so \(W=\tfrac12\sqrt{504}\)
M1
\(W^{2}=\tfrac{504}{4}=126\) and \(X^{2}+Y^{2}+Z^{2}=81+36+9=126\)
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