Original questions in the style of Papers 1, 2 and 3, each with a full worked mark scheme. Tick questions to build a printable test. Topic 3 at Standard Level →
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Original questions in the style of each exam boardFull worked mark schemes with method and accuracy marksCreated by Evan Kotronis with AI assistanceUpdated 14 October 2026 · new questions added continuouslyFree access until 30 November 2026How every question is checked →
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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 \(B\hat{G}C=C\hat{G}D=D\hat{G}B=\dfrac{2\pi}{3}\).
(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)
Show mark scheme
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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Working
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
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 \(O\).
(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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Working
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)
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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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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
24AAHL3-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 \(AB\gt AC\).
(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)
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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) \(\rho=\sqrt3\)
25AAHL3-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.
(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) \(\dfrac{20}{\pi+2}\le r\lt10\)
26AAHL3-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)
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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 \(\dfrac{1}{2}\lt\dfrac{\pi}{6}\lt\dfrac{1}{\sqrt3}\)
27AAHL3-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 \(k=0,1,\ldots,11\).
(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)
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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 \(\dfrac{1}{3}\)
29AAHL3-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)
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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.
30AAHL3-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.
34AAHL3-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)
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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) \(x=\frac{\pi}{3},\ \frac{5\pi}{3}\)
35AAHL3-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.
37AAHL3-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 \(0\lt\beta\lt\alpha\lt\dfrac{\pi}{2}\).
(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\((-1,-2,-3)\)
42AAHL3-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 \(\overrightarrow{OG}=\dfrac23\overrightarrow{OM}\).
(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.
44AAHL3-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 \(\boldsymbol{a}\).
45AAHL3-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.
46AAHL3-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.
47AAHL3-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) \(p=1\)
48AAHL3-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 \(L_{2}\).
49AAHL3-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 \(k=-1\))
51AAHL3-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.
52AAHL3-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: \(\begin{pmatrix}2\\3\\2\end{pmatrix}\)
53AAHL3-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.
55AAHL3-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) \(\dfrac{3}{8}\)
56AAHL3-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 \(=\dfrac{1}{2}a^{2}\sin\dfrac{\pi}{3}=\dfrac{\sqrt3a^{2}}{4}\)
60AAHL3-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) \(\tan\frac{\pi}{12}=2-\sqrt3\)
62AAHL3-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
63AAHL3-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 \(2\pi-a\)
64AAHL3-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.
65AAHL3-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
66AAHL3-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) \(12\sqrt3\)
67AAHL3-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)
68AAHL3-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 \(\Pi\).
71AAHL3-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 \(\dfrac{AF}{\sin50^{\circ}}=\dfrac{250}{\sin100^{\circ}}\)
72AAHL3-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.
(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 \(AB=t\sqrt{18^{2}+24^{2}-2(18)(24)\cos75^{\circ}}=26.0t\)
73AAHL3-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
75AAHL3-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 \(-3\lt t\lt3\).
(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
76AAHL3-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.
(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 \(\theta=2.48\)
77AAHL3-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)
80AAHL3-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
81AAHL3-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
83AAHL3-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) \(60.2^{\circ}\)
84AAHL3-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)
Show mark scheme
Working
Mark
Notes
(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
85AAHL3-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 \(35^{\circ}\).
(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)
Show mark scheme
Working
Mark
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°
86AAHL3-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)
Show mark scheme
Working
Mark
Notes
(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 \(\alpha=0.3398\)
87AAHL3-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
88AAHL3-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)
Show mark scheme
Working
Mark
Notes
(a)
\(\theta=T\hat{E}H-B\hat{E}H\), where \(\tan T\hat{E}H=\dfrac{5}{x}\) and \(\tan B\hat{E}H=\dfrac{2}{x}\)
89AAHL3-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)
Show mark scheme
Working
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
90AAHL3-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.
91AAHL3-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 \(\alpha\)
92AAHL3-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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