IB Maths AA HL 3.2 Geometry and Trigonometry Ahl Content Questions

Practise IB Mathematics AA HL 3.2 by solving advanced vectors, lines, planes, 3D geometry, identities and trigonometric proof problems.

Exam points

Transform and solve reciprocal, inverse, compound and multiple-angle trigonometric equations, selecting admissible roots on a stated interval.
Use vectors to represent positions, displacements, magnitudes, unit directions and geometric relations, including motion and spatial models.
Apply scalar products to calculate angles, test perpendicularity or parallelism, and determine unknown parameters or lengths.
Construct and use vector or parametric equations of lines, including direction-vector angles, intersections and kinematic time relationships.
Classify and solve 3D line relationships, distinguishing coincident, parallel, intersecting and skew configurations.
Use vector products to construct perpendicular directions, test relationships and calculate areas in spatial geometry.
Construct plane equations and use normal vectors to test membership, relationships and unknown parameters.
Solve line-plane and plane-plane systems, calculate 3D angles, and interpret solution sets geometrically.

Covered syllabus points

AHL 3.9 (HL)—Reciprocal and inverse trigonometric functions AHL 3.10 (HL)—Compound-angle identities AHL 3.11 (HL)—Further trigonometric equations AHL 3.12 (HL)—Vectors AHL 3.13 (HL)—Scalar product AHL 3.14 (HL)—Vector equations of lines AHL 3.15 (HL)—Line relationships in 3D AHL 3.16 (HL)—Vector product AHL 3.17 (HL)—Planes AHL 3.18 (HL)—Intersections and angles in 3D

[Maximum number: 1]

Two boats A and B travel due north.
Initially, boat B is positioned 50 metres due east of boat A .
The distances travelled by boat A and boat B , after t seconds, are x metres and y metres respectively. The angle $\theta$ is the radian measure of the bearing of boat B from boat A . This information is shown on the following diagram.

Show that $y=x+50 \cot \theta$ .

At time T, the following conditions are true.
Boat B has travelled 10 metres further than boat A .
Boat B is travelling at double the speed of boat A .
The rate of change of the angle $\theta$ is -0.1 radians per second.

$\tan \theta=\frac{50}{y-x}$ OR $\cot \theta=\frac{y-x}{50}$ $y=x+50 \cot \theta$
AG
Note: y-x may be identified as a length on a diagram, and not written explicitly.

[Maximum number: 3]

Show that $\sin 105^{\circ}+\cos 105^{\circ}=\frac{1}{\sqrt{2}}$ .

(b) EITHER
choosing two appropriate angles, for example $60^{\circ}$ and $45^{\circ}$ $\sin 105^{\circ}=\sin 60^{\circ} \cos 45^{\circ}+\cos 60^{\circ} \sin 45^{\circ}$ and
$\cos 105^{\circ}=\cos 60^{\circ} \cos 45^{\circ}-\sin 60^{\circ} \sin 45^{\circ}$ $\sin 105^{\circ}+\cos 105^{\circ}=\frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{2}}+\frac{1}{2} \times \frac{1}{\sqrt{2}}+\frac{1}{2} \times \frac{1}{\sqrt{2}}-\frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{2}}$ $=\frac{1}{\sqrt{2}}$
OR
attempt to square the expression
$\left(\sin 105^{\circ}+\cos 105^{\circ}\right)^{2}=\sin ^{2} 105^{\circ}+2 \sin 105^{\circ} \cos 105^{\circ}+\cos ^{2} 105^{\circ}$ $\left(\sin 105^{\circ}+\cos 105^{\circ}\right)^{2}=1+\sin 210^{\circ}$ $=\frac{1}{2}$ $\sin 105^{\circ}+\cos 105^{\circ}=\frac{1}{\sqrt{2}}$

[Maximum number: 6]

Hence or otherwise solve the equation $\sin x+\sin 3 x=\cos x$ in the interval $0<x<\pi$ .

EITHER
$\sin x+\sin 3 x=\cos x \Rightarrow \frac{1-\cos 4 x}{2 \sin x}=\cos x$ $\Rightarrow 1-\cos 4 x=2 \sin x \cos x,(\sin x \neq 0)$ $\Rightarrow 1-\left(1-2 \sin ^{2} 2 x\right)=\sin 2 x$ $\Rightarrow \sin 2 x(2 \sin 2 x-1)=0$ $\Rightarrow \sin 2 x=0$ or $\sin 2 x=\frac{1}{2}$ $2 x=\pi, 2 x=\frac{\pi}{6}$ and $2 x=\frac{5 \pi}{6}$

OR
$\sin x+\sin 3 x=\cos x \Rightarrow 2 \sin 2 x \cos x=\cos x$ $\Rightarrow(2 \sin 2 x-1) \cos x=0,(\sin x \neq 0)$ $\Rightarrow \sin 2 x=\frac{1}{2}$ or $\cos x=0$ $2 x=\frac{\pi}{6}, 2 x=\frac{5 \pi}{6}$ and $x=\frac{\pi}{2}$

THEN
$\therefore x=\frac{\pi}{2}, x=\frac{\pi}{12}$ and $x=\frac{5 \pi}{12}$

[Maximum number: 3]

Consider the points A(1,2,3), B(k,-2,1) and C(5,0,2), where $k \in \mathbb{R}$ .

(a)

Write down $\overrightarrow{\mathrm{AB}}$ and $\overrightarrow{\mathrm{AC}}$ .

[ 2 ]

$\quad \overrightarrow{\mathrm{AB}}=\left(\begin{array}{c}k-1 \\ -4 \\ -2\end{array}\right), \overrightarrow{\mathrm{AC}}=\left(\begin{array}{c}4 \\ -2 \\ -1\end{array}\right)$

(b)

Given that the points A, B and C lie on a straight line, show that k=9.

[ 1 ]

METHOD 1
$k-1=2 \times 4$ M1
k=9
METHOD 2
in order by y or z-ordinate, the points are (k,-2,1),(5,0,2)(1,2,3)
k-5=5-1
k=9
AG

(c)

For $k \neq 9$ , let $\Pi$ be the plane containing A, B and C .

[ 9 ]

(i)

Find the Cartesian equation of the plane $\Pi$ .

[ 4 ]

(d) (i) METHOD 1
attempt to find cross product of two of $\overrightarrow{\mathrm{AB}}, \overrightarrow{\mathrm{AC}}$ and $\overrightarrow{\mathrm{BC}}$ or their opposites
eg $\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}=\left(\begin{array}{c}0 \\ k-9 \\ 18-2 k\end{array}\right)\left(=(k-9)\left(\begin{array}{c}0 \\ 1 \\ -2\end{array}\right)\right)$
attempt to substitute their cross product and a point into the equation of a plane
(k-9) y+2(9-k) z=2(k-9)+6(9-k)
$(k-9) y+2(9-k) z=36-4 k(\Rightarrow y-2 z=-4$ since $k \neq 9)$
METHOD 2
attempt to find vector equation of $\Pi$ and write x, y and z in parametric form
$\left(\boldsymbol{r}=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)+\lambda\left(\begin{array}{c}k-1 \\ -4 \\ -2\end{array}\right)+\mu\left(\begin{array}{c}4 \\ -2 \\ -1\end{array}\right) \Rightarrow x=1+\lambda(k-1)+4 \mu, y=2-4 \lambda-2 \mu\right.$ ,
$z=3-2 \lambda-\mu$ or equivalent
attempt to eliminate both parameters to work towards Cartesian form
$(k-9) y+2(9-k) z=36-4 k(\Rightarrow y-2 z=-4$ since $k \neq 9)$

(ii)

Find the coordinates of the point on the plane $\Pi$ which is closest to the origin (0,0,0).

[ 5 ]

METHOD 1
attempt to find the equation of the line through (0,0,0) perpendicular to the plane

EITHER

(r=) t\left(\begin{array}{c} 0 \\ 1 \\ -2 \end{array}\right)

attempt to find the point where the line and plane intersect

t+4 t+4=0

t=-\frac{4}{5}

(r=) t(k-9)\left(\begin{array}{c} 0 \\ 1 \\ -2 \end{array}\right)

attempt to find the point where the line and plane intersect

\begin{aligned} & t(k-9)^{2}+4 t(k-9)^{2}+4(k-9)=0 \\ & t=-\frac{4}{5(k-9)} \end{aligned}

THEN
so the point on the plane closest to the origin is (0,-0.8,1.6)

METHOD 2
choose a point on the plane (p, q, r)

q-2 r+4=0 \text { OR } q(k-9)-2 r(k-9)+4(k-9)=0 \Rightarrow q=2 r-4

distance to the origin is $\sqrt{p^{2}+(2 r-4)^{2}+r^{2}}$
since p is independent of r, distance is minimised when p=0
attempt to find the value of r for which their $\sqrt{(2 r-4)^{2}+r^{2}}$ is minimised

r=1.6

so the point on the plane closest to the origin is (0,-0.8,1.6)

METHOD 3
attempt to find a vector from the origin to the closest point on the plane
EITHER
$(\boldsymbol{r}=) t\left(\begin{array}{c}0 \\ 1 \\ -2\end{array}\right)$
distance to the origin $=\left(\frac{4}{\sqrt{1^{2}+(-2)^{2}}}=\frac{4}{\sqrt{5}}\right)=\frac{4 \sqrt{5}}{5}$ $t= \pm \frac{4}{5}$
check in equation of plane y-2 z=-4 to get $t=-\frac{4}{5}$
OR
$(\boldsymbol{r}=) t(k-9)\left(\begin{array}{c}0 \\ 1 \\ -2\end{array}\right)$
distance to the origin $=\left(\frac{4}{\sqrt{1^{2}+(-2)^{2}}}=\frac{4}{\sqrt{5}}\right)=\frac{4 \sqrt{5}}{5}$ $t= \pm \frac{4}{5(k-9)}$
check in equation of plane y-2 z=-4 to get $t=-\frac{4}{5(k-9)}$
THEN
so the point on the plane closest to the origin is (0,-0.8,1.6)

[Maximum number: 13]

The following question compares the distance and direction between cities on a flat surface to the distance and direction between cities on a sphere.
Consider a model where the cities of Bogotá, Moscow, and Nairobi lie on a flat surface. In this model, Nairobi is 6000 km due south of Moscow and Bogotá is 12500 km due west of Nairobi, as shown in the following diagram.

(a)

Use the scalar product to find the angle between p and n.

[ 2 ]

$\mathbf{p}\cdot\mathbf{n}=0$ , so the angle is $90^\circ$ .

(b)

Find the shortest distance from Bogotá to Moscow on the sphere.

The bearing from B to M is defined as the angle at vertex B in the spherical triangle containing B, M and P . It is given that $\boldsymbol{b} \times \boldsymbol{p}=\left(\begin{array}{c}36 \cos 120^{\circ} \\ -36 \sin 120^{\circ} \\ 0\end{array}\right)$ .

[ 5 ]

METHOD 1

EITHER (Scalar product)
attempt to use scalar product to find BÔM $=\arccos \left(\frac{\boldsymbol{b} . \boldsymbol{m}}{|\boldsymbol{b}||\boldsymbol{m}|}\right)$

Note: This may be written as $\cos ($ BÔM $)=\frac{\boldsymbol{b} . \boldsymbol{m}}{|\boldsymbol{b}||\boldsymbol{m}|}$ .
$=\arccos \left(\frac{\left(6 \cos 120^{\circ}\right)\left(6 \cos 57.3^{\circ}\right)}{6^{2}}\right)$

Note: This (A1) can be awarded for $\cos \theta=\frac{6 \cos 120^{\circ} \times 6 \cos 57.3^{\circ}}{6^{2}}$ .
$=106^{\circ}\left(=105.673 \ldots{ }^{\circ}\right)$ OR 1.84 radians $(1.84434 \ldots)$

OR (Cosine Rule)
$|\mathrm{BM}|=\sqrt{(6 \sin 120-0)^{2}+(6 \cos 120-6 \sin 57.3)^{2}+(0-6 \sin 57.3)^{2}}$ $=9.56(9.56299 \ldots)$ OR 9560 km
attempt to use the cosine rule to find angle BOM
$|\mathrm{BM}|^{2}=6^{2}+6^{2}-2 \times 6 \times 6 \times \cos \mathrm{BOM}$ $\mathrm{BOM}=106^{\circ}(=105.673 \ldots)$ OR 1.84 radians $(1.84434 \ldots)$

THEN
attempt to use $\frac{\theta}{360} \times 2 \pi r$ OR $s=r \theta$ $=\frac{105.671^{\circ}}{360} \times 2 \pi \times 6$ $=11.1(11.0660 \ldots$ thousand km)

Note: Accept an answer of 11100 OR $11066.0 \ldots$ without units (i.e. "km" is implicit).

METHOD 2 (Cross product)
attempt to use cross product to find BÔM $=\arcsin \frac{|\boldsymbol{b} \times \boldsymbol{m}|}{|\boldsymbol{b}||\boldsymbol{m}|}$

Note: This may be written as $\sin$ BÔM $=\frac{|\boldsymbol{b} \times \boldsymbol{m}|}{|\boldsymbol{b}||\boldsymbol{m}|}$ .

\begin{aligned} & =\frac{\left|\left(\begin{array}{c} 36 \cos 120 \sin \theta \\ -36 \sin 120 \sin \theta \\ 36 \sin 120 \cos \theta \end{array}\right)\right|}{6 \times 6} \\ & =\frac{\sqrt{\sin ^{2} \theta+3}}{2}=0.962818 \ldots \\ & \text { BÔM }=180-\arcsin 0.962818 \ldots \\ & \quad=106^{\circ}(105.673 \ldots, 1.844346 \mathrm{rads}) \end{aligned}

Note: Award at most (M1)(A1)(A0)(M1)A0 for an answer of 7.78 thousand km from finding the principal root of the arcsin.

(c)

Using the method from part (c), find the bearing from Bogotá to Moscow.
Disclaimer:
Content used in IB assessments is taken from authentic, third-party sources. The views expressed within them belong to their individual authors and/or publishers and do not necessarily reflect the views of the IB.

[ 6 ]

attempt to calculate $\boldsymbol{b} \times \boldsymbol{m}$ $\boldsymbol{b} \times \boldsymbol{m}=\left(\begin{array}{c}36 \cos 120^{\circ} \sin 57.3^{\circ} \\ -36 \sin 120^{\circ} \sin 57.3^{\circ} \\ 36 \sin 120^{\circ} \cos 57.3^{\circ}\end{array}\right)\left(=\left(\begin{array}{c}-15.1464 \ldots \\ -26.2344 \ldots \\ 16.8449 \ldots\end{array}\right)\right)$
attempt to use scalar product to find angle between vectors

\begin{aligned} & (\cos \theta=)\left(\frac{(\boldsymbol{b} \times \boldsymbol{m}) \cdot(\boldsymbol{b} \times \boldsymbol{p})}{|\boldsymbol{b} \times \boldsymbol{m} \| \boldsymbol{b} \times \boldsymbol{p}|}\right) \\ & (\cos \theta=)\left(\frac{1090.54 \ldots}{34.6614 \ldots \times 36}\right) \end{aligned}

Note: Award (A1) for numerator, (A1) for denominator.

\text { bearing is }(0) 29.1^{\circ}\left(29.0770 \ldots{ }^{\circ}\right)

Note: Do not penalise absence of degree symbol.
Award full marks for answer in radians =0.507.
Award full marks for $029^{\circ}$ .
Special Case: If a candidate calculates $\boldsymbol{m} \times \boldsymbol{b}$ instead of $\boldsymbol{b} \times \boldsymbol{m}$ then you can award at most (M1)(A0)(M1)(A1)(A1)A0.

(d)

Find the vector $\boldsymbol{a} \times \boldsymbol{p}$ .

[ 2 ]

attempt to find cross product (e.g. correct except for sign error)
$\boldsymbol{a} \times \boldsymbol{p}=\left(\begin{array}{c}0 \\ -36 \\ 0\end{array}\right)$

(e)

Show that the angle at vertex A in the spherical triangle is $90^{\circ}$ .

Moscow, M , has position vector $\overrightarrow{\mathrm{OM}}=\boldsymbol{m}=\left(\begin{array}{c}0 \\ 6 \cos \theta \\ 6 \sin \theta\end{array}\right)$ , as shown in the following diagram.

Figure for Question (e) — IB Maths AA HL

The shortest distance between two points on the sphere lies along an arc of a circle on the sphere with centre O . In this model the shortest distance from Moscow to Nairobi is 6000 km .

[ 3 ]

METHOD 1

\boldsymbol{a} \times \boldsymbol{n}=\left(\begin{array}{c} 0 \\ 0 \\ 36 \end{array}\right)

choosing to use scalar product for their $\boldsymbol{a} \times \boldsymbol{p}$ and $\boldsymbol{a} \times \boldsymbol{n}$ $(\boldsymbol{a} \times \boldsymbol{p}) \cdot(\boldsymbol{a} \times \boldsymbol{n})=\mathbf{0}$
so angle is $90^{\circ}$

METHOD 2
$\boldsymbol{a} \times \boldsymbol{n}=\left(\begin{array}{c}0 \\ 0 \\ 36\end{array}\right)$

Note: The final A1 can be awarded for seeing the dot product equals zero without incorrect working. Follow through from (i) is permitted and if they explicitly state that they are using $\boldsymbol{a} \times \boldsymbol{n}$ then allow internal follow through. \\ Note: The final A1 can be awarded for seeing the dot product equals zero
without incorrect working. Follow through from (i) is permitted and if they
explicitly state that they are using $\boldsymbol{a} \times \boldsymbol{n}$ then allow internal follow through.
o angle is $90^{\circ}$
attempt to use cross product for their $\boldsymbol{a} \times \boldsymbol{p}$ and $\boldsymbol{a} \times \boldsymbol{n}$ . This must include both an attempt to find the cross product and knowledge that $\sin \theta=\frac{|x \times y|}{|x||y|}$ $(\boldsymbol{a} \times \boldsymbol{n}) \times(\boldsymbol{a} \times \boldsymbol{p})=\left(\begin{array}{c}1296 \\ 0 \\ 0\end{array}\right)$ OR $(\boldsymbol{a} \times \boldsymbol{p}) \times(\boldsymbol{a} \times \boldsymbol{n})=\left(\begin{array}{c}-1296 \\ 0 \\ 0\end{array}\right)$ $\sin \theta=\frac{1296}{36 \times 36}=1$
so angle is $90^{\circ}$
AG

[Maximum number: 2]

Consider the points A(1,2,3), B(k,-2,1) and C(5,0,2), where $k \in \mathbb{R}$ .

For k=9, let $L_{1}$ be the line passing through A, B and C .

Find a vector equation of the line $L_{1}$ .

attempt to set up a vector equation using a point, a parameter and a direction vector

r=\left(\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right)+\lambda\left(\begin{array}{c} 4 \\ -2 \\ -1 \end{array}\right) \text { (or equivalent) }

Note: " r= " or equivalent must be seen for A1.

[Maximum number: 8]

Consider the points A(1,2,3), B(k,-2,1) and C(5,0,2), where $k \in \mathbb{R}$ .

For k=9, let $L_{1}$ be the line passing through A, B and C .

Line $L_{2}$ has the equation $\frac{x-1}{2}=\frac{y}{3}=1-z$ . Show that the lines $L_{1}$ and $L_{2}$
are skew.

METHOD 1
point on line $L_{1}$ has coordinates $(1+4 \lambda, 2-2 \lambda, 3-\lambda)$
attempt to use a different parameter for $L_{2}$ $\frac{x-1}{2}=\frac{y}{3}=1-z=\mu$ or $r=\left(\begin{array}{l}1 \\ 0 \\ 1\end{array}\right)+\mu\left(\begin{array}{c}2 \\ 3 \\ -1\end{array}\right)$
point on line $L_{2}$ has coordinates $(1+2 \mu, 3 \mu, 1-\mu)$
Note: This A1 may be implied by $\boldsymbol{r}=\left(\begin{array}{l}1 \\ 0 \\ 1\end{array}\right)+\mu\left(\begin{array}{c}2 \\ 3 \\ -1\end{array}\right)$ .
$1+4 \lambda=1+2 \mu$ $2-2 \lambda=3 \mu$ $3-\lambda=1-\mu$
any two of the above equations
attempt to solve two simultaneous equations with two parameters
eg $\lambda=0.25, \mu=0.5$ or $\lambda=1.6, \mu=-0.4$ or $\lambda=-2, \mu=-4$
substitute into third equation or solve a different pair of simultaneous equations
obtain contradiction eg $3-0.25 \neq 1-0.5$ or $1+4(1.6) \neq 1+2(-0.4)$ or $2-2(-2) \neq 3(-4)$ (so the lines do not intersect)
Note: Do not award this R1 if it is based on incorrect values.
lines are not parallel
so lines are skew

METHOD 2
point on line $L_{1}$ has coordinates $(1+4 \lambda, 2-2 \lambda, 3-\lambda)$
attempt to use the equation of $L_{2}$ to generate at least two equations in $\lambda$
if the two lines intersect,
$\frac{(1+4 \lambda)-1}{2}=\frac{2-2 \lambda}{3}\left(\Rightarrow 2 \lambda=\frac{2-2 \lambda}{3}\right)$ $\frac{(1+4 \lambda)-1}{2}=1-(3-\lambda)(\Rightarrow 2 \lambda=\lambda-2)$ $\frac{2-2 \lambda}{3}=1-(3-\lambda) \Rightarrow\left(\frac{2-2 \lambda}{3}=\lambda-2\right)$
any two of the above equations
attempt to solve at least one equation in $\lambda$
one of $\lambda=\frac{1}{4}, \lambda=-2, \lambda=\frac{8}{5}$ seen
substitute into second equation or solve second equation
obtain contradiction eg $\lambda=\frac{1}{4} \neq-2$ or $2\left(\frac{1}{4}\right) \neq \frac{1}{4}-2$ (so the lines do not
intersect)
Note: Do not award this R1 if it is based on incorrect values.
lines are not parallel
so lines are skew

METHOD 3
attempt to find a Cartesian equation for $L_{1}$ $\frac{x-1}{4}=\frac{y-2}{-2}=\frac{z-3}{-1}$
attempt to isolate one variable in both equations
$L_{1}: z=\frac{1-x}{4}+3=\frac{y-2}{2}+3 \quad L_{2}: z=\frac{1-x}{2}+1=\frac{-y}{3}+1 \quad$ OR
$L_{1}: y=\frac{1-x}{2}+2=2(z-3)+2 \quad L_{2}: y=\frac{3(x-1)}{2}=3(1-z) \quad$ OR
$L_{1}: x=1-2(y-2)=1-4(z-3) \quad L_{2}: x=\frac{2 y}{3}+1=1-2(z-1)$
attempt to solve for each of the other two variables
e.g. $\frac{1-x}{2}+1=\frac{1-x}{4}+3$ and $\frac{-y}{3}+1=\frac{y-2}{2}+3$
x=-7, y=-1.2 OR x=2, z=1.4 OR y=1.5, z=5
obtain contradiction eg $z=5 \neq 1.4$ OR $y=1.5 \neq-1.2$ OR $x=2 \neq-7$
(so the lines do not intersect)
Note: Do not award this R1 if it is based on incorrect values.
lines are not parallel
so lines are skew

[Maximum number: 7]

(a)

The z-axis meets the plane $\Pi$ at the point P. Find the coordinates of P.

[ 2 ]

On the z-axis, x=y=0. (M1) Thus -11z=64, so
$\boxed{P=\left(0,0,-\frac{64}{11}\right)}.\quad (A1)$

(b)

Find the angle between the line $\frac{x-2}{3}=\frac{y+5}{4}=\frac{z}{2}$ and the plane $\Pi$ .

[ 5 ]

For line direction (3,4,2) and plane normal (16,24,-11),
$122=\sqrt{29}\sqrt{953}\cos\theta.\quad (M1)(A1)(A1)$
Thus $\theta=42.8^\circ$ is the angle with the normal. (A1) The angle with the plane is therefore
$\boxed{90^\circ-42.8^\circ=47.2^\circ}\quad(0.824\text{ rad}).\quad (A1)$
Note: The equivalent sine formula is accepted.

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Covered syllabus points

Question 1

Question 2

Question 3

Question 4

Question (a)

Question (b)

Question (c)

Question (i)

Question (ii)

Question 5

Question (a)

Question (b)

Question (c)

Question (d)

Question (e)

Question 6

Question 7

Question 8

Question (a)

Question (b)