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Vectors

Practice exam-style IB Math AI questions for Vectors, aligned with the syllabus and grouped by topic.

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Question 1
HL • Paper 1
Easy
Calculator Permitted
HL • Paper 1
Easy
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A rescue drone is at the point A(2,−1,4)A(2,-1,4) and travels directly towards the point B(8,3,1)B(8,3,1). Coordinates are measured in metres. The drone travels at a constant speed of 12 m s−112\ \text{m s}^{-1}.

A

the vector AB→\overrightarrow{AB}.

[1]
B

the unit vector in the direction from AA to BB.

[2]
C

the velocity vector of the drone.

[2]
D

the position of the drone after 0.50.5 seconds.

[1]
Question 2
HL • Paper 1
Easy
Calculator Permitted
HL • Paper 1
Easy
Calculator Permitted

A line LL passes through the points P(1,−2,3)P(1,-2,3) and Q(5,4,−1)Q(5,4,-1).

A

a vector equation of LL.

[2]
B

parametric equations for LL.

[1]
C

whether each of the points R(7,7,−3)R(7,7,-3) and S(3,1,2)S(3,1,2) lies on LL.

[2]
Question 3
HL • Paper 1
Easy
Calculator Permitted
HL • Paper 1
Easy
Calculator Permitted

The positions of two boats, AA and BB, at time tt hours are modelled by

rA(t)=(25)+t(3−1),rB(t)=(148)+t(−1−2)\boldsymbol{r}_A(t)=\begin{pmatrix}2\\5\end{pmatrix}+t\begin{pmatrix}3\\-1\end{pmatrix},\qquad \boldsymbol{r}_B(t)=\begin{pmatrix}14\\8\end{pmatrix}+t\begin{pmatrix}-1\\-2\end{pmatrix}

Distances are measured in kilometres.

A

the speed of boat AA.

[1]
B

when and where the two boats meet.

[3]
C

the distance travelled by boat AA before the boats meet.

[2]
Question 4
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

A particle moves in a vertical plane. Its velocity at time tt seconds is

v(t)=(46−2t) m s−1\boldsymbol{v}(t)=\begin{pmatrix}4\\6-2t\end{pmatrix}\ \text{m s}^{-1}

At t=0t=0, its position vector is (1−3)\begin{pmatrix}1\\-3\end{pmatrix} metres.

A

the position vector r(t)\boldsymbol{r}(t) of the particle.

[3]
B

the time at which the particle reaches its maximum vertical position.

[1]
C

the maximum vertical coordinate of the particle.

[1]
D

the position of the particle when it reaches its maximum vertical position.

[1]

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Question 5
HL • Paper 1
Medium
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HL • Paper 1
Medium
Calculator Permitted

A ball is projected from ground level. Its velocity after tt seconds is modelled by

v(t)=(812−9.8t) m s−1\boldsymbol{v}(t)=\begin{pmatrix}8\\12-9.8t\end{pmatrix}\ \text{m s}^{-1}

At t=0t=0, its position is (00)\begin{pmatrix}0\\0\end{pmatrix}. Air resistance is neglected.

A

the position vector of the ball at time tt.

[2]
B

the positive value of tt at which the ball returns to ground level.

[2]
C

the horizontal distance travelled before the ball returns to ground level.

[1]
D

the speed of the ball immediately before it reaches the ground.

[1]
Question 6
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

A constant force F=(120−4030)\boldsymbol{F}=\begin{pmatrix}120\\-40\\30\end{pmatrix} newtons moves an object through a displacement d=(62−1)\boldsymbol{d}=\begin{pmatrix}6\\2\\-1\end{pmatrix} metres.

A

the work done by the force.

[2]
B

the angle between the force and the displacement. Give your answer in degrees.

[3]
C

what the sign of the work done indicates about the force.

[1]
Question 7
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

The points A(1,0,2)A(1,0,2), B(4,2,1)B(4,2,1) and C(−1,3,5)C(-1,3,5) form a triangle.

A

the vectors AB→\overrightarrow{AB} and AC→\overrightarrow{AC}.

[2]
B

AB→×AC→\overrightarrow{AB}\times\overrightarrow{AC}.

[2]
C

the area of triangle ABCABC.

[1]
D

a unit vector perpendicular to the plane containing AA, BB and CC.

[1]
Question 8
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

A force F=(63−2)\boldsymbol{F}=\begin{pmatrix}6\\3\\-2\end{pmatrix} newtons acts on a straight beam with direction vector b=(2−12)\boldsymbol{b}=\begin{pmatrix}2\\-1\\2\end{pmatrix}.

A

the scalar component of F\boldsymbol{F} acting in the direction of the beam.

[2]
B

the magnitude of the component of F\boldsymbol{F} perpendicular to the beam.

[2]
C

whether the component found in part (a) acts with or against the stated direction of b\boldsymbol{b}. Give a reason.

[1]

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Question 9
HL • Paper 2
Medium
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HL • Paper 2
Medium
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Two cables exert forces

F1=(4−25) kN,F2=(−162) kN\boldsymbol{F}_1=\begin{pmatrix}4\\-2\\5\end{pmatrix}\ \text{kN}, \qquad \boldsymbol{F}_2=\begin{pmatrix}-1\\6\\2\end{pmatrix}\ \text{kN}

on a joint.

A

the resultant of the two forces.

[1]
B

the magnitude of the resultant.

[2]
C

A third cable keeps the joint in equilibrium. the force exerted by this cable.

[1]
D

the angle between F1\boldsymbol{F}_1 and F2\boldsymbol{F}_2. Give your answer in degrees.

[2]
Question 10
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

Two autonomous vehicles move for 0≤t≤50\le t\le5, where tt is measured in minutes. Their position vectors, in kilometres, are

rA(t)=(04)+t(21),rB(t)=(100)+t(−12)\boldsymbol{r}_A(t)=\begin{pmatrix}0\\4\end{pmatrix}+t\begin{pmatrix}2\\1\end{pmatrix},\qquad \boldsymbol{r}_B(t)=\begin{pmatrix}10\\0\end{pmatrix}+t\begin{pmatrix}-1\\2\end{pmatrix}
A

the relative position vector of BB from AA at time tt.

[2]
B

an expression for the square of the distance between the vehicles.

[1]
C

the time at which the distance between the vehicles is a minimum.

[2]
D

the minimum distance between the vehicles.

[1]
Question 11
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
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Two lines have vector equations

L1: r=(12−1)+λ(2−12),L2: r=(4−23)+μ(12−2)L_1:\ \boldsymbol{r}=\begin{pmatrix}1\\2\\-1\end{pmatrix}+\lambda\begin{pmatrix}2\\-1\\2\end{pmatrix}, \qquad L_2:\ \boldsymbol{r}=\begin{pmatrix}4\\-2\\3\end{pmatrix}+\mu\begin{pmatrix}1\\2\\-2\end{pmatrix}
A

the acute angle between L1L_1 and L2L_2. Give your answer in degrees.

[3]
B

whether the two lines intersect.

[3]
Question 12
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

Two vehicles start moving at the same time. Their positions after tt hours are

rA(t)=(00)+t(21),rB(t)=(6−2)+t(−11)\boldsymbol{r}_A(t)=\begin{pmatrix}0\\0\end{pmatrix}+t\begin{pmatrix}2\\1\end{pmatrix}, \qquad \boldsymbol{r}_B(t)=\begin{pmatrix}6\\-2\end{pmatrix}+t\begin{pmatrix}-1\\1\end{pmatrix}

Distances are measured in kilometres.

A

Using independent parameters λ\lambda and μ\mu, the point at which the geometric paths of the vehicles intersect.

[3]
B

the time at which each vehicle reaches the intersection point.

[2]
C

whether the vehicles collide at the intersection point. Give a reason.

[1]

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Question 13
HL • Paper 1
Medium
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HL • Paper 1
Medium
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The position vector of a particle moving in a horizontal plane is

r(t)=(5cos⁡(0.4t)5sin⁡(0.4t)) m\boldsymbol{r}(t)=\begin{pmatrix}5\cos(0.4t)\\5\sin(0.4t)\end{pmatrix}\ \text{m}

where tt is measured in seconds.

A

the velocity vector of the particle.

[2]
B

the speed of the particle is constant.

[1]
C

the acceleration vector of the particle.

[2]
D

the position vector is perpendicular to the velocity vector for all tt.

[1]
Question 14
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

A force F=(040−30)\boldsymbol{F}=\begin{pmatrix}0\\40\\-30\end{pmatrix} newtons is applied at a point whose position relative to a pivot is r=(0.250.100)\boldsymbol{r}=\begin{pmatrix}0.25\\0.10\\0\end{pmatrix} metres. The torque vector is defined by τ=r×F\boldsymbol{\tau}=\boldsymbol{r}\times\boldsymbol{F}.

A

the torque vector.

[2]
B

the magnitude of the torque.

[2]
C

the angle between r\boldsymbol{r} and F\boldsymbol{F}. Give your answer in degrees.

[2]
Question 15
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

A straight pipeline is modelled by the line

L: r=(1−12)+λ(21−2)L:\ \boldsymbol{r}=\begin{pmatrix}1\\-1\\2\end{pmatrix}+\lambda\begin{pmatrix}2\\1\\-2\end{pmatrix}

A monitoring station is located at P(7,3,1)P(7,3,1).

A

Let QQ be the point on LL closest to PP. Form an equation in λ\lambda using the fact that PQ→\overrightarrow{PQ} is perpendicular to LL.

[2]
B

the coordinates of QQ.

[2]
C

the shortest distance from the monitoring station to the pipeline.

[2]
Question 16
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

A particle AA moves according to

rA(t)=(3cos⁡t3sin⁡t)\boldsymbol{r}_A(t)=\begin{pmatrix}3\cos t\\3\sin t\end{pmatrix}

where tt is measured in seconds. A second particle BB follows the same motion exactly 22 seconds later.

A

the position vector rB(t)\boldsymbol{r}_B(t).

[1]
B

the velocity vector of BB at t=2t=2.

[2]
C

the distance between AA and BB is constant, and this distance.

[3]

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Question 17
HL • Paper 2
Medium
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HL • Paper 2
Medium
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An aircraft flies in a horizontal plane. East and north are represented by the positive xx- and yy-directions respectively. The aircraft has an airspeed of 240 km h−1240\ \text{km h}^{-1} in the direction 3i+4j3\boldsymbol{i}+4\boldsymbol{j}. A constant wind has velocity −30i+20j km h−1-30\boldsymbol{i}+20\boldsymbol{j}\ \text{km h}^{-1}.

A
I.

the velocity vector of the aircraft relative to the ground.

[3]
II.

the ground speed of the aircraft.

[2]
B
I.

the displacement of the aircraft after 1.751.75 hours.

[2]
II.

The aircraft then returns directly to its starting point with a ground speed of 260 km h−1260\ \text{km h}^{-1}. its return velocity vector and the return time.

[3]
Question 18
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

A straight railway tunnel passes through the points A(−2,1,4)A(-2,1,4) and B(4,5,−2)B(4,5,-2). A maintenance shaft is modelled by the line

M: r=(77−5)+μ(1−22)M:\ \boldsymbol{r}=\begin{pmatrix}7\\7\\-5\end{pmatrix}+\mu\begin{pmatrix}1\\-2\\2\end{pmatrix}
A
I.

a vector equation of the railway tunnel.

[2]
II.

the point C(7,7,−5)C(7,7,-5) lies on the railway tunnel.

[3]
B

the acute angle at which the maintenance shaft meets the railway tunnel.

[4]
Question 19
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

A towing cable exerts a force of magnitude 500 N500\ \text{N} in the direction 4i+j4\boldsymbol{i}+\boldsymbol{j}. During one stage of the tow, the load has displacement 24i+10j24\boldsymbol{i}+10\boldsymbol{j} metres.

A
I.

the force vector.

[2]
II.

the work done by the cable.

[2]
B
I.

the scalar component of the force parallel to the displacement.

[2]
II.

the magnitude of the component of the force perpendicular to the displacement and its size.

[3]
Question 20
HL • Paper 3
Medium
Calculator Permitted
HL • Paper 3
Medium
Calculator Permitted

Three environmental sensors are placed at A(2,−1,3)A(2,-1,3), B(8,2,0)B(8,2,0) and C(−1,5,6)C(-1,5,6). A communications hub is to be positioned using the medians of triangle ABCABC.

Point

x

y

z

A

2

-1

3

B

8

2

0

C

-1

5

6

A
I.

AB→\overrightarrow{AB} and AC→\overrightarrow{AC}.

[2]
II.

the midpoint MM of BCBC and a vector equation of the median AMAM.

[2]
B

The hub GG is placed two-thirds of the way from AA to MM. GG and its position vector is 13(a+b+c)\frac13(\boldsymbol{a}+\boldsymbol{b}+\boldsymbol{c}). why the same point lies two-thirds of the way along each median.

[4]

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Question 21
HL • Paper 3
Medium
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HL • Paper 3
Medium
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A robotic tool applies the constant force F=(3,−2,1)\boldsymbol{F}=(3,-2,1) newtons while moving from O=(0,0,0)O=(0,0,0) to B=(4,1,3)B=(4,1,3) metres. It may move directly or via A=(1,3,−1)A=(1,3,-1).

A
I.

the work done along the direct route from OO to BB.

[2]
II.

the total work done along the route OABOAB.

[2]
B

For a constant force F\boldsymbol{F} and any sequence of points P0,P1,…,PnP_0,P_1,\ldots,P_n, the total work from P0P_0 to PnP_n is independent of the intermediate points. why this conclusion need not hold if the force changes with position.

[4]
Question 22
HL • Paper 2
Hard
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HL • Paper 2
Hard
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Two survey vessels move for 0≤t≤60\leq t\leq6, where tt is measured in hours and positions are measured in kilometres:

rA(t)=(012)+t(51),rB(t)=(180)+t(−14)\boldsymbol{r}_A(t)=\begin{pmatrix}0\\12\end{pmatrix}+t\begin{pmatrix}5\\1\end{pmatrix},\qquad \boldsymbol{r}_B(t)=\begin{pmatrix}18\\0\end{pmatrix}+t\begin{pmatrix}-1\\4\end{pmatrix}
A
AI.

the relative position vector of BB from AA at time tt.

[2]
AII.

the square of their separation is 45t2−288t+46845t^2-288t+468.

[2]
B
BI.

the time at which the vessels are closest and their minimum separation.

[4]
BII.

Safety regulations require the vessels to remain at least 3 km3\ \text{km} apart. whether the regulation is satisfied, with a reason.

[1]
Question 23
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
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A water droplet moves in a vertical plane. Its velocity at time tt seconds is

v(t)=(68−1.6t) m s−1\boldsymbol{v}(t)=\begin{pmatrix}6\\8-1.6t\end{pmatrix}\ \text{m s}^{-1}

At t=0t=0, its position is (−3,2)(-3,2) metres.

A
I.

the position vector r(t)\boldsymbol{r}(t).

[3]
II.

the maximum vertical coordinate reached by the droplet.

[3]
B

The droplet passes through the horizontal level y=10y=10 twice.

I.

the two horizontal coordinates at which the droplet passes through y=10y=10.

[3]
II.

the speed of the droplet when it passes through y=10y=10 for the second time. If you did not obtain a second time, use t=8.873t=8.873.

[2]
Question 24
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
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A rectangular solar panel has adjacent vertices A(1,2,0)A(1,2,0), B(7,3,−2)B(7,3,-2) and D(2,6,5)D(2,6,5). Coordinates are measured in metres. The direction from the panel towards the Sun is represented by s=(2−1−2)\boldsymbol{s}=\begin{pmatrix}2\\-1\\-2\end{pmatrix}.

A
I.

AB→×AD→\overrightarrow{AB}\times\overrightarrow{AD}.

[3]
II.

the area of the panel and a unit normal vector to it.

[3]
B
I.

the acute angle between the Sun's direction and a normal to the panel.

[2]
II.

The effective collecting area is the area of the panel multiplied by the cosine of this angle. the effective collecting area.

[2]

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Question 25
HL • Paper 2
Hard
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HL • Paper 2
Hard
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A mechanic applies a force of magnitude 180 N180\ \text{N} in the direction 3j−4k3\boldsymbol{j}-4\boldsymbol{k}. The position vector from the centre of a bolt to the point of application is

r=0.25i+0.08j m\boldsymbol{r}=0.25\boldsymbol{i}+0.08\boldsymbol{j}\ \text{m}

The torque vector is τ=r×F\boldsymbol{\tau}=\boldsymbol{r}\times\boldsymbol{F}.

A
I.

the force vector.

[2]
II.

the torque vector and its magnitude.

[4]
B

a unit vector along the axis of rotation indicated by the torque and the effect of reversing the force.

[3]
Question 26
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
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In a computer-graphics model, a surface has outward normal vector n=2i−j+2k\boldsymbol{n}=2\boldsymbol{i}-\boldsymbol{j}+2\boldsymbol{k}. The direction from the surface towards a light source is s=i−2j+2k\boldsymbol{s}=\boldsymbol{i}-2\boldsymbol{j}+2\boldsymbol{k}. The illumination factor is modelled by the cosine of the acute angle between the two vectors.

A
I.

unit vectors in the directions of n\boldsymbol{n} and s\boldsymbol{s}.

[2]
II.

the illumination factor and the angle between the normal and the light direction.

[3]
B

A second surface has normal vector m=ki+j+2k\boldsymbol{m}=k\boldsymbol{i}+\boldsymbol{j}+2\boldsymbol{k}. the value of kk for which this surface receives no direct illumination, and your answer.

[3]
Question 27
HL • Paper 2
Hard
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HL • Paper 2
Hard
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A river is 600 m600\ \text{m} wide and flows due east at 1.8 m s−11.8\ \text{m s}^{-1}. A boat moves through the water at a constant speed of 4.5 m s−14.5\ \text{m s}^{-1}. The boat starts on the south bank directly opposite a landing point on the north bank. The diagram is schematic and not to scale; the boat's heading is specified in each part.

A
I.

the boat's velocity relative to the water if it is to travel directly to the landing point.

[3]
II.

the time taken to reach the landing point.

[2]
B

The boat instead points due north while maintaining the same speed through the water.

I.

If the boat is pointed due north relative to the water, the crossing time and the distance east of the landing point at which the boat reaches the north bank.

[3]
II.

the direct-to-landing strategy in part (a) with the due-north strategy in part (b)(i) in terms of crossing time and arrival position.

[2]
Question 28
HL • Paper 2
Hard
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HL • Paper 2
Hard
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A particle AA moves around a circle according to

rA(t)=(8cos⁡(0.3t)8sin⁡(0.3t)) m\boldsymbol{r}_A(t)=\begin{pmatrix}8\cos(0.3t)\\8\sin(0.3t)\end{pmatrix}\ \text{m}

A particle BB follows the same motion exactly 44 seconds later.

A
I.

the position vector of BB at time tt.

[1]
II.

the velocity vector and speed of BB at t=4t=4.

[3]
B
I.

the distance between AA and BB is constant, and this distance.

[3]
II.

the first time t≥4t\geq4 at which BB is at the point (0,8)(0,8).

[2]

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Question 29
HL • Paper 2
Hard
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HL • Paper 2
Hard
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Two inspection pods move through a three-dimensional storage facility. Their positions after tt seconds are

rA(t)=(103)+t(21−1),rB(t)=(9−50)+t(−121)\boldsymbol{r}_A(t)=\begin{pmatrix}1\\0\\3\end{pmatrix}+t\begin{pmatrix}2\\1\\-1\end{pmatrix},\qquad \boldsymbol{r}_B(t)=\begin{pmatrix}9\\-5\\0\end{pmatrix}+t\begin{pmatrix}-1\\2\\1\end{pmatrix}

Distances are measured in metres.

A
I.

the angle between the velocity vectors of the pods.

[3]
II.

the acute angle between their geometric paths.

[1]
B
I.

the relative position of BB from AA is (8−3t−5+t−3+2t)\begin{pmatrix}8-3t\\-5+t\\-3+2t\end{pmatrix}.

[2]
II.

when the pods are closest and their positions and separation at that time.

[4]
Question 30
HL • Paper 2
Hard
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HL • Paper 2
Hard
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Two cables exert forces on a joint. Cable 1 exerts a force of magnitude 8 kN8\ \text{kN} in the direction 2i−j+2k2\boldsymbol{i}-\boldsymbol{j}+2\boldsymbol{k}. Cable 2 exerts a force of magnitude 10 kN10\ \text{kN} in the direction −3i+4j-3\boldsymbol{i}+4\boldsymbol{j}. A mast through the joint has direction j+k\boldsymbol{j}+\boldsymbol{k}.

A
I.

the two force vectors.

[3]
II.

the resultant force and its magnitude.

[3]
B
I.

the scalar component of the resultant acting in the direction of the mast.

[2]
II.

the magnitude of the component perpendicular to the mast.

[2]
Question 31
HL • Paper 2
Hard
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HL • Paper 2
Hard
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A stationary camera drone is at D(4,−3,12)D(4,-3,12) metres. Its camera is initially directed towards a ground marker M(10,5,0)M(10,5,0). The ground is represented by the plane z=0z=0.

A
I.

a vector equation of the initial line of sight.

[2]
II.

A transparent vertical screen is represented by x=7x=7. where the line of sight passes through the screen and its distance from the drone.

[3]
B

A ground vehicle has position

rV(t)=(0−50)+t(220),t≥0\boldsymbol{r}_V(t)=\begin{pmatrix}0\\-5\\0\end{pmatrix}+t\begin{pmatrix}2\\2\\0\end{pmatrix},\qquad t\geq0

where tt is measured in seconds.

I.

the time when the vehicle is closest to the drone.

[3]
II.

the minimum distance from the drone to the vehicle. If you did not obtain a time, use t=1.50t=1.50.

[2]
Question 32
HL • Paper 3
Hard
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HL • Paper 3
Hard
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An underwater vehicle travels from A(1,2,−1)A(1,2,-1) to B(13,8,7)B(13,8,7). Coordinates are measured in metres. The journey takes 66 minutes. The water current has velocity

c=(1−10.5)\boldsymbol{c}=\begin{pmatrix}1\\-1\\0.5\end{pmatrix}

in metres per minute. The vehicle maintains a constant velocity relative to the water.

A
I.

the displacement vector AB→\overrightarrow{AB} and the velocity of the vehicle relative to the ground.

[2]
II.

the velocity and speed of the vehicle relative to the water.

[2]
B

Suppose instead that the current is kck\boldsymbol{c}, where k≥0k\geq0, while the journey time and route remain unchanged. the value of kk for which the magnitude of the required velocity relative to the water is minimized, and this minimum speed.

[4]

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Question 33
HL • Paper 3
Hard
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HL • Paper 3
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Two straight mining tunnels are modelled by

L1: r=(120)+λ(2−11),L2: r=(7−13)+μ(−120)L_1:\ \boldsymbol{r}=\begin{pmatrix}1\\2\\0\end{pmatrix}+\lambda\begin{pmatrix}2\\-1\\1\end{pmatrix},\qquad L_2:\ \boldsymbol{r}=\begin{pmatrix}7\\-1\\3\end{pmatrix}+\mu\begin{pmatrix}-1\\2\\0\end{pmatrix}

Tunnel

Point

Direction

Vector equation

L1

(1, 2, 0)

(2, -1, 1)

r = (1, 2, 0) + λ(2, -1, 1)

L2

(7, -1, 3)

(-1, 2, 0)

r = (7, -1, 3) + μ(-1, 2, 0)

A
I.

the point at which the tunnels intersect.

[2]
II.

the acute angle between the tunnels. Give your answer in degrees.

[2]
B

A third tunnel through the intersection is to bisect the acute angle between L1L_1 and L2L_2. a vector equation for this tunnel and that its direction bisects the angle.

[4]
Question 34
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

Two aircraft fly at constant altitude. Their positions, in kilometres, are

rA(t)=(00)+t(41),rB(t)=(10−2)+t(−13)\boldsymbol{r}_A(t)=\begin{pmatrix}0\\0\end{pmatrix}+t\begin{pmatrix}4\\1\end{pmatrix},\qquad \boldsymbol{r}_B(t)=\begin{pmatrix}10\\-2\end{pmatrix}+t\begin{pmatrix}-1\\3\end{pmatrix}

where tt is measured in minutes and 0≤t≤40\leq t\leq4. A separation below 22 km requires an alert.

A
I.

an expression for the relative position of BB from AA and for the square of their separation.

[2]
II.

when the aircraft are closest and whether an alert is required.

[2]
B

For two objects with relative position p+tv\boldsymbol{p}+t\boldsymbol{v}, derive a formula for the unrestricted time of closest approach. the additional checks needed when the model applies only for 0≤t≤T0\leq t\leq T, and how an actual collision can be distinguished from a close approach.

[4]
Question 35
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

Two inspection robots follow straight routes across a factory floor. Robot AA starts at time t=0t=0 and has position

rA(t)=t(21)\boldsymbol{r}_A(t)=t\begin{pmatrix}2\\1\end{pmatrix}

Robot BB begins at (8,0)(8,0) and moves with velocity (−12)\begin{pmatrix}-1\\2\end{pmatrix} once activated.

A
I.

Using independent route parameters, the intersection point of the two geometric paths.

[2]
II.

If both robots are activated at t=0t=0, whether they collide at the intersection.

[2]
B

Robot BB is instead activated hh minutes after robot AA. Write a time-shifted position model for BB and the value of hh that causes a collision. Generalize your result in terms of the two travel times to an intersection.

[4]
Question 36
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A camera moves around an elliptical track. Its position after tt seconds is

r(t)=(4cos⁡(0.5t)3sin⁡(0.5t))\boldsymbol{r}(t)=\begin{pmatrix}4\cos(0.5t)\\3\sin(0.5t)\end{pmatrix}

in metres. For part (b), take the first circuit to mean 0≤t<4π0\leq t<4\pi, with tt measured in seconds.

A
I.

the velocity vector and the speed of the camera.

[2]
II.

the minimum and maximum speeds during one complete circuit.

[2]
B

During the first complete circuit, all times when the position vector is perpendicular to the velocity vector. these times geometrically and why the speed is not constant even though the path is closed and periodic.

[4]

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Question 37
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A parallelogram-shaped solar panel has adjacent edge vectors

u=(401),v=(131)\boldsymbol{u}=\begin{pmatrix}4\\0\\1\end{pmatrix},\qquad \boldsymbol{v}=\begin{pmatrix}1\\3\\1\end{pmatrix}

in metres. Sunlight travels parallel to s=(2,−1,2)\boldsymbol{s}=(2,-1,2).

A
I.

a vector perpendicular to the panel and its area.

[2]
II.

a unit normal to the panel with a positive vertical component.

[2]
B

The effective collecting area is the area of the panel projected perpendicular to the sunlight. this projected area and the greatest projected area attainable by rotating the panel without changing its shape.

[4]
Question 38
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

At an airport, the wind velocity is w=(8,−6,2) m s−1\boldsymbol{w}=(8,-6,2)\ \text{m s}^{-1}. A sloping runway has direction vector d=(2,1,2)\boldsymbol{d}=(2,1,2) in the stated take-off direction.

A
I.

the signed scalar component of the wind in the runway direction and its sign.

[2]
II.

the magnitude of the component perpendicular to the runway.

[2]
B

A new runway must be horizontal and may have any unit direction q=(cos⁡θ,sin⁡θ,0)\boldsymbol{q}=(\cos\theta,\sin\theta,0). the direction that maximizes the tailwind component. this maximum component and the remaining perpendicular component of the wind.

[4]
Question 39
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A triangular fabric panel has vertices A=(0,0,0)A=(0,0,0), B=(4,0,0)B=(4,0,0) and C=(1,k,2)C=(1,k,2), where kk is a real design parameter. Coordinates are measured in metres.

Point

x [m]

y [m]

z [m]

A

0

0

0

B

4

0

0

C

1

k

2

A
I.

AB→×AC→\overrightarrow{AB}\times\overrightarrow{AC} in terms of kk.

[2]
II.

obtain an expression for the area of the panel.

[2]
B

The required panel area is 10 m210\ \text{m}^2. the possible values of kk and the angle BACBAC for either design. Then the minimum possible area over all real values of kk and the geometry of the minimizing design.

[4]
Question 40
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

Robot AA starts at time t=0t=0 minutes and has position

rA(t)=(2−4)+t(32)\boldsymbol{r}_A(t)=\begin{pmatrix}2\\-4\end{pmatrix}+t\begin{pmatrix}3\\2\end{pmatrix}

Robot BB starts at time t=2t=2 minutes from (20,0)(20,0) and then moves with velocity (−21)\begin{pmatrix}-2\\1\end{pmatrix}. Positions are measured in kilometres.

A
I.

Using independent parameters, the point where the geometric paths of the robots intersect.

[4]
II.

whether the robots collide at this point.

[2]
B

For t≥2t\geq2, when the robots are closest and their minimum separation.

[5]

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Question 41
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

A triangular sail has vertices A(1,0,2)A(1,0,2), B(5,2,1)B(5,2,1) and C(2,k,6)C(2,k,6), where kk is a real number. Coordinates are measured in metres.

A
I.

AB→\overrightarrow{AB} and AC→\overrightarrow{AC} in terms of kk.

[2]
II.

AB→×AC→=(k+8−174k−2)\overrightarrow{AB}\times\overrightarrow{AC}=\begin{pmatrix}k+8\\-17\\4k-2\end{pmatrix}.

[3]
B

The area of the sail is 4252 m2\dfrac{\sqrt{425}}{2}\ \text{m}^2. Therefore, ∣AB→×AC→∣2=425\left\lvert\overrightarrow{AB}\times\overrightarrow{AC}\right\rvert^2=425.

I.

the possible values of kk.

[3]
II.

The chosen orientation requires the k\boldsymbol{k}-component of AB→×AC→\overrightarrow{AB}\times\overrightarrow{AC} to be positive. kk and your choice.

[2]
Question 42
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

Two pipes have centre-lines

L1: r=λ(110),L2: r=(102)+μ(011)L_1:\ \boldsymbol{r}=\lambda\begin{pmatrix}1\\1\\0\end{pmatrix},\qquad L_2:\ \boldsymbol{r}=\begin{pmatrix}1\\0\\2\end{pmatrix}+\mu\begin{pmatrix}0\\1\\1\end{pmatrix}

A shortest connecting pipe is to be installed between them.

A
I.

Let PP be a point on L1L_1 and QQ a point on L2L_2. Write PQ→\overrightarrow{PQ} in terms of λ\lambda and μ\mu.

[2]
II.

Use perpendicularity to the endpoints of the shortest connecting pipe.

[2]
B

the length of the connecting pipe. Then the result using the cross product of the direction vectors and the displacement between any one point on each line.

[4]
Question 43
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A package is launched from ground level with initial velocity (c10) m s−1\begin{pmatrix}c\\10\end{pmatrix}\ \text{m s}^{-1}, where c>0c>0. Its acceleration is (0−10) m s−2\begin{pmatrix}0\\-10\end{pmatrix}\ \text{m s}^{-2}. A vertical barrier is 99 m from the launch point and has height 33 m.

A
I.

the position of the package is

r(t)=(ct10t−5t2)\boldsymbol{r}(t)=\begin{pmatrix}ct\\10t-5t^2\end{pmatrix}
[2]
II.

For c=6c=6, whether the package clears the barrier.

[2]
B

the complete interval of values of cc for which the package reaches the barrier and passes strictly above its 3 m height before returning to ground level.

[4]
Question 44
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

In a computer graphics model, a surface has normal vector n=(2,−1,2)\boldsymbol{n}=(2,-1,2). A light source lies in the direction s=(1,−2,2)\boldsymbol{s}=(1,-2,2) from the surface. The diffuse illumination is modelled by

I=I0max⁡(0,n^⋅s^)I=I_0\max\left(0,\widehat{\boldsymbol{n}}\cdot\widehat{\boldsymbol{s}}\right)
A
I.

the unit vectors n^\widehat{\boldsymbol{n}} and s^\widehat{\boldsymbol{s}}.

[2]
II.

II as a fraction of I0I_0.

[2]
B

The light source is moved so that its direction is s(k)=(1,−2,k)\boldsymbol{s}(k)=(1,-2,k), where kk is real. the value of kk that maximizes the illumination and the maximum value of I/I0I/I_0.

[4]

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Question 45
HL • Paper 3
Hard
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HL • Paper 3
Hard
Calculator Permitted

A force F=(0,40,−30)\boldsymbol{F}=(0,40,-30) newtons is applied to a handle. The point of application is initially at position r=(0.3,0,0)\boldsymbol{r}=(0.3,0,0) metres relative to the pivot. The torque is τ=r×F\boldsymbol{\tau}=\boldsymbol{r}\times\boldsymbol{F}.

A
I.

the torque vector.

[2]
II.

the torque magnitude and why it is as large as possible for a handle of length 0.30.3 m and a force of magnitude 5050 N.

[2]
B

The handle is rotated in the xyxy-plane so that r(ϕ)=0.3(cos⁡ϕ,sin⁡ϕ,0)\boldsymbol r(\phi)=0.3(\cos\phi,\sin\phi,0). the maximum and minimum possible torque magnitudes and the corresponding orientations of the handle.

[4]
Question 46
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

Two satellites are modelled over a short time interval by

rA(t)=t(120),rB(t)=(6−34)+t(−11−1)\boldsymbol{r}_A(t)=t\begin{pmatrix}1\\2\\0\end{pmatrix},\qquad \boldsymbol{r}_B(t)=\begin{pmatrix}6\\-3\\4\end{pmatrix}+t\begin{pmatrix}-1\\1\\-1\end{pmatrix}

where distance is measured in kilometres and time in minutes. A warning is active whenever their separation is at most 66 km.

A
I.

the relative position of satellite BB from satellite AA and an expression for the square of their separation.

[2]
II.

the time and distance of closest approach.

[2]
B

the interval of time for which the warning is active and its duration. why the midpoint of this interval is the time of closest approach.

[4]
Question 47
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A cable supporting an aerial platform has direction vector d=(3,4,12)\boldsymbol{d}=(3,4,12). The platform's weight is represented by the force W=(0,0,−120)\boldsymbol{W}=(0,0,-120) newtons.

A
I.

the signed scalar component of the weight in the stated cable direction.

[2]
II.

the magnitude of the component of the weight perpendicular to the cable.

[2]
B

A redesigned cable has direction vector (3,4,h)(3,4,h), where h>0h>0. Safety regulations require the component of the weight perpendicular to the cable to be at most 3030 N. the least possible value of hh and this design requirement geometrically.

[4]
Question 48
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

Two particles move in a horizontal plane for 0≤t≤40\leq t\leq4, where positions are measured in metres and time in seconds:

rA(t)=(t24t),rB(t)=(12−2tt2)\boldsymbol{r}_A(t)=\begin{pmatrix}t^2\\4t\end{pmatrix},\qquad \boldsymbol{r}_B(t)=\begin{pmatrix}12-2t\\t^2\end{pmatrix}
A
I.

the velocity vectors of the two particles.

[2]
II.

the square of the distance between the particles is

D2(t)=(12−2t−t2)2+(t2−4t)2D^2(t)=(12-2t-t^2)^2+(t^2-4t)^2
[3]
B
I.

the time at which the particles are closest. You should any stationary point with the endpoints of the interval.

[4]
II.

the minimum distance between the particles and that the relative position is perpendicular to the relative velocity at this time. If you did not obtain a time, use the unrounded value t=2.69623 st=2.69623\ \text{s}.

[2]

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Trigonometry

Voronoi Diagrams