Free worked solutions · 2026 Edition

IB Maths AI HL Chapter 10: Vectors worked solutions

Worked solutions to the practice questions in Chapter 10 of Photon Guidebook: Mathematics Applications and Interpretation HL for the IB Diploma (2026 Edition). Each solution shows every step with the mark it earns, the way an IB mark scheme does.

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Basics

One idea at a time, in the order the chapter teaches them.

Question 1

[4 marks]
A sensor is at P(2,-4,9) and a receiver at Q(-5,8,3), in metres.
Find \overrightarrow{PQ} and the distance PQ.
[4]
Show solutionHide solution
\overrightarrow{PQ}=(-7,12,-6). M1A1
PQ=\sqrt{49+144+36}=\sqrt{229}=15.1 m. M1A1

Final answer

\overrightarrow{PQ}=(-7,12,-6); PQ=15.1 m.

Question 2

[4 marks]
A boat displacement is (-6,9) km, with components east and north. Bearings are clockwise from north.
Find its magnitude and bearing.
[4]
Show solutionHide solution
|\mathbf d|=\sqrt{117}=10.8 km. M1A1
Angle west of north =\tan^{-1}(6/9)=33.6901^\circ.
M1
Bearing =326^\circ (3 s.f.).
A1

Final answer

10.8 km; bearing 326^\circ.

Question 3

[3 marks]
Point P(7,2,3) lies on the segment AB, where A(-2,5,9) and B(13,0,-1).
Determine AP:PB.
[3]
Show solutionHide solution
\overrightarrow{AB}=(15,-5,-10) and \overrightarrow{AP}=(9,-3,-6).
M1
\overrightarrow{AP}=\frac35\overrightarrow{AB}.
A1
AP:PB=3:2.
A1

Final answer

AP:PB=3:2.

Question 4

[4 marks]
The line \mathbf r=(1,-4,7)+t(2,5,-3) contains a point (9,k,-5).
Find t and k, checking the third coordinate.
[4]
Show solutionHide solution
1+2t=9 gives t=4. M1A1
k=-4+5(4)=16; 7-3(4)=-5 confirms consistency. A1R1

Final answer

t=4, k=16.

Question 5

[3 marks]
A drone starts at (2,5,1) m and flies at 14 m/s in direction (2,-3,6), with constant velocity.
Write its position vector at time t\ge0 seconds.
[3]
Show solutionHide solution
Direction magnitude 7; velocity =2(2,-3,6)=(4,-6,12) m/s. M1A1
\mathbf r=(2,5,1)+t(4,-6,12), t\ge0.
A1

Final answer

\mathbf r=(2,5,1)+t(4,-6,12), t\ge0.

Question 6

[5 marks]
L:\mathbf r=(2,1,3)+s(3,-2,1) and M:\mathbf r=(9,-4,8)+t(1,-1,3).
Find the intersection or show that none exists.
[5]
Show solutionHide solution
2+3s=9+t, 1-2s=-4-t give s=2, t=-1. M1A2
Both third coordinates are 5, so intersection (8,-3,5). A1R1

Final answer

The lines meet at (8,-3,5), with s=2 and t=-1.

Question 7

[3 marks]
L_1:\mathbf r=(0,0,1)+s(1,0,0) and L_2:\mathbf r=(2,0,4)+t(0,1,0).
Explain why these lines are skew.
[3]
Show solutionHide solution
The directions are not proportional, so the lines are not parallel.
R1
z=1 on L_1 and z=4 on L_2, so no intersection is possible.
R1
Non-parallel, non-intersecting spatial lines are skew.
R1

Final answer

Not parallel; z=1 on one line and z=4 on the other, so there is no common point: skew.

Question 8

[5 marks]
Lines are \mathbf r=(0, 2, 5)+s(1, 3, -2) and \mathbf r=(1, -4, 3)+t(4, -1, 2).
Calculate their smaller angle between 0^\circ and 90^\circ, explaining why base points do not enter the calculation.
[5]
Show solutionHide solution
Use direction vectors: dot product −3 and magnitude product \sqrt{294}. M1A1
\cos\phi=|-3|/17.146428199; \phi=79.923462871^\circ=79.9^\circ. M1A1
Translating a line does not change its direction.
R1

Final answer

79.9^\circ.

Question 9

[4 marks]
Two sides from the same corner of a triangular sail are \mathbf u=(3,-2,5) m and \mathbf v=(1,4,0) m. A parallelogram area is |\mathbf u\times\mathbf v|; a triangle uses half this.
Find the area of the triangle.
[4]
Show solutionHide solution
\mathbf u\times\mathbf v=(-20,5,14). M1A1
Area =1/2\sqrt{621}=12.45993579=12.5\text{ m}^2. M1A1

Final answer

\mathbf u\times\mathbf v=(-20,5,14); area 12.5 m^2.

Question 10

[4 marks]
A gate handle contact is at position \mathbf r=(0.5,0.2,0) m relative to the pivot. Force \mathbf F=(-10,25,6) N acts there. The moment is \boldsymbol\tau=\mathbf r\times\mathbf F.
Find the moment vector and its magnitude.
[4]
Show solutionHide solution
\boldsymbol\tau=(6/5,-3,29/2) N m. M1A1
|\boldsymbol\tau|=\sqrt{22\,069/100}=14.85563866=14.9 N m. M1A1

Final answer

\boldsymbol\tau=(1.2,-3,14.5) N m; |\boldsymbol\tau|=14.9 N m.

Tutorial

Exam-style questions on each question type, with the marks the IB would give.

Question 11

[6 marks]
A cable-car cabin leaves the lower station \mathrm{S}, which is the origin, and travels along a straight cable. After one minute its displacement from \mathrm{S} is \mathbf{d} = 120\mathbf{i} - 90\mathbf{j} + 80\mathbf{k} metres, where \mathbf{i} points east, \mathbf{j} north and \mathbf{k} vertically upwards.
(a) Find the distance the cabin has travelled along the cable in this minute.
[2]
(b) Find the unit vector in the direction of the cable.
[2]
(c) Find the position vector of the cabin when it has travelled 500 m along the cable from \mathrm{S}. Give the components to the nearest metre.
[2]
Show solutionHide solution
(a) |\mathbf{d}| = \sqrt{120^2 + (-90)^2 + 80^2} = \sqrt{28\,900}
[1]
= 170 m
[1]
(b) \hat{\mathbf{d}} = \dfrac{1}{170}\mathbf{d}
[1]
= \begin{pmatrix}0.7059\\-0.5294\\0.4706\end{pmatrix} (exact: \frac{1}{17}(12,-9,8); the components square and sum to 1)
[1]
(c) 500\,\hat{\mathbf{d}} = \dfrac{500}{170}\,\mathbf{d}
[1]
= \begin{pmatrix}352.9\\-264.7\\235.3\end{pmatrix}, i.e. about \begin{pmatrix}353\\-265\\235\end{pmatrix} m
[1]

Final answer

(a) 170 m. (b) \frac1{17}(12,-9,8)=(0.706,-0.529,0.471). (c) (353,-265\,235) m.

Question 12

[7 marks]
A triangular roof panel has corners \mathrm{A}(2,1,4), \mathrm{B}(6,4,4) and \mathrm{C}(2,7,12), measured in metres.
(a) Find the angles \mathrm{B\hat{A}C} and \mathrm{A\hat{B}C}, to 1 decimal place.
[4]
(b) Hence find the angle \mathrm{A\hat{C}B}.
[1]
(c) Check your answer to part (b) by calculating \mathrm{A\hat{C}B} directly from a scalar product.
[2]
Show solutionHide solution
(a) \overrightarrow{\mathrm{AB}} = \begin{pmatrix}4\\3\\0\end{pmatrix}, \overrightarrow{\mathrm{AC}} = \begin{pmatrix}0\\6\\8\end{pmatrix}
[1]
\cos\mathrm{B\hat{A}C} = \dfrac{0 + 18 + 0}{5 \times 10} = 0.36 \Rightarrow \mathrm{B\hat{A}C} = 68.9^\circ [1]
[1]
At \mathrm{B}: \overrightarrow{\mathrm{BA}} = \begin{pmatrix}-4\\-3\\0\end{pmatrix}, \overrightarrow{\mathrm{BC}} = \begin{pmatrix}-4\\3\\8\end{pmatrix}, so \cos\mathrm{A\hat{B}C} = \dfrac{16 - 9}{5\sqrt{89}} = 0.1484 and \mathrm{A\hat{B}C} = 81.5^\circ
[1]
(b) Angles of a triangle sum to 180^\circ: \mathrm{A\hat{C}B} = 180 - 68.9 - 81.5 = 29.6^\circ
[1]
(c) \overrightarrow{\mathrm{CA}} = \begin{pmatrix}0\\-6\\-8\end{pmatrix}, \overrightarrow{\mathrm{CB}} = \begin{pmatrix}4\\-3\\-8\end{pmatrix}, so \overrightarrow{\mathrm{CA}}\cdot\overrightarrow{\mathrm{CB}} = 18 + 64 = 82
[1]
\cos\mathrm{A\hat{C}B} = \dfrac{82}{10\sqrt{89}} = 0.8692, so \mathrm{A\hat{C}B} = 29.6^\circ, which agrees.
[1]

Final answer

(a) 68.9^\circ and 81.5^\circ. (b) 29.6^\circ. (c) 29.6^\circ, in agreement.

Question 13

[9 marks]
A flat triangular solar roof panel has corners \mathrm{A}(0,0,3), \mathrm{B}(6,0,3) and \mathrm{C}(2,5,6), in metres, with z the height above the ground.
(a) Find \overrightarrow{\mathrm{AB}} and \overrightarrow{\mathrm{AC}}.
[1]
(b) Find \overrightarrow{\mathrm{AB}}\times\overrightarrow{\mathrm{AC}}.
[2]
(c) Find the area of the panel, to 3 significant figures.
[2]
(d) Find the unit vector perpendicular to the panel that points upwards (positive z-component).
[2]
(e) An aerial of length 4.5 m is fixed at \mathrm{A} perpendicular to the panel and pointing upwards. Find the coordinates of its top, to 3 significant figures.
[2]
Show solutionHide solution
(a) \overrightarrow{\mathrm{AB}} = \begin{pmatrix}6\\0\\0\end{pmatrix}, \overrightarrow{\mathrm{AC}} = \begin{pmatrix}2\\5\\3\end{pmatrix}
[1]
(b) \overrightarrow{\mathrm{AB}}\times\overrightarrow{\mathrm{AC}} = \begin{pmatrix}0(3)-0(5)\\0(2)-6(3)\\6(5)-0(2)\end{pmatrix}
[1]
= \begin{pmatrix}0\\-18\\30\end{pmatrix}
[1]
(c) Area = \tfrac12\sqrt{0 + 324 + 900} = \tfrac12\sqrt{1224}
[1]
= 17.5 m^2
[1]
(d) The cross product is perpendicular to the panel and has z-component 30 > 0, so it points upwards. Its magnitude is \sqrt{1224}.
[1]
\hat{\mathbf{n}} = \dfrac{1}{\sqrt{1224}}\begin{pmatrix}0\\-18\\30\end{pmatrix} = \begin{pmatrix}0\\-0.5145\\0.8575\end{pmatrix}
[1]
(e) Top of aerial: \overrightarrow{\mathrm{OA}} + 4.5\,\hat{\mathbf{n}} = \begin{pmatrix}0\\0\\3\end{pmatrix} + \begin{pmatrix}0\\-2.3152\\3.8587\end{pmatrix}
[1]
= \begin{pmatrix}0\\-2.3152\\6.8587\end{pmatrix}, so the top is at (0,\ -2.32,\ 6.86).
[1]

Final answer

(a) (6,0,0), (2,5,3). (b) (0,-18,30). (c) 17.5 m^2. (d) (0,-0.514,0.857). (e) (0,-2.32,6.86).

Question 14

[6 marks]
A delivery drone flies along the line \mathbf{r} = \begin{pmatrix}12\\-8\\25\end{pmatrix} + \lambda\begin{pmatrix}4\\5\\-1\end{pmatrix}, with distances in metres.
(a) Show that the drone passes through the beacon \mathrm{B}(36,22,19).
[2]
(b) Show that the drone does not pass through the beacon \mathrm{C}(20,2,22).
[2]
(c) The point \mathrm{D}(k,17,20) lies on the line. Find k.
[2]
Show solutionHide solution
(a) x: 12 + 4\lambda = 36 \Rightarrow \lambda = 6
[1]
y: -8 + 30 = 22 ✓, z: 25 - 6 = 19 ✓, so \mathrm{B} is on the line.
[1]
(b) x: 12 + 4\lambda = 20 \Rightarrow \lambda = 2. Then y = -8 + 10 = 2 ✓ but z = 25 - 2 = 23 ≠ 22.
[1]
The same \lambda does not fit every component, so \mathrm{C} is not on the line.
[1]
(c) y: -8 + 5\lambda = 17 \Rightarrow \lambda = 5 (check z: 25 - 5 = 20 ✓)
[1]
k = 12 + 4(5) = 32
[1]

Final answer

(a) \lambda=6 and the other components agree. (b) \lambda=2 gives z=23\ne22. (c) k=32.

Question 15

[7 marks]
In a theatre, a laser at \mathrm{A}(1,2,3) is aimed at the point \mathrm{T}(5,4,2), in metres. The beam continues in a straight line. The floor is at height z = 0 and one wall is the vertical surface x = 9.
(a) Find a vector equation of the line of the beam.
[2]
(b) Find the value of \lambda at which the beam would reach the wall, and the value at which it would reach the floor.
[3]
(c) Hence state which surface the beam hits first, and find the coordinates of that point.
[2]
Show solutionHide solution
(a) \overrightarrow{\mathrm{AT}} = \begin{pmatrix}4\\2\\-1\end{pmatrix}
[1]
\mathbf{r} = \begin{pmatrix}1\\2\\3\end{pmatrix} + \lambda\begin{pmatrix}4\\2\\-1\end{pmatrix}
[1]
(b) Wall: 1 + 4\lambda = 9 \Rightarrow \lambda = 2 [1]
[1]
Floor: 3 - \lambda = 0 \Rightarrow \lambda = 3
[1]
(c) The smaller positive \lambda is reached first, so the beam hits the wall first (at \lambda = 2, before reaching the floor at \lambda = 3).
[1]
Point: \begin{pmatrix}9\\6\\1\end{pmatrix} = \begin{pmatrix}9\\6\\1\end{pmatrix}, that is (9,6,1) (1 m above the floor).
[1]

Final answer

(a) \mathbf r=(1,2,3)+\lambda(4,2,-1). (b) Wall \lambda=2, floor \lambda=3. (c) The wall, at (9,6,1).

Question 16

[7 marks]
Two underground pipes are modelled by the lines L_1: \mathbf{r} = \begin{pmatrix}2\\1\\0\end{pmatrix} + \lambda\begin{pmatrix}1\\2\\1\end{pmatrix} and L_2: \mathbf{r} = \begin{pmatrix}0\\7\\-4\end{pmatrix} + \mu\begin{pmatrix}2\\-1\\3\end{pmatrix}, in metres.
(a) Show that the pipes meet, and find the coordinates of the meeting point \mathrm{M}.
[4]
(b) Find the acute angle between the pipes, to the nearest degree.
[3]
Show solutionHide solution
(a) x: 2 + \lambda = 2\mu and y: 1 + 2\lambda = 7 - \mu
[1]
Solving: \lambda = 2\mu - 2, so 1 + 4\mu - 4 = 7 - \mu \Rightarrow \mu = 2, \lambda = 2
[1]
z: L_1 gives 0 + 2 = 2, L_2 gives -4 + 3(2) = 2 ✓, so the pipes meet.
[1]
\mathrm{M} = (4,\ 5,\ 2)
[1]
(b) \mathbf{d}_1\cdot\mathbf{d}_2 = 2 - 2 + 3 = 3, |\mathbf{d}_1| = \sqrt6, |\mathbf{d}_2| = \sqrt{14}
[1]
\cos\theta = \dfrac{3}{\sqrt{84}} = 0.3273
[1]
\theta = 70.9^\circ \approx 71^\circ
[1]

Final answer

(a) M(4,5,2). (b) 71^\circ.

Question 17

[8 marks]
Two cyclists ride along straight paths in a large park. Relative to the park gate, cyclist \mathrm{A} has position \mathbf{r}_A = \begin{pmatrix}2\\1\end{pmatrix} + t\begin{pmatrix}9\\12\end{pmatrix} and cyclist \mathrm{B} has position \mathbf{r}_B = \begin{pmatrix}14\\-4\end{pmatrix} + t\begin{pmatrix}-3\\16\end{pmatrix}, where distances are in kilometres and t is the time in hours after 14:00.
(a) Find the vector \mathbf{D}(t) from \mathrm{A} to \mathrm{B} at time t.
[2]
(b) Find the distance between the cyclists at 14:30, to 3 significant figures.
[2]
(c) The cyclists can see each other when they are within 10 km of one another. Find, to 3 significant figures, the times t at which they are exactly 10 km apart.
[4]
Show solutionHide solution
(a) \mathbf{D}(t) = \mathbf{r}_B - \mathbf{r}_A
[1]
= \begin{pmatrix}12-12t\\-5+4t\end{pmatrix}
[1]
(b) t = 0.5: \mathbf{D} = \begin{pmatrix}6\\-3\end{pmatrix}
[1]
|\mathbf{D}| = \sqrt{36 + 9} = 6.71 km
[1]
(c) |\mathbf{D}|^2 = (12 - 12t)^2 + (4t - 5)^2 = 160t^2 - 328t + 169
[1]
Set equal to 10^2: 160t^2 - 328t + 69 = 0
[1]
GDC solver (or quadratic formula):
[1]
t = 0.238 or t = 1.81
[1]

Final answer

(a) \mathbf D(t)=(12-12t,\,-5+4t). (b) 6.71 km. (c) t=0.238 and t=1.81.

Question 18

[6 marks]
A straight road is modelled by the line \mathbf{r} = \begin{pmatrix}0\\3\end{pmatrix} + \lambda\begin{pmatrix}4\\3\end{pmatrix} and a hikers' camp is at \mathrm{C}(10,9), with distances in kilometres. The hikers want to reach the road by the shortest path.
(a) Find the coordinates of the point \mathrm{F} on the road that is nearest to the camp.
[4]
(b) Find the shortest distance from the camp to the road.
[2]
Show solutionHide solution
(a) A general point on the road is \mathrm{Q} = (4\lambda, 3 + 3\lambda), so \overrightarrow{\mathrm{CQ}} = \begin{pmatrix}4\lambda-10\\3\lambda-6\end{pmatrix}
[1]
The shortest path is perpendicular to the road: \overrightarrow{\mathrm{CQ}}\cdot\begin{pmatrix}4\\3\end{pmatrix} = 0
[1]
4(4\lambda - 10) + 3(3\lambda - 6) = 0 \Rightarrow 25\lambda = 58 \Rightarrow \lambda = 2.32
[1]
\mathrm{F} = (9.28, 9.96)
[1]
(b) \overrightarrow{\mathrm{CF}} = \begin{pmatrix}-0.72\\0.96\end{pmatrix}
[1]
\mathrm{CF} = \sqrt{0.72^2 + 0.96^2} = 1.2 km
[1]

Final answer

(a) F(9.28,\,9.96). (b) 1.2 km.

Question 19

[6 marks]
A ship leaves a port \mathrm{P} (the origin) at 06:00 and sails at 18 km h^{-1} on a bearing of 135^\circ. A lighthouse \mathrm{L} stands at (20,-10), with distances in kilometres east and north of the port.
(a) Find the velocity of the ship, to 3 significant figures.
[2]
(b) Show that the lighthouse is not on the ship's path.
[2]
(c) Find the shortest distance between the lighthouse and the ship's path, to 3 significant figures.
[2]
Show solutionHide solution
(a) \mathbf{v} = \begin{pmatrix}18\sin 135^\circ\\18\cos 135^\circ\end{pmatrix}
[1]
= \begin{pmatrix}12.73\\-12.73\end{pmatrix} km h^{-1}
[1]
(b) The path is \mathbf{r} = t\begin{pmatrix}1\\-1\end{pmatrix} (direction proportional to \mathbf{v})
[1]
x = 20 gives y = -20 ≠ -10, so \mathrm{L} is not on the path.
[1]
(c) For \mathrm{Q} = (s, -s) on the path, \overrightarrow{\mathrm{LQ}} = \begin{pmatrix}s-20\\-s+10\end{pmatrix}; perpendicular to \begin{pmatrix}1\\-1\end{pmatrix}: (s - 20) + (s - 10) = 0 \Rightarrow s = 15
[1]
\overrightarrow{\mathrm{LQ}} = \begin{pmatrix}-5\\-5\end{pmatrix}, so the distance is \sqrt{50} = 7.07 km.
[1]

Final answer

(a) (12.7,\,-12.7) km h^{-1}. (b) x=20 gives y=-20\ne-10. (c) 7.07 km.

Question 20

[8 marks]
Two paragliders \mathrm{P} and \mathrm{Q} fly with constant velocities. Relative to a take-off marker at the origin, in metres, with time t in seconds, \mathbf{r}_P = \begin{pmatrix}10\\-20\\40\end{pmatrix} + t\begin{pmatrix}5\\4\\-1\end{pmatrix} and \mathbf{r}_Q = \begin{pmatrix}-30\\20\\10\end{pmatrix} + t\begin{pmatrix}8\\-2\\2\end{pmatrix}, where the third component is height.
(a) Find the vector \overrightarrow{\mathrm{PQ}} from \mathrm{P} to \mathrm{Q} at time t.
[2]
(b) Find the time at which the two paragliders are at the same height, and the horizontal distance between them then, to 3 significant figures.
[3]
(c) Use your GDC to find the least distance between the paragliders and the time at which it occurs, to 3 significant figures.
[3]
Show solutionHide solution
(a) \overrightarrow{\mathrm{PQ}} = \mathbf{r}_Q - \mathbf{r}_P
[1]
= \begin{pmatrix}-40+3t\\40-6t\\-30+3t\end{pmatrix}
[1]
(b) Same height: the third component is zero, -30 + 3t = 0 \Rightarrow t = 10 s [1]
[1]
\overrightarrow{\mathrm{PQ}}(10) = \begin{pmatrix}-10\\-20\\0\end{pmatrix}, so the horizontal distance is \sqrt{(-10)^2 + (-20)^2} = \sqrt{500} = 22.4 m
[1]
(c) |\overrightarrow{\mathrm{PQ}}|^2 = (3t - 40)^2 + (40 - 6t)^2 + (3t - 30)^2 = 54t^2 - 900t + 4100; graph on the GDC and find the minimum
[1]
t = \dfrac{900}{108} = 8.33 s
[1]
Least |\overrightarrow{\mathrm{PQ}}|^2 = 350, so the least distance is \sqrt{350} = 18.7 m
[1]

Final answer

(a) (-40+3t,\,40-6t,\,-30+3t). (b) t=10 s; 22.4 m. (c) 18.7 m at t=8.33 s.

Mastery

Full exam questions that mix question types, at Paper 1 and Paper 2 standard.

Question 21

[10 marks]
A patrol boat starts at the point (-3,4) and moves with constant velocity \begin{pmatrix}2\\-1\end{pmatrix} (km per hour). Its position after t hours is \mathbf{r}(t)=\begin{pmatrix}-3\\4\end{pmatrix}+t\begin{pmatrix}2\\-1\end{pmatrix}. A lighthouse is at the origin O.
(a) Find the position of the boat after 2 hours.
[2]
(b) Find the speed of the boat.
[2]
(c) Show that the square of the distance from O at time t is D^{2}=5t^{2}-20t+25.
[3]
(d) Hence find the time at which the boat is closest to the lighthouse, and the least distance.
[3]
Show solutionHide solution
(a) \mathbf{r}(2)=\begin{pmatrix}-3\\4\end{pmatrix}+2\begin{pmatrix}2\\-1\end{pmatrix}=\begin{pmatrix}1\\2\end{pmatrix}.
[2]
(b) Speed =\sqrt{2^{2}+(-1)^{2}}=\sqrt5=2.24 km h^{-1}.
[2]
(c) Position =(-3+2t,\;4-t), so
D^{2}=(-3+2t)^{2}+(4-t)^{2}=9-12t+4t^{2}+16-8t+t^{2}=5t^{2}-20t+25.
[3]
(d) Minimum of D^{2} at t=-\dfrac{-20}{2(5)}=2 hours.
D^{2}=5(4)-20(2)+25=5, so least distance =\sqrt5=2.24 km.
[3]

Final answer

(a) (1,2). (b) 2.24 km h^{-1}. (c) Shown. (d) t=2 h; 2.24 km.

Question 22

[12 marks]
Two ships move with constant velocity. At time t=0 hours, ship A is at (0,10) and ship B is at (12,0), with distances in km. Their velocities are \mathbf{v}_A=\begin{pmatrix}3\\1\end{pmatrix} and \mathbf{v}_B=\begin{pmatrix}-1\\3\end{pmatrix} km h^{-1}.
(a) Write down the position vector of each ship at time t.
[2]
(b) Show that the displacement of A relative to B is \begin{pmatrix}-12\\10\end{pmatrix}+t\begin{pmatrix}4\\-2\end{pmatrix}.
[3]
(c) Show that the square of the distance between the ships is S^{2}=20t^{2}-136t+244.
[3]
(d) Find the time at which the ships are closest together, and the minimum distance between them, giving your answer to 3 significant figures.
[4]
Show solutionHide solution
(a) \mathbf{r}_A=\begin{pmatrix}0\\10\end{pmatrix}+t\begin{pmatrix}3\\1\end{pmatrix}, \quad\mathbf{r}_B=\begin{pmatrix}12\\0\end{pmatrix}+t\begin{pmatrix}-1\\3\end{pmatrix}.
[2]
(b) \mathbf{r}_A-\mathbf{r}_B=\begin{pmatrix}0-12\\10-0\end{pmatrix}+t\begin{pmatrix}3-(-1)\\1-3\end{pmatrix}=\begin{pmatrix}-12\\10\end{pmatrix}+t\begin{pmatrix}4\\-2\end{pmatrix}.
[3]
(c) Components: (-12+4t,\;10-2t).
S^{2}=(-12+4t)^{2}+(10-2t)^{2}=144-96t+16t^{2}+100-40t+4t^{2}=20t^{2}-136t+244.
[3]
(d) Minimum at t=\dfrac{136}{2(20)}=3.4 hours.
S^{2}=20(3.4)^{2}-136(3.4)+244=12.8, so S=\sqrt{12.8}=3.58 km.
[4]

Final answer

(a), (b), (c) Shown. (d) t=3.4 h; 3.58 km.

Question 23

[8 marks]
Three telecommunication towers are placed at A(1,0,2), B(3,1,0) and C(2,-1,3), with coordinates in kilometres. Engineers wish to find the area of the flat triangular land parcel ABC.
(a) Find the vectors \overrightarrow{AB} and \overrightarrow{AC}.
[2]
(b) Find the vector (cross) product \overrightarrow{AB}\times\overrightarrow{AC}.
[3]
(c) Hence find the area of triangle ABC.
[3]
Show solutionHide solution
(a) \overrightarrow{AB}=B-A=\begin{pmatrix}2\\1\\-2\end{pmatrix}, \quad\overrightarrow{AC}=C-A=\begin{pmatrix}1\\-1\\1\end{pmatrix}.
[2]
(b) \overrightarrow{AB}\times\overrightarrow{AC}=\begin{pmatrix}(1)(1)-(-2)(-1)\\(-2)(1)-(2)(1)\\(2)(-1)-(1)(1)\end{pmatrix}=\begin{pmatrix}-1\\-4\\-3\end{pmatrix}.
[3]
(c) |\overrightarrow{AB}\times\overrightarrow{AC}|=\sqrt{(-1)^{2}+(-4)^{2}+(-3)^{2}}=\sqrt{26}.
Area =\tfrac12\sqrt{26}=2.55 km^{2}.
[3]

Final answer

(a) (2,1,-2), (1,-1,1). (b) (-1,-4,-3). (c) 2.55 km^2.

Question 24

[16 marks]
At an air show, a searchlight beam leaves a lamp at \mathrm{A}(0,0,6) in the direction \begin{pmatrix}2\\1\\-1\end{pmatrix} and a second beam leaves a lamp at \mathrm{B}(12,-2,0) in the direction \begin{pmatrix}-2\\3\\1\end{pmatrix}, with coordinates in metres.
(a) Write down vector equations L_1 and L_2 for the two beams.
[2]
(b) Show that the beams meet, and find the coordinates of the meeting point \mathrm{X}.
[4]
(c) Find the acute angle between the beams, to the nearest degree.
[3]
(d) Beam L_2 continues until it hits the wall y = 10. Find the coordinates of the point on the wall that it hits.
[3]
(e) The lamps and the point \mathrm{X} form a triangle \mathrm{ABX}. Find the area of this triangle, to 3 significant figures.
[4]
Show solutionHide solution
(a) L_1: \mathbf{r} = \begin{pmatrix}0\\0\\6\end{pmatrix} + \lambda\begin{pmatrix}2\\1\\-1\end{pmatrix}
[1]
L_2: \mathbf{r} = \begin{pmatrix}12\\-2\\0\end{pmatrix} + \mu\begin{pmatrix}-2\\3\\1\end{pmatrix}
[1]
(b) x: 2\lambda = 12 - 2\mu and y: \lambda = -2 + 3\mu
[1]
Substituting: 2(-2 + 3\mu) = 12 - 2\mu \Rightarrow 8\mu = 16 \Rightarrow \mu = 2, \lambda = 4
[1]
z: L_1 gives 6 - 4 = 2 and L_2 gives 0 + 2 = 2 ✓, so the beams meet.
[1]
\mathrm{X} = (8,\ 4,\ 2)
[1]
(c) \mathbf{d}_1\cdot\mathbf{d}_2 = -4 + 3 - 1 = -2, |\mathbf{d}_1| = \sqrt6, |\mathbf{d}_2| = \sqrt{14}
[1]
\cos\theta = \dfrac{|-2|}{\sqrt{84}} = 0.2182
[1]
\theta = 77.4^\circ \approx 77^\circ (the acute angle; the angle between the direction vectors is 103^\circ)
[1]
(d) y: -2 + 3\mu = 10 \Rightarrow \mu = 4 [1]
[1]
Point: \begin{pmatrix}4\\10\\4\end{pmatrix}, that is (4, 10, 4).
[1]
(e) \overrightarrow{\mathrm{AB}} = \begin{pmatrix}12\\-2\\-6\end{pmatrix}, \overrightarrow{\mathrm{AX}} = \begin{pmatrix}8\\4\\-4\end{pmatrix}
[1]
\overrightarrow{\mathrm{AB}}\times\overrightarrow{\mathrm{AX}} = \begin{pmatrix}32\\0\\64\end{pmatrix} [1]
[1]
Area = \tfrac12\sqrt{32^2 + 0 + 64^2} = \tfrac12\sqrt{5120} = 35.8 m^2
[1]

Final answer

(a) Lines written. (b) X(8,4,2). (c) 77^\circ. (d) (4,10,4). (e) 35.8 m^2.

Question 25

[14 marks]
A rescue helicopter flies in a straight line at constant velocity. Relative to a hiker's position on a ridge, in hundreds of metres, and with t in minutes after take-off, the helicopter is at \mathbf{r} = \begin{pmatrix}-60\\-45\\15\end{pmatrix} + t\begin{pmatrix}12\\9\\-1\end{pmatrix}, where the third coordinate is height above sea level. The hiker is at \mathrm{H}(0,0,3).
(a) Find the speed of the helicopter in m s^{-1}, to 3 significant figures.
[3]
(b) Find the position of the helicopter 3 minutes after take-off, and its height above sea level in metres.
[2]
(c) Show that the square of the distance from the helicopter to the hiker is 226t^2 - 2274t + 5769 (in units of hundreds of metres squared).
[3]
(d) Find the least distance between the helicopter and the hiker, in metres to 3 significant figures.
[3]
(e) The helicopter's winch cable is 600 m long. Determine whether the hiker can be reached at any time during the flight.
[3]
Show solutionHide solution
(a) Speed in hundreds of metres per minute: \sqrt{12^2 + 9^2 + 1^2} = \sqrt{226} = 15.03 [1]
[1]
15.03 \times 100 \text{ m} \div 60 \text{ s} = 25.1 m s^{-1}
[1]
(b) \mathbf{r}(3) = \begin{pmatrix}-24\\-18\\12\end{pmatrix}
[1]
Height = 12 \times 100 = 1200 m.
[1]
(c) \overrightarrow{\mathrm{HR}} = \mathbf{r} - \mathbf{h} = \begin{pmatrix}-60+12t\\-45+9t\\12-t\end{pmatrix}
[1]
|\overrightarrow{\mathrm{HR}}|^2 = (12t - 60)^2 + (9t - 45)^2 + (12 - t)^2 = 144t^2 - 1440t + 3600 + 81t^2 - 810t + 2025 + 144 - 24t + t^2
[1]
= 226t^2 - 2274t + 5769
AG[1]
(d) GDC minimum: t = \dfrac{2274}{452} = 5.031 min [1]
[1]
Least |\overrightarrow{\mathrm{HR}}|^2 = 48.78, so the least distance is 6.98 hundred metres = 698 m
[1]
(e) The least distance is 698 m, which is greater than 600 m [1]
[1]
so the winch can never reach the hiker (the helicopter would need to change course).
[1]

Final answer

(a) 25.1 m s^{-1}. (b) (-24,-18,12); 1200 m. (c) Shown. (d) 698 m. (e) No: 698 m is more than 600 m.

Question 26

[18 marks]
In this question x is east, y is north and z is vertically up. Distances are in km and t is the time in hours after 08:00. A survey drone D flies with position \mathbf{r}_D=\begin{pmatrix}-4\\2\\0.1\end{pmatrix}+t\begin{pmatrix}24\\18\\1.2\end{pmatrix}.
(a) Find the speed of D.
[2]
(b) Find the bearing on which D flies.
[2]
(c) Find the angle between the path of D and the horizontal.
[2]
(d) Find the time at which D reaches a height of 0.7 km, and its position then.
[2]
A second drone E flies at a constant height. At 08:00 it is at (17,-1,0.7) and it flies at 25 km h^{-1} in the direction of \begin{pmatrix}-3\\4\\0\end{pmatrix}.
(e) Find the velocity of E.
[2]
(f) Show that the flight paths of D and E cross, and determine whether the drones collide.
[4]
(g) Find the least distance between the drones and the time at which it occurs.
[3]
(h) The drones must stay at least 2 km apart. Comment on this rule.
[1]
Show solutionHide solution
(a) \sqrt{24^2+18^2+1.2^2}=\sqrt{901.44}.
M1
=30.0 km h^{-1}.
A1
(b) \tan^{-1}\dfrac{24}{18}.
M1
Bearing 053.1^\circ.
A1
(c) \sin^{-1}\dfrac{1.2}{\sqrt{901.44}}, or the angle between the path and (24,18,0).
M1
2.29^\circ.
A1
(d) 0.1+1.2t=0.7, so t=0.5, which is 08:30.
M1
Position (8,11,0.7).
A1
(e) \dfrac{25}{5}\begin{pmatrix}-3\\4\\0\end{pmatrix}=\begin{pmatrix}-15\\20\\0\end{pmatrix} km h^{-1}.
M1A1
(f) The path of E is (17-15\mu,\,-1+20\mu,\,0.7). At \mu=0.6 it is at (8,11,0.7).
M1A1
This is the point D reaches at t=0.5, so the paths cross.
R1
D is there at 08:30 and E at 08:36, so the drones do not collide.
R1
(g) \mathbf{r}_E-\mathbf{r}_D=\begin{pmatrix}21-39t\\-3+2t\\0.6-1.2t\end{pmatrix}.
M1
Least value of its length at t=0.541 h, which is 08:32.
A1
Least distance 1.92 km.
A1
(h) 1.92<2, so the rule is broken: one drone must change course or height.
R1

Final answer

(a) 30.0 km h^{-1}. (b) 053.1^\circ. (c) 2.29^\circ. (d) 08:30; (8,11,0.7). (e) (-15,20,0) km h^{-1}. (f) The paths cross at (8,11,0.7); D is there at 08:30, E at 08:36: no collision. (g) 1.92 km at 08:32. (h) 1.92<2: the rule is broken.

Going further

Beyond the syllabus: the worked examples and investigations of the chapter's Going further pages.

G1 Closest approach without a graph

G1 Worked

[6 marks]
Two boats have relative position \mathbf{D}(t)=\begin{pmatrix}30-30t\\-5+10t\end{pmatrix} km, where t is in hours after 12:00.
(a) Write down \mathbf{p} and \mathbf{w}, where \mathbf{D}(t)=\mathbf{p}+t\mathbf{w}.
[1]
(b) Show that |\mathbf{D}|^2=1000t^2-1900t+925.
[2]
(c) Use the vertex of this quadratic to find the time of closest approach and the least distance.
[3]
Show solutionHide solution
(a) \mathbf{p}=(30,-5) and \mathbf{w}=(-30,10) (A1). (b) |\mathbf{D}|^2=(30-30t)^2+(-5+10t)^2 (M1) =900-1800t+900t^2+25-100t+100t^2 (A1) =1000t^2-1900t+925 (AG). (c) \mathbf{p}\cdot\mathbf{w}=-950 and |\mathbf{w}|^2=1000 (M1), so t^*=0.95, which is 12:57 (A1), and d^2=925-\frac{950^2}{1000}=22.5, so d=4.74 km (A1).

Final answer

\mathbf{p}=(30,-5), \mathbf{w}=(-30,10); shown; t=0.95 (12:57), 4.74 km.

G2 The distance from a point to a line by the vector product

G2 Worked

[6 marks]
A buried cable follows the line \mathbf{r}=\begin{pmatrix}2\\-1\\4\end{pmatrix}+\lambda\begin{pmatrix}1\\2\\2\end{pmatrix}, in metres. A manhole cover is at Q(6,3,1).
(a) Find \overrightarrow{AQ}\times\mathbf{b}, where A(2,-1,4) is on the cable and \mathbf{b}=(1,2,2).
[2]
(b) Hence find the shortest distance from the manhole to the cable.
[2]
(c) Check your answer by finding the foot X of the perpendicular from Q and the length QX.
[2]
Show solutionHide solution
(a) \overrightarrow{AQ}=(4,4,-3) (M1) and \overrightarrow{AQ}\times\mathbf{b}=(14,-11,4) (A1). (b) d=\frac{\sqrt{196+121+16}}{\sqrt{9}}=\frac{\sqrt{333}}{3} (M1) =6.08 m (A1). (c) \overrightarrow{QX}=(-4+\lambda,-4+2\lambda,3+2\lambda) and \overrightarrow{QX}\cdot\mathbf{b}=-6+9\lambda=0, so \lambda=\frac23 (M1). Then \overrightarrow{QX}=(-\frac{10}{3},-\frac83,\frac{13}{3}) and QX=\sqrt{37}=6.08 m (A1).

Final answer

(14,-11,4); 6.08 m; X=(\frac83,\frac13,\frac{16}{3}) and QX=6.08 m.

G3 Investigation

[20 marks]
A yacht has lost power and drifts with constant velocity. Take x east and y north, with distances in km and t in hours after the alarm sounds. The yacht is at \mathbf{p}=(12,9) when the alarm sounds and drifts with velocity \mathbf{w}=(3,-4) km h^{-1}. A lifeboat leaves the origin at the moment of the alarm and travels in a straight line at a constant 25 km h^{-1}.
(a) Write down the position of the yacht at time t.
[1]
(b) The lifeboat meets the yacht at time T. Explain why |\mathbf{p}+T\mathbf{w}|=25T.
[2]
(c) Show that 600T^2=225, and hence find T in minutes.
[3]
(d) Find the meeting point and the bearing on which the lifeboat must steer.
[3]
(e) For a target starting at \mathbf{p} with velocity \mathbf{w} and a chaser of speed s leaving the origin, show that (|\mathbf{w}|^2-s^2)T^2+2(\mathbf{p}\cdot\mathbf{w})T+|\mathbf{p}|^2=0.
[3]
(f) Explain why an interception is always possible when s>|\mathbf{w}|.
[3]
(g) A small inflatable with top speed 4 km h^{-1} also leaves the origin. Show that it cannot reach the yacht.
[2]
(h) A second yacht starts at (12,9) and drifts with velocity (-3,-4) km h^{-1}. Find the earliest time at which the inflatable, at 4 km h^{-1}, can reach it.
[3]
Show solutionHide solution
(a) (12+3t,\,9-4t) (A1). (b) The lifeboat travels in a straight line from the origin at 25 km h^{-1}, so after time T it is 25T km from the origin (R1). At the meeting its position is the yacht's position \mathbf{p}+T\mathbf{w}, so the distance of that point from the origin is 25T (R1). (c) (12+3T)^2+(9-4T)^2=625T^2 (M1), which simplifies to 225+25T^2=625T^2 (A1), so 600T^2=225 and T=0.612 h, which is 36.7 minutes (A1). (d) Meeting point (12+3T,\,9-4T)=(13.8,\,6.55) (M1 A1); the bearing is \tan^{-1}\frac{13.84}{6.55}=064.7^\circ (A1). (e) Square |\mathbf{p}+T\mathbf{w}|=sT to get |\mathbf{p}|^2+2T\,\mathbf{p}\cdot\mathbf{w}+T^2|\mathbf{w}|^2=s^2T^2 (M1 A1), and rearrange to the printed form (A1). (f) If s>|\mathbf{w}| the coefficient of T^2 is negative (R1) and the constant term |\mathbf{p}|^2 is positive (R1), so the product of the roots is negative: exactly one root is positive, which is the interception time (R1). (g) Here \mathbf{p}\cdot\mathbf{w}=0 and s=4<5=|\mathbf{w}|, so the equation is 9T^2+225=0 (M1), which has no real root, so there is no interception (R1). (h) Now \mathbf{p}\cdot\mathbf{w}=-72 and |\mathbf{w}|=5, so 9T^2-144T+225=0 (M1), with roots T=8\pm\sqrt{39} (A1). The earliest is T=8-\sqrt{39}=1.76 h, which is 1 h 45 min (A1).

Final answer

(12+3t,\,9-4t); explained; T=0.612 h (36.7 minutes); (13.8,\,6.55) and 064.7^\circ; shown; explained; shown; T=1.76 h (1 h 45 min).

From the book

Photon Guidebook: Mathematics Applications and Interpretation HL for the IB Diploma

Chapter 10 teaches vectors from the basics to full exam questions: how the IB asks it, every question type with a marked example, and these practice questions in three steps.

About the book

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