Free worked solutions · 2026 Edition

Chapter 11: Vectors worked solutions

Worked solutions to the practice questions in Chapter 11 of The Ultimate Guide for the IB: Mathematics: Analysis and Approaches HL (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

[8 marks]

The points A(1, -2, 3), B(4, 0, 2) and C(10, 4, 0) are given.

(a) Show that A, B and C are collinear.

[3]

(b) Find the ratio AB : BC.

[2]

(c) The point D(p, 8, q) is also collinear with A and B. Find the value of p and the value of q.

[3]

Show solutionHide solution

(a) \vec{AB} = \begin{pmatrix} 3 \\ 2 \\ -1 \end{pmatrix}, \vec{AC} = \begin{pmatrix} 9 \\ 6 \\ -3 \end{pmatrix}

\vec{AC} = 3\vec{AB}, so \vec{AB} and \vec{AC} are parallel

and they share the point A, so A, B and C lie on one straight line

Note: the R1 needs the common point; parallel vectors alone do not give collinear points.

(b) \vec{BC} = \begin{pmatrix} 6 \\ 4 \\ -2 \end{pmatrix} = 2\vec{AB} (M1)

AB : BC = 1 : 2

(c) \vec{AD} = s\vec{AB}: \begin{pmatrix} p - 1 \\ 10 \\ q - 3 \end{pmatrix} = s\begin{pmatrix} 3 \\ 2 \\ -1 \end{pmatrix}

y-components: 10 = 2s \Rightarrow s = 5

p = 16, q = -2

Final answer

(a) \vec{AC}=3\vec{AB}, shared point A
(b) 1:2
(c) p=16, q=-2

Question 2

[6 marks]

The points A(2, 1, -1), B(4, -1, 0) and C(5, -5, -1) are the vertices of a triangle.

(a) Find \vec{BA} and \vec{BC}.

[2]

(b) Find the exact size of the angle A\hat{B}C.

[3]

(c) A student uses \vec{AB} and \vec{BC} instead and obtains \frac{\pi}{4}. Explain why this is not the size of A\hat{B}C.

[1]

Show solutionHide solution

(a) \vec{BA} = \begin{pmatrix} -2 \\ 2 \\ -1 \end{pmatrix}

\vec{BC} = \begin{pmatrix} 1 \\ -4 \\ -1 \end{pmatrix}

(b) \vec{BA} \cdot \vec{BC} = -2 - 8 + 1 = -9

\cos A\hat{B}C = \dfrac{-9}{3 \times 3\sqrt{2}} = -\dfrac{1}{\sqrt{2}}

A\hat{B}C = \dfrac{3\pi}{4}

(c) \vec{AB} points towards B, not away from it; \vec{AB} = -\vec{BA}, so this gives the supplement \pi - A\hat{B}C

Note: for the angle at a vertex both vectors must point away from that vertex.

Final answer

(a) \vec{BA}=(-2,2,-1), \vec{BC}=(1,-4,-1)
(b) \frac{3\pi}{4}
(c) the vectors do not both start at B; the answer is the supplement

Question 3

[5 marks]

The line L has Cartesian equation \dfrac{x - 3}{4} = \dfrac{y + 2}{-1} = \dfrac{z - 1}{2}.

(a) The point P(p, 1, q) lies on L. Find the value of p and the value of q.

[3]

(b) Find the coordinates of the point on L whose x-coordinate is equal to its z-coordinate.

[2]

Show solutionHide solution

(a) \dfrac{1 + 2}{-1} = -3, so each fraction equals -3

p = 3 + 4(-3) = -9

q = 1 + 2(-3) = -5

(b) x = 3 + 4\lambda, z = 1 + 2\lambda, so 3 + 4\lambda = 1 + 2\lambda, \lambda = -1

the point (-1, -1, -1)

Final answer

(a) p=-9, q=-5
(b) (-1,-1,-1)

Question 4

[6 marks]

The lines L_1 and L_2 have equations

\begin{gathered}L_1 : \mathbf{r} = \begin{pmatrix} 2 \\ -3 \\ 7 \end{pmatrix} + \lambda \begin{pmatrix} 5 \\ 2 \\ -4 \end{pmatrix},\\ L_2 : \mathbf{r} = \begin{pmatrix} 5 \\ 7 \\ 9 \end{pmatrix} + \mu \begin{pmatrix} -1 \\ 4 \\ 3 \end{pmatrix}.\end{gathered}

Show that L_1 and L_2 intersect, and find the coordinates of the point of intersection.

[6]

Show solutionHide solution

2 + 5\lambda = 5 - \mu, -3 + 2\lambda = 7 + 4\mu, 7 - 4\lambda = 9 + 3\mu

from the first equation \mu = 3 - 5\lambda; substituting into the second: -3 + 2\lambda = 7 + 12 - 20\lambda

\lambda = 1

\mu = -2

third equation: 7 - 4(1) = 3 and 9 + 3(-2) = 3, both sides equal, so the lines intersect

point of intersection (7, -1, 3)

Note: any two equations may be solved first; the R1 requires the remaining equation to be checked with BOTH values.

Final answer

\lambda=1, \mu=-2; (7,-1,3)

Question 5

[6 marks]

The points A(3, 1, -3), B(5, -1, 6) and C(4, -1, -1) are the vertices of a triangle.

(a) Find \vec{AB} and \vec{AC}.

[2]

(b) Find \vec{AB} \times \vec{AC}.

[2]

(c) Hence find the area of triangle ABC.

[2]

Show solutionHide solution

(a) \vec{AB} = \begin{pmatrix} 2 \\ -2 \\ 9 \end{pmatrix}

\vec{AC} = \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix}

(b) \begin{pmatrix} (-2)(2) - 9(-2) \\ 9(1) - 2(2) \\ 2(-2) - (-2)(1) \end{pmatrix}

= \begin{pmatrix} 14 \\ 5 \\ -2 \end{pmatrix}

(c) area = \frac{1}{2}\sqrt{196 + 25 + 4}

= \frac{1}{2}\sqrt{225} = \frac{15}{2}

Note: accept any two sides from one vertex (e.g. \vec{BA} \times \vec{BC}). Do not award the final A1 for 15 (the parallelogram).

Final answer

(a) (2,-2,9), (1,-2,2)
(b) (14,5,-2)
(c) \frac{15}{2}

Question 6

[7 marks]

Find the Cartesian equation of the plane

(a) through P(-3, 2, 4) and parallel to the plane 5x + y - 4z = 2;

[2]

(b) through Q(4, -3, 1) and perpendicular to the line \dfrac{x - 1}{3} = \dfrac{5 - y}{4} = \dfrac{z}{2};

[3]

(c) through the origin and perpendicular to the line with parametric equations x = 3 - t, y = 2t, z = 7.

[2]

Show solutionHide solution

(a) parallel planes share the normal \begin{pmatrix} 5 \\ 1 \\ -4 \end{pmatrix}: 5x + y - 4z = 5(-3) + 2 - 4(4)

5x + y - 4z = -29

(b) \dfrac{5 - y}{4} = \dfrac{y - 5}{-4}, so the direction of the line is \begin{pmatrix} 3 \\ -4 \\ 2 \end{pmatrix}, which is a normal of the plane

3x - 4y + 2z = 3(4) - 4(-3) + 2(1)

3x - 4y + 2z = 26

(c) direction \begin{pmatrix} -1 \\ 2 \\ 0 \end{pmatrix} is normal to the plane, and the plane passes through O

-x + 2y = 0 (or x - 2y = 0)

Note: in (b) a normal read as (3, \ldots) with the sign of the y-component not reversed earns M1 A0 for the equation. In (c) the equation has no z-term; do not award the A1 for -x + 2y + 7z = 0.

Final answer

(a) 5x+y-4z=-29
(b) 3x-4y+2z=26
(c) x-2y=0

Question 7

[7 marks]

The points P(4, -5, 1) and Q(-4, 3, -1) lie on the line L. The plane \Pi has equation 3x + 5y - z = 1.

(a) Find \vec{PQ}.

[1]

(b) Find the acute angle between L and \Pi, giving your answer in degrees.

[4]

(c) Find the coordinates of the point where L meets \Pi.

[2]

Show solutionHide solution

(a) \vec{PQ} = \begin{pmatrix} -8 \\ 8 \\ -2 \end{pmatrix}

(b) \begin{pmatrix} -8 \\ 8 \\ -2 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ 5 \\ -1 \end{pmatrix} = -24 + 40 + 2 = 18

\left| \vec{PQ} \right| = \sqrt{132}, | \mathbf{n} | = \sqrt{35} (M1)

\sin\alpha = \frac{18}{\sqrt{132}\sqrt{35}} = 0.26482\ldots

\alpha = 15.4^{\circ}

(c) 3(4 - 8t) + 5(-5 + 8t) - (1 - 2t) = 1 \Rightarrow -14 + 18t = 1, t = \frac{5}{6}

(-2.67, 1.67, -0.667)

Note: in (b) award M1 for using \sin (or 90^{\circ} - \theta with \cos\theta = 0.265); 74.6^{\circ} is the angle with the normal and earns M1(M1)M0A0. Exact point in (c): \left(-\frac{8}{3}, \frac{5}{3}, -\frac{2}{3}\right).

Final answer

(a) (-8,8,-2)
(b) 15.4^\circ
(c) \left(-\frac83,\frac53,-\frac23\right)

Question 8

[8 marks]

The line L has equation \mathbf{r} = \begin{pmatrix} -3 \\ 2 \\ 6 \end{pmatrix} + \lambda \begin{pmatrix} 4 \\ 7 \\ -4 \end{pmatrix} and P is the point (4, 9, 5).

(a) Show that P does not lie on L.

[2]

(b) Find the coordinates of the foot of the perpendicular from P to L.

[4]

(c) Hence find the shortest distance from P to L.

[2]

Show solutionHide solution

(a) -3 + 4\lambda = 4 gives \lambda = \frac{7}{4}, but 2 + 7\lambda = 9 gives \lambda = 1

the values differ, so P is not on L

(b) \vec{PN} = \begin{pmatrix} -7 + 4\lambda \\ -7 + 7\lambda \\ 1 - 4\lambda \end{pmatrix}

\vec{PN} \cdot \begin{pmatrix} 4 \\ 7 \\ -4 \end{pmatrix} = 4(-7 + 4\lambda) + 7(-7 + 7\lambda) - 4(1 - 4\lambda) = 0

81\lambda - 81 = 0, \lambda = 1

N(1, 9, 2)

(c) \vec{PN} = \begin{pmatrix} -3 \\ 0 \\ -3 \end{pmatrix}

PN = 3\sqrt{2}

Final answer

(a) the \lambda values differ
(b) (1,9,2)
(c) 3\sqrt2

Question 9

[6 marks]

The lines L_1 and L_2 are given by

\begin{gathered}L_1:\ \mathbf{r}=\begin{pmatrix}1\\-2\\3\end{pmatrix}+\lambda\begin{pmatrix}2\\1\\-1\end{pmatrix},\\ L_2:\ \mathbf{r}=\begin{pmatrix}5\\0\\1\end{pmatrix}+\mu\begin{pmatrix}-4\\-2\\2\end{pmatrix}.\end{gathered}

(a) Show that L_1 and L_2 are the same line.

[3]

(b) The plane \Pi has equation

\begin{gathered}\mathbf{r}=\begin{pmatrix}1\\-2\\3\end{pmatrix}+s\begin{pmatrix}2\\1\\-1\end{pmatrix}\\ {}+t\begin{pmatrix}0\\1\\1\end{pmatrix}.\end{gathered}

Determine whether the point D(3,0,3) lies in \Pi.

[3]

Show solutionHide solution

(a) \begin{pmatrix}-4\\-2\\2\end{pmatrix}=-2\begin{pmatrix}2\\1\\-1\end{pmatrix}, so the direction vectors are parallel.

The point (5,0,1) is on L_1 when \lambda=2: (1+4,\,-2+2,\,3-2)=(5,0,1).

The lines are parallel and share a point, so they coincide.

(b) \vec{AD}=(3,0,3)-(1,-2,3)=(2,2,0), so solve (2,2,0)=s(2,1,-1)+t(0,1,1).

x: 2=2s gives s=1. z: 0=-s+t gives t=1.

y: s+t=2, which matches the y-component, so D lies in \Pi.

Final answer

(a) parallel directions and a shared point
(b) s=t=1, so D lies in \Pi

Tutorial

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

Question 10

[6 marks]

The points A, B and C have coordinates A(2,\,-1,\,3), B(5,\,k,\,9) and C(11,\,8,\,m), where k, m \in \mathbb{R}.

(a) Given that A, B and C are collinear, find the value of k and the value of m.

[4]

(b) The point D is such that A is the midpoint of [DB]. Find the coordinates of D.

[2]

Show solutionHide solution

(a) \vec{AB} = \begin{pmatrix}3\\ k+1\\6\end{pmatrix} and \vec{AC} = \begin{pmatrix}9\\9\\ m-3\end{pmatrix}.

[1]

Collinear means \vec{AC} = \lambda\,\vec{AB} (they share the point A).

[1]

The x-components give 9 = 3\lambda, so \lambda = 3.

[1]

y: 9 = 3(k+1), so k = 2. z: m - 3 = 18, so m = 21.

[1]

(b) A is the midpoint of [DB], so \vec{OA} = \tfrac12(\vec{OD} + \vec{OB}), that is \vec{OD} = 2\vec{OA} - \vec{OB}.

[1]

With B(5,\,2,\,9): \vec{OD} = \begin{pmatrix}4\\-2\\6\end{pmatrix} - \begin{pmatrix}5\\2\\9\end{pmatrix} = \begin{pmatrix}-1\\-4\\-3\end{pmatrix}, so D(-1,\,-4,\,-3).

[1]

Final answer

(a) k=2, m=21
(b) D(-1,-4,-3)

Question 11

[7 marks]

The points O(0,\,0,\,0), A(2,\,-1,\,1) and B(1,\,k,\,-2) form a triangle, where k > 0.

(a) Given that the area of triangle OAB is \dfrac{5\sqrt3}{2}, find the value of k.

[5]

(b) Taking [OA] as the base, find the exact height of the triangle.

[2]

Show solutionHide solution

(a) \vec{OA}\times\vec{OB} = \begin{pmatrix}(-1)(-2) - (1)(k)\\(1)(1) - (2)(-2)\\(2)(k) - (-1)(1)\end{pmatrix}

[1]

= \begin{pmatrix}2-k\\5\\2k+1\end{pmatrix}.

[1]

|\vec{OA}\times\vec{OB}|^2 = (2-k)^2 + 25 + (2k+1)^2 = 5k^2 + 30.

[1]

Area = \tfrac12\sqrt{5k^2+30} = \dfrac{5\sqrt3}{2}, so 5k^2 + 30 = 75.

[1]

k^2 = 9, and since k > 0, k = 3.

[1]

(b) |\vec{OA}| = \sqrt{4+1+1} = \sqrt6.

[1]

\tfrac12\sqrt6\,h = \dfrac{5\sqrt3}{2}, so h = \dfrac{5\sqrt3}{\sqrt6} = \dfrac{5}{\sqrt2} = \dfrac{5\sqrt2}{2}.

[1]

Final answer

(a) k=3
(b) \frac{5\sqrt2}{2}

Question 12

[5 marks]

The lines L_1 and L_2 have equations

\begin{gathered}L_1:\ \mathbf{r} = \begin{pmatrix}1\\0\\2\end{pmatrix} + \lambda\begin{pmatrix}1\\2\\2\end{pmatrix},\\ L_2:\ \frac{x-3}{-2} = y + 1 = \frac{z}{-3}.\end{gathered}

(a) Write down a direction vector of L_2.

[1]

(b) Find the acute angle between L_1 and L_2, in degrees to three significant figures.

[3]

(c) A student's calculator gives 122.3^\circ for the angle between the two direction vectors. Explain why this is not the answer to part (b).

[1]

Show solutionHide solution

(a) \mathbf{b}_2 = \begin{pmatrix}-2\\1\\-3\end{pmatrix}.

[1]

(b) \mathbf{b}_1 = \begin{pmatrix}1\\2\\2\end{pmatrix}, so \mathbf{b}_1\cdot\mathbf{b}_2 = -2 + 2 - 6 = -6, |\mathbf{b}_1| = 3, |\mathbf{b}_2| = \sqrt{14}.

[1]

For the acute angle use the modulus: \cos\theta = \dfrac{|-6|}{3\sqrt{14}} = \dfrac{2}{\sqrt{14}}.

[1]

\theta = 57.6884\ldots \approx 57.7^\circ.

[1]

(c) Without the modulus, \cos\theta = -\tfrac{2}{\sqrt{14}} gives the obtuse angle between the direction vectors; the angle between the lines is defined to be the acute one, 180^\circ - 122.3^\circ = 57.7^\circ.

[1]

Final answer

(a) (-2,1,-3)
(b) 57.7^\circ
(c) 122.3^\circ is the obtuse angle between the directions

Question 13

[8 marks]

The lines L_1 and L_2 have equations

\begin{gathered}L_1:\ \mathbf{r} = \begin{pmatrix}2\\-1\\3\end{pmatrix} + \lambda\begin{pmatrix}1\\2\\-1\end{pmatrix},\\ L_2:\ \mathbf{r} = \begin{pmatrix}6\\ k\\2\end{pmatrix} + \mu\begin{pmatrix}2\\-1\\1\end{pmatrix},\end{gathered}

where k \in \mathbb{R}. The lines intersect.

(a) Find the value of k.

[4]

(b) Find the coordinates of the point of intersection.

[1]

(c) Find the acute angle between the lines, in degrees to three significant figures.

[3]

Show solutionHide solution

(a) Equate the x- and z-components (the equations that do not contain k):

[1]

2 + \lambda = 6 + 2\mu and 3 - \lambda = 2 + \mu.

Adding: 5 = 8 + 3\mu, so \mu = -1 and \lambda = 2.

[1]

y-components: -1 + 2\lambda = k - \mu, so -1 + 4 = k + 1.

[1]

k = 2.

[1]

(b) \lambda = 2: \begin{pmatrix}2\\-1\\3\end{pmatrix} + 2\begin{pmatrix}1\\2\\-1\end{pmatrix} = \begin{pmatrix}4\\3\\1\end{pmatrix}, so the point is (4,\,3,\,1).

[1]

(c) \mathbf{b}_1\cdot\mathbf{b}_2 = 2 - 2 - 1 = -1 and |\mathbf{b}_1| = |\mathbf{b}_2| = \sqrt6.

[1]

\cos\theta = \dfrac{|-1|}{6}.

[1]

\theta = 80.4059\ldots \approx 80.4^\circ.

[1]

Final answer

(a) k=2
(b) (4,3,1)
(c) 80.4^\circ

Question 14

[6 marks]

The line L has equation \mathbf{r} = \begin{pmatrix}1\\2\\0\end{pmatrix} + \lambda\begin{pmatrix}1\\-1\\2\end{pmatrix}. The point P has coordinates (3,\,0,\,1) and does not lie on L.

(a) Find the Cartesian equation of the plane \Pi that contains both L and P.

[4]

(b) The point Q has coordinates (k,\,5,\,2) and lies in \Pi. Find the value of k.

[1]

(c) Explain, by referring to a normal vector of \Pi, why the z-coordinate of Q played no part in your answer to part (b).

[1]

Show solutionHide solution

(a) Let A_0(1,2,0) be the point of L used in its equation. Two vectors in \Pi: the direction of L, \mathbf{d} = \begin{pmatrix}1\\-1\\2\end{pmatrix}, and \vec{A_0P} = \begin{pmatrix}2\\-2\\1\end{pmatrix}.

[1]

\mathbf{n} = \mathbf{d}\times\vec{A_0P} = \begin{pmatrix}(-1)(1) - (2)(-2)\\(2)(2) - (1)(1)\\(1)(-2) - (-1)(2)\end{pmatrix}

[1]

= \begin{pmatrix}3\\3\\0\end{pmatrix}, which may be simplified to \begin{pmatrix}1\\1\\0\end{pmatrix}.

[1]

d = \begin{pmatrix}1\\1\\0\end{pmatrix}\cdot\begin{pmatrix}1\\2\\0\end{pmatrix} = 3, so \Pi:\ x + y = 3.

[1]

(Check P: 3 + 0 = 3 ✓.)

(b) k + 5 = 3, so k = -2.

[1]

(c) The normal \begin{pmatrix}1\\1\\0\end{pmatrix} has zero z-component, so the plane is parallel to the z-axis and its equation does not involve z.

[1]

Final answer

(a) x+y=3
(b) k=-2
(c) the normal has zero z-component

Question 15

[6 marks]

The line L has equation \mathbf{r} = \begin{pmatrix}1\\-2\\4\end{pmatrix} + \lambda\begin{pmatrix}2\\-1\\2\end{pmatrix} and the plane \Pi has equation x + 2y - 2z = -15.

(a) Find the coordinates of the point where L meets \Pi.

[3]

(b) Find the acute angle between L and \Pi, in degrees to three significant figures.

[3]

Show solutionHide solution

(a) x = 1 + 2\lambda, y = -2 - \lambda, z = 4 + 2\lambda. Substitute: (1 + 2\lambda) + 2(-2 - \lambda) - 2(4 + 2\lambda) = -15.

[1]

-11 - 4\lambda = -15, so \lambda = 1.

[1]

Point: (3,\,-3,\,6).

[1]

(b) Direction \mathbf{b} = \begin{pmatrix}2\\-1\\2\end{pmatrix} and normal \mathbf{n} = \begin{pmatrix}1\\2\\-2\end{pmatrix}; \mathbf{b}\cdot\mathbf{n} = 2 - 2 - 4 = -4, |\mathbf{b}| = |\mathbf{n}| = 3.

[1]

The angle between a line and a plane uses sine: \sin\alpha = \dfrac{|\mathbf{b}\cdot\mathbf{n}|}{|\mathbf{b}||\mathbf{n}|} = \dfrac49.

[1]

\alpha = \arcsin\tfrac49 = 26.3878\ldots \approx 26.4^\circ.

[1]

Final answer

(a) (3,-3,6)
(b) 26.4^\circ

Question 16

[8 marks]

A cable is modelled by the line L with equation \mathbf{r} = \begin{pmatrix}2\\-1\\4\end{pmatrix} + \lambda\begin{pmatrix}1\\1\\-1\end{pmatrix}. A flat glass roof is modelled by the plane \Pi with equation 3x - y + 2z = -6. All units are metres.

(a) Show that L is parallel to \Pi and does not lie in it.

[2]

(b) Find the shortest distance from the cable to the roof, in metres to three significant figures.

[4]

(c) Find the coordinates of the point on \Pi that is closest to the point (2,\,-1,\,4) of the cable.

[2]

Show solutionHide solution

(a) \mathbf{d}\cdot\mathbf{n} = \begin{pmatrix}1\\1\\-1\end{pmatrix}\cdot\begin{pmatrix}3\\-1\\2\end{pmatrix} = 3 - 1 - 2 = 0, so L is parallel to \Pi.

[1]

The point (2,-1,4) gives 3(2) - (-1) + 2(4) = 15 \ne -6, so L is not in \Pi.

[1]

(b) Because L is parallel to \Pi, every point of L is the same distance from \Pi. Take A(2,-1,4) and the perpendicular line \mathbf{r} = \begin{pmatrix}2\\-1\\4\end{pmatrix} + \mu\begin{pmatrix}3\\-1\\2\end{pmatrix}.

[1]

Substitute: 3(2 + 3\mu) - (-1 - \mu) + 2(4 + 2\mu) = -6, so 15 + 14\mu = -6.

[1]

\mu = -\tfrac32.

[1]

Distance = |\mu|\,|\mathbf{n}| = \tfrac32\sqrt{14} = 5.6124\ldots \approx 5.61 m.

[1]

(c) F = \begin{pmatrix}2\\-1\\4\end{pmatrix} - \tfrac32\begin{pmatrix}3\\-1\\2\end{pmatrix} = \begin{pmatrix}-\tfrac52\\\ tfrac12\\1\end{pmatrix}.

[1]

The point is \left(-\tfrac52,\,\tfrac12,\,1\right) (check: -7.5 - 0.5 + 2 = -6 ✓).

[1]

Final answer

(a) direction \cdot\,\mathbf n=0 and 15\neq-6
(b) \frac{3\sqrt{14}}{2}\approx5.61 m
(c) \left(-\frac52,\frac12,1\right)

Question 17

[8 marks]

The line L has equation \mathbf{r} = \begin{pmatrix}0\\2\\1\end{pmatrix} + \lambda\begin{pmatrix}0\\1\\2\end{pmatrix} and the plane \Pi has equation 2x + y - 2z = -9. The line L' is the mirror image of L in \Pi, so that \Pi acts as a plane mirror.

Find a vector equation of L'.

Show solutionHide solution

The mirror image passes through the point X where L meets \Pi (it does not move). Points of L: (0,\,2 + \lambda,\,1 + 2\lambda). In \Pi: 0 + (2 + \lambda) - 2(1 + 2\lambda) = -9.

[1]

-3\lambda = -9, so \lambda = 3.

[1]

X = (0,\,5,\,7).

[1]

Take A(0,2,1) on L and reflect it. The line through A perpendicular to \Pi is \begin{pmatrix}0\\2\\1\end{pmatrix} + \mu\begin{pmatrix}2\\1\\-2\end{pmatrix}. In \Pi: 2(2\mu) + (2 + \mu) - 2(1 - 2\mu) = -9.

[1]

9\mu = -9, so \mu = -1 and the foot is \begin{pmatrix}0\\2\\1\end{pmatrix} - \begin{pmatrix}2\\1\\-2\end{pmatrix} = (-2,\,1,\,3).

[1]

A' = 2F - A = (-4,\,2,\,6) - (0,\,2,\,1) = (-4,\,0,\,5).

[1]

X is on the mirror so it is on L', and A' is also on L'. Direction \vec{XA'} = \begin{pmatrix}-4\\-5\\-2\end{pmatrix}.

[1]

L':\ \mathbf{r} = \begin{pmatrix}0\\5\\7\end{pmatrix} + t\begin{pmatrix}4\\5\\2\end{pmatrix}.

[1]

(Check: |XA| = |XA'| = \sqrt{45}.)

Final answer

\mathbf r=\begin{pmatrix}0\\5\\7\end{pmatrix}+t\begin{pmatrix}4\\5\\2\end{pmatrix}

Question 18

[10 marks]

A pyramid has base ABC and apex V, where A(2,\,0,\,1), B(-1,\,3,\,2), C(0,\,1,\,-2) and V(4,\,4,\,5). The units are centimetres.

(a) Find the area of triangle ABC, in \text{cm}^2 to three significant figures.

[3]

(b) Find the Cartesian equation of the plane containing A, B and C.

[2]

(c) Find the perpendicular height of the pyramid, in cm to three significant figures.

[3]

(d) Find the volume of the pyramid, in \text{cm}^3.

[2]

Show solutionHide solution

(a) \vec{AB} = \begin{pmatrix}-3\\3\\1\end{pmatrix} and \vec{AC} = \begin{pmatrix}-2\\1\\-3\end{pmatrix}.

[1]

\vec{AB}\times\vec{AC} = \begin{pmatrix}(3)(-3) - (1)(1)\\(1)(-2) - (-3)(-3)\\(-3)(1) - (3)(-2)\end{pmatrix} = \begin{pmatrix}-10\\-11\\3\end{pmatrix}.

[1]

Area = \tfrac12\sqrt{100 + 121 + 9} = \tfrac12\sqrt{230} = 7.5829\ldots \approx 7.58 \text{cm}^2.

[1]

(b) Normal \begin{pmatrix}10\\11\\-3\end{pmatrix} (reversed sign), and d = 10(2) + 0 - 3(1) = 17.

[1]

\Pi:\ 10x + 11y - 3z = 17 (check B: -10 + 33 - 6 = 17 ✓; C: 11 + 6 = 17 ✓).

[1]

(c) The height is the distance from V to \Pi. Perpendicular line through V: (4 + 10\lambda,\,4 + 11\lambda,\,5 - 3\lambda).

[1]

In \Pi: 10(4 + 10\lambda) + 11(4 + 11\lambda) - 3(5 - 3\lambda) = 17, so 69 + 230\lambda = 17 and \lambda = -\tfrac{26}{115}.

[1]

h = |\lambda|\sqrt{100 + 121 + 9} = \tfrac{26}{115}\sqrt{230} = 3.4287\ldots \approx 3.43 cm.

[1]

(d) V_{\text{pyramid}} = \tfrac13\times(\text{base area})\times h.

[1]

= \tfrac13(7.5829)(3.4287) = 8.6667\ldots = \dfrac{26}{3} \approx 8.67 \text{cm}^3.

[1]

Final answer

(a) 7.58 cm^2
(b) 10x+11y-3z=17
(c) 3.43 cm
(d) \frac{26}{3} cm^3

Question 19

[6 marks]

Two particles A and B move so that, t seconds after they start, their position vectors in metres are

\begin{gathered}\mathbf{r}_A = \begin{pmatrix}2\\-1\\5\end{pmatrix} + t\begin{pmatrix}1\\2\\-1\end{pmatrix},\\ \mathbf{r}_B = \begin{pmatrix}11\\ k\\ m\end{pmatrix} + t\begin{pmatrix}-2\\2\\2\end{pmatrix},\end{gathered}

where k and m are constants.

The particles collide.

(a) Find the time at which they collide.

[2]

(b) Find the value of k and the value of m.

[3]

(c) Find the coordinates of the point of collision.

[1]

Show solutionHide solution

(a) At the collision the two positions are equal, with the same t in every component. The x-components do not involve k or m: 2 + t = 11 - 2t.

[1]

t = 3 s.

[1]

(b) y-components: -1 + 2t = k + 2t gives -1 + 6 = k + 6, so k = -1. [1]

[1]

z-components: 5 - t = m + 2t gives 5 - 3 = m + 6, so m = -4.

[1]

(c) \mathbf{r}_A(3) = \begin{pmatrix}2+3\\-1+6\\5-3\end{pmatrix} = \begin{pmatrix}5\\5\\2\end{pmatrix}, so the collision is at (5,\,5,\,2).

[1]

Final answer

(a) t=3
(b) k=-1, m=-4
(c) (5,5,2)

Mastery

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

Question 20

[8 marks]

The points A and B have position vectors \mathbf{a} and \mathbf{b} relative to an origin O. The point P lies on the line AB such that \vec{AP} = \lambda\vec{AB}, where 0 < \lambda < 1.

(a) Show that \vec{OP} = (1 - \lambda)\mathbf{a} + \lambda\mathbf{b}.

[1]

It is also given that |\mathbf{a}| = 3, |\mathbf{b}| = 5 and \mathbf{a}\cdot\mathbf{b} = -4.

(b) Find the value of \lambda for which \vec{OP} is perpendicular to \vec{AB}.

[7]

Show solutionHide solution

(a) \vec{AB} = \mathbf{b} - \mathbf{a}, so

\vec{OP} = \vec{OA} + \vec{AP} = \mathbf{a} + \lambda(\mathbf{b} - \mathbf{a}) = (1-\lambda)\mathbf{a} + \lambda\mathbf{b} \qquad \textbf{AG}

A1[1]

(b) \vec{OP}\cdot\vec{AB} = 0:

\left[(1-\lambda)\mathbf{a} + \lambda\mathbf{b}\right]\cdot(\mathbf{b} - \mathbf{a}) = 0

Expanding:

(1-\lambda)\,\mathbf{a}\cdot\mathbf{b} - (1-\lambda)|\mathbf{a}|^2 + \lambda|\mathbf{b}|^2 - \lambda\,\mathbf{a}\cdot\mathbf{b} = 0

Substituting \mathbf{a}\cdot\mathbf{b} = -4, |\mathbf{a}|^2 = 9, |\mathbf{b}|^2 = 25:

(1-\lambda)(-4) - 9(1-\lambda) + 25\lambda + 4\lambda = 0

-13 + 13\lambda + 29\lambda = 0 \;\Rightarrow\; 42\lambda = 13

\lambda = \frac{13}{42}

A1[7]

Final answer

(b) \lambda=\frac{13}{42}

Question 21

[8 marks]

The vectors \boldsymbol{a}, \boldsymbol{b} and \boldsymbol{c} are given by \boldsymbol{a} = \begin{pmatrix} -6 \\ 8 \end{pmatrix}, \boldsymbol{b} = \begin{pmatrix} -5 \\ 12 \end{pmatrix} and \boldsymbol{c} = \begin{pmatrix} p \\ -3 \end{pmatrix}, where p \in \mathbb{R}.

(a) Write \boldsymbol{a}\cdot\boldsymbol{c} and \boldsymbol{b}\cdot\boldsymbol{c} in terms of p.

[3]

The angle between \boldsymbol{a} and \boldsymbol{c} equals the angle between \boldsymbol{b} and \boldsymbol{c}.

(b) Determine the value of p.

[5]

Show solutionHide solution

(a) \boldsymbol{a}\cdot\boldsymbol{c} = -6p - 24

M1A1

\boldsymbol{b}\cdot\boldsymbol{c} = -5p - 36

A1[3]

(b) |\boldsymbol{a}| = \sqrt{36 + 64} = 10 and |\boldsymbol{b}| = \sqrt{25 + 144} = 13.

Equal angles mean equal cosines; the common factor |\boldsymbol{c}| cancels:

\frac{\boldsymbol{a}\cdot\boldsymbol{c}}{|\boldsymbol{a}||\boldsymbol{c}|} = \frac{\boldsymbol{b}\cdot\boldsymbol{c}}{|\boldsymbol{b}||\boldsymbol{c}|} \;\Rightarrow\; \frac{-6p - 24}{10} = \frac{-5p - 36}{13}

13(-6p - 24) = 10(-5p - 36) \;\Rightarrow\; -78p - 312 = -50p - 360

-28p = -48 \;\Rightarrow\; p = \frac{12}{7}

A1[5]

Final answer

(a) -6p-24, -5p-36
(b) p=\frac{12}{7}

Question 22

[7 marks]

The plane \Pi has equation x + 2y - 2z = 7. The line L passes through the point (1,\,3,\,-2) and has direction vector 2\mathbf{i} + p\mathbf{k}, where p \in \mathbb{R}.

The line L is inclined to the plane \Pi at an acute angle of 45^\circ.

Find the two possible values of p.

[7]

Show solutionHide solution

Normal to the plane: \mathbf{n} = \begin{pmatrix} 1 \\ 2 \\ -2 \end{pmatrix}, |\mathbf{n}| = 3. Direction of L: \mathbf{d} = \begin{pmatrix} 2 \\ 0 \\ p \end{pmatrix}, |\mathbf{d}| = \sqrt{4 + p^2}.

[2]

The line is inclined at 45^\circ to the plane, so the angle between \mathbf{d} and \mathbf{n} is 45^\circ or 135^\circ:

\frac{|\mathbf{n} \cdot \mathbf{d}|}{|\mathbf{n}||\mathbf{d}|} = \sin 45^\circ \quad\Rightarrow\quad \frac{|2 - 2p|}{3\sqrt{4 + p^2}} = \frac{1}{\sqrt{2}}

[2]

Squaring: 2(2 - 2p)^2 = 9(4 + p^2)

8 - 16p + 8p^2 = 36 + 9p^2 \;\Rightarrow\; p^2 + 16p + 28 = 0

[2]

(p + 2)(p + 14) = 0 \quad\Rightarrow\quad p = -2 \ \text{ or } \ p = -14

[1]

Final answer

p=-2 or p=-14

Question 23

[8 marks]

Two planes are given by

\begin{gathered}\Pi_1:\ x + 2y + 2z = 6,\\ \Pi_2:\ 2x - 2y + z = 6,\end{gathered}

and L is their line of intersection.

(a) Show that a vector equation of L is \mathbf{r} = \begin{pmatrix} 4 \\ 1 \\ 0 \end{pmatrix} + \lambda\begin{pmatrix} 2 \\ 1 \\ -2 \end{pmatrix}, \lambda \in \mathbb{R}.

[3]

(b) Determine the point P of L nearest to the origin O, and state the length OP.

[5]

Show solutionHide solution

(a) The point (4,1,0) is on both planes: 4 + 2 + 0 = 6 and 8 - 2 + 0 = 6.

Direction perpendicular to both normals:

\begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}\times\begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix} = \begin{pmatrix} 2(1) - 2(-2) \\ 2(2) - 1(1) \\ 1(-2) - 2(2) \end{pmatrix} = \begin{pmatrix} 6 \\ 3 \\ -6 \end{pmatrix} = 3\begin{pmatrix} 2 \\ 1 \\ -2 \end{pmatrix}

M1A1

So L: \mathbf{r} = \begin{pmatrix} 4 \\ 1 \\ 0 \end{pmatrix} + \lambda\begin{pmatrix} 2 \\ 1 \\ -2 \end{pmatrix}

AG[3]

(b) A general point of L is (4 + 2\lambda,\ 1 + \lambda,\ -2\lambda). At the closest point, \vec{OP} is perpendicular to the direction of L:

2(4 + 2\lambda) + (1 + \lambda) - 2(-2\lambda) = 0 \;\Rightarrow\; 9 + 9\lambda = 0 \;\Rightarrow\; \lambda = -1

M1A1

P = (2,\ 0,\ 2)

OP = \sqrt{4 + 0 + 4} = 2\sqrt{2}

A1[5]

Final answer

(b) P(2,0,2), OP=2\sqrt2

Question 24

[9 marks]

The points P(3,\, p,\, 1), Q(1,\, 2,\, 0) and R(p,\, 0,\, 2) are given, where p \in \mathbb{R}.

(a) Show that \vec{PQ} \times \vec{PR} = \begin{pmatrix} 2 - 2p \\ 5 - p \\ p^{2} - 3p + 6 \end{pmatrix}.

[4]

(b) Find the smallest possible value of \left|\vec{PQ} \times \vec{PR}\right|^{2}, and the value of p at which it occurs.

[3]

(c) Deduce the smallest possible area of triangle PQR.

[2]

Show solutionHide solution

(a)

Position differences:

\vec{PQ} = \begin{pmatrix} -2 \\ 2-p \\ -1 \end{pmatrix}, \qquad \vec{PR} = \begin{pmatrix} p-3 \\ -p \\ 1 \end{pmatrix}

Components of the cross product:

\begin{aligned} \text{first} &= (2-p)(1) - (-1)(-p) = 2 - p - p = 2-2p \\ \text{second} &= (-1)(p-3) - (-2)(1) = 3 - p + 2 = 5 - p \\ \text{third} &= (-2)(-p) - (2-p)(p-3) = 2p - (-p^2 + 5p - 6) = p^2 - 3p + 6 \end{aligned}

So \vec{PQ} \times \vec{PR} = \begin{pmatrix} 2-2p \\ 5-p \\ p^2-3p+6 \end{pmatrix}, as required.

[4]

(b)

Squaring and adding the components:

S(p) = (2-2p)^2 + (5-p)^2 + (p^2-3p+6)^2 = p^4 - 6p^3 + 26p^2 - 54p + 65

Setting S'(p) = 4p^3 - 18p^2 + 52p - 54 = 0 has a single real root (a GDC shows this is the only turning point of the quartic), p = 1.6197\ldots, and then

S_{\min} = 27.1327\ldots

Smallest value \approx 27.1 when p \approx 1.62.

[3]

(c)

Area = \tfrac{1}{2}\left|\vec{PQ} \times \vec{PR}\right|, which is least when S is least:

\text{Area}_{\min} = \tfrac{1}{2}\sqrt{27.1327\ldots} = 2.6044\ldots

Smallest area \approx 2.60 square units.

[2]

Final answer

(b) least value \approx27.1 at p\approx1.62
(c) \approx2.60

Question 25

[5 marks]

From an origin O, two ships A and B are tracked. At t hours after midday their position vectors, in kilometres, are

\begin{gathered}\mathbf{r}_{A} = \begin{pmatrix} 3 \\ 2 \end{pmatrix} + t\begin{pmatrix} 2 \\ 5 \end{pmatrix},\\ \mathbf{r}_{B} = \begin{pmatrix} 9 \\ -1 \end{pmatrix} + t\begin{pmatrix} 1 \\ 9 \end{pmatrix}.\end{gathered}

Determine the least distance between the two ships.

[5]

Show solutionHide solution

The displacement from A to B at time t is

\mathbf{r}_{B} - \mathbf{r}_{A} = \begin{pmatrix} 6 \\ -3 \end{pmatrix} + t\begin{pmatrix} -1 \\ 4 \end{pmatrix} = \begin{pmatrix} 6 - t \\ 4t - 3 \end{pmatrix}

Its magnitude is

d(t) = \sqrt{(6 - t)^{2} + (4t - 3)^{2}} = \sqrt{17t^{2} - 36t + 45}

The square root is increasing, so d is least where the quadratic under it is least, at

t = \frac{36}{2(17)} = \frac{18}{17}

d\left(\frac{18}{17}\right) = \sqrt{17\left(\frac{18}{17}\right)^{2} - 36\left(\frac{18}{17}\right) + 45} = \frac{21}{\sqrt{17}} = 5.09324\ldots

minimum distance= 5.09 km \left(= \dfrac{21\sqrt{17}}{17}\right)

[5]

Final answer

\frac{21}{\sqrt{17}}\approx5.09 km

Question 26

[15 marks]

In a survey of a mine, all distances are in kilometres. A straight tunnel T has equation \mathbf{r} = \begin{pmatrix}-8\\4\\-1\end{pmatrix} + \lambda\begin{pmatrix}3\\1\\2\end{pmatrix}, where the z-axis points vertically upwards. A flat ore seam S lies in the plane x + 2y + 3z = 30.

(a) Find the coordinates of the point P where the tunnel meets the seam.

[3]

(b) Find the acute angle between the tunnel and the seam, in degrees to three significant figures.

[3]

A ventilation shaft is to be drilled in a straight line from the surface point Q(2,\,6,\,12) to the seam, by the shortest possible route.

(c) Find the coordinates of the point F where the shaft meets the seam, and the length of the shaft, to three significant figures.

[4]

A second tunnel U passes through P with direction vector \begin{pmatrix}1\\1\\-1\end{pmatrix}.

(d) Show that U lies entirely in the seam S.

[2]

(e) Find the acute angle between the tunnels T and U, in degrees to three significant figures.

[3]

Show solutionHide solution

(a) (-8 + 3\lambda) + 2(4 + \lambda) + 3(-1 + 2\lambda) = 30.

[1]

11\lambda - 3 = 30, so \lambda = 3.

[1]

P = (1,\,7,\,5).

[1]

(b) Tunnel direction \mathbf{b} = \begin{pmatrix}3\\1\\2\end{pmatrix}, seam normal \mathbf{n} = \begin{pmatrix}1\\2\\3\end{pmatrix}: \mathbf{b}\cdot\mathbf{n} = 3 + 2 + 6 = 11 and |\mathbf{b}| = |\mathbf{n}| = \sqrt{14}.

[1]

Line and plane: \sin\alpha = \dfrac{11}{14}.

[1]

\alpha = 51.787\ldots \approx 51.8^\circ.

[1]

(c) The shortest route is along the normal: (2 + \mu,\,6 + 2\mu,\,12 + 3\mu).

[1]

In the seam: (2 + \mu) + 2(6 + 2\mu) + 3(12 + 3\mu) = 50 + 14\mu = 30, so \mu = -\tfrac{10}{7}.

[1]

F = \left(\tfrac47,\,\tfrac{22}{7},\,\tfrac{54}{7}\right) = (0.571,\,3.14,\,7.71).

[1]

Length QF = \tfrac{10}{7}\sqrt{14} = 5.3452\ldots \approx 5.35 km.

[1]

(d) The direction of U is perpendicular to the normal of S: \begin{pmatrix}1\\1\\-1\end{pmatrix}\cdot\begin{pmatrix}1\\2\\3\end{pmatrix} = 1 + 2 - 3 = 0.

[1]

So U is parallel to S, and since U passes through P, which is in S, the whole of U lies in S.

[1]

(e) \begin{pmatrix}3\\1\\2\end{pmatrix}\cdot\begin{pmatrix}1\\1\\-1\end{pmatrix} = 3 + 1 - 2 = 2, with |\mathbf{b}_T| = \sqrt{14} and |\mathbf{b}_U| = \sqrt3.

[1]

\cos\theta = \dfrac{2}{\sqrt{14}\sqrt{3}} = 0.3086\ldots

[1]

\theta = 72.02\ldots \approx 72.0^\circ.

[1]

Final answer

(a) P(1,7,5)
(b) 51.8^\circ
(c) F\left(\frac47,\frac{22}{7},\frac{54}{7}\right), 5.35 km
(d) \mathbf u\cdot\mathbf n=0 and P is in S
(e) 72.0^\circ

Going further

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

G1 Vectors that share one angle

G1 Worked example

[9 marks]

The points A(1,1,1), B(1,-1,-1), C(-1,1,-1) and D(-1,-1,1) are four corners of a cube centred at the origin O.

(a) Show that the six edges of the tetrahedron ABCD all have the same length.

[2]

(b) Show that \overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}=\mathbf 0, and find the exact value of \cos A\hat OB.

[3]

(c) Explain, without further calculation, why the six angles A\hat OB, A\hat OC, A\hat OD, B\hat OC, B\hat OD and C\hat OD are equal.

[1]

(d) Show that \overrightarrow{OA} is perpendicular to the plane BCD. The line OA meets this plane at F. Find the ratio AF:OF.

[3]

Show solutionHide solution

(a) \overrightarrow{AB}=\begin{pmatrix}0\\-2\\-2\end{pmatrix}, so AB=\sqrt8=2\sqrt2

each of the six edges joins two corners that differ in exactly two coordinates, each by 2, so every edge is 2\sqrt2

Note: listing all six lengths also earns A1.

(b) the four x-coordinates 1+1-1-1, and likewise the y- and z-coordinates, add to 0

\overrightarrow{OA}\cdot\overrightarrow{OB}=1-1-1=-1 and |\overrightarrow{OA}|=|\overrightarrow{OB}|=\sqrt3

\cos A\hat OB=\dfrac{-1}{3}, so A\hat OB=109.5^\circ

(c) OA=OB=OC=OD=\sqrt3 and all edges are equal, so the six triangles such as OAB are congruent (three equal sides), and their angles at O are equal

(d) \overrightarrow{BC}\times\overrightarrow{BD}=\begin{pmatrix}-2\\2\\0\end{pmatrix}\times\begin{pmatrix}-2\\0\\2\end{pmatrix}=\begin{pmatrix}4\\4\\4\end{pmatrix}, parallel to \overrightarrow{OA}=\begin{pmatrix}1\\1\\1\end{pmatrix}

M1A1

plane BCD: x+y+z=-1. On OA, \mathbf r=t\begin{pmatrix}1\\1\\1\end{pmatrix} meets it when 3t=-1, so F=\left(-\frac13,-\frac13,-\frac13\right); OF=\dfrac{1}{\sqrt3}, AF=\dfrac{4}{\sqrt3}, so AF:OF=4:1

Final answer

(b) \cos A\hat OB=-\frac13 (109.5^\circ)
(d) AF:OF=4:1

G1 Investigation

[13 marks]

In this investigation all vectors are unit vectors, and "the vectors make equal angles" means that every pair of them makes the same angle.

(a) Three vectors \mathbf u, \mathbf v, \mathbf w satisfy \mathbf u+\mathbf v+\mathbf w=\mathbf 0. By considering |\mathbf u+\mathbf v|^2, show that \mathbf u\cdot\mathbf v=-\frac12, and write down the angle between \mathbf u and \mathbf v.

[2]

(b) The vectors \mathbf v_1,\mathbf v_2,\dots,\mathbf v_n, where n\ge2, satisfy \mathbf v_1+\mathbf v_2+\cdots+\mathbf v_n=\mathbf 0 and \mathbf v_i\cdot\mathbf v_j=c whenever i\ne j. By expanding |\mathbf v_1+\mathbf v_2+\cdots+\mathbf v_n|^2, show that c=-\dfrac{1}{n-1}.

[3]

(c) Now remove the condition that the sum is \mathbf 0. Show that if n vectors make equal angles, with \mathbf v_i\cdot\mathbf v_j=c for i\ne j, then c\ge-\dfrac{1}{n-1}.

[2]

(d) Three different vectors in one plane make equal angles \theta. Show that \theta=120^\circ. Deduce that no four vectors in one plane make equal angles.

[3]

(e) Suppose five vectors in three dimensions make equal angles, with \mathbf v_i\cdot\mathbf v_j=c for i\ne j. Take \mathbf v_1=\mathbf k and write \mathbf v_i=\mathbf h_i+c\,\mathbf k for i=2,3,4,5, where each \mathbf h_i is perpendicular to \mathbf k. By finding \mathbf h_i\cdot\mathbf h_j and |\mathbf h_i|, obtain a contradiction.

[3]

Show solutionHide solution

(a) \mathbf u+\mathbf v=-\mathbf w, so |\mathbf u+\mathbf v|^2=1; also |\mathbf u+\mathbf v|^2=1+2\,\mathbf u\cdot\mathbf v+1, so \mathbf u\cdot\mathbf v=-\frac12

A1AG

the angle is 120^\circ

(b) |\mathbf v_1+\cdots+\mathbf v_n|^2=\sum_i \mathbf v_i\cdot\mathbf v_i+\sum_{i\ne j}\mathbf v_i\cdot\mathbf v_j

=n+n(n-1)c, since there are n(n-1) ordered pairs with i\ne j

this equals |\mathbf 0|^2=0, so c=-\dfrac{1}{n-1}

A1AG

Note: n=3 gives -\frac12, as in (a); n=4 gives -\frac13, the tetrahedron angle 109.5^\circ.

(c) the same expansion gives n+n(n-1)c=|\mathbf v_1+\cdots+\mathbf v_n|^2\ge0

so c\ge-\dfrac{1}{n-1}, with equality exactly when the sum is \mathbf 0

(d) measure directions from \mathbf u: \mathbf v and \mathbf w are each at \theta to \mathbf u, on opposite sides of it since \mathbf v\ne\mathbf w, so the angle between \mathbf v and \mathbf w is 2\theta or 360^\circ-2\theta

2\theta=\theta is impossible, and 360^\circ-2\theta=\theta gives \theta=120^\circ

with four vectors, any three of them are at 120^\circ to each other, so they point in three directions 120^\circ apart; the fourth would have to be at 120^\circ to \mathbf u, which is the direction of \mathbf v or of \mathbf w: it is not a different vector

(e) \mathbf v_i\cdot\mathbf k=c, so the \mathbf k-part of \mathbf v_i is c\,\mathbf k; then \mathbf v_i\cdot\mathbf v_j=\mathbf h_i\cdot\mathbf h_j+c^2=c, so \mathbf h_i\cdot\mathbf h_j=c-c^2, and |\mathbf h_i|^2=1-c^2

M1A1

c\ne\pm1 (the vectors are different, and c=-1 is impossible for three vectors), so \mathbf h_2,\dots,\mathbf h_5 are non-zero, of equal length, with equal scalar products: four vectors in the plane perpendicular to \mathbf k making equal angles, which (d) rules out

Final answer

(a) 120^\circ
(b) c=-\frac1{n-1}
(c)–(e) proofs

G2 Signed areas

G2 Worked example

[9 marks]

The plane \dfrac x2+\dfrac y3+\dfrac z6=1 meets the coordinate axes at A(2,0,0), B(0,3,0) and C(0,0,6). O is the origin.

(a) Find \overrightarrow{AB}\times\overrightarrow{AC} and hence the area of triangle ABC.

[3]

(b) Find the areas of the triangles OAB, OBC and OCA.

[2]

(c) Show that the square of the area of ABC equals the sum of the squares of these three areas.

[1]

(d) Explain why the \mathbf k-component of \frac12\overrightarrow{AB}\times\overrightarrow{AC} is the area of the triangle that ABC projects onto in the xy-plane, and hence why the result in (c) holds.

[3]

Show solutionHide solution

(a) \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}

M1A1

area =\frac12\sqrt{18^2+12^2+6^2}=\frac12\sqrt{504}=3\sqrt{14}

(b) right angles at O: OAB=\frac12(2)(3)=3, OBC=\frac12(3)(6)=9, OCA=\frac12(6)(2)=6

A1A1

Note: A1 for two correct areas, A1 for the third.

(c) (3\sqrt{14})^2=126=3^2+9^2+6^2

(d) the \mathbf k-component of \overrightarrow{AB}\times\overrightarrow{AC} uses only the x- and y-components of the two vectors

so it equals the \mathbf k-component for the projected vectors, whose vector product is \pm2\times(area of the projected triangle) \mathbf k; here ABC projects onto OAB, and \frac12\times6=3

the same holds for the \mathbf i- and \mathbf j-components and the yz- and zx-planes, so |\tfrac12\overrightarrow{AB}\times\overrightarrow{AC}|^2=9^2+6^2+3^2, the sum of the squared projected areas

Final answer

(a) \begin{pmatrix}18\\12\\6\end{pmatrix}, 3\sqrt{14}
(b) 3, 9, 6

G2 Investigation

[11 marks]

(a) Use a vector product to find the area of the triangle with vertices P(1,1), Q(5,2) and R(2,6).

[2]

(b) Find the signed areas of the triangles OPQ, OQR and ORP, where O is the origin, and show that they add to the area in (a). With a sketch, explain why two of them are negative.

[2]

(c) The quadrilateral with vertices A(3,-1), B(4,3), C(-1,4), D(-2,-2), taken in that order, contains O. By splitting it into the triangles OAB, OBC, OCD and ODA, find its area.

[2]

(d) Conjecture a formula for the area of a convex polygon with vertices (x_1,y_1),\dots,(x_n,y_n) listed anticlockwise. Prove it when O lies inside the polygon.

[3]

(e) Use your formula to find the area of the pentagon with vertices (0,0), (4,0), (5,3), (2,5), (-1,2). State what changes if the same vertices are listed clockwise.

[2]

Show solutionHide solution

(a) \overrightarrow{PQ}\times\overrightarrow{PR}=\begin{pmatrix}4\\1\\0\end{pmatrix}\times\begin{pmatrix}1\\5\\0\end{pmatrix}=\begin{pmatrix}0\\0\\19\end{pmatrix}

area =\dfrac{19}{2}

(b) \frac12(1\times2-5\times1)=-\frac32, ; \frac12(5\times6-2\times2)=13, ; \frac12(2\times1-1\times6)=-2; the sum is \frac{19}{2}

O lies outside PQR and P lies inside triangle OQR. Triangle OQR (anticlockwise, positive) is made of PQR together with OPQ and ORP; going round O\to P\to Q and O\to R\to P turns clockwise, so those two areas come with a minus sign and are taken away

(c) the four values of x_iy_{i+1}-x_{i+1}y_i are 13, 19, 10 and 8, all positive

area =\frac12(13+19+10+8)=25

(d) conjecture: area =\dfrac12\left|\displaystyle\sum_{i=1}^{n}(x_iy_{i+1}-x_{i+1}y_i)\right|, where (x_{n+1},y_{n+1})=(x_1,y_1)

with O inside, the segments from O to the vertices cut the polygon into n triangles OV_iV_{i+1} that do not overlap

going round anticlockwise, each turn from \overrightarrow{OV_i} to \overrightarrow{OV_{i+1}} is anticlockwise, so by (b) each triangle has area \frac12(x_iy_{i+1}-x_{i+1}y_i) with a positive bracket; adding gives the formula

(e) the terms are 0, 12, 19, 9, 0: area =\frac12\times40=20

listed clockwise, every bracket changes sign and the sum is -40; the modulus still gives 20

Note: here O is a vertex, not inside. The two triangles at O have zero area and the formula still holds; the proof in (d) can be run from any point inside.

Final answer

(a) \frac{19}{2}
(b) -\frac32, 13, -2
(c) 25
(e) 20; the sum becomes -40

TI-84 Plus CE keystrokes

The GDC panels in the book give keystrokes for the TI-Nspire CX II and the Casio fx-CG50. These are the TI-84 Plus CE keystrokes for the same calculator jobs. The full list of TI-84 keystrokes is in Chapter 19.

Job	TI-84 Plus CE keystrokes
Linear system	`2nd x⁻¹ (MATRIX), enter the augmented matrix, then MATH, rref(`

From the book

The Ultimate Guide for the IB: Mathematics: Analysis and Approaches HL

Chapter 11 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.

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