Vectors in Three Dimensions – A Level Maths Year 2

1 Vectors in Three Dimensions

Everything from Vectors in Two Dimensions carries straight over to 3-D: you just add a third component. The axes $x$, $y$ and $z$ are at right angles to each other, with unit vectors $\mathbf{i}$, $\mathbf{j}$ and $\mathbf{k}$ along them.

Two ways to write a 3-D vector

$\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k} = \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix}$

The point $P(2, 3, 4)$ has position vector $\overrightarrow{OP} = 2\mathbf{i} + 3\mathbf{j} + 4\mathbf{k}$: go $2$ along $x$, $3$ along $y$, then $4$ up in the $z$ direction.

The axes are right-handed. Point the fingers of your right hand along the $x$-axis and curl them towards the $y$-axis: your thumb points along the $z$-axis. All diagrams and textbooks use this convention.

Adding, subtracting and multiplying by a scalar all work component by component, exactly as in 2-D:

Arithmetic

$\begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix} + \begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix} = \begin{pmatrix} a_1 + b_1 \\ a_2 + b_2 \\ a_3 + b_3 \end{pmatrix}$ · $\lambda\begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix} = \begin{pmatrix} \lambda a_1 \\ \lambda a_2 \\ \lambda a_3 \end{pmatrix}$

Parallel vectors are scalar multiples of each other. $\begin{pmatrix} 3 \\ -6 \\ 9 \end{pmatrix} = -3\begin{pmatrix} -1 \\ 2 \\ -3 \end{pmatrix}$, so these are parallel (in opposite directions). If the ratios of the components are not all equal, the vectors are not parallel.

2 Magnitude, Distance and Unit Vectors

Magnitude (Pythagoras in 3-D)

$|\mathbf{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}$

Distance between $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$

$AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

Unit vector in the direction of a

$\hat{\mathbf{a}} = \dfrac{\mathbf{a}}{|\mathbf{a}|}$

Worked Example 1 — Magnitude and unit vector

$\mathbf{a} = 2\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}$. Find $|\mathbf{a}|$, a unit vector in the direction of $\mathbf{a}$, and a vector of magnitude $21$ in the same direction.

①$|\mathbf{a}| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$

②$\hat{\mathbf{a}} = \tfrac{1}{7}(2\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}) = \tfrac{2}{7}\mathbf{i} - \tfrac{3}{7}\mathbf{j} + \tfrac{6}{7}\mathbf{k}$

③Magnitude $21$: $21\hat{\mathbf{a}} = 3\mathbf{a} = 6\mathbf{i} - 9\mathbf{j} + 18\mathbf{k}$

Square the negative component with its sign: $(-3)^2 = 9$, never $-9$.

Worked Example 2 — Distance between two points

Find the distance between $A(1, 2, -1)$ and $B(4, -2, 11)$.

①$\overrightarrow{AB} = \begin{pmatrix} 4 - 1 \\ -2 - 2 \\ 11 - (-1) \end{pmatrix} = \begin{pmatrix} 3 \\ -4 \\ 12 \end{pmatrix}$

②$AB = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$

Worked Example 3 — An unknown component

The vector $k\mathbf{i} + 2k\mathbf{j} + 2\mathbf{k}$ has magnitude $7$. Find the possible values of $k$.

①$k^2 + (2k)^2 + 2^2 = 49$

②$5k^2 + 4 = 49 \Rightarrow k^2 = 9 \Rightarrow k = \pm 3$

Square both sides before doing anything else — it removes the square root and usually leaves a quadratic.

Worked Example 4 — The angle a vector makes with an axis

Find the angle between $\mathbf{a} = 2\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}$ and each of the coordinate axes.

①The component along an axis is the adjacent side of a right-angled triangle whose hypotenuse is $|\mathbf{a}| = 7$.

②$x$-axis: $\cos\theta_x = \tfrac{2}{7} \Rightarrow \theta_x = 73.4^\circ$

③$y$-axis: $\cos\theta_y = -\tfrac{3}{7} \Rightarrow \theta_y = 115.4^\circ$

④$z$-axis: $\cos\theta_z = \tfrac{6}{7} \Rightarrow \theta_z = 31.0^\circ$

The angle is measured to the positive direction of the axis, so a negative component gives an obtuse angle.

3 Position Vectors

The position vector of a point $A$ is $\mathbf{a} = \overrightarrow{OA}$, the vector from the origin to $A$. Its components are just the coordinates of $A$. The vector from $A$ to $B$ is "end minus start".

Key results

$\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$ · midpoint of $AB$: $\tfrac{1}{2}(\mathbf{a} + \mathbf{b})$

Worked Example 5 — Vector, length and midpoint

$A$ and $B$ have position vectors $2\mathbf{i} + \mathbf{j} + 3\mathbf{k}$ and $5\mathbf{i} + 7\mathbf{j} - 3\mathbf{k}$. Find $\overrightarrow{AB}$, $|\overrightarrow{AB}|$ and the midpoint of $AB$.

①$\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = 3\mathbf{i} + 6\mathbf{j} - 6\mathbf{k}$

②$|\overrightarrow{AB}| = \sqrt{9 + 36 + 36} = \sqrt{81} = 9$

③Midpoint: $\tfrac{1}{2}(7\mathbf{i} + 8\mathbf{j} + 0\mathbf{k}) = 3.5\mathbf{i} + 4\mathbf{j}$, i.e. the point $(3.5, 4, 0)$.

4 Geometric Problems in 3-D

The methods from Vector Geometry work unchanged in three dimensions.

To show…	Show that…
$A$, $B$, $C$ are collinear	$\overrightarrow{AC} = \lambda\overrightarrow{AB}$ — parallel and sharing the point $A$
$ABCD$ is a parallelogram	$\overrightarrow{AB} = \overrightarrow{DC}$ (equivalently $\overrightarrow{AD} = \overrightarrow{BC}$)
$P$ divides $AB$ in the ratio $m : n$	$\mathbf{p} = \mathbf{a} + \dfrac{m}{m+n}(\mathbf{b} - \mathbf{a})$

Worked Example 6 — Collinear points

Show that $A(1, 2, 3)$, $B(3, 5, 4)$ and $C(7, 11, 6)$ are collinear, and find the ratio $AB : BC$.

①$\overrightarrow{AB} = \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix}$, $\overrightarrow{BC} = \begin{pmatrix} 4 \\ 6 \\ 2 \end{pmatrix} = 2\overrightarrow{AB}$

②$\overrightarrow{AB}$ and $\overrightarrow{BC}$ are parallel and share the point $B$, so $A$, $B$, $C$ lie on one straight line.

③$AB : BC = 1 : 2$

Parallel alone is not enough — two parallel vectors could lie on different lines. Always state the common point.

Worked Example 7 — Completing a parallelogram

$A(1, 0, 2)$, $B(4, 1, 3)$ and $C(6, 4, 1)$ are three vertices of parallelogram $ABCD$. Find $D$.

①In $ABCD$, $\overrightarrow{AD} = \overrightarrow{BC} = \mathbf{c} - \mathbf{b} = \begin{pmatrix} 2 \\ 3 \\ -2 \end{pmatrix}$.

②$\mathbf{d} = \mathbf{a} + \overrightarrow{AD} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} + \begin{pmatrix} 2 \\ 3 \\ -2 \end{pmatrix} = \begin{pmatrix} 3 \\ 3 \\ 0 \end{pmatrix}$

③$D$ is $(3, 3, 0)$. Check: $\overrightarrow{DC} = (3, 1, 1)$ and $\overrightarrow{AB} = (3, 1, 1)$ ✓

The order of the letters matters: in $ABCD$ the vertices go round the shape, so $A$ is opposite $C$. A quick sketch avoids putting $D$ in the wrong place.

Worked Example 8 — Dividing a line in a ratio

$A(1, -2, 4)$ and $B(6, 8, -1)$. Find the point $P$ on $AB$ with $AP : PB = 2 : 3$.

①$P$ is $\tfrac{2}{5}$ of the way from $A$ to $B$.

②$\overrightarrow{AB} = \begin{pmatrix} 5 \\ 10 \\ -5 \end{pmatrix}$, so $\tfrac{2}{5}\overrightarrow{AB} = \begin{pmatrix} 2 \\ 4 \\ -2 \end{pmatrix}$.

③$\mathbf{p} = \begin{pmatrix} 1 \\ -2 \\ 4 \end{pmatrix} + \begin{pmatrix} 2 \\ 4 \\ -2 \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \\ 2 \end{pmatrix}$, so $P$ is $(3, 2, 2)$.

5 Forces as 3-D Vectors

Forces, velocities and accelerations are vectors, so in 3-D they are written with $\mathbf{i}$, $\mathbf{j}$ and $\mathbf{k}$ components. The mechanics is the same as in 2-D:

Resultant and Newton's second law

$\mathbf{R} = \mathbf{F}_1 + \mathbf{F}_2 + \cdots$ · $\mathbf{R} = m\mathbf{a}$ · equilibrium: $\mathbf{R} = \mathbf{0}$

Worked Example 9 — Resultant, acceleration and equilibrium

Forces $\mathbf{F}_1 = (2\mathbf{i} - \mathbf{j} + 3\mathbf{k})$ N and $\mathbf{F}_2 = (-4\mathbf{i} + 3\mathbf{j} + \mathbf{k})$ N act on a particle of mass $2$ kg. (a) Find the magnitude of the acceleration. (b) Find the third force $\mathbf{F}_3$ that would keep the particle in equilibrium.

①$\mathbf{R} = \mathbf{F}_1 + \mathbf{F}_2 = (-2\mathbf{i} + 2\mathbf{j} + 4\mathbf{k})$ N

②$\mathbf{a} = \dfrac{\mathbf{R}}{m} = (-\mathbf{i} + \mathbf{j} + 2\mathbf{k})$ m s$^{-2}$

③$|\mathbf{a}| = \sqrt{1 + 1 + 4} = \sqrt{6} \approx 2.45$ m s$^{-2}$

④(b) Equilibrium needs $\mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_3 = \mathbf{0}$, so $\mathbf{F}_3 = -\mathbf{R} = (2\mathbf{i} - 2\mathbf{j} - 4\mathbf{k})$ N.

⑤$|\mathbf{F}_3| = \sqrt{4 + 4 + 16} = \sqrt{24} = 2\sqrt{6} \approx 4.90$ N

Not on these pages: the scalar (dot) product and vector equations of lines are Further Mathematics topics. A Level Mathematics only needs the ideas above.

6 Quick Reference

Notation

$a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}$ or a column of three.

Magnitude

$\sqrt{a_1^2 + a_2^2 + a_3^2}$

Unit vector

$\hat{\mathbf{a}} = \mathbf{a} \div |\mathbf{a}|$

Between points

$\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$; distance $= |\mathbf{b} - \mathbf{a}|$.

Midpoint

$\tfrac{1}{2}(\mathbf{a} + \mathbf{b})$

Parallel

One vector is a scalar multiple of the other.

Collinear

Parallel and a common point.

Parallelogram

$\overrightarrow{AB} = \overrightarrow{DC}$

Ratio m : n

$\mathbf{a} + \tfrac{m}{m+n}(\mathbf{b} - \mathbf{a})$

Angle with axis

$\cos\theta_x = \dfrac{a_1}{|\mathbf{a}|}$, and similarly for $y$, $z$.

7 Practice Questions

Question 1

Find the magnitude of $4\mathbf{i} - 4\mathbf{j} + 2\mathbf{k}$.

▶ Show solution

$\sqrt{16 + 16 + 4} = \sqrt{36} = 6$

Question 2

Find a unit vector in the direction of $\begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix}$.

▶ Show solution

Magnitude $\sqrt{1 + 4 + 4} = 3$.

Unit vector $\tfrac{1}{3}\begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix} = \begin{pmatrix} 1/3 \\ -2/3 \\ 2/3 \end{pmatrix}$

Question 3

Find the distance between the points $(3, -1, 2)$ and $(-1, 2, 14)$.

▶ Show solution

Difference: $(-4, 3, 12)$.

Distance $= \sqrt{16 + 9 + 144} = \sqrt{169} = 13$

Question 4

$A$ is $(2, 0, -1)$ and $B$ is $(5, 4, -1)$. Find $|\overrightarrow{AB}|$ and a vector of magnitude $15$ in the direction of $\overrightarrow{AB}$.

▶ Show solution

$\overrightarrow{AB} = 3\mathbf{i} + 4\mathbf{j} + 0\mathbf{k}$, $|\overrightarrow{AB}| = \sqrt{9 + 16} = 5$.

Magnitude $15$ is $3$ times as long: $9\mathbf{i} + 12\mathbf{j}$.

Question 5

The vector $t\mathbf{i} + 2\mathbf{j} + (t - 1)\mathbf{k}$ has magnitude $3$. Find the possible values of $t$.

▶ Show solution

$t^2 + 4 + (t-1)^2 = 9$

$2t^2 - 2t + 5 = 9 \Rightarrow 2t^2 - 2t - 4 = 0 \Rightarrow t^2 - t - 2 = 0$

$(t - 2)(t + 1) = 0$, so $t = 2$ or $t = -1$.

Question 6

$A(1, 1, 1)$, $B(2, 3, 0)$ and $C(5, 9, p)$ are collinear. Find $p$.

▶ Show solution

$\overrightarrow{AB} = (1, 2, -1)$ and $\overrightarrow{AC} = (4, 8, p - 1)$.

The $x$ and $y$ components show $\overrightarrow{AC} = 4\overrightarrow{AB}$, so $p - 1 = -4$ and $p = -3$.

Question 7

$ABCD$ is a parallelogram with $A(2, 1, 0)$, $B(5, 3, 1)$ and $D(1, 4, 2)$. Find $C$, and the lengths of both diagonals in exact form.

▶ Show solution

$\overrightarrow{BC} = \overrightarrow{AD} = (-1, 3, 2)$, so $\mathbf{c} = (5, 3, 1) + (-1, 3, 2) = (4, 6, 3)$.

$\overrightarrow{AC} = (2, 5, 3)$, $|\overrightarrow{AC}| = \sqrt{4 + 25 + 9} = \sqrt{38}$.

$\overrightarrow{BD} = (-4, 1, 1)$, $|\overrightarrow{BD}| = \sqrt{16 + 1 + 1} = \sqrt{18} = 3\sqrt{2}$.

Question 8

$A(1, 2, 3)$, $B(4, 6, 3)$ and $C(1, 2, 15)$. Show that triangle $ABC$ is right-angled, and find its area.

▶ Show solution

$\overrightarrow{AB} = (3, 4, 0)$, $AB = 5$. $\overrightarrow{AC} = (0, 0, 12)$, $AC = 12$. $\overrightarrow{BC} = (-3, -4, 12)$, $BC = \sqrt{9 + 16 + 144} = 13$.

$5^2 + 12^2 = 25 + 144 = 169 = 13^2$, so by the converse of Pythagoras the triangle is right-angled at $A$.

Area $= \tfrac{1}{2} \times 5 \times 12 = 30$ square units.

Question 9

Three forces $\mathbf{F}_1 = (3\mathbf{i} + 2\mathbf{j} - \mathbf{k})$ N, $\mathbf{F}_2 = (-\mathbf{i} + 4\mathbf{j} + 5\mathbf{k})$ N and $\mathbf{F}_3$ act on a particle of mass $0.5$ kg, which accelerates at $(2\mathbf{i} - 2\mathbf{j} + 4\mathbf{k})$ m s$^{-2}$. Find $\mathbf{F}_3$ and its magnitude.

▶ Show solution

Resultant $= m\mathbf{a} = 0.5(2\mathbf{i} - 2\mathbf{j} + 4\mathbf{k}) = \mathbf{i} - \mathbf{j} + 2\mathbf{k}$.

$\mathbf{F}_1 + \mathbf{F}_2 = 2\mathbf{i} + 6\mathbf{j} + 4\mathbf{k}$.

$\mathbf{F}_3 = (\mathbf{i} - \mathbf{j} + 2\mathbf{k}) - (2\mathbf{i} + 6\mathbf{j} + 4\mathbf{k}) = (-\mathbf{i} - 7\mathbf{j} - 2\mathbf{k})$ N

$|\mathbf{F}_3| = \sqrt{1 + 49 + 4} = \sqrt{54} = 3\sqrt{6} \approx 7.35$ N

Question 10

Triangle $ABC$ has vertices $A(2, -1, 3)$, $B(6, 1, -1)$ and $C(-2, 5, 3)$. $M$ is the midpoint of $AB$ and $N$ is the midpoint of $BC$.

(a) Find the length $AB$. (b) Find the coordinates of $M$ and $N$. (c) The point $G$ lies on $CM$ with $CG : GM = 2 : 1$. Find $G$. (d) Show that $A$, $G$ and $N$ are collinear, and find the ratio $AG : GN$.

▶ Show solution

(a) $\overrightarrow{AB} = (4, 2, -4)$, $AB = \sqrt{16 + 4 + 16} = 6$.

(b) $\mathbf{m} = \tfrac{1}{2}(\mathbf{a} + \mathbf{b}) = (4, 0, 1)$; $\mathbf{n} = \tfrac{1}{2}(\mathbf{b} + \mathbf{c}) = (2, 3, 1)$.

(c) $\overrightarrow{CM} = (6, -5, -2)$, and $G$ is $\tfrac{2}{3}$ of the way from $C$ to $M$:

$\mathbf{g} = (-2, 5, 3) + \tfrac{2}{3}(6, -5, -2) = \left(2, \tfrac{5}{3}, \tfrac{5}{3}\right)$

(d) $\overrightarrow{AG} = \left(0, \tfrac{8}{3}, -\tfrac{4}{3}\right)$ and $\overrightarrow{AN} = (0, 4, -2)$, so $\overrightarrow{AG} = \tfrac{2}{3}\overrightarrow{AN}$.

They are parallel and share the point $A$, so $A$, $G$, $N$ are collinear. $G$ is $\tfrac{2}{3}$ of the way along $AN$, so $AG : GN = 2 : 1$.

($G$ is the centroid: all three medians pass through it, and $\mathbf{g} = \tfrac{1}{3}(\mathbf{a} + \mathbf{b} + \mathbf{c})$ — check: $\tfrac{1}{3}(6, 5, 5)$ ✓.)

Vectors in Three Dimensions · A Level Maths Year 2 · Created with MathJax