Vectors and Translations

A vector has both a size (magnitude) and a direction.
A scalar has size only.

Vector quantities	Scalar quantities
Displacement (5 km north)	Distance (5 km)
Velocity (30 m/s east)	Speed (30 m/s)
Force (10 N downwards)	Mass (10 kg)

Column vector

$\begin{pmatrix} x \\ y \end{pmatrix}$ — $x$ is the movement across, $y$ is the movement up

Signs: positive $x$ means right, negative $x$ means left. Positive $y$ means up, negative $y$ means down.
So $\begin{pmatrix} -3 \\ 5 \end{pmatrix}$ means $3$ left and $5$ up.

Notation	Meaning
$\overrightarrow{AB}$	The vector from point $A$ to point $B$ — note the arrow direction
$\mathbf{a}$ (bold)	A named vector, in print
$\underline{a}$ (underlined)	The same thing, written by hand — always underline
$\|\mathbf{a}\|$ or $\|\overrightarrow{AB}\|$	The magnitude (length) of the vector
$-\mathbf{a}$	The same length but the opposite direction

Always underline your vectors when writing by hand. Bold cannot be written with a pen, so the underline is how you show that $\underline{a}$ is a vector rather than an ordinary number $a$.

The reversal rule

$\overrightarrow{BA} = -\overrightarrow{AB}$

Worked Example 1 — Reversing a vector

$\overrightarrow{PQ} = \begin{pmatrix} 5 \\ -2 \end{pmatrix}$. Write down $\overrightarrow{QP}$.

①Going backwards means reversing both components.

②$\overrightarrow{QP} = \begin{pmatrix} -5 \\ 2 \end{pmatrix}$

The rule

$\overrightarrow{AB} = \begin{pmatrix} x_B - x_A \\ y_B - y_A \end{pmatrix}$ — finish minus start

Worked Example 2 — Vector between two points

$A = (2,\ 7)$ and $B = (9,\ 3)$. Find $\overrightarrow{AB}$ and $\overrightarrow{BA}$.

①$x$: $9 - 2 = 7$

②$y$: $3 - 7 = -4$

③$\overrightarrow{AB} = \begin{pmatrix} 7 \\ -4 \end{pmatrix}$

④$\overrightarrow{BA} = \begin{pmatrix} -7 \\ 4 \end{pmatrix}$

Finish minus start, not the other way round. Doing start minus finish gives the vector pointing backwards.

A translation slides every point of a shape by the same vector. It is the transformation that vectors were made for.

Translating a point

image $=$ object $+$ vector — add the components

Worked Example 3 — Applying a translation

Translate the triangle with vertices $(1,\ 2)$, $(4,\ 2)$ and $(1,\ 6)$ by $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$.

①$(1,\ 2) \to (1-2,\ 2+3) = (-1,\ 5)$

②$(4,\ 2) \to (4-2,\ 2+3) = (2,\ 5)$

③$(1,\ 6) \to (1-2,\ 6+3) = (-1,\ 9)$

The image is the same size and shape — a translation preserves everything except position.

Worked Example 4 — Finding the translation vector

A shape is translated so that the vertex at $(5,\ 1)$ moves to $(-2,\ 8)$. Find the translation vector.

①Image minus object.

②$x$: $-2 - 5 = -7$

③$y$: $8 - 1 = 7$

④Vector $= \begin{pmatrix} -7 \\ 7 \end{pmatrix}$

Use corresponding vertices. Comparing the top-left corner of one shape with the bottom-right of the other gives the wrong answer.

Worked Example 5 — Reversing a translation

Shape $B$ is the image of shape $A$ after a translation of $\begin{pmatrix} 4 \\ -6 \end{pmatrix}$. Describe the translation that maps $B$ back onto $A$.

①To undo a translation, reverse the vector.

②A translation by $\begin{pmatrix} -4 \\ 6 \end{pmatrix}$.

Magnitude (length)

$\left| \begin{pmatrix} x \\ y \end{pmatrix} \right| = \sqrt{x^2 + y^2}$

This is just Pythagoras. The vector is the hypotenuse of a right-angled triangle whose legs are $x$ and $y$.
The magnitude is always positive, whatever the signs of $x$ and $y$ — squaring removes them.

Worked Example 6 — Magnitude

Find the magnitude of $\mathbf{a} = \begin{pmatrix} 5 \\ 12 \end{pmatrix}$ and of $\mathbf{b} = \begin{pmatrix} -3 \\ 4 \end{pmatrix}$.

①$|\mathbf{a}| = \sqrt{5^2 + 12^2} = \sqrt{169} = 13$

②$|\mathbf{b}| = \sqrt{(-3)^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$

Note that $(-3)^2 = 9$, not $-9$.

Worked Example 7 — Distance between two points

$P = (1,\ 4)$ and $Q = (7,\ 12)$. Find the distance $PQ$.

①$\overrightarrow{PQ} = \begin{pmatrix} 6 \\ 8 \end{pmatrix}$

②$|\overrightarrow{PQ}| = \sqrt{36 + 64} = \sqrt{100} = 10$

Finding a distance is exactly the same as finding a magnitude.

Two vectors are equal if they have the same magnitude and the same direction. Their position on the page does not matter — a vector records a journey, not a place.

Two vectors are parallel if one is a multiple of the other.

Worked Example 8 — Testing for parallel vectors

Are $\begin{pmatrix} 3 \\ 2 \end{pmatrix}$ and $\begin{pmatrix} 9 \\ 6 \end{pmatrix}$ parallel?

①Is one a multiple of the other? Try $\times 3$: $\;3 \times 3 = 9$ ✓ and $3 \times 2 = 6$ ✓

②$\begin{pmatrix} 9 \\ 6 \end{pmatrix} = 3\begin{pmatrix} 3 \\ 2 \end{pmatrix}$

Yes, they are parallel, and the second is three times as long.

Both components must scale by the same factor. $\begin{pmatrix} 3 \\ 2 \end{pmatrix}$ and $\begin{pmatrix} 9 \\ 5 \end{pmatrix}$ are not parallel.

Worked Example 9 — Equal vectors in a shape

$ABCD$ is a parallelogram. Explain why $\overrightarrow{AB} = \overrightarrow{DC}$.

①In a parallelogram, $AB$ and $DC$ are opposite sides, so they are equal in length.

②They are also parallel, and going $A \to B$ points the same way as going $D \to C$.

③Same magnitude and same direction, so the vectors are equal. ∎

Note that $\overrightarrow{AB} = -\overrightarrow{CD}$, because $C \to D$ points the opposite way.

Vector vs scalar

A vector has size and direction; a scalar has size only.

Column vector

$\begin{pmatrix} x \\ y \end{pmatrix}$: $x$ across, $y$ up. Left and down are negative.

Notation

$\overrightarrow{AB}$, bold $\mathbf{a}$, or underlined $\underline{a}$ by hand.

Reversing

$\overrightarrow{BA} = -\overrightarrow{AB}$ — change both signs.

Between points

$\overrightarrow{AB} = $ finish $-$ start.

Translation

Image $=$ object $+$ vector.

Magnitude

$\sqrt{x^2 + y^2}$ — always positive.

Parallel

One vector is a scalar multiple of the other.

Equal

Same magnitude and direction; position is irrelevant.

Question 1

Describe in words the movement given by (a) $\begin{pmatrix} 4 \\ -3 \end{pmatrix}$, (b) $\begin{pmatrix} -6 \\ 0 \end{pmatrix}$.

▶ Show solution

(a) $4$ units right and $3$ units down.

(b) $6$ units left, with no vertical movement.

Question 2

$A = (3,\ 1)$ and $B = (8,\ 9)$. Find $\overrightarrow{AB}$ and $\overrightarrow{BA}$.

▶ Show solution

$\overrightarrow{AB} = \begin{pmatrix} 8-3 \\ 9-1 \end{pmatrix} = \begin{pmatrix} 5 \\ 8 \end{pmatrix}$

$\overrightarrow{BA} = \begin{pmatrix} -5 \\ -8 \end{pmatrix}$

Question 3

Translate the point $(6,\ -2)$ by the vector $\begin{pmatrix} -9 \\ 5 \end{pmatrix}$.

▶ Show solution

$x$: $6 + (-9) = -3$

$y$: $-2 + 5 = 3$

Image $= (-3,\ 3)$

Question 4

Find the magnitude of (a) $\begin{pmatrix} 8 \\ 15 \end{pmatrix}$, (b) $\begin{pmatrix} -7 \\ -24 \end{pmatrix}$.

▶ Show solution

(a) $\sqrt{64 + 225} = \sqrt{289} = 17$

(b) $\sqrt{49 + 576} = \sqrt{625} = 25$

(Both are Pythagorean triples: $8,15,17$ and $7,24,25$.)

Question 5

A shape is translated so that $(2,\ 5)$ maps to $(11,\ -1)$. Find the translation vector, and the vector that maps the image back.

▶ Show solution

Forward: $\begin{pmatrix} 11-2 \\ -1-5 \end{pmatrix} = \begin{pmatrix} 9 \\ -6 \end{pmatrix}$

Back: $\begin{pmatrix} -9 \\ 6 \end{pmatrix}$

Question 6

Which of these vectors are parallel to $\begin{pmatrix} 2 \\ -5 \end{pmatrix}$?

$\begin{pmatrix} 4 \\ -10 \end{pmatrix}$, $\begin{pmatrix} -6 \\ 15 \end{pmatrix}$, $\begin{pmatrix} 4 \\ -5 \end{pmatrix}$

▶ Show solution

$\begin{pmatrix} 4 \\ -10 \end{pmatrix} = 2\begin{pmatrix} 2 \\ -5 \end{pmatrix}$ ✓ parallel

$\begin{pmatrix} -6 \\ 15 \end{pmatrix} = -3\begin{pmatrix} 2 \\ -5 \end{pmatrix}$ ✓ parallel (pointing the opposite way)

$\begin{pmatrix} 4 \\ -5 \end{pmatrix}$: the $x$ doubled but the $y$ did not ✗ not parallel

Question 7

$P = (-4,\ 6)$ and $Q = (8,\ 1)$. Find the exact distance $PQ$.

▶ Show solution

$\overrightarrow{PQ} = \begin{pmatrix} 12 \\ -5 \end{pmatrix}$

$|PQ| = \sqrt{144 + 25} = \sqrt{169} = 13$

Question 8

$\overrightarrow{AB} = \begin{pmatrix} 3 \\ 7 \end{pmatrix}$ and $A = (2,\ -1)$. Find the coordinates of $B$.

▶ Show solution

$B = A + \overrightarrow{AB}$

$x$: $2 + 3 = 5$; $y$: $-1 + 7 = 6$

$B = (5,\ 6)$

Question 9

$ABCD$ is a parallelogram with $A = (1,\ 2)$, $B = (5,\ 3)$ and $D = (2,\ 7)$.

(a) Find $\overrightarrow{AB}$. (b) Use the fact that $\overrightarrow{DC} = \overrightarrow{AB}$ to find $C$. (c) Find the length $AB$, to 2 d.p.

▶ Show solution

(a) $\overrightarrow{AB} = \begin{pmatrix} 5-1 \\ 3-2 \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \end{pmatrix}$

(b) $C = D + \overrightarrow{DC} = (2 + 4,\ 7 + 1) = (6,\ 8)$

(c) $|AB| = \sqrt{16 + 1} = \sqrt{17} = 4.12$

Question 10

A boat's journey is described by two successive translations: $\begin{pmatrix} 8 \\ 6 \end{pmatrix}$ km, then $\begin{pmatrix} -3 \\ 9 \end{pmatrix}$ km.

(a) Find the single vector describing the whole journey. (b) Find the direct distance from start to finish, to 2 d.p. (c) Find the total distance actually travelled, to 2 d.p. (d) Explain why your answers to (b) and (c) differ.

▶ Show solution

(a) Add the components: $\begin{pmatrix} 8 + (-3) \\ 6 + 9 \end{pmatrix} = \begin{pmatrix} 5 \\ 15 \end{pmatrix}$

(b) $\sqrt{25 + 225} = \sqrt{250} = 15.81$ km

(c) First leg: $\sqrt{64 + 36} = \sqrt{100} = 10$ km

Second leg: $\sqrt{9 + 81} = \sqrt{90} = 9.4868$ km

Total travelled $= 10 + 9.4868 = 19.49$ km

(d) The displacement (b) is the straight-line vector from start to finish, while the distance travelled (c) follows the actual path. Because the boat changed direction, the path is longer than the straight line.