E-Math notes · Chapter 16

Vectors in Two Dimensions: O-Level E-Math Notes

A vector is a journey: how far, and which way. In this chapter you learn to write vectors as columns, move points with them, find lengths, add and subtract vectors, multiply by a number, and use all of that to solve geometry problems about parallel lines, midpoints and collinear points.

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Chapter
16

Figure 1 · The two legs of the flight drawn as arrows from

O

. One grid square is 10 metres.

8 outcomes · 25 worked examples · every step shown ✦

Maths in real life

When something moves in two straight legs, the single move from start to finish is the sum of the two leg vectors. That is the triangle law.

Notation and directed segments

A vector is a journey, not a place.

Translation and position vectors

Translation by $\begin{pmatrix}p\\q\end{pmatrix}$ adds $p$ to every $x$ -coordinate and $q$ to every $y$ -coordinate.

By the end of this chapter

Write and read vectors in column form, as directed line segments, and with the notation $\overrightarrow{AB}$ , $\mathbf{a}$ , $|\overrightarrow{AB}|$ and $|\mathbf{a}|$ .
Describe a translation by a vector, and use position vectors, including $\overrightarrow{AB}=\mathbf{b}-\mathbf{a}$ .
Find the magnitude of a vector.
Add and subtract vectors with the triangle and parallelogram laws, and multiply a vector by a scalar.
Decide when vectors are parallel and when points are collinear.
Solve geometry problems by expressing vectors in terms of two others.

Supports G7.6 · G7.5 · G7.4 · G7.7 · G7.8

Maths in real life: where vectors are used

Vectors are how we describe movement when direction matters. Three real situations come first, so you can see where the chapter is going. Each one is solved in full, with a Your Turn that has its own numbers. Come back to them when the ideas are new.

A drone flies two legs

When something moves in two straight legs, the single move from start to finish is the sum of the two leg vectors. That is the triangle law.

Example 1.1

The delivery drone

A delivery drone leaves its launch pad $O$ and flies $\begin{pmatrix}90\\40\end{pmatrix}$ metres to point $A$ , then $\begin{pmatrix}-50\\80\end{pmatrix}$ metres to point $B$ . The top number is metres east and the bottom number is metres north.

(i)Find the position vector of $B$ from $O$ .
(ii)Find the straight-line distance $OB$ to three significant figures.
(iii)Write the vector for the flight straight back from $B$ to $O$ .

Show solutionHide solution

Colour keyadd, subtract, multiply, divide or simplifyexpand, rewrite, substitute or solveround, estimate or compare

(i)

\displaystyle \overrightarrow{OB}\;=\overrightarrow{OA}+\overrightarrow{AB}

follow the path, head to tail

\displaystyle =\begin{pmatrix}90\\40\end{pmatrix}+\begin{pmatrix}-50\\80\end{pmatrix}

\displaystyle =\begin{pmatrix}40\\120\end{pmatrix}

add all the top numbers, then all the bottom numbers

(ii)

\displaystyle |\overrightarrow{OB}|\;=\sqrt{40^{2}+120^{2}}

square each component, add, then take the root

\displaystyle =\sqrt{1600+14400}

\displaystyle =\sqrt{16000}

\displaystyle \approx126

three significant figures

(iii)

\displaystyle \overrightarrow{BO}\;=-\overrightarrow{OB}=-\begin{pmatrix}40\\120\end{pmatrix}=\begin{pmatrix}-40\\-120\end{pmatrix}

reverse the arrow: change the sign of both numbers

The two legs of the flight drawn as arrows from

O

. One grid square is 10 metres.

Final answer

(i)

\overrightarrow{OB}=\begin{pmatrix}40\\120\end{pmatrix}

metres

(ii)

|\overrightarrow{OB}|=40 \sqrt{10}\approx126

metres

(iii)

\overrightarrow{BO}=\begin{pmatrix}-40\\-120\end{pmatrix}

metres

Your Turn 1.1

A delivery drone leaves its launch pad $O$ and flies $\begin{pmatrix}30\\70\end{pmatrix}$ metres to point $A$ , then $\begin{pmatrix}-10\\-40\end{pmatrix}$ metres to point $B$ . The top number is metres east and the bottom number is metres north.

(i)Find the position vector of $B$ from $O$ .
(ii)Find the straight-line distance $OB$ to three significant figures.
(iii)Write the vector for the flight straight back from $B$ to $O$ .

Show worked solutionHide worked solution

Colour keyadd, subtract, multiply, divide or simplifyexpand, rewrite, substitute or solveround, estimate or compare

(i)

\displaystyle \overrightarrow{OB}\;=\overrightarrow{OA}+\overrightarrow{AB}

follow the path, head to tail

\displaystyle =\begin{pmatrix}30\\70\end{pmatrix}+\begin{pmatrix}-10\\-40\end{pmatrix}

\displaystyle =\begin{pmatrix}20\\30\end{pmatrix}

add all the top numbers, then all the bottom numbers

(ii)

\displaystyle |\overrightarrow{OB}|\;=\sqrt{20^{2}+30^{2}}

square each component, add, then take the root

\displaystyle =\sqrt{400+900}

\displaystyle =\sqrt{1300}

\displaystyle \approx36.1

three significant figures

(iii)

\displaystyle \overrightarrow{BO}\;=-\overrightarrow{OB}=-\begin{pmatrix}20\\30\end{pmatrix}=\begin{pmatrix}-20\\-30\end{pmatrix}

reverse the arrow: change the sign of both numbers

Final answer

(i)

\overrightarrow{OB}=\begin{pmatrix}20\\30\end{pmatrix}

metres

(ii)

|\overrightarrow{OB}|=10 \sqrt{13}\approx36.1

metres

(iii)

\overrightarrow{BO}=\begin{pmatrix}-20\\-30\end{pmatrix}

metres

A ferry in a current

Two movements that happen at the same time combine by adding their vectors, so a ferry's true velocity is its own velocity plus the current.

Example 1.2

The ferry and the current

A ferry steers directly north through water that flows east. In column form the ferry's own velocity is $\begin{pmatrix}0\\12\end{pmatrix}$ km/h and the current is $\begin{pmatrix}5\\0\end{pmatrix}$ km/h (east on top, north underneath). The true velocity is the sum of the two.

(i)Find the true velocity as a column vector.
(ii)Find the true speed.
(iii)Find the displacement after 2 hours.
(iv)How far is the ferry from where it started?

Show solutionHide solution

Colour keyadd, subtract, multiply, divide or simplify

(i)

\displaystyle \mathbf{f}+\mathbf{c}\;=\begin{pmatrix}0\\12\end{pmatrix}+\begin{pmatrix}5\\0\end{pmatrix}

add across, add up

\displaystyle =\begin{pmatrix}5\\12\end{pmatrix}

(ii)

\displaystyle |\mathbf{v}|\;=\sqrt{5^{2}+12^{2}}

square each component, add, then take the root

\displaystyle =\sqrt{25+144}

\displaystyle =\sqrt{169}

\displaystyle =13

a perfect square, so the magnitude is exact

(iii)

\displaystyle 2\mathbf{v}\;=2\begin{pmatrix}5\\12\end{pmatrix}

multiply both components by the number

\displaystyle =\begin{pmatrix}10\\24\end{pmatrix}

(iv)

\displaystyle |\mathbf{d}|\;=\sqrt{10^{2}+24^{2}}

square each component, add, then take the root

\displaystyle =\sqrt{100+576}

\displaystyle =\sqrt{676}

\displaystyle =26

a perfect square, so the magnitude is exact

Ferry velocity (up), current (right) and the true velocity (diagonal): the triangle law.

Final answer

(i)

\begin{pmatrix}5\\12\end{pmatrix}

km/h

(ii)

|\mathbf{v}|=13

km/h

(iii)

\begin{pmatrix}10\\24\end{pmatrix}

(iv)

|\mathbf{d}|=26

Your Turn 1.2

A ferry steers directly north through water that flows east. In column form the ferry's own velocity is $\begin{pmatrix}0\\15\end{pmatrix}$ km/h and the current is $\begin{pmatrix}8\\0\end{pmatrix}$ km/h (east on top, north underneath). The true velocity is the sum of the two.

(i)Find the true velocity as a column vector.
(ii)Find the true speed.
(iii)Find the displacement after 3 hours.
(iv)How far is the ferry from where it started?

Show worked solutionHide worked solution

Colour keyadd, subtract, multiply, divide or simplify

(i)

\displaystyle \mathbf{f}+\mathbf{c}\;=\begin{pmatrix}0\\15\end{pmatrix}+\begin{pmatrix}8\\0\end{pmatrix}

add across, add up

\displaystyle =\begin{pmatrix}8\\15\end{pmatrix}

(ii)

\displaystyle |\mathbf{v}|\;=\sqrt{8^{2}+15^{2}}

square each component, add, then take the root

\displaystyle =\sqrt{64+225}

\displaystyle =\sqrt{289}

\displaystyle =17

a perfect square, so the magnitude is exact

(iii)

\displaystyle 3\mathbf{v}\;=3\begin{pmatrix}8\\15\end{pmatrix}

multiply both components by the number

\displaystyle =\begin{pmatrix}24\\45\end{pmatrix}

(iv)

\displaystyle |\mathbf{d}|\;=\sqrt{24^{2}+45^{2}}

square each component, add, then take the root

\displaystyle =\sqrt{576+2025}

\displaystyle =\sqrt{2601}

\displaystyle =51

a perfect square, so the magnitude is exact

Final answer

(i)

\begin{pmatrix}8\\15\end{pmatrix}

km/h

(ii)

|\mathbf{v}|=17

km/h

(iii)

\begin{pmatrix}24\\45\end{pmatrix}

(iv)

|\mathbf{d}|=51

A point that divides a string

A point that divides a line in a ratio is found by taking that fraction of the line's vector from the starting end.

Example 1.3

Lanterns on a string

A string of lanterns is hung from the top of pole $A=(2,\ 3)$ to the top of pole $B=(12,\ 8)$ (coordinates in metres). The red lantern is at $P$ on the string, where $AP:PB=2:3$ .

(i)Find $\overrightarrow{AB}$ .
(ii)Find the coordinates of $P$ .
(iii)Find the length $AP$ to three significant figures.

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Colour keyadd, subtract, multiply, divide or simplifyround, estimate or compare

(i)

\displaystyle \overrightarrow{AB}\;=\overrightarrow{OB}-\overrightarrow{OA}=\begin{pmatrix}12\\8\end{pmatrix}-\begin{pmatrix}2\\3\end{pmatrix}

end minus start

\displaystyle =\begin{pmatrix}10\\5\end{pmatrix}

subtract the top numbers, then the bottom numbers

(ii)

\displaystyle \overrightarrow{AP}\;=\frac{2}{5}\overrightarrow{AB}=\frac{2}{5}\begin{pmatrix}10\\5\end{pmatrix}=\begin{pmatrix}4\\2\end{pmatrix}

AP:PB=2:3

, so

AP

is 2 of 5 equal parts of

AB

\displaystyle \overrightarrow{OP}\;=\overrightarrow{OA}+\overrightarrow{AP}=\begin{pmatrix}2\\3\end{pmatrix}+\begin{pmatrix}4\\2\end{pmatrix}=\begin{pmatrix}6\\5\end{pmatrix}

start from a vector you know

(iii)

\displaystyle |\overrightarrow{AP}|\;=\sqrt{4^{2}+2^{2}}

square each component, add, then take the root

\displaystyle =\sqrt{16+4}

\displaystyle =\sqrt{20}

\displaystyle \approx4.47

three significant figures

The string

AB

with the lantern at

P

Final answer

(i)

\overrightarrow{AB}=\begin{pmatrix}10\\5\end{pmatrix}

metres

(ii)

P=(6,\ 5)

(iii)

|\overrightarrow{AP}|=2 \sqrt{5}\approx4.47

metres

Your Turn 1.3

A string of lanterns is hung from the top of pole $A=(1,\ 2)$ to the top of pole $B=(11,\ 12)$ (coordinates in metres). The red lantern is at $P$ on the string, where $AP:PB=3:2$ .

(i)Find $\overrightarrow{AB}$ .
(ii)Find the coordinates of $P$ .
(iii)Find the length $AP$ to three significant figures.

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Colour keyadd, subtract, multiply, divide or simplifyround, estimate or compare

(i)

\displaystyle \overrightarrow{AB}\;=\overrightarrow{OB}-\overrightarrow{OA}=\begin{pmatrix}11\\12\end{pmatrix}-\begin{pmatrix}1\\2\end{pmatrix}

end minus start

\displaystyle =\begin{pmatrix}10\\10\end{pmatrix}

subtract the top numbers, then the bottom numbers

(ii)

\displaystyle \overrightarrow{AP}\;=\frac{3}{5}\overrightarrow{AB}=\frac{3}{5}\begin{pmatrix}10\\10\end{pmatrix}=\begin{pmatrix}6\\6\end{pmatrix}

AP:PB=3:2

, so

AP

is 3 of 5 equal parts of

AB

\displaystyle \overrightarrow{OP}\;=\overrightarrow{OA}+\overrightarrow{AP}=\begin{pmatrix}1\\2\end{pmatrix}+\begin{pmatrix}6\\6\end{pmatrix}=\begin{pmatrix}7\\8\end{pmatrix}

start from a vector you know

(iii)

\displaystyle |\overrightarrow{AP}|\;=\sqrt{6^{2}+6^{2}}

square each component, add, then take the root

\displaystyle =\sqrt{36+36}

\displaystyle =\sqrt{72}

\displaystyle \approx8.49

three significant figures

Final answer

(i)

\overrightarrow{AB}=\begin{pmatrix}10\\10\end{pmatrix}

metres

(ii)

P=(7,\ 8)

(iii)

|\overrightarrow{AP}|=6 \sqrt{2}\approx8.49

metres

Think About It

Why does the drone's straight-line distance come out shorter than the total length of its two legs?

Show the worked answerHide the worked answer

\displaystyle |\overrightarrow{OA}|\;=\sqrt{90^{2}+40^{2}}=\sqrt{9700}\approx98.5

first leg

\displaystyle |\overrightarrow{AB}|\;=\sqrt{(-50)^{2}+80^{2}}=\sqrt{8900}\approx94.3

second leg

\displaystyle |\overrightarrow{OB}|\approx126<98.5+94.3

the straight line is shorter, as the triangle law predicts

Final answerThe total flown is the length of

OA

plus the length of

AB

. The straight line

OB

is one side of the triangle

OAB

, and one side of a triangle is always shorter than the other two sides together (when the legs are not in the same direction). With the numbers in the first problem the legs are 98.5 and 94.3 long (to three significant figures) but

OB

is only 126.

Supports G7.1 · G7.2

Vectors, scalars and the notation

Quick revision. A scalar has size only, like the mass of a bag of rice or the time a bus takes. A vector has size and direction: 'move 3 steps right and 2 steps up' is a vector, because both the amount and the way matter.

The Core Idea

A vector is a journey, not a place.

It says how far and which way to move. Two arrows of the same length pointing the same way are the same vector, wherever they are drawn.

Reading and writing vectors

Table 1: the four pieces of vector notation
Written as	Read as	Example
$\begin{pmatrix}x\\y\end{pmatrix}$	a column vector: $x$ units across (right if positive, left if negative) and $y$ units up (down if negative)	$\begin{pmatrix}3\\2\end{pmatrix}$ means 3 right, 2 up
$\overrightarrow{AB}$	the vector from $A$ to $B$ . The letters are in travel order	$\overrightarrow{AB}$ and $\overrightarrow{BA}$ are different vectors
$\mathbf{a}$	a vector given one letter. Bold in print. In your own writing, underline it	$\overrightarrow{AB}=\mathbf{a}$
$\|\overrightarrow{AB}\|$ or $\|\mathbf{a}\|$	the magnitude (length). It is a plain number, not an arrow	found in a later section

The vector

\mathbf{a}

drawn as a directed line segment: 3 units right and 2 units up, written

\begin{pmatrix}3\\2\end{pmatrix}

A vector drawn as an arrow is a directed line segment: a straight line with a start point, an end point and an arrowhead showing the direction. A vector only records the move, so you can slide the arrow anywhere without changing the vector.

Example 2.1

Reading a vector from a grid

On a square grid, $P$ is at $(1,\ 1)$ and $Q$ is at $(5,\ 4)$ .

(i)Write $\overrightarrow{PQ}$ as a column vector.
(ii)Write $\overrightarrow{QP}$ as a column vector.

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Colour keyadd, subtract, multiply, divide or simplifyexpand, rewrite, substitute or solve

(i)

\displaystyle \overrightarrow{PQ}\;=\overrightarrow{OQ}-\overrightarrow{OP}=\begin{pmatrix}5\\4\end{pmatrix}-\begin{pmatrix}1\\1\end{pmatrix}

count across, then up (or end minus start)

\displaystyle =\begin{pmatrix}4\\3\end{pmatrix}

subtract the top numbers, then the bottom numbers

(ii)

\displaystyle \overrightarrow{QP}\;=-\overrightarrow{PQ}=-\begin{pmatrix}4\\3\end{pmatrix}=\begin{pmatrix}-4\\-3\end{pmatrix}

reverse the arrow: change the sign of both numbers

The arrow from

P

Q

Final answer

(i)

\overrightarrow{PQ}=\begin{pmatrix}4\\3\end{pmatrix}

(ii)

\overrightarrow{QP}=\begin{pmatrix}-4\\-3\end{pmatrix}

Your Turn 2.1

On a square grid, $P$ is at $(2,\ 5)$ and $Q$ is at $(6,\ 2)$ .

(i)Write $\overrightarrow{PQ}$ as a column vector.
(ii)Write $\overrightarrow{QP}$ as a column vector.

Show worked solutionHide worked solution

Colour keyadd, subtract, multiply, divide or simplifyexpand, rewrite, substitute or solve

(i)

\displaystyle \overrightarrow{PQ}\;=\overrightarrow{OQ}-\overrightarrow{OP}=\begin{pmatrix}6\\2\end{pmatrix}-\begin{pmatrix}2\\5\end{pmatrix}

count across, then up (or end minus start)

\displaystyle =\begin{pmatrix}4\\-3\end{pmatrix}

subtract the top numbers, then the bottom numbers

(ii)

\displaystyle \overrightarrow{QP}\;=-\overrightarrow{PQ}=-\begin{pmatrix}4\\-3\end{pmatrix}=\begin{pmatrix}-4\\3\end{pmatrix}

reverse the arrow: change the sign of both numbers

Final answer

(i)

\overrightarrow{PQ}=\begin{pmatrix}4\\-3\end{pmatrix}

(ii)

\overrightarrow{QP}=\begin{pmatrix}-4\\3\end{pmatrix}

Think About It

The vector from

A

B

and the vector from

B

A

: do they have the same length? Do they have the same direction?

Show the worked answerHide the worked answer

Colour keyadd, subtract, multiply, divide or simplify

\displaystyle \overrightarrow{AB}\;=\begin{pmatrix}3\\2\end{pmatrix},\qquad \overrightarrow{BA}=\begin{pmatrix}-3\\-2\end{pmatrix}

subtract in each order

\displaystyle \overrightarrow{BA}\;=-\overrightarrow{AB}

all the signs flip: opposite direction

\displaystyle |\overrightarrow{AB}|\;=|\overrightarrow{BA}|=\sqrt{13}

squaring removes the signs: same length

Final answerThey have the same length but opposite directions, so

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

. Check with

A=(1,\ 1)

and

B=(4,\ 3)

Equal vectors and negative vectors

The first two arrows are the same vector

\mathbf{a}=\begin{pmatrix}2\\1\end{pmatrix}

drawn in different places. The third arrow is

-\mathbf{a}

: same length, opposite direction.

SGSchoolKaki Tip

Two vectors are equal only when both components match. If one component matches and the other does not, they are different vectors.

Example 2.2

Equal, negative or neither?

Given $\mathbf{a}=\begin{pmatrix}4\\-1\end{pmatrix}$ and the four vectors: $\mathbf{p}=\begin{pmatrix}4\\-1\end{pmatrix}$ , $\mathbf{q}=\begin{pmatrix}-4\\1\end{pmatrix}$ , $\mathbf{r}=\begin{pmatrix}1\\-4\end{pmatrix}$ , $\mathbf{s}=\begin{pmatrix}8\\-2\end{pmatrix}$ .

(i)Which vector is equal to $\mathbf{a}$ ?
(ii)Which vector is $-\mathbf{a}$ ?
(iii)Which vector points the same way as $\mathbf{a}$ but is twice as long?

Show solutionHide solution

Colour keyadd, subtract, multiply, divide or simplifyexpand, rewrite, substitute or solve

(i)

\displaystyle \mathbf{p}\;=\begin{pmatrix}4\\-1\end{pmatrix}=\mathbf{a}

equal vectors have both components the same

It is

\mathbf{p}

(ii)

\displaystyle -\mathbf{a}\;=-\begin{pmatrix}4\\-1\end{pmatrix}=\begin{pmatrix}-4\\1\end{pmatrix}

change the sign of both components

It is

\mathbf{q}

(iii)

\displaystyle 2\mathbf{a}\;=2\begin{pmatrix}4\\-1\end{pmatrix}=\begin{pmatrix}8\\-2\end{pmatrix}

multiply both components by two

It is

\mathbf{s}

Final answer

(i)

\mathbf{p}

(ii)

\mathbf{q}

(iii)

\mathbf{s}

Your Turn 2.2

Given $\mathbf{a}=\begin{pmatrix}-3\\2\end{pmatrix}$ and the four vectors: $\mathbf{p}=\begin{pmatrix}2\\-3\end{pmatrix}$ , $\mathbf{q}=\begin{pmatrix}3\\-2\end{pmatrix}$ , $\mathbf{r}=\begin{pmatrix}-3\\2\end{pmatrix}$ , $\mathbf{s}=\begin{pmatrix}-6\\4\end{pmatrix}$ .

(i)Which vector is equal to $\mathbf{a}$ ?
(ii)Which vector is $-\mathbf{a}$ ?
(iii)Which vector points the same way as $\mathbf{a}$ but is twice as long?

Show worked solutionHide worked solution

Colour keyadd, subtract, multiply, divide or simplifyexpand, rewrite, substitute or solve

(i)

\displaystyle \mathbf{r}\;=\begin{pmatrix}-3\\2\end{pmatrix}=\mathbf{a}

equal vectors have both components the same

It is

\mathbf{r}

(ii)

\displaystyle -\mathbf{a}\;=-\begin{pmatrix}-3\\2\end{pmatrix}=\begin{pmatrix}3\\-2\end{pmatrix}

change the sign of both components

It is

\mathbf{q}

(iii)

\displaystyle 2\mathbf{a}\;=2\begin{pmatrix}-3\\2\end{pmatrix}=\begin{pmatrix}-6\\4\end{pmatrix}

multiply both components by two

It is

\mathbf{s}

Final answer

(i)

\mathbf{r}

(ii)

\mathbf{q}

(iii)

\mathbf{s}

Supports G7.3 · G7.4

Translation and position vectors

A translation slides every point of a shape the same distance in the same direction, so the shape keeps its size and its direction. A translation is described by one column vector.

The Core Idea

Translation by $\begin{pmatrix}p\\q\end{pmatrix}$ adds $p$ to every $x$ -coordinate and $q$ to every $y$ -coordinate.

The image of a point is its original position plus the translation vector.

Triangle

ABC

is translated by

\mathbf{t}=\begin{pmatrix}3\\-2\end{pmatrix}

: every point moves 3 right and 2 down, so the three arrows

AA'

BB'

and

CC'

are the same vector.

Example 3.1

Translating a point

$P$ is the point $(2,\ -1)$ . It is translated by the vector $\begin{pmatrix}-3\\4\end{pmatrix}$ .

(i)Find the coordinates of the image of $P$ .
(ii)The point $Q'=(5,\ 3)$ is the image of a point $Q$ under the same translation. Find $Q$ .

Show solutionHide solution

Colour keyadd, subtract, multiply, divide or simplify

(i)

\displaystyle \overrightarrow{OP'}\;=\begin{pmatrix}2\\-1\end{pmatrix}+\begin{pmatrix}-3\\4\end{pmatrix}

add the translation vector to the position vector of

P

\displaystyle =\begin{pmatrix}-1\\3\end{pmatrix}

add across, then add up

(ii)

\displaystyle \overrightarrow{OQ}\;=\begin{pmatrix}5\\3\end{pmatrix}-\begin{pmatrix}-3\\4\end{pmatrix}

go back: subtract the translation

\displaystyle =\begin{pmatrix}8\\-1\end{pmatrix}

Final answer

(i)The image of

P

P'=(-1,\ 3)

(ii)

Q=(8,\ -1)

Your Turn 3.1

$P$ is the point $(-4,\ 3)$ . It is translated by the vector $\begin{pmatrix}5\\-6\end{pmatrix}$ .

(i)Find the coordinates of the image of $P$ .
(ii)The point $Q'=(1,\ -2)$ is the image of a point $Q$ under the same translation. Find $Q$ .

Show worked solutionHide worked solution

Colour keyadd, subtract, multiply, divide or simplify

(i)

\displaystyle \overrightarrow{OP'}\;=\begin{pmatrix}-4\\3\end{pmatrix}+\begin{pmatrix}5\\-6\end{pmatrix}

add the translation vector to the position vector of

P

\displaystyle =\begin{pmatrix}1\\-3\end{pmatrix}

add across, then add up

(ii)

\displaystyle \overrightarrow{OQ}\;=\begin{pmatrix}1\\-2\end{pmatrix}-\begin{pmatrix}5\\-6\end{pmatrix}

go back: subtract the translation

\displaystyle =\begin{pmatrix}-4\\4\end{pmatrix}

Final answer

(i)The image of

P

P'=(1,\ -3)

(ii)

Q=(-4,\ 4)

Example 3.2

Finding the translation vector

A translation maps $A=(1,\ 2)$ to $A'=(6,\ -1)$ .

(i)Write down the translation as a column vector.
(ii)The same translation maps $B=(-2,\ 4)$ to $B'$ . Find $B'$ .

Show solutionHide solution

Colour keyadd, subtract, multiply, divide or simplify

(i)

\displaystyle \overrightarrow{AA'}\;=\begin{pmatrix}6\\-1\end{pmatrix}-\begin{pmatrix}1\\2\end{pmatrix}

image minus original: end minus start

\displaystyle =\begin{pmatrix}5\\-3\end{pmatrix}

(ii)

\displaystyle \overrightarrow{OB'}\;=\begin{pmatrix}-2\\4\end{pmatrix}+\begin{pmatrix}5\\-3\end{pmatrix}

add the translation vector to the position vector of

B

\displaystyle =\begin{pmatrix}3\\1\end{pmatrix}

add across, then add up

Final answer

(i)

\begin{pmatrix}5\\-3\end{pmatrix}

(ii)The image of

B

B'=(3,\ 1)

Your Turn 3.2

A translation maps $A=(-4,\ 3)$ to $A'=(2,\ -1)$ .

(i)Write down the translation as a column vector.
(ii)The same translation maps $B=(5,\ -2)$ to $B'$ . Find $B'$ .

Show worked solutionHide worked solution

Colour keyadd, subtract, multiply, divide or simplify

(i)

\displaystyle \overrightarrow{AA'}\;=\begin{pmatrix}2\\-1\end{pmatrix}-\begin{pmatrix}-4\\3\end{pmatrix}

image minus original: end minus start

\displaystyle =\begin{pmatrix}6\\-4\end{pmatrix}

(ii)

\displaystyle \overrightarrow{OB'}\;=\begin{pmatrix}5\\-2\end{pmatrix}+\begin{pmatrix}6\\-4\end{pmatrix}

add the translation vector to the position vector of

B

\displaystyle =\begin{pmatrix}11\\-6\end{pmatrix}

add across, then add up

Final answer

(i)

\begin{pmatrix}6\\-4\end{pmatrix}

(ii)The image of

B

B'=(11,\ -6)

Position vectors

Choose a fixed point $O$ , the origin. The position vector of a point $A$ is the vector $\overrightarrow{OA}$ , usually written $\mathbf{a}$ . If $A$ has coordinates $(x, y)$ then $\overrightarrow{OA}=\begin{pmatrix}x\\y\end{pmatrix}$ : a position vector is the point's coordinates written as a column.

The Core Idea

$\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=\mathbf{b}-\mathbf{a}$

End minus start. To get from $A$ to $B$ , go back to $O$ along $-\mathbf{a}$ and then out to $B$ along $\mathbf{b}$ .

A

and

B

have position vectors: the vector from

O

A

\mathbf{a}=\begin{pmatrix}2\\4\end{pmatrix}

and the vector from

O

B

\mathbf{b}=\begin{pmatrix}6\\2\end{pmatrix}

. The vector from

A

B

\mathbf{b}-\mathbf{a}=\begin{pmatrix}4\\-2\end{pmatrix}

SGSchoolKaki Tip

Do not write $\overrightarrow{AB}=\mathbf{a}-\mathbf{b}$ . The vector from $A$ to $B$ is end minus start: $\mathbf{b}-\mathbf{a}$ .

Example 3.3

Position vectors and the vector between two points

$A$ is the point $(2,\ 5)$ and $B$ is the point $(7,\ 1)$ . $O$ is the origin.

(i)Write down the position vector $\overrightarrow{OA}$ .
(ii)Write down the position vector $\overrightarrow{OB}$ .
(iii)Find $\overrightarrow{AB}$ .

Show solutionHide solution

Colour keyadd, subtract, multiply, divide or simplify

(i)

\displaystyle \overrightarrow{OA}\;=\begin{pmatrix}2\\5\end{pmatrix}

a position vector is just the coordinates written as a column

(ii)

\displaystyle \overrightarrow{OB}\;=\begin{pmatrix}7\\1\end{pmatrix}

the same idea for

B

(iii)

\displaystyle \overrightarrow{AB}\;=\overrightarrow{AO}+\overrightarrow{OB}=\overrightarrow{OB}-\overrightarrow{OA}

go from

A

back to

O

, then on to

B

: end minus start

\displaystyle =\begin{pmatrix}7\\1\end{pmatrix}-\begin{pmatrix}2\\5\end{pmatrix}

\displaystyle =\begin{pmatrix}5\\-4\end{pmatrix}

subtract across, subtract up

Final answer

(i)

\overrightarrow{OA}=\begin{pmatrix}2\\5\end{pmatrix}

(ii)

\overrightarrow{OB}=\begin{pmatrix}7\\1\end{pmatrix}

(iii)

\overrightarrow{AB}=\begin{pmatrix}5\\-4\end{pmatrix}

Your Turn 3.3

$A$ is the point $(-3,\ 2)$ and $B$ is the point $(4,\ -6)$ . $O$ is the origin.

(i)Write down the position vector $\overrightarrow{OA}$ .
(ii)Write down the position vector $\overrightarrow{OB}$ .
(iii)Find $\overrightarrow{AB}$ .

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Colour keyadd, subtract, multiply, divide or simplify

(i)

\displaystyle \overrightarrow{OA}\;=\begin{pmatrix}-3\\2\end{pmatrix}

a position vector is just the coordinates written as a column

(ii)

\displaystyle \overrightarrow{OB}\;=\begin{pmatrix}4\\-6\end{pmatrix}

the same idea for

B

(iii)

\displaystyle \overrightarrow{AB}\;=\overrightarrow{AO}+\overrightarrow{OB}=\overrightarrow{OB}-\overrightarrow{OA}

go from

A

back to

O

, then on to

B

: end minus start

\displaystyle =\begin{pmatrix}4\\-6\end{pmatrix}-\begin{pmatrix}-3\\2\end{pmatrix}

\displaystyle =\begin{pmatrix}7\\-8\end{pmatrix}

subtract across, subtract up

Final answer

(i)

\overrightarrow{OA}=\begin{pmatrix}-3\\2\end{pmatrix}

(ii)

\overrightarrow{OB}=\begin{pmatrix}4\\-6\end{pmatrix}

(iii)

\overrightarrow{AB}=\begin{pmatrix}7\\-8\end{pmatrix}

Think About It

The point

A

has position vector

\begin{pmatrix}3\\1\end{pmatrix}

. What is

\overrightarrow{AO}

Show the worked answerHide the worked answer

Colour keyadd, subtract, multiply, divide or simplify

\displaystyle \overrightarrow{AO}\;=\overrightarrow{OO}-\overrightarrow{OA}=\mathbf{0}-\begin{pmatrix}3\\1\end{pmatrix}

end minus start, with

O

as the end

\displaystyle =\begin{pmatrix}-3\\-1\end{pmatrix}

the origin has position vector zero

Final answer

\overrightarrow{AO}

runs from

A

back to

O

, so it is the negative of

\overrightarrow{OA}

\begin{pmatrix}-3\\-1\end{pmatrix}

Example 3.4

Chaining vectors from a position vector

The position vector of $P$ is $\begin{pmatrix}3\\-2\end{pmatrix}$ . The vectors $\overrightarrow{PQ}=\begin{pmatrix}-5\\6\end{pmatrix}$ and $\overrightarrow{QR}=\begin{pmatrix}4\\-3\end{pmatrix}$ are given.

(i)Find the coordinates of $Q$ .
(ii)Find the position vector of $R$ .

Show solutionHide solution

(i)

\displaystyle \overrightarrow{OQ}\;=\overrightarrow{OP}+\overrightarrow{PQ}=\begin{pmatrix}3\\-2\end{pmatrix}+\begin{pmatrix}-5\\6\end{pmatrix}

path from

O

P

, then on to

Q

\displaystyle =\begin{pmatrix}-2\\4\end{pmatrix}

(ii)

\displaystyle \overrightarrow{OR}\;=\overrightarrow{OQ}+\overrightarrow{QR}=\begin{pmatrix}-2\\4\end{pmatrix}+\begin{pmatrix}4\\-3\end{pmatrix}

extend the path one more step

\displaystyle =\begin{pmatrix}2\\1\end{pmatrix}

Final answer

(i)

Q=(-2,\ 4)

(ii)

\overrightarrow{OR}=\begin{pmatrix}2\\1\end{pmatrix}

Your Turn 3.4

The position vector of $P$ is $\begin{pmatrix}-1\\5\end{pmatrix}$ . The vectors $\overrightarrow{PQ}=\begin{pmatrix}4\\-7\end{pmatrix}$ and $\overrightarrow{QR}=\begin{pmatrix}-6\\2\end{pmatrix}$ are given.

(i)Find the coordinates of $Q$ .
(ii)Find the position vector of $R$ .

Show worked solutionHide worked solution

(i)

\displaystyle \overrightarrow{OQ}\;=\overrightarrow{OP}+\overrightarrow{PQ}=\begin{pmatrix}-1\\5\end{pmatrix}+\begin{pmatrix}4\\-7\end{pmatrix}

path from

O

P

, then on to

Q

\displaystyle =\begin{pmatrix}3\\-2\end{pmatrix}

(ii)

\displaystyle \overrightarrow{OR}\;=\overrightarrow{OQ}+\overrightarrow{QR}=\begin{pmatrix}3\\-2\end{pmatrix}+\begin{pmatrix}-6\\2\end{pmatrix}

extend the path one more step

\displaystyle =\begin{pmatrix}-3\\0\end{pmatrix}

Final answer

(i)

Q=(3,\ -2)

(ii)

\overrightarrow{OR}=\begin{pmatrix}-3\\0\end{pmatrix}

Links To

Coordinates of points and gradients are used throughout, as in Chapter 15, Coordinate geometry

Supports G7.5 · G7.4

The magnitude of a vector

The magnitude of a vector is its length. A column vector $\begin{pmatrix}x\\y\end{pmatrix}$ is the hypotenuse of a right-angled triangle whose other two sides are $|x|$ and $|y|$ , so Pythagoras' theorem gives its length.

The Core Idea

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

A magnitude is never negative, because it is a square root. Square the components first, so the signs disappear.

The vector

\begin{pmatrix}4\\3\end{pmatrix}

is the hypotenuse of a right-angled triangle with sides 4 and 3, so its length is 5.

In general

|\mathbf{a}|^{2}=x^{2}+y^{2}

for

\mathbf{a}=\begin{pmatrix}x\\y\end{pmatrix}

SGSchoolKaki Tip

Magnitude is a number, so write $|\mathbf{a}|=$ a number, not an arrow. If the answer is not a whole number, leave it as a surd for exact working and give three significant figures at the end.

Example 4.1

Finding the magnitude

Find the magnitude of each vector. Give exact answers, and a value to three significant figures when the answer is not a whole number.

(i) $\mathbf{a}=\begin{pmatrix}7\\-24\end{pmatrix}$ . Find $|\mathbf{a}|$ .
(ii) $\mathbf{b}=\begin{pmatrix}-3\\5\end{pmatrix}$ . Find $|\mathbf{b}|$ .
(iii) $\mathbf{c}=\begin{pmatrix}3\\-6\end{pmatrix}$ . Find $|\mathbf{c}|$ .

Show solutionHide solution

Colour keyadd, subtract, multiply, divide or simplifyround, estimate or compare

(i)

\displaystyle |\mathbf{a}|\;=\sqrt{7^{2}+(-24)^{2}}

square each component, add, then take the root

\displaystyle =\sqrt{49+576}

\displaystyle =\sqrt{625}

\displaystyle =25

a perfect square, so the magnitude is exact

(ii)

\displaystyle |\mathbf{b}|\;=\sqrt{(-3)^{2}+5^{2}}

square each component, add, then take the root

\displaystyle =\sqrt{9+25}

\displaystyle =\sqrt{34}

\displaystyle \approx5.83

three significant figures

(iii)

\displaystyle |\mathbf{c}|\;=\sqrt{3^{2}+(-6)^{2}}

square each component, add, then take the root

\displaystyle =\sqrt{9+36}

\displaystyle =\sqrt{45}

\displaystyle \approx6.71

three significant figures

Final answer

(i)

|\mathbf{a}|=25

(ii)

|\mathbf{b}|=\sqrt{34}\approx5.83

(iii)

|\mathbf{c}|=3 \sqrt{5}\approx6.71

Your Turn 4.1

Find the magnitude of each vector. Give exact answers, and a value to three significant figures when the answer is not a whole number.

(i) $\mathbf{a}=\begin{pmatrix}12\\-35\end{pmatrix}$ . Find $|\mathbf{a}|$ .
(ii) $\mathbf{b}=\begin{pmatrix}-4\\7\end{pmatrix}$ . Find $|\mathbf{b}|$ .
(iii) $\mathbf{c}=\begin{pmatrix}4\\-10\end{pmatrix}$ . Find $|\mathbf{c}|$ .

Show worked solutionHide worked solution

Colour keyadd, subtract, multiply, divide or simplifyround, estimate or compare

(i)

\displaystyle |\mathbf{a}|\;=\sqrt{12^{2}+(-35)^{2}}

square each component, add, then take the root

\displaystyle =\sqrt{144+1225}

\displaystyle =\sqrt{1369}

\displaystyle =37

a perfect square, so the magnitude is exact

(ii)

\displaystyle |\mathbf{b}|\;=\sqrt{(-4)^{2}+7^{2}}

square each component, add, then take the root

\displaystyle =\sqrt{16+49}

\displaystyle =\sqrt{65}

\displaystyle \approx8.06

three significant figures

(iii)

\displaystyle |\mathbf{c}|\;=\sqrt{4^{2}+(-10)^{2}}

square each component, add, then take the root

\displaystyle =\sqrt{16+100}

\displaystyle =\sqrt{116}

\displaystyle \approx10.8

three significant figures

Final answer

(i)

|\mathbf{a}|=37

(ii)

|\mathbf{b}|=\sqrt{65}\approx8.06

(iii)

|\mathbf{c}|=2 \sqrt{29}\approx10.8

Distance as a magnitude

The distance between points $A$ and $B$ is the magnitude of $\overrightarrow{AB}$ . Find the column vector first (end minus start), then find its magnitude. This agrees with the distance formula you met in coordinate geometry.

Links To

The same length formula appears there as the distance between two points: Chapter 15, Coordinate geometry

Example 4.2

The distance between two points

$A=(-2,\ 3)$ and $B=(4,\ -5)$ . Find the distance $AB$ .

Write $\overrightarrow{AB}$ as a column vector, then find $|\overrightarrow{AB}|$ .

Show solutionHide solution

Colour keyadd, subtract, multiply, divide or simplify

\displaystyle \overrightarrow{AB}\;=\overrightarrow{OB}-\overrightarrow{OA}=\begin{pmatrix}4\\-5\end{pmatrix}-\begin{pmatrix}-2\\3\end{pmatrix}

end minus start

\displaystyle =\begin{pmatrix}6\\-8\end{pmatrix}

subtract the top numbers, then the bottom numbers

\displaystyle |\overrightarrow{AB}|\;=\sqrt{6^{2}+(-8)^{2}}

square each component, add, then take the root

\displaystyle =\sqrt{36+64}

\displaystyle =\sqrt{100}

\displaystyle =10

a perfect square, so the magnitude is exact

Final answer

|\overrightarrow{AB}|=10

units

Your Turn 4.2

$A=(1,\ -4)$ and $B=(-11,\ 5)$ . Find the distance $AB$ .

Write $\overrightarrow{AB}$ as a column vector, then find $|\overrightarrow{AB}|$ .

Show worked solutionHide worked solution

Colour keyadd, subtract, multiply, divide or simplify

\displaystyle \overrightarrow{AB}\;=\overrightarrow{OB}-\overrightarrow{OA}=\begin{pmatrix}-11\\5\end{pmatrix}-\begin{pmatrix}1\\-4\end{pmatrix}

end minus start

\displaystyle =\begin{pmatrix}-12\\9\end{pmatrix}

subtract the top numbers, then the bottom numbers

\displaystyle |\overrightarrow{AB}|\;=\sqrt{(-12)^{2}+9^{2}}

square each component, add, then take the root

\displaystyle =\sqrt{144+81}

\displaystyle =\sqrt{225}

\displaystyle =15

a perfect square, so the magnitude is exact

Final answer

|\overrightarrow{AB}|=15

units

Think About It

The vectors

\begin{pmatrix}3\\4\end{pmatrix}

and

\begin{pmatrix}-4\\3\end{pmatrix}

are different. Do they have the same magnitude?

Show the worked answerHide the worked answer

\displaystyle \left|\begin{pmatrix}3\\4\end{pmatrix}\right|\;=\sqrt{3^{2}+4^{2}}=\sqrt{25}=5

first vector

\displaystyle \left|\begin{pmatrix}-4\\3\end{pmatrix}\right|\;=\sqrt{(-4)^{2}+3^{2}}=\sqrt{25}=5

second vector: the same sum of squares

Final answerYes: both have magnitude 5. Different vectors can have the same magnitude, because squaring removes the signs and only the sizes of the components matter here.

Example 4.3

Working backwards from a magnitude

Find the possible values of the unknown in each vector.

(i) $\begin{pmatrix}k\\4\end{pmatrix}$ has magnitude 5.
(ii) $\begin{pmatrix}3\\m\end{pmatrix}$ has magnitude $\sqrt{34}$ .

Show solutionHide solution

Colour keyadd, subtract, multiply, divide or simplifyexpand, rewrite, substitute or solve

(i)

\displaystyle \sqrt{k^{2}+16}\;=5

write the magnitude equation

\displaystyle k^{2}+16\;=25

square both sides: the root disappears

\displaystyle k^{2}\;=9

\displaystyle k\;=3\ \text{or}\ k=-3

a square has two roots: positive and negative

(ii)

\displaystyle \sqrt{9+m^{2}}\;=\sqrt{34}

write the magnitude equation

\displaystyle 9+m^{2}\;=34

square both sides: the root disappears

\displaystyle m^{2}\;=25

\displaystyle m\;=5\ \text{or}\ m=-5

a square has two roots: positive and negative

Final answer

(i)

k=3

k=-3

(ii)

m=5

m=-5

Your Turn 4.3

Find the possible values of the unknown in each vector.

(i) $\begin{pmatrix}k\\12\end{pmatrix}$ has magnitude 13.
(ii) $\begin{pmatrix}2\\m\end{pmatrix}$ has magnitude $\sqrt{20}$ .

Show worked solutionHide worked solution

Colour keyadd, subtract, multiply, divide or simplifyexpand, rewrite, substitute or solve

(i)

\displaystyle \sqrt{k^{2}+144}\;=13

write the magnitude equation

\displaystyle k^{2}+144\;=169

square both sides: the root disappears

\displaystyle k^{2}\;=25

\displaystyle k\;=5\ \text{or}\ k=-5

a square has two roots: positive and negative

(ii)

\displaystyle \sqrt{4+m^{2}}\;=2 \sqrt{5}

write the magnitude equation

\displaystyle 4+m^{2}\;=20

square both sides: the root disappears

\displaystyle m^{2}\;=16

\displaystyle m\;=4\ \text{or}\ m=-4

a square has two roots: positive and negative

Final answer

(i)

k=5

k=-5

(ii)

m=4

m=-4

SGSchoolKaki Tip

Do not add the components first. $\left|\begin{pmatrix}3\\4\end{pmatrix}\right|$ is $\sqrt{3^{2}+4^{2}}$ , not $3+4$ .

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Notes version 1.0 · Last updated 3 Nov 2026 · Chapter 16 of 19