Maths · 8 Transformations and vectors

8.2 Vectors in two dimensions

Before you start

Plotting and reading coordinates on a grid, including negative coordinates.
Translation from 8.1: a slide in which every point of a shape moves the same distance in the same direction, with no turning and no change of size.
Adding, subtracting and multiplying negative numbers, especially subtracting a negative: $2 - (- 3) = 5$ .
Solving a linear equation such as $3 q + 5 = - 1$ , and a pair of simultaneous equations.

Syllabus objectives covered (3)

Describe a translation using a vector represented by \binom{x}{y}, \overrightarrow{AB} or a.
Add and subtract vectors.
Multiply a vector by a scalar.

Part 1 of 7

What a vector is, and the three ways it is written

A vector is a movement with a size and a direction. 'Three to the right and two up' is a vector: it tells you how far to go and which way.
The column vector $(\binom{x}{y})$ records that movement as two numbers, one above the other. The top number $x$ is the movement across: positive to the right, negative to the left.
The bottom number $y$ is the movement up or down: positive up, negative down. The two numbers are called the components of the vector.
$(\binom{3}{2}) is 3 right, 2 up (\binom{- 4}{- 1}) is 4 left, 1 down$
graph diagram — not drawn yet
A square grid with x from 0 to 6 and y from 0 to 5. A solid arrow from (1, 1) to (4, 3) labelled $(\binom{3}{2})$ , with a dashed horizontal line from (1, 1) to (4, 1) labelled 3 and a dashed vertical line from (4, 1) to (4, 3) labelled 2. A second solid arrow from (0, 3) to (3, 5), also labelled $(\binom{3}{2})$ .
The top number is the movement across and the bottom number the movement up. The two arrows are in different places but are the same vector.
Write the two numbers in brackets with no comma and no fraction line. $(\binom{3}{2})$ is not the point $(3, 2)$ and not the fraction $\frac{3}{2}$ . A coordinate says where something is; a vector says how it moves.
A vector can also be named by the two points it joins. $\vec{A B}$ is the movement from $A$ to $B$ . The order of the letters gives the direction: start at the first letter, finish at the second.
Or a vector can be named by a single letter. The syllabus says vectors will be printed as $\vec{A B}$ or as a bold letter such as $𝐚$ . You cannot write bold by hand, so underline the letter instead: $a$ .
Two vectors are equal when their column vectors are the same. Where they are drawn does not matter: every arrow that goes 3 right and 2 up is the vector $(\binom{3}{2})$ .

Part 2 of 7

Describing a translation with a vector

In a translation, every point of the shape moves by the same vector. So one vector describes the whole movement, and that vector is what you give.
Pick one corner of the original shape (the object) and find the matching corner of the moved shape (the image). Matching means the same position in the shape: the right-angle corner goes to the right-angle corner.
Count across from the object corner to the image corner. That is the top number, with a minus sign if you went left. Then count up or down. That is the bottom number, with a minus sign if you went down.
Check with a second pair of corners. If it gives a different vector, you matched the wrong corners, or the transformation is not a translation.
To describe the transformation fully, give two things: the word translation, and the column vector. For example: 'a translation by the vector $(\binom{- 4}{2})$ '. The vector alone does not name the transformation, and the word alone does not say where the shape went.
Direction matters. The vector goes from object to image. If $(\binom{- 4}{2})$ maps shape $P$ onto shape $Q$ , then the vector that maps $Q$ back onto $P$ is $(\binom{4}{- 2})$ . Both signs change, because you go back the way you came.

Part 3 of 7

Finding $\vec{A B}$ from coordinates

When you know the coordinates of $A$ and $B$ , you do not need a grid. The movement across is the change in $x$ , and the movement up is the change in $y$ .
$\vec{A B} = (\binom{x_{B} - x_{A}}{y_{B} - y_{A}})$
Read it as end minus start. $B$ is where you finish, so its coordinate comes first in each subtraction.
For $A (2, 5)$ and $B (- 1, 3)$ : across $- 1 - 2 = - 3$ , and up $3 - 5 = - 2$ . So $\vec{A B} = (\binom{- 3}{- 2})$ .
Check the signs against a quick picture. $B$ is to the left of $A$ and lower down, so both numbers must be negative, and they are.
Going the other way reverses every sign: $\vec{B A} = (\binom{3}{2})$ . In general, $\vec{B A} = - \vec{A B}$ .
It works backwards too. If you know $A$ and the vector $\vec{A B}$ , add the vector to the coordinates of $A$ to land on $B$ . From $A (2, 5)$ , the vector $(\binom{- 3}{- 2})$ takes you to $(2 - 3, 5 - 2) = (- 1, 3)$ , which is $B$ .

Part 4 of 7

Adding vectors: one movement after another

To add two column vectors, add the top numbers and add the bottom numbers. Each row is its own sum; the rows never mix.
$(\binom{3}{1}) + (\binom{- 1}{3}) = (\binom{3 + (- 1)}{1 + 3}) = (\binom{2}{4})$
The meaning: $𝐚 + 𝐛$ is the single movement that has the same effect as doing $𝐚$ and then $𝐛$ . Going 3 right and then 1 left leaves you 2 right; going 1 up and then 3 up leaves you 4 up.
On a diagram, draw $𝐛$ starting where $𝐚$ finishes, nose to tail. The arrow from the start of $𝐚$ to the end of $𝐛$ is $𝐚 + 𝐛$ , called the resultant.
graph diagram — not drawn yet
A square grid with x from -2 to 4 and y from 0 to 5. A solid arrow from (0, 0) to (3, 1) labelled $𝐚$ , then a solid arrow from (3, 1) to (2, 4) labelled $𝐛$ . A thicker arrow from (0, 0) to (2, 4) labelled $𝐚 + 𝐛$ . Dashed arrows complete a parallelogram: one from (0, 0) to (-1, 3) labelled $𝐛$ and one from (-1, 3) to (2, 4) labelled $𝐚$ .
Doing $𝐚$ then $𝐛$ ends at the same point as doing $𝐛$ then $𝐚$ . The arrow from start to finish is $𝐚 + 𝐛 = (\binom{2}{4})$ .
With named points, adding is a route. Going from $A$ to $B$ and then from $B$ to $C$ takes you from $A$ to $C$ . The middle letters match, and the answer runs from the first letter to the last.
$\vec{A B} + \vec{B C} = \vec{A C}$
Order does not matter in addition: $𝐚 + 𝐛 = 𝐛 + 𝐚$ , because $3 + (- 1)$ and $(- 1) + 3$ are the same number. On the diagram, doing $𝐛$ first takes the other way round the parallelogram and ends at the same point.

Part 5 of 7

Subtracting vectors, and the negative of a vector

To subtract, subtract the top numbers and subtract the bottom numbers. Keep the order: the first vector minus the second.
$(\binom{2}{3}) - (\binom{- 4}{1}) = (\binom{2 - (- 4)}{3 - 1}) = (\binom{6}{2})$
The trap here is the double negative. $2 - (- 4)$ is $2 + 4 = 6$ , not $- 2$ . Write each row out in full with brackets, as above, before you work it out.
The negative of a vector, $- 𝐛$ , is the same movement in the opposite direction. Change the sign of both numbers: if $𝐛 = (\binom{- 4}{1})$ , then $- 𝐛 = (\binom{4}{- 1})$ .
So subtracting a vector is the same as adding its negative: $𝐚 - 𝐛 = 𝐚 + (- 𝐛)$ . On a diagram, go along $𝐚$ and then go backwards along $𝐛$ .
Order matters in subtraction. $𝐛 - 𝐚$ is the negative of $𝐚 - 𝐛$ . With the vectors above, $(\binom{- 4}{1}) - (\binom{2}{3}) = (\binom{- 6}{- 2})$ , which is $(\binom{6}{2})$ with both signs changed.
Routes use the same idea. To get from $B$ to $C$ when you know $\vec{A B}$ and $\vec{A C}$ , go back from $B$ to $A$ , then on to $C$ .
$\vec{B C} = - \vec{A B} + \vec{A C} = \vec{A C} - \vec{A B}$

Part 6 of 7

Multiplying a vector by a scalar

A scalar is an ordinary number: it has a size but no direction. Multiplying a vector by a scalar $k$ multiplies both components by $k$ .
$3 (\binom{2}{- 1}) = (\binom{3 \times 2}{3 \times (- 1)}) = (\binom{6}{- 3})$
The effect is to stretch the arrow. $3 𝐚$ means $𝐚 + 𝐚 + 𝐚$ , the movement done three times in a row. So it points the same way as $𝐚$ and is three times as long.
A fraction shrinks it. $\frac{1}{2} (\binom{6}{- 4}) = (\binom{3}{- 2})$ : the same direction, half as long.
A negative scalar also reverses the direction. $- 2 𝐚$ is twice as long as $𝐚$ and points the opposite way.
graph diagram — not drawn yet
One square grid with x from 0 to 22 and y from 0 to 5, carrying four separate arrows: $𝐚$ from (0, 1) to (4, 3); $2 𝐚$ from (5, 0) to (13, 4); $\frac{1}{2} 𝐚$ from (14, 1) to (16, 2); $- 𝐚$ from (22, 3) to (18, 1). Each arrow is labelled with its name and its column vector: $(\binom{4}{2})$ , $(\binom{8}{4})$ , $(\binom{2}{1})$ and $(\binom{- 4}{- 2})$ .
All four arrows lie at the same slope. A positive scalar keeps the direction of $𝐚$ , a negative one reverses it, and the size of the scalar sets the length.
Because $k 𝐚$ lies along the same line as $𝐚$ , forwards or backwards, $k 𝐚$ is parallel to $𝐚$ . Two vectors are parallel when one is a scalar multiple of the other.
To test two column vectors, look for one number $k$ that turns one into the other in both rows. $(\binom{6}{- 9}) = - 3 (\binom{- 2}{3})$ , so those two are parallel. For $(\binom{4}{2})$ and $(\binom{2}{3})$ , the top row needs $k = 2$ but the bottom row needs $k = \frac{2}{3}$ , so they are not parallel.

Part 7 of 7

Putting the operations together

In an expression such as $3 𝐚 - 2 𝐛$ , do the scalar multiplications first, then add or subtract. It is the same order as $3 \times 4 - 2 \times 5$ : multiply before you subtract.
Write each scalar multiple as its own column vector before you combine them. With $𝐚 = (\binom{1}{3})$ and $𝐛 = (\binom{2}{- 1})$ :
$3 𝐚 - 2 𝐛 = (\binom{3}{9}) - (\binom{4}{- 2}) = (\binom{3 - 4}{9 - (- 2)}) = (\binom{- 1}{11})$
When a vector equation contains unknown letters, use this fact: two vectors are equal only when both rows match. So one vector equation gives two ordinary equations, one from the top row and one from the bottom row.
For example, $k (\binom{2}{q}) = (\binom{8}{- 12})$ . The top row gives $2 k = 8$ , so $k = 4$ . The bottom row gives $k q = - 12$ , so $4 q = - 12$ and $q = - 3$ .
When there are two unknown scalars, as in $m 𝐚 + n 𝐛 = 𝐜$ , the two row equations are simultaneous equations. Solve them together by substitution or elimination.

Worked examples

Example 1. Triangle $P$ has vertices $(1, 1)$ , $(3, 1)$ and $(1, 4)$ . Triangle $Q$ has vertices $(- 3, 3)$ , $(- 1, 3)$ and $(- 3, 6)$ . Describe fully the single transformation that maps $P$ onto $Q$ . Then write down the vector that maps $Q$ onto $P$ .

Show the working

Plot the triangles, or picture them. They are the same shape and size and face the same way, with no turning and no flip. That points to a translation.
graph diagram — not drawn yet
A square grid with x from -4 to 4 and y from 0 to 7, with labelled axes. Triangle P with vertices (1, 1), (3, 1) and (1, 4), shaded and labelled P. Triangle Q with vertices (-3, 3), (-1, 3) and (-3, 6), unshaded and labelled Q. A dashed arrow from (1, 1) to (-3, 3) labelled $(\binom{- 4}{2})$ .
Every vertex of P moves 4 left and 2 up to its matching vertex of Q, so one vector describes the whole translation.
Match corresponding vertices. The right angle of $P$ is at $(1, 1)$ ; the right angle of $Q$ is at $(- 3, 3)$ .
Across is the change in $x$ , end minus start. Up is the change in $y$ .
$(\binom{- 3 - 1}{3 - 1}) = (\binom{- 4}{2})$
Check with a second pair: $(3, 1)$ goes to $(- 1, 3)$ , and $(\binom{- 1 - 3}{3 - 1}) = (\binom{- 4}{2})$ again. The third pair, $(1, 4)$ to $(- 3, 6)$ , agrees too.
Give both parts of the description: the word and the vector. It is a translation by $(\binom{- 4}{2})$ .
To map $Q$ back onto $P$ , go back the way you came. Change both signs: $(\binom{4}{- 2})$ .

Answer: A translation by the vector $(\binom{- 4}{2})$ . The vector that maps $Q$ onto $P$ is $(\binom{4}{- 2})$ .

Example 2. The points $A (2, - 1)$ , $B (5, 3)$ and $C (- 1, 4)$ are given. Write $\vec{A B}$ and $\vec{B C}$ as column vectors. Then show that $\vec{A B} + \vec{B C} = \vec{A C}$ .

Show the working

For $\vec{A B}$ , use end minus start: the end is $B$ , the start is $A$ .
$\vec{A B} = (\binom{5 - 2}{3 - (- 1)}) = (\binom{3}{4})$
Sign check: $B$ is to the right of $A$ and above it, so both numbers should be positive. They are.
For $\vec{B C}$ , the end is $C$ and the start is $B$ .
$\vec{B C} = (\binom{- 1 - 5}{4 - 3}) = (\binom{- 6}{1})$
Add the two vectors row by row.
$(\binom{3}{4}) + (\binom{- 6}{1}) = (\binom{- 3}{5})$
Now find $\vec{A C}$ separately, straight from the coordinates. A 'show that' needs both sides worked out on their own, so that the match is evidence and not an assumption.
$\vec{A C} = (\binom{- 1 - 2}{4 - (- 1)}) = (\binom{- 3}{5})$
Both results are $(\binom{- 3}{5})$ . Going from $A$ to $B$ and then from $B$ to $C$ is the same movement as going straight from $A$ to $C$ .

Answer: $\vec{A B} = (\binom{3}{4})$ and $\vec{B C} = (\binom{- 6}{1})$ . Their sum and $\vec{A C}$ are both $(\binom{- 3}{5})$ , so $\vec{A B} + \vec{B C} = \vec{A C}$ .

Example 3. $𝐚 = (\binom{3}{- 2})$ and $𝐛 = (\binom{- 1}{4})$ . Work out $2 𝐚 - 3 𝐛$ as a column vector.

Show the working

Scalars first. Multiply both components of $𝐚$ by 2.
$2 𝐚 = (\binom{6}{- 4})$
Multiply both components of $𝐛$ by 3.
$3 𝐛 = (\binom{- 3}{12})$
Subtract row by row. Write each row out with brackets, because the top row subtracts a negative number.
$(\binom{6}{- 4}) - (\binom{- 3}{12}) = (\binom{6 - (- 3)}{- 4 - 12}) = (\binom{9}{- 16})$
Check the top row another way: $2 \times 3 - 3 \times (- 1) = 6 + 3 = 9$ . The bottom row: $2 \times (- 2) - 3 \times 4 = - 4 - 12 = - 16$ .

Answer: $2 𝐚 - 3 𝐛 = (\binom{9}{- 16})$

Example 4. Find the values of $m$ and $n$ such that $m (\binom{1}{2}) + n (\binom{3}{- 1}) = (\binom{9}{4})$ .

Show the working

Multiply out the scalars and add, so the left side becomes one column vector.
$(\binom{m + 3 n}{2 m - n}) = (\binom{9}{4})$
Two vectors are equal only when both rows match. That gives two equations.
$m + 3 n = 9 2 m - n = 4$
These are simultaneous equations. The second rearranges to $n = 2 m - 4$ . Substitute that into the first.
$m + 3 (2 m - 4) = 9 \Rightarrow 7 m - 12 = 9 \Rightarrow m = 3$
Then $n = 2 \times 3 - 4 = 2$ .
Check in the original equation.
$3 (\binom{1}{2}) + 2 (\binom{3}{- 1}) = (\binom{3}{6}) + (\binom{6}{- 2}) = (\binom{9}{4})$

Answer: $m = 3$ and $n = 2$

Where people go wrong

Mistake: Writing the vector with the up-down movement on top, or as a coordinate pair such as $(5, - 3)$ .

Instead: The top number is the movement across (right is positive) and the bottom number is the movement up or down (up is positive): $(\binom{5}{- 3})$ .

The column vector keeps the coordinate order, $x$ then $y$ . It is written as a column, which keeps a movement visibly different from a position.

Mistake: Working out $\vec{A B}$ as start minus end, subtracting $B$ 's coordinates from $A$ 's.

Instead: Use end minus start: $\vec{A B} = (\binom{x_{B} - x_{A}}{y_{B} - y_{A}})$ .

Start minus end gives $\vec{B A}$ , the reverse movement, so both signs come out wrong. A quick sketch shows which way $B$ is from $A$ and so what the signs must be.

Mistake: Describing the translation the wrong way round, from the image back to the object.

Instead: The vector goes from the shape you start with to the shape it maps onto. Read 'maps $A$ onto $B$ ' as 'from $A$ to $B$ '.

The reverse translation has both signs changed, so a vector in the wrong direction is wrong in both rows.

Mistake: $(\binom{2}{3}) - (\binom{- 4}{1}) = (\binom{- 2}{2})$

Instead: $2 - (- 4) = 2 + 4 = 6$ , so the answer is $(\binom{6}{2})$ .

Subtracting a negative number adds it. Writing the row as $2 - (- 4)$ , with brackets, before you work it out keeps the sign in view.

Mistake: Multiplying only the top number by the scalar: $3 (\binom{2}{- 1}) = (\binom{6}{- 1})$ .

Instead: Multiply both components: $3 (\binom{2}{- 1}) = (\binom{6}{- 3})$ .

$3 𝐚$ means $𝐚 + 𝐚 + 𝐚$ . Adding three copies triples the up-down movement as well as the movement across.

Check yourself

Write down the column vector for a movement of 5 units to the left and 3 units up.
Answer
$(\binom{- 5}{3})$
$P$ is the point $(4, - 2)$ and $Q$ is the point $(- 1, 6)$ . Write $\vec{P Q}$ and $\vec{Q P}$ as column vectors.
Answer
$\vec{P Q} = (\binom{- 1 - 4}{6 - (- 2)}) = (\binom{- 5}{8})$ , and $\vec{Q P} = (\binom{5}{- 8})$ .
Shape $F$ is translated by $(\binom{3}{- 7})$ to give shape $G$ . Write down the vector of the translation that maps $G$ onto $F$ .
Answer
$(\binom{- 3}{7})$ : the reverse movement, with both signs changed.
$𝐚 = (\binom{4}{1})$ and $𝐛 = (\binom{- 2}{5})$ . Work out $𝐚 + 𝐛$ and $𝐚 - 𝐛$ .
Answer
$𝐚 + 𝐛 = (\binom{2}{6})$ and $𝐚 - 𝐛 = (\binom{4 - (- 2)}{1 - 5}) = (\binom{6}{- 4})$ .
$\vec{A B} = (\binom{2}{5})$ and $\vec{A C} = (\binom{- 1}{3})$ . Find $\vec{B C}$ .
Answer
$\vec{B C} = \vec{A C} - \vec{A B} = (\binom{- 1 - 2}{3 - 5}) = (\binom{- 3}{- 2})$
Work out $- 3 (\binom{2}{- 5})$ .
Answer
$(\binom{- 6}{15})$
Explain whether $(\binom{6}{- 4})$ and $(\binom{- 3}{2})$ are parallel.
Answer
They are parallel, because $(\binom{6}{- 4}) = - 2 (\binom{- 3}{2})$ : one is a scalar multiple of the other. The negative scalar means they point in opposite directions.
Find $p$ and $q$ if $(\binom{p}{5}) + 3 (\binom{2}{q}) = (\binom{4}{- 1})$ .
Answer
Top row: $p + 6 = 4$ , so $p = - 2$ . Bottom row: $5 + 3 q = - 1$ , so $q = - 2$ .

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. Triangle $A$ has vertices $(- 3, 1)$ , $(- 1, 1)$ and $(- 1, 4)$ . Triangle $B$ has vertices $(2, - 2)$ , $(4, - 2)$ and $(4, 1)$ .

(a) Describe fully the single transformation that maps triangle $A$ onto triangle $B$ . 2 marks No calculator

Mark scheme

B1 Translation
B1 $(\binom{5}{- 3})$

(b) Write down the column vector of the translation that maps triangle $B$ onto triangle $A$ . 1 mark No calculator

Mark scheme

B1 $(\binom{- 5}{3})$ ; FT the negative of their vector in (a)

Question 2. The points $A (1, 2)$ , $B (4, - 2)$ and $C (- 3, 0)$ are marked on a grid.

(a) Write $\vec{A B}$ as a column vector. 1 mark No calculator

Mark scheme

B1 $(\binom{3}{- 4})$

(b) Write $\vec{B C}$ as a column vector. 1 mark No calculator

Mark scheme

B1 $(\binom{- 7}{2})$

(c) Show that $\vec{A B} + \vec{B C} = \vec{A C}$ . 2 marks No calculator

Mark scheme

M1 $(\binom{3}{- 4}) + (\binom{- 7}{2}) = (\binom{- 4}{- 2})$ ; FT their (a) and (b)
A1 $\vec{A C} = (\binom{- 3 - 1}{0 - 2}) = (\binom{- 4}{- 2})$ found from the coordinates, and the two results stated to be equal

(d) The point $D$ is such that $\vec{C D} = \vec{A B}$ . Find the coordinates of $D$ . 2 marks No calculator

Mark scheme

M1 $(- 3 + 3, 0 + (- 4))$ oe; FT their (a)
A1 $(0, - 4)$

Question 3. $𝐚 = (\binom{2}{- 3})$ and $𝐛 = (\binom{- 4}{1})$ .

(a) Work out $3 𝐚 + 𝐛$ as a column vector. 2 marks No calculator

Mark scheme

M1 $3 𝐚 = (\binom{6}{- 9})$ soi
A1 $(\binom{2}{- 8})$

(b) Work out $𝐚 - 2 𝐛$ as a column vector. 2 marks No calculator

Mark scheme

M1 $2 𝐛 = (\binom{- 8}{2})$ soi
A1 $(\binom{10}{- 5})$

(c) The vector $𝐝 = (\binom{- 6}{9})$ . Explain why $𝐝$ is parallel to $𝐚$ . 2 marks No calculator

Mark scheme

B1 $𝐝 = - 3 𝐚$ oe, e.g. $- 3 (\binom{2}{- 3}) = (\binom{- 6}{9})$ shown
B1 states that $𝐝$ is a scalar multiple of $𝐚$ , so the two vectors are parallel

(d) Find the values of $m$ and $n$ such that $m 𝐚 + n 𝐛 = (\binom{- 8}{- 3})$ . 3 marks No calculator

Mark scheme

M1 $2 m - 4 n = - 8$ and $- 3 m + n = - 3$ oe
M1 correct method to eliminate one unknown, by elimination or substitution; FT their equations
A1 $m = 2$ and $n = 3$

How this is examined

Turning a vector equation with two unknowns into two ordinary equations, one for each row: a Find that gives two column vectors and asks for the scalars $m$ and $n$ in $m 𝐚 + n 𝐛 = 𝐜$ M1 for both row equations written down, e.g. $2 m - 4 n = - 8$ from the top row and $- 3 m + n = - 3$ from the bottom row. M1 for a correct method to solve them, by substitution or elimination. A1 for both values. Write the two equations on separate lines before you solve, so the first method mark can be seen.
Describing a translation fully: the word and the vector: a two-mark Describe: 'Describe fully the single transformation that maps shape A onto shape B', with the two shapes on a grid or given by their vertices One B mark for the word 'translation'. One B mark for the correct column vector from object to image, e.g. $(\binom{5}{- 3})$ . The objective asks for the vector, so give it as a column vector, not as a coordinate pair or in words.
Working out a combination such as $3 𝐚 + 𝐛$ or $𝐚 - 2 𝐛$ : a two-mark Work out with two column vectors given in the stem M1 for the scalar multiple written as a column vector, e.g. $2 𝐛 = (\binom{- 8}{2})$ . A1 for the final column vector with both rows correct. Write the scalar multiple on its own line first: it carries the method mark even if a sign slips later.
Explaining why two vectors are parallel: a two-mark Explain or Show (that) that gives two column vectors Write one vector as a number times the other, e.g. $(\binom{- 6}{9}) = - 3 (\binom{2}{- 3})$ , then say in words that one is a scalar multiple of the other, so they are parallel. 'They look parallel' gives no reason.
Getting the direction of $\vec{A B}$ right: end minus start: a one-mark Write that gives two points and asks for $\vec{A B}$ as a column vector, which may lead into a Find that uses the vector to reach a new point B1 for the column vector with both signs right: $\vec{A B} = (\binom{x_{B} - x_{A}}{y_{B} - y_{A}})$ . With a single mark for the answer, one wrong sign scores nothing, so check the signs against a sketch: if $B$ is to the left of $A$ , the top number is negative.

Words to know

Vector: A quantity with both a size and a direction, such as a movement on a grid.
Column vector: $(\binom{x}{y})$ : $x$ is the movement across (right is positive), $y$ is the movement up or down (up is positive).
Component: One of the two numbers in a column vector: the $x$ -component (across) or the $y$ -component (up or down).
Translation: A transformation that moves every point of a shape by the same vector, with no turning and no change of size.
Scalar: An ordinary number, with a size but no direction. Multiplying a vector by a scalar multiplies both of its components.
Resultant: The single vector with the same effect as two or more vectors done one after another, e.g. $𝐚 + 𝐛$ .
Negative of a vector: $- 𝐚$ : the same size as $𝐚$ in the opposite direction. $\vec{B A} = - \vec{A B}$ .
Parallel vectors: Vectors where one is a scalar multiple of the other. They point in the same direction or in opposite directions.

All Maths lessons Next: 8.3 Magnitude of a vector

8.2 Vectors in two dimensions

What a vector is, and the three ways it is written

Describing a translation with a vector

Finding AB→ from coordinates

Adding vectors: one movement after another

Subtracting vectors, and the negative of a vector

Multiplying a vector by a scalar

Putting the operations together

Worked examples

Where people go wrong

Check yourself

Exam questions

How this is examined

Words to know

Finding $\vec{A B}$ from coordinates