Vectors (incl. vector proof) – GCSE Maths exam questions with mark scheme

1Grade 1–3FoundationNon-calculator3 marks

$\mathbf{a}=\begin{pmatrix}4\\1\end{pmatrix}$ and $\mathbf{b}=\begin{pmatrix}2\\-5\end{pmatrix}$.

Write as column vectors:

(a) $3\mathbf{a}$ [1]

(b) $\mathbf{a}-\mathbf{b}$ [1]

(a) $\begin{pmatrix}12\\3\end{pmatrix}$B1
(b) $\begin{pmatrix}2\\6\end{pmatrix}$B1
(c) $\begin{pmatrix}8\\-9\end{pmatrix}$B1
Answer: (a) $\begin{pmatrix}12\\3\end{pmatrix}$ (b) $\begin{pmatrix}2\\6\end{pmatrix}$ (c) $\begin{pmatrix}8\\-9\end{pmatrix}$
Examiner tip: Take care with subtracting a negative: $1-(-5)=6$.

2Grade 4–5Non-calculator3 marks

$\mathbf{a}=\begin{pmatrix}3\\-2\end{pmatrix}$ and $\mathbf{b}=\begin{pmatrix}-1\\4\end{pmatrix}$.

(a) Work out $\mathbf{a}+\mathbf{b}$ as a column vector. [1]

(b) Work out $2\mathbf{a}-3\mathbf{b}$ as a column vector. [2]

(a) $\begin{pmatrix}2\\2\end{pmatrix}$B1
(b) $\begin{pmatrix}6\\-4\end{pmatrix}$ and $\begin{pmatrix}-3\\12\end{pmatrix}$ seen, or one correct component in the final answerM1
$\begin{pmatrix}9\\-16\end{pmatrix}$A1
Answer: (a) $\begin{pmatrix}2\\2\end{pmatrix}$ (b) $\begin{pmatrix}9\\-16\end{pmatrix}$
Examiner tip: Multiply both components by the scalar before subtracting – write out $2\mathbf{a}$ and $3\mathbf{b}$ separately.

3Grade 4–5Non-calculator3 marks

Point A has coordinates $(1,2)$ and point B has coordinates $(5,-1)$.

(a) Write $\overrightarrow{AB}$ as a column vector. [1]

$\overrightarrow{BC}=\begin{pmatrix}-2\\6\end{pmatrix}$.

(b) Work out the coordinates of point C. [2]

(a) $\begin{pmatrix}4\\-3\end{pmatrix}$B1
(b) $(5+(-2),\;-1+6)$ or one correct coordinateM1
$(3,5)$A1
Answer: (a) $\begin{pmatrix}4\\-3\end{pmatrix}$ (b) $(3,5)$
Examiner tip: $\overrightarrow{AB}$ is 'end minus start' – B minus A, not A minus B.

4Grade 6–7HigherNon-calculator5 marks

(a) $\mathbf{p}=\begin{pmatrix}5\\-12\end{pmatrix}$ and $\mathbf{p}+\mathbf{r}=\begin{pmatrix}1\\3\end{pmatrix}$. Work out $\mathbf{r}$ as a column vector. [2]

(b) $m$ and $n$ are numbers such that

$$m\begin{pmatrix}2\\1\end{pmatrix}+n\begin{pmatrix}1\\3\end{pmatrix}=\begin{pmatrix}7\\11\end{pmatrix}$$

Work out the values of $m$ and $n$. [3]

(a) $\mathbf{r}=\begin{pmatrix}1\\3\end{pmatrix}-\begin{pmatrix}5\\-12\end{pmatrix}$ or one correct componentM1
$\begin{pmatrix}-4\\15\end{pmatrix}$A1
(b) forms both equations $2m+n=7$ and $m+3n=11$M1
correct method to eliminate one variableM1
$m=2,\;n=3$A1
Answer: (a) $\begin{pmatrix}-4\\15\end{pmatrix}$ (b) $m=2,\;n=3$
Examiner tip: Take care with the signs: $3-(-12)=15$. In (b), write one equation for the top components and one for the bottom, then solve them simultaneously.

5Grade 6–7HigherNon-calculator4 marks

OABC is a parallelogram. $\overrightarrow{OA}=\mathbf{a}$ and $\overrightarrow{OC}=\mathbf{c}$. M is the midpoint of AB.

Write each vector in terms of $\mathbf{a}$ and $\mathbf{c}$. Give your answers in their simplest form.

(a) $\overrightarrow{AC}$ [1]

(b) $\overrightarrow{OM}$ [2]

(a) $\mathbf{c}-\mathbf{a}$ (or $-\mathbf{a}+\mathbf{c}$)B1
(b) $\overrightarrow{OA}+\tfrac12\overrightarrow{AB}$ with $\overrightarrow{AB}=\mathbf{c}$M1
$\mathbf{a}+\tfrac12\mathbf{c}$A1
(c) $\mathbf{a}-\tfrac12\mathbf{c}$ (ft $-\mathbf{c}+$ their $\overrightarrow{OM}$)B1
Answer: (a) $\mathbf{c}-\mathbf{a}$ (b) $\mathbf{a}+\tfrac12\mathbf{c}$ (c) $\mathbf{a}-\tfrac12\mathbf{c}$
Examiner tip: In a parallelogram, opposite sides are equal vectors: $\overrightarrow{AB}=\overrightarrow{OC}=\mathbf{c}$.

6Grade 8–9HigherNon-calculator5 marks

OAB is a triangle. $\overrightarrow{OA}=\mathbf{a}$ and $\overrightarrow{OB}=\mathbf{b}$.

M is the midpoint of OB. P is the point on AB such that AP : PB = 1 : 2. OA is extended to the point Q so that OQ = 2OA.

Use a vector method to prove that M, P and Q lie on a straight line. [5]

$\overrightarrow{AB}=\mathbf{b}-\mathbf{a}$ or $\overrightarrow{BA}=\mathbf{a}-\mathbf{b}$ (may be implied by a correct $\overrightarrow{AP}=\tfrac13(\mathbf{b}-\mathbf{a})$ or $\overrightarrow{OP}$)M1
$\overrightarrow{OP}=\mathbf{a}+\tfrac13(\mathbf{b}-\mathbf{a})\;\left(=\tfrac23\mathbf{a}+\tfrac13\mathbf{b}\right)$M1
$\overrightarrow{MP}=\tfrac23\mathbf{a}-\tfrac16\mathbf{b}$ or $\overrightarrow{MQ}=2\mathbf{a}-\tfrac12\mathbf{b}$ (or another correct vector joining two of M, P, Q, e.g. $\overrightarrow{PQ}=\tfrac43\mathbf{a}-\tfrac13\mathbf{b}$, or its reverse)M1
two correct vectors joining pairs of M, P, Q, e.g. $\overrightarrow{MP}=\tfrac16(4\mathbf{a}-\mathbf{b})$ and $\overrightarrow{MQ}=\tfrac12(4\mathbf{a}-\mathbf{b})$ (or $\overrightarrow{PQ}=\tfrac13(4\mathbf{a}-\mathbf{b})$)A1
concludes e.g. $\overrightarrow{MQ}=3\overrightarrow{MP}$ (or $\overrightarrow{PQ}=2\overrightarrow{MP}$, or equivalent), so the vectors are parallel and share a common point (e.g. M or P), hence M, P and Q are collinearC1
Answer: $\overrightarrow{MP}=\tfrac16(4\mathbf{a}-\mathbf{b})$, $\overrightarrow{MQ}=\tfrac12(4\mathbf{a}-\mathbf{b})=3\overrightarrow{MP}$; parallel with common point M, so collinear.
Examiner tip: Showing the vectors are multiples is not enough – the final mark needs 'parallel' AND 'a common point' to conclude they're on a straight line.

7Grade 8–9HigherNon-calculator5 marks

OAB is a triangle. $\overrightarrow{OA}=\mathbf{a}$ and $\overrightarrow{OB}=\mathbf{b}$.

M is the midpoint of AB. N is the point on OB such that ON : NB = 3 : 1. The lines OM and AN meet at X.

Work out the ratio OX : XM. [5]

$\overrightarrow{OM}=\tfrac12(\mathbf{a}+\mathbf{b})$M1
$\overrightarrow{AN}=-\mathbf{a}+\tfrac34\mathbf{b}$M1
two expressions for $\overrightarrow{OX}$: $\lambda\cdot\tfrac12(\mathbf{a}+\mathbf{b})$ and $\mathbf{a}+\mu\left(\tfrac34\mathbf{b}-\mathbf{a}\right)$M1
equates coefficients, $\tfrac{\lambda}{2}=1-\mu$ and $\tfrac{\lambda}{2}=\tfrac{3\mu}{4}$, and solves ($\mu=\tfrac47$, $\lambda=\tfrac67$)M1
6 : 1A1
Answer: 6 : 1
Examiner tip: Use a different letter for each line's multiplier, then compare the coefficients of $\mathbf{a}$ and of $\mathbf{b}$ separately.

8Grade 1–3FoundationNon-calculator3 marks

Triangle P and triangle Q are drawn on the grid.

(a) Describe fully the single transformation that maps triangle P onto triangle Q. [2]

Triangle Q is then translated by the vector $\begin{pmatrix}-2\\-6\end{pmatrix}$ to give triangle R.

(b) Write down the column vector of the single translation that maps triangle P onto triangle R. [1]

(a) Translation (accept translate, translated; do not accept move, shift, slide or transformation alone)B1
(a) by $\begin{pmatrix}5\\4\end{pmatrix}$ (accept "5 right and 4 up"; do not accept coordinates $(5,4)$ or an upside-down vector). B0 for both marks if a second transformation is also givenB1
(b) $\begin{pmatrix}3\\-2\end{pmatrix}$ (ft their vector from (a) $+\begin{pmatrix}-2\\-6\end{pmatrix}$)B1
Answer: (a) Translation by $\begin{pmatrix}5\\4\end{pmatrix}$ (b) $\begin{pmatrix}3\\-2\end{pmatrix}$
Examiner tip: Write the vector as a column with the horizontal move on top – a coordinate like (5, 4) or "move it 5 and 4" does not get the second mark.

9Grade 1–3FoundationCalculator4 marks

Three vectors $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ are drawn on the centimetre grid.

(a) Write $\mathbf{a}$ as a column vector. [1]

(b) Write $\mathbf{c}$ in terms of $\mathbf{a}$. [1]

(a) $\begin{pmatrix}3\\2\end{pmatrix}$B1
(b) $-2\mathbf{a}$ (accept $\mathbf{c}=-2\mathbf{a}$; do not accept $2\mathbf{a}$)B1
(c) reads $\mathbf{b}=\begin{pmatrix}2\\-4\end{pmatrix}$ and finds $2\mathbf{b}=\begin{pmatrix}4\\-8\end{pmatrix}$, or correct method to subtract their $\mathbf{a}$ from their $2\mathbf{b}$ (may be seen in one component only)M1
$\begin{pmatrix}1\\-10\end{pmatrix}$ (ft their $\mathbf{a}$ and $\mathbf{b}$ if M1 earned)A1
Answer: (a) $\begin{pmatrix}3\\2\end{pmatrix}$ (b) $-2\mathbf{a}$ (c) $\begin{pmatrix}1\\-10\end{pmatrix}$
Examiner tip: Check the direction of the arrow: c is twice as long as a but points the opposite way, so it is $-2\mathbf{a}$, not $2\mathbf{a}$.

10Grade 4–5Non-calculator3 marks

$\mathbf{p}=\begin{pmatrix}6\\-9\end{pmatrix}$, $\mathbf{q}=\begin{pmatrix}-4\\6\end{pmatrix}$ and $\mathbf{r}=\begin{pmatrix}9\\6\end{pmatrix}$.

(a) Show that $\mathbf{p}$ is parallel to $\mathbf{q}$. [2]

(b) Jake says,

"$\mathbf{r}$ is parallel to $\mathbf{p}$ because both vectors have a 9 in them."

Explain why Jake is wrong. [1]

(a) writes one vector as a multiple of the other, e.g. $\mathbf{p}=-\tfrac32\mathbf{q}$ or $\mathbf{q}=-\tfrac23\mathbf{p}$, or both as multiples of the same vector, e.g. $\mathbf{p}=3\begin{pmatrix}2\\-3\end{pmatrix}$ and $\mathbf{q}=-2\begin{pmatrix}2\\-3\end{pmatrix}$M1
fully correct working with conclusion that $\mathbf{p}$ is a (scalar) multiple of $\mathbf{q}$, so they are parallel (accept "same ratio $6:-9=-4:6$" with conclusion; do not accept "same gradient" with no working)A1
(b) explanation that $\mathbf{r}$ is not a multiple of $\mathbf{p}$, e.g. $9\div6=1.5$ but $6\div(-9)\ne1.5$, or "the 9 is in a different position / to be parallel both components must be multiplied by the same number" (ignore any reference to the sign alone if the multiple idea is present)C1
Answer: (a) $\mathbf{p}=-\tfrac32\mathbf{q}$, so parallel (b) $\mathbf{r}$ is not a scalar multiple of $\mathbf{p}$ (top ×1.5, bottom ×$-\tfrac23$).
Examiner tip: Parallel means "a multiple of": you must show the same number (here $-\tfrac32$) multiplies both components – a negative multiple still counts as parallel.

11Grade 4–5Non-calculator4 marks

ABCD is a quadrilateral.

$\overrightarrow{AB}=\begin{pmatrix}5\\2\end{pmatrix}$, $\overrightarrow{BC}=\begin{pmatrix}1\\-4\end{pmatrix}$, $\overrightarrow{CD}=\begin{pmatrix}-5\\-2\end{pmatrix}$

(a) Work out $\overrightarrow{AD}$ as a column vector. [2]

(b) What type of quadrilateral is ABCD? Give a reason for your answer, using vectors. [2]

(a) $\overrightarrow{AD}=\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}$ or adds all three vectors (at least one component correct)M1
$\begin{pmatrix}1\\-4\end{pmatrix}$A1
(b) parallelogramB1
reason, e.g. $\overrightarrow{AD}=\overrightarrow{BC}$ (or $\overrightarrow{AB}=-\overrightarrow{CD}$ i.e. $\overrightarrow{AB}=\overrightarrow{DC}$), so a pair of opposite sides is parallel and equal in length (accept "both pairs of opposite sides are parallel"; dependent on B1)C1
Answer: (a) $\begin{pmatrix}1\\-4\end{pmatrix}$ (b) parallelogram, since $\overrightarrow{AD}=\overrightarrow{BC}$ (and $\overrightarrow{AB}=\overrightarrow{DC}$)
Examiner tip: A route round the shape is the sum of the vectors along it; and "equal vectors" tells you both parallel AND the same length – say both.

12Grade 6–7HigherCalculator5 marks

ABCDEF is a regular hexagon.

$\overrightarrow{AB}=\mathbf{p}$ and $\overrightarrow{BC}=\mathbf{q}$.

M is the midpoint of DE.

Write each vector in terms of $\mathbf{p}$ and $\mathbf{q}$. Give each answer in its simplest form.

(a) $\overrightarrow{AD}$ [1]

(b) $\overrightarrow{CD}$ [2]

(a) $2\mathbf{q}$B1
(b) correct route, e.g. $\overrightarrow{CD}=\overrightarrow{CB}+\overrightarrow{BA}+\overrightarrow{AD}=-\mathbf{q}-\mathbf{p}+2\mathbf{q}$ (ft their $\overrightarrow{AD}$), or uses $\overrightarrow{CD}=\overrightarrow{BO}$ where O is the centre with $\overrightarrow{BO}=\mathbf{q}-\mathbf{p}$ via $\overrightarrow{AO}=\mathbf{q}$M1
$\mathbf{q}-\mathbf{p}$ (accept $-\mathbf{p}+\mathbf{q}$)A1
(c) $\overrightarrow{DE}=-\mathbf{p}$ seen or used, or a correct route e.g. $\overrightarrow{AD}+\tfrac12\overrightarrow{DE}$ (ft their $\overrightarrow{AD}$)M1
$2\mathbf{q}-\tfrac12\mathbf{p}$ (accept $-\tfrac12\mathbf{p}+2\mathbf{q}$ or $\tfrac12(4\mathbf{q}-\mathbf{p})$; not in simplest form, e.g. $2\mathbf{q}-\mathbf{p}+\tfrac12\mathbf{p}$, is A0)A1
Answer: (a) $2\mathbf{q}$ (b) $\mathbf{q}-\mathbf{p}$ (c) $2\mathbf{q}-\tfrac12\mathbf{p}$
Examiner tip: In a regular hexagon, opposite sides are equal and opposite ($\overrightarrow{DE}=-\mathbf{p}$) and the long diagonal AD is parallel to BC and twice as long.

13Grade 6–7HigherNon-calculator5 marks

OABC is a parallelogram. $\overrightarrow{OA}=\mathbf{a}$ and $\overrightarrow{OC}=\mathbf{c}$.

P is the point on AB such that AP : PB = 1 : 2.
Q is the point on CB such that CQ : QB = 1 : 2.

(a) Use a vector method to prove that PQ is parallel to AC. [4]

(b) Write down the ratio PQ : AC. [1]

(a) $\overrightarrow{OP}=\mathbf{a}+\tfrac13\mathbf{c}$ (or $\overrightarrow{AP}=\tfrac13\mathbf{c}$, $\overrightarrow{PB}=\tfrac23\mathbf{c}$)M1
$\overrightarrow{OQ}=\mathbf{c}+\tfrac13\mathbf{a}$ (or $\overrightarrow{CQ}=\tfrac13\mathbf{a}$, $\overrightarrow{BQ}=-\tfrac23\mathbf{a}$)M1
$\overrightarrow{PQ}=\tfrac23\mathbf{c}-\tfrac23\mathbf{a}$ (from any correct route, e.g. $\overrightarrow{PB}+\overrightarrow{BQ}$)A1
$\overrightarrow{PQ}=\tfrac23(\mathbf{c}-\mathbf{a})$ and $\overrightarrow{AC}=\mathbf{c}-\mathbf{a}$, so $\overrightarrow{PQ}=\tfrac23\overrightarrow{AC}$, a multiple of $\overrightarrow{AC}$, hence parallel (needs $\overrightarrow{AC}$ stated and a conclusion)C1
(b) 2 : 3 (ft their scalar multiple from (a))B1
Answer: (a) $\overrightarrow{PQ}=\tfrac23(\mathbf{c}-\mathbf{a})=\tfrac23\overrightarrow{AC}$, so PQ is parallel to AC (b) 2 : 3
Examiner tip: Factorise your answer to show the multiple clearly – writing $\tfrac23\mathbf{c}-\tfrac23\mathbf{a}$ without linking it to $\overrightarrow{AC}=\mathbf{c}-\mathbf{a}$ loses the final mark.

14Grade 8–9HigherNon-calculator5 marks

OAB is a triangle. $\overrightarrow{OA}=\mathbf{a}$ and $\overrightarrow{OB}=\mathbf{b}$.

M is the midpoint of OA.
N is the point on AB such that AN : NB = 3 : 1.

The line MN is extended to meet the line OB, extended beyond B, at the point X.

Work out the ratio OB : BX. [5]

$\overrightarrow{ON}=\mathbf{a}+\tfrac34(\mathbf{b}-\mathbf{a})\;\left(=\tfrac14\mathbf{a}+\tfrac34\mathbf{b}\right)$M1
$\overrightarrow{MN}=-\tfrac14\mathbf{a}+\tfrac34\mathbf{b}$ (ft their $\overrightarrow{ON}$)M1
forms an equation for X, e.g. $\overrightarrow{OX}=\tfrac12\mathbf{a}+k\left(-\tfrac14\mathbf{a}+\tfrac34\mathbf{b}\right)=\lambda\mathbf{b}$ and equates coefficients of $\mathbf{a}$: $\tfrac12-\tfrac14k=0$M1
$k=2$ so $\overrightarrow{OX}=\tfrac32\mathbf{b}$ (accept $\lambda=\tfrac32$)A1
OB : BX = 2 : 1 (accept 1 : 0.5)A1
Answer: OB : BX = 2 : 1 (since $\overrightarrow{OX}=\tfrac32\mathbf{b}$)
Examiner tip: Use "X is on line MN" and "X is on line OB" to get two expressions for $\overrightarrow{OX}$, then match the $\mathbf{a}$ parts – there is no $\mathbf{a}$ in a vector along OB.

15Grade 8–9HigherNon-calculator4 marks

$\overrightarrow{AB}=\begin{pmatrix}k\\4\end{pmatrix}$ and $\overrightarrow{CD}=\begin{pmatrix}3\\k+4\end{pmatrix}$, where $k$ is a constant.

$\overrightarrow{AB}$ is parallel to $\overrightarrow{CD}$ and the two vectors point in the same direction.

Find the value of $k$. You must show your working. [4]

sets up a correct equation using parallel vectors, e.g. $\dfrac{k}{3}=\dfrac{4}{k+4}$, or $k=3\lambda$ and $4=\lambda(k+4)$M1
rearranges to a correct quadratic, e.g. $k^2+4k-12=0$ (or $3\lambda^2+4\lambda-4=0$), and attempts to solve (e.g. $(k+6)(k-2)=0$ or formula)M1
$k=2$ and $k=-6$ (or $\lambda=\tfrac23$ and $\lambda=-2$)A1
selects $k=2$ with a reason for rejecting $k=-6$, e.g. when $k=-6$, $\overrightarrow{AB}=\begin{pmatrix}-6\\4\end{pmatrix}=-2\begin{pmatrix}3\\-2\end{pmatrix}=-2\overrightarrow{CD}$, which points in the opposite direction (accept "the multiple is negative", or rejects $\lambda=-2$ because it is negative and gives $k=3\times\tfrac23=2$); $k=2$ alone with no consideration of the second root scores M1M1A0C0 at mostC1
Answer: $k=2$ (then $\overrightarrow{CD}=\tfrac32\overrightarrow{AB}$)
Examiner tip: Both roots make the vectors parallel – the "same direction" condition is there to make you test each one and reject the negative multiple.

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