GCSE · Vectors

Vectors on a shapeExam-style questions

Expressing any journey around a figure in terms of two given vectors — why every route gives the same answer, why going against an arrow flips the sign, and why a point splitting a side in the ratio 2:3 sits two fifths along, not two thirds.

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12 questions · GCSE (Higher) level · mark scheme and worked solution on each · no sign-up to try

DifficultyAll Level 2 Level 3 Level 4 Level 5

1
$OABC$ is a parallelogram. $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OC} = \mathbf{b}$ .

$D$ is the point on $AB$ such that $AD:DB = 3:1$ .

$E$ is the midpoint of $OC$ .The diagram is a sketch and is not accurately drawn.

(a) Find $\overrightarrow{OD}$ in terms of $\mathbf{a}$ and $\mathbf{b}$ .

(b) Find $\overrightarrow{ED}$ in terms of $\mathbf{a}$ and $\mathbf{b}$ , giving your answer in its simplest form.
Level 5 · 6 marksTry it
2
$OAB$ 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 $BN:NA = 1:3$ .The diagram is a sketch and is not accurately drawn.

(a) Find $\overrightarrow{ON}$ in terms of $\mathbf{a}$ and $\mathbf{b}$ , giving your answer in its simplest form.

(b) Find $\overrightarrow{MN}$ in terms of $\mathbf{a}$ and $\mathbf{b}$ , giving your answer in its simplest form.
Level 5 · 6 marksTry it
3
$OAB$ is a triangle.

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

$P$ is the point on $AB$ such that $AP : PB = 1 : 2$ .Diagram NOT accurately drawn
(a) Write $\overrightarrow{AB}$ in terms of $\mathbf{a}$ and $\mathbf{b}$ . (1 mark)
(b) Show that $\overrightarrow{OP} = \dfrac{2}{3}\mathbf{a} + \dfrac{1}{3}\mathbf{b}$ (3 marks)
Level 5 · 4 marksTry it
4
$OAB$ is a triangle. $\overrightarrow{OA} = 2\mathbf{a}$ $\qquad$ $\overrightarrow{OB} = 6\mathbf{b}$

$D$ is the point on $OA$ extended such that $\overrightarrow{OD} = k\overrightarrow{OA}$

$X$ is the point on $OB$ such that $OX : XB = 1 : 2$

$Y$ is the point on $BD$ such that $BY : YD = 3 : 1$
(a) Find $\overrightarrow{XY}$ in terms of $k$ , $\mathbf{a}$ and $\mathbf{b}$ .

Give your answer in its simplest form.

(4 marks)
(b) Is $XY$ parallel to $OA$ ?

Give a reason for your answer.

(1 mark)
Level 5 · 5 marksTry it