Vector geometry

From the diagram, $A$ lies midway along $OC$, and $B$ is the point on $OD$ for which $OB = \dfrac{1}{3}OD$. Also, $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$.

May/June 2015

$ABCDE$ is a pentagon. $AFB$, $AHE$ and $BGC$ lie on straight lines. $F$ lies halfway along $AB$. $H$ lies halfway along $AE$. $G$ divides $BC$ in the ratio $1:2$. $\overrightarrow{AH}=a$, $\overrightarrow{AF}=a-b$, $\overrightarrow{BG}=\overrightarrow{ED}=c$.

May/June 2015

The diagram shows $X$ on $AB$ with $AX = \frac{1}{4}AB$, $Y$ on $AC$ with $AY = \frac{1}{3}AC$, and $Z$ on the extension of $BC$ with $CZ = 2BC$. Also, $\overrightarrow{AY} = p$ and $\overrightarrow{AX} = q$.

May/June 2016

$OPRQ$ forms a parallelogram, and $S$ lies on $PR$ so that $PS : SR = 1 : 3$. $\vec{OP} = \mathbf{p}$ and $\vec{OQ} = \mathbf{q}$.

May/June 2017

$OACB$ forms a parallelogram. $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$. $P$ and $Q$ are points on $OC$ such that $OP = PQ = QC$.

May/June 2017

The diagram shows triangle $OYC$. Point $A$ lies on $OY$ and point $B$ lies on $CY$. $AB$ is parallel to $OC$. The lines $AC$ and $OB$ intersect at $X$.

May/June 2018

The diagram shows that $\overrightarrow{PQ} = 4\mathbf{p}$, $\overrightarrow{QR} = 3\mathbf{q}$ and $\overrightarrow{PT} = \mathbf{p} + 2\mathbf{q}$. Also, $\overrightarrow{QU} = \frac{2}{3}\overrightarrow{QR}$ and $\overrightarrow{PT} = \frac{2}{3}\overrightarrow{PS}$.

May/June 2019

$\mathbf{f} = \begin{pmatrix}4\\-3\end{pmatrix}$ and $\mathbf{g} = \begin{pmatrix}1\\-5\end{pmatrix}$. Points $O, A$ and $B$ are given with $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$. Point $P$ lies on $OA$ so that $OP = \frac{1}{3}OA$. The points $O, Q$ and $R$ are collinear, and $Q$ is the midpoint of $PB$.

May/June 2019

In triangle $ACD$, $B$ lies at the midpoint of $AC$, and $E$ lies at the midpoint of $AD$. $\overrightarrow{AB}=6\mathbf{a}+3\mathbf{b}$ and $\overrightarrow{DC}=5\mathbf{a}+2\mathbf{b}$.

May/June 2022

$OABC$ and $OPQR$ are parallelograms. $A$ lies on $OP$ and $C$ lies on $OR$. Also, $\overrightarrow{OA}=\mathbf{a}$ and $\overrightarrow{OC}=\mathbf{c}$. The ratio $OA:OP=1:4$ and $OC:CR=2:3$.

May/June 2022

In the parallelogram $OABC$, $\overrightarrow{OA} = 2a$ and $\overrightarrow{OC} = 3c$. $M$ is the midpoint of $BC$, and $T$ lies on $OB$ with $OT : TB = 2 : 1$.

May/June 2023

$F$ is located at $(6,1)$, $G$ is located at $(-2,4)$ and $\overrightarrow{GH} = \begin{pmatrix} -1 \\ -8 \end{pmatrix}$. Calculate $|\overrightarrow{FH}|$.

May/June 2023

$OCB$ forms a triangle. Point $A$ lies on $OC$ such that $OA : AC = 1 : 3$. $X$ is the midpoint of $BC$. $\vec{OA} = \mathbf{a}$ and $\vec{OB} = \mathbf{b}$.

May/June 2024

$\vec{OA}=\vec{a}$ and $\vec{OB}=\vec{b}$. Point $X$ lies on $AB$ with $AX:XB=3:2$. $OXC$ forms a straight line. $AC$ is parallel to $OB$.

May/June 2025

The diagram depicts quadrilateral $OACB$. $\vec{OA} = \vec{a}$ and $\vec{OB} = \vec{b}$. $OB$ runs parallel to $AC$, and $AC = 2OB$.

May/June 2025

The diagram shows that $ADB$ and $ACF$ are straight lines. $BC$ meets $DF$ at $E$. The ratio $AC : CF = 2 : 1$. Also, $\overrightarrow{DB} = \mathbf{p}$, $\overrightarrow{BE} = 3\mathbf{q}$, $\overrightarrow{EC} = 2\mathbf{q}$ and $\overrightarrow{AC} = 3\mathbf{p} + 5\mathbf{q}$.

Oct/Nov 2018

In the sketch, $B$ lies at the midpoint of $OD$, and the ratio $OA : AC = 1 : 3$. Also, $\overrightarrow{OA} = \vec{a}$ and $\overrightarrow{OB} = \vec{b}$.

Oct/Nov 2019

$OAC$ forms a triangle, and $B$ lies on $AC$ with $AB : BC = 3 : 2$. $\overrightarrow{OA} = \vec{a}$ and $\overrightarrow{OB} = \vec{b}$.

Oct/Nov 2022

The diagram shows parallelogram $OABC$. The vector $\vec{OA}$ is $\vec{a}$ and $\vec{OC}$ is $\vec{c}$. $X$ lies halfway along $AC$. $Y$ is on $AB$ such that $AY : YB = 2 : 1$.

Oct/Nov 2023

$OAB$ forms a triangle. $P$ is on $AB$ and $AP : PB = 2:3$. $\overrightarrow{OA} = 4a$ and $\overrightarrow{OP} = 3a + 2b$.

Oct/Nov 2023

The diagram shows $\overrightarrow{OA}=a$, $\overrightarrow{OB}=2b$ and $AB:BC=1:3$. $OBD$ is a straight line.

Oct/Nov 2024