Vectors (Edexcel GCSE Maths: Higher): Exam Questions

$APB$ is a triangle.
$N$ is a point on $AP$.

$\overrightarrow{AB}=\mathbf{a}\overrightarrow{AN}\mathbf{=}2\mathbf{b}\overrightarrow{NP}\mathbf{=}\mathbf{b}$

Find the vector $\overrightarrow{PB}$, in terms of $\mathbf{a}$ and $\mathbf{b}$.

$B$ is the midpoint of $AC$.
$M$ is the midpoint of $PB$.

Show that $NMC$ is a straight line.

$PQRS$ is a quadrilateral.

$\overrightarrow{QR}=8\mathbf{a}$
$\overrightarrow{QP}=4\mathbf{b}$
$\overrightarrow{RS}=4\mathbf{a}+2\mathbf{b}$

$X$ is the point on $PR$ such that $PX:XR=3:1$.

Prove that $X$ is the midpoint of $QS$.

$OPTR$is a trapezium.

$\begin{array}{l}\overrightarrow{OP}=\mathbf{a}\\ \overrightarrow{PT}=\mathbf{b}\\ \overrightarrow{OR}=3\mathbf{b}\end{array}$

i) Find $\overrightarrow{OT}$ in terms of $\mathbf{a}$ and$\mathbf{b}$

[1]

ii) Find $\overrightarrow{PR}$ in terms of $\mathbf{a}$ and $\mathbf{b}$ Give your answer in its simplest form.

[1]

$S$ is the point on $PR$ such that $PS:SR$ = 1 : 3

Find $\overrightarrow{OS}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.
Give your answer in its simplest form.

What does your answer to part (b) tell you about the position of point $S$?

$OAB$is a triangle.

$\begin{array}{l}\overrightarrow{OA}=5\mathbf{a}\\ \overrightarrow{OB}=2\mathbf{b}\end{array}$

$T$ is the point on $AB$such that $AT:TB=5:1$

Show that $OT$is parallel to the vector $\mathbf{a}+2\mathbf{b}$

$\overrightarrow{OA}=6\mathbf{a}$ and $\overrightarrow{OB}=6\mathbf{b}$
$M$ is the midpoint of $AB$.

Write $\overrightarrow{OM}$in terms of $\mathbf{a}$ and $\mathbf{b}$.
Give your answer in its simplest form.

$N$ is the midpoint of $OB$.
$G$ is the point on $OM$ such that $OG:GM$= 2 : 1

Show that $AGN$ is a straight line.

$CAYB$ is a quadrilateral.

$\begin{array}{l}\overrightarrow{CA}=3\mathbf{a}\\ \overrightarrow{CB}=6\mathbf{b}\\ \overrightarrow{BY}=5\mathbf{a}\mathbf{-}\mathbf{b}\end{array}$

$X$ is the point on $AB$such that $AX:XB=1:2$

Prove that $\overrightarrow{CX}=\frac{2}{5}\overrightarrow{CY}$

$OAB$is a triangle.
$P$ is the point on $AB$such that $AP:PB=5:3$

$\begin{array}{l}\overrightarrow{OA}=2\mathbf{a}\\ \overrightarrow{OB}=2\mathbf{b}\end{array}$

$\overrightarrow{OP}=k(3\mathbf{a}+5\mathbf{b})$ where $k$ is a scalar quantity.

Find the value of $k$.

$ABC$is a triangle.
The midpoint of $BC$is $M$.
$P$ is a point on $AM$.

$\begin{array}{rcl}\overrightarrow{AB}& =& 4\mathbf{a}\\ \overrightarrow{AC}& =& 2\mathbf{b}\\ \overrightarrow{AP}& =& \frac{3}{2}\mathbf{a}+\frac{3}{4}\mathbf{b}\end{array}$

Find the ratio $AP:PM$

$OAB$ is a triangle.

$\overrightarrow{OA}=\mathbf{a}$

$\overrightarrow{OB}=\mathbf{b}$

The point $C$ lies on $OA$ such that $OC:CA=1:2$
The point $D$ lies on $OB$ such that $OD:DB=1:2$

Using a vector method, prove that $ABDC$ is a trapezium.

The diagram shows parallelogram $ABCD$.

$\overrightarrow{AB}=\left(\begin{array}{c}2\\ 7\end{array}\right)$       $\overrightarrow{AC}=\left(\begin{array}{c}10\\ 11\end{array}\right)$

The point $B$ has coordinates (5, 8)
Work out the coordinates of the point $C$.

The point $E$ has coordinates (63, 211)
Use a vector method to prove that $ABE$ is a straight line.

The diagram shows triangle $ABD.$

$N$ is the midpoint of $BC$.

$C$ is the midpoint of $AD$.

$M$ is the point on $AB$ such that $AM:MB=3:1$

$\overrightarrow{AB}=\mathbf{p}$     and    $\overrightarrow{AC}=\mathbf{q}$

Express, in terms of $\mathbf{p}$ and $\mathbf{q}$,

i) $\overrightarrow{BD}$

ii) $\overrightarrow{MN}$

State, giving reasons, two different geometric facts relating $MN$ and $BD.$

In triangle $JKL$

$M$ is the midpoint of
$JK$ $JN:NL=3:2$
$\overrightarrow{KL}=7\mathbf{a}$      $\overrightarrow{NL}=4\mathbf{b}$

Work out $\overrightarrow{JM}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.

Give your answer in its simplest form.

$PQR$ is a straight line.

$PQ:QR=3:1$

$\overrightarrow{PQ}=\mathbf{a}$

Choose the vector $\overrightarrow{RQ}$

$\mathbf{a}=\left(\begin{array}{c}6\\ -10\end{array}\right)$       $\mathbf{b}=\left(\begin{array}{c}-1\\ 2\end{array}\right)$      $\mathbf{c}=\left(\begin{array}{c}-4\\ 7\end{array}\right)$

Work out $\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{+}\mathbf{c}$

Show that $\mathbf{a}+2\mathbf{c}$ is parallel to $\mathbf{b}$

Is $BCD$ a straight line?
Show working to support your answer.

Given that

      $m\left(\begin{array}{c}4\\ 1\end{array}\right)+n\left(\begin{array}{c}5\\ 2\end{array}\right)=\left(\begin{array}{c}12\\ 6\end{array}\right)$

find the value of $m$ and the value of $n$.

$m$ = .......................

$n$ = ....................... 

Two vectors, a and b, are shown on the 1 centimetre grid below.

Show that the vector a + 2b has length 7cm.
You may use the grid below.

Vector $\mathbf{a}=\left(\begin{array}{c}2\\ 1\end{array}\right)$ , vector $\mathbf{b}=\left(\begin{array}{c}-2\\ 1\end{array}\right)$ and vector $\mathbf{c}=\left(\begin{array}{c}-12\\ 0\end{array}\right)$

Find the value $k$ so that $k\left(\mathbf{a}-\mathbf{b}\right)=\mathbf{c}$

$k$ = .....................................................

P, Q, R and S are the midpoints of OX, XY, YZ and OZ respectively.

$\overrightarrow{\mathrm{OP}}=\mathbf{a}$ , $\overrightarrow{\mathrm{XQ}}=\mathbf{b}$ and $\overrightarrow{\mathrm{OS}}=\mathbf{c}$.

Show that $\mathrm{PQ}$ is parallel to $\mathrm{SR}$.

$STUV$ is a quadrilateral.

$\stackrel{\rightharpoonup }{TS}=10\mathbf{a}-2\mathbf{b}$

$\stackrel{\rightharpoonup }{UT}=12\mathbf{b}$

$\stackrel{\rightharpoonup }{UV}=6\mathbf{a}$

$W$ is the point on $VT$ such that $VW:WT=1:2$.

Prove that $\stackrel{\rightharpoonup }{UW}=\frac{2}{3}\stackrel{\rightharpoonup }{WS}$

You must show all your working.

State two different facts about the geometric relationship between the point $U$, $W$ and $S$.

Let $\mathit{v}=\left(\begin{array}{c}-2\\ 3\end{array}\right)$, $\mathit{w}=\left(\begin{array}{c}-1\\ 4\end{array}\right)$ and $\mathit{x}=2\mathit{v}-\mathit{w}$.

Write the vector $\mathit{x}$ as a single column vector.

Show that $\mathit{v}+\mathit{x}$ is parallel to $\left(\begin{array}{c}-1\\ 1\end{array}\right)$.

$PQRS$ is a parallelogram.

$\begin{array}{l}\overrightarrow{PQ}=\mathit{a}\\ \overrightarrow{SP}=\mathit{b}\end{array}$

$M$ is the midpoint of $PR$.

$N$ is the midpoint of $PM$.

Determine the vector $\overrightarrow{NS}$ in terms of $\mathit{a}$ and $\mathit{b}$.

$OACB$ is a parallelogram.

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

$D$is the point such that $\overrightarrow{AC}=\overrightarrow{CD}$
The point $N$ divides $AB$in the ratio $2:1$

Write an expression for $\overrightarrow{ON}$ in terms of $\mathbf{a}$and $\mathbf{b}$.

Prove that $OND$is a straight line.

$ABC$is a straight line.

$AB:BC=2:5$

$\begin{array}{l}\overrightarrow{OA}=2\mathbf{a}+\mathbf{b}\\ \overrightarrow{OB}=3\mathbf{a}+2\mathbf{b}\end{array}$

Express $\overrightarrow{OC}$ in terms of $\mathbf{a}$and $\mathbf{b}$.

Give your answer in its simplest form.

$ACEF$is a parallelogram.
$B$ is the midpoint of $AC$.
$M$ is the midpoint of $BE$.

$\begin{array}{l}\overrightarrow{CB}=\mathbf{a}\\ \overrightarrow{ED}=\mathbf{b}\\ \overrightarrow{DC}=\mathbf{2}\mathbf{b}\end{array}$

Show that $AMD$ is a straight line.

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

$X$ is the midpoint of the line $AC$.

$OCD$is a straight line so that $OC:CD=k:1$

Given that $\overrightarrow{XD}=3\mathbf{c}-\frac{1}{2}\mathbf{a}$ find the value of$k.$

$OAN,OMB$ and $APB$are straight lines.
$AN=2OA.$
$M$ is the midpoint of $OB$.

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

$\overrightarrow{AP}=k\overrightarrow{AB}$ where $k$ is a scalar quantity.

Given that $MPN$is a straight line, find the value of $k$.

$OMA,ONB$ and $ABC$are straight lines.
$M$ is the midpoint of $OA$.
$B$ is the midpoint of $AC$.

$\overrightarrow{OA}=6\mathbf{a}\overrightarrow{OB}=6\mathbf{b}\overrightarrow{ON}=k\mathbf{b}$ 
where $k$ is a scalar quantity.

Given that $MNC$is a straight line, find the value of $k$.

$CDEF$is a quadrilateral.

$\overrightarrow{CD}=\mathbf{a},\overrightarrow{DE}=\mathbf{b}\mathrm{and}\overrightarrow{FC}=\mathbf{a}\mathbf{-}\mathbf{b}.$

Express $\overrightarrow{FE}$ in terms of $\mathbf{a}$ and/or $\mathbf{b}$.
Give your answer in its simplest form.

$M$ is the midpoint of $DE$.
$X$ is the point on $FM$such that $FX:XM=n:1$
$CXE$ is a straight line.

Work out the value of $n$.

The diagram shows triangle $OAB$

$\overrightarrow{OA}=8\mathbf{a}$     $\overrightarrow{OB}=6\mathbf{b}$
$M$ is the point on $OB$ such that $OM:MB=1:2$
$N$ is the midpoint of $AB$
$P$ is the point of intersection of $ON$ and $AM$

Using a vector method, find $\overrightarrow{OP}$ as a simplified expression in terms of $\mathbf{a}$ and $\mathbf{b}$
Show your working clearly.

$\overrightarrow{OP}$= ...................................

$OXYZ$ is a trapezium.

$\overrightarrow{OX}=\mathbf{a}$

$\overrightarrow{XY}=\mathbf{b}$

$\overrightarrow{OZ}=3\mathbf{b}$

$M$is the midpoint of $OX$
$N$ is the point such that  $MNZ$ and  $ONY$ are straight lines.
Given that $ON:OY=\lambda :1$
use a vector method to find the value of $\lambda$

$\lambda =$ ....................................................

In the diagram

$\begin{array}{rcl}\overrightarrow{DE}& =& \mathbf{a}\\ \overrightarrow{DH}& =& \mathbf{b}\\ \overrightarrow{HG}& =& 8\mathbf{b}\\ EX:XH& =& 3:1\\ EF:FG& =& 1:3\\ & & \end{array}$

Show that  $\overrightarrow{DX}=\frac{1}{4}\mathbf{a}+\frac{3}{4}\mathbf{b}$