Vectors in Two Dimensions (Cambridge (CIE) IGCSE Additional Maths): Exam Questions

The parallelogram $OABC$ is such that $\overrightarrow{OA}=\mathbf{a}$ and $\overrightarrow{OC}=\mathbf{c}$. The point $D$ lies on $OC$ such that $OD:DC$ = $1:2$. The point $E$ lies on $AC$ such that $AE:EC$ =$2:1$.

Show that $\overrightarrow{OB}=k\overrightarrow{DE}$ , where $k$ is an integer to be found.

In the diagram $\overrightarrow{OP}=2\mathbf{b}$, $\overrightarrow{OS}=3\mathbf{a}$, $\overrightarrow{SR}=\mathbf{b}$ and $\overrightarrow{PQ}=\mathbf{a}$. The lines $OR$ and $QS$ intersect at $X$.

Find $\overrightarrow{OQ}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.

Find $\overrightarrow{QS}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.

Given that $\overrightarrow{QX}=\mu \overrightarrow{QS}$ , find $\overrightarrow{OX}$ in terms of $\mathbf{a},\mathbf{b}\mathrm{and}\mu$.

Given that $\overrightarrow{OX}=\lambda \overrightarrow{OR}$, find $\overrightarrow{OX}$ in terms of $\mathbf{a},\mathbf{b}\mathrm{and}\lambda$.

Find the value of $\lambda$ and of $\mu$.

Find the value of $\frac{QX}{XS}$.

Find the value of $\frac{OR}{OX}$.

The diagram shows the triangle $OAC$. The point $B$ is the midpoint of $OC$. The point $Y$ lies on $AC$ such that $OY$ intersects $AB$ at the point $X$ where $AX:XB=3:1$. It is given that $\overrightarrow{OA}=a$ and $\overrightarrow{OB}=b$.

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

Find AC in terms of a and b.

Given that $\overrightarrow{OY}=h\overrightarrow{OX}$ , find $\overrightarrow{AY}$ in terms of $a$, $b$ and $h$.

Given that $\overrightarrow{AY}=m\overrightarrow{AC}$, find the value of $h$ and of $m$.

The vectors $\mathbf{a}$ and $\mathbf{b}$ are such that $\mathbf{a}=\alpha \mathbf{i}+\mathbf{j}$ and $\mathbf{b}=12\mathbf{i}+\beta \mathbf{j}$.

Find the value of each of the constants $\alpha$ and $\beta$ such that $4\mathbf{a}-\mathbf{b}=(\alpha +3)\mathbf{i}-2\mathbf{j}$.

Hence find the unit vector in the direction of $\mathbf{b}-4\mathbf{a}$.

The kite $ABCD$ has diagonals $CA$ and $BD$ which intersect at $O$, where $\overrightarrow{OA}=\mathbf{a}$, $\overrightarrow{OB}=\mathbf{b}$, $\overrightarrow{OC}=k\mathbf{a}$ and $\overrightarrow{OD}=-\mathbf{b}$.

The point $X$ is the midpoint of $AB$, as shown.

Find

(i) $\overrightarrow{AB}$ in terms of $\mathbf{a}$ and $\mathbf{b}$

(ii) $\overrightarrow{OX}$ in terms of $\mathbf{a}$ and $\mathbf{b}$

(iii) $\overrightarrow{CD}$ in terms of $k$, $\mathbf{a}$ and $\mathbf{b}$

The point $Y$ lies on $CD$ such that $CY:YD=1:2$.

If $\overrightarrow{YX}$ is parallel to $\mathbf{a}+\mathbf{b}$, find $k$.

A particle $P$ is initially at the point with position vector $\left(\begin{array}{c}30\\ 10\end{array}\right)$ and moves with a constant speed of $10{\mathrm{ms}}^{-1}$ in the same direction as $\left(\begin{array}{c}-4\\ 3\end{array}\right)$.

Find the position vector of $P$ after $t\mathrm{s}$.

As $P$ starts moving, a particle $Q$ starts to move such that its position vector after$t\mathrm{s}$ is given by  $\left(\begin{array}{c}-80\\ 90\end{array}\right)+t\left(\begin{array}{c}5\\ 12\end{array}\right)$.

Write down the speed of $Q$.

Find the exact distance between $P$ and $Q$ when $t=10$, giving your answer in its simplest surd form.

Relative to an origin $O$, the position vectors of the points $A,B,C$ and $D$ are

$\overrightarrow{OA}=\left(\begin{array}{c}6\\ -5\end{array}\right),\overrightarrow{OB}=\left(\begin{array}{c}10\\ 3\end{array}\right),\overrightarrow{OC}=\left(\begin{array}{c}x\\ y\end{array}\right)and\overrightarrow{OD}=\left(\begin{array}{c}12\\ 7\end{array}\right)$

Find the unit vector in the direction of $\overrightarrow{AB}$.

The point $A$ is the mid‑point of $BC$. Find the value of $x$ and of $y$.

The point $E$ lies on $OD$ such that $OE:OD$ is $1:1+\lambda$. Find the value of $\lambda$ such that $\overrightarrow{BE}$ is parallel to the $x$‑axis.

In this question all distances are in km.
A ship $P$ sails from a point $A$, which has position vector $\left(\begin{array}{c}0\\ 0\end{array}\right)$, with a speed of $52{\mathrm{kmh}}^{-1}$ in the direction of $\left(\begin{array}{c}-5\\ 12\end{array}\right)$

Find the velocity vector of the ship.

Write down the position vector of $P$ at a time $t$ hours after leaving $A$.

At the same time that ship $P$ sails from $A$, a ship $Q$ sails from a point $B$, which has position vector $\left(\begin{array}{c}12\\ 8\end{array}\right)$, with velocity vector $\left(\begin{array}{c}-25\\ 45\end{array}\right){\mathrm{kmh}}^{-1}$.

Write down the position vector of $Q$ at a time $t$ hours after leaving $B$.

Using your answers to parts (b) and (c), find the displacement vector $\overrightarrow{PQ}$ at time$t$ hours.

Hence show that $PQ=\sqrt{34t^2-168t+208}$. 

Find the value of $t$ when $P$ and $Q$ are first $2\mathrm{km}$ apart.

The position vectors of three points, $A,B\mathrm{and}C$, relative to an origin $O$, are $\left(\begin{array}{c}-5\\ -7\end{array}\right),\left(\begin{array}{c}10\\ -4\end{array}\right)\mathrm{and}\left(\begin{array}{c}x\\ y\end{array}\right)$ respectively. Given that $\overrightarrow{AC}$ = $4\overrightarrow{BC}$, find the unit vector in the direction of $\overrightarrow{OC}$.

The diagram shows a triangle $OAB$ such that $\overrightarrow{OA}=\mathbf{a}$ and $\overrightarrow{OB}=\mathbf{b}$. The point $P$ lies on $OA$ such that $OP=\frac{3}{4}OA$. The point $Q$ is the mid-point of $AB$. The lines $OB$ and $PQ$ are extended to meet at the point $R$. Find, in terms of $\mathbf{a}$ and $\mathbf{b}$,

Find $\overrightarrow{AB}$

Find $\overrightarrow{PQ}$. Give your answer in its simplest form.

It is given that $n\overrightarrow{PQ}=\overrightarrow{QR}\mathrm{and}\overrightarrow{BR}=k\mathbf{b}$, where $n\mathrm{and}k$ are positive constants.

Find $\overrightarrow{QR}$ in terms of $n,\mathbf{a}\mathrm{and}\mathbf{b}$.

Find $\overrightarrow{QR}$ in terms of $k,\mathbf{a}\mathrm{and}\mathbf{b}$.

Hence find the value of $n$ and of $k$.