The position vector of point Q is $\begin{pmatrix}-7\\-3\end{pmatrix}$ and the position vector of point R is $\begin{pmatrix}5\\1\end{pmatrix}$. Find $|\overrightarrow{QR}|$. S is the point on QR with coordinates (k, -2). Find the position vector of S. OAB is a triangle. $\overrightarrow{OA} = a$ and $\overrightarrow{OB} = b$. P is the point on AB such that AP : PB = 2 : 3. M and N are the midpoints of OP and OB respectively. Express $\overrightarrow{AP}$ in terms of a and b, as simply as possible. Express $\overrightarrow{OM}$ in terms of a and b, as simply as possible. What type of quadrilateral is PBNM? Justify your answer using vectors.

1. Find the components of $\overrightarrow{QR}$.

$\overrightarrow{QR}=\begin{pmatrix}5\\1\end{pmatrix}-\begin{pmatrix}-7\\-3\end{pmatrix}=\begin{pmatrix}12\\4\end{pmatrix}$

2. Find the magnitude of the vector.

$|\overrightarrow{QR}|=\sqrt{12^2+4^2}=\sqrt{160}=4\sqrt{10}$

1. Write a general point on QR using a parameter.

Since $Q=(-7,-3)$ and $R=(5,1)$,

$\overrightarrow{QR}=\begin{pmatrix}5-(-7)\\1-(-3)\end{pmatrix}=\begin{pmatrix}12\\4\end{pmatrix}.$

A point on QR can be written as

$\begin{pmatrix}-7\\-3\end{pmatrix}+t\begin{pmatrix}12\\4\end{pmatrix}=\begin{pmatrix}-7+12t\\-3+4t\end{pmatrix}.$

2. Use the given y-coordinate of S to find the parameter and hence k.

For S, the y-coordinate is $-2$.

$-3+4t=-2$

$4t=1$

$t=\frac14$

So

$k=-7+12\left(\frac14\right)=-7+3=-4.$

1. Express $\overrightarrow{AB}$ in terms of $a$ and $b$.

Since $\overrightarrow{OA}=a$ and $\overrightarrow{OB}=b$,
$\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=b-a.$

2. Use the ratio $AP:PB=2:3$ to find $\overrightarrow{AP}$.

Since $AP:PB=2:3$, point $P$ divides $AB$ in the ratio $2:3$ from $A$ to $B$.
$\overrightarrow{AP}=\frac{2}{2+3}\overrightarrow{AB}$
$\overrightarrow{AP}=\frac{2}{5}(b-a).$

1. Find $\overrightarrow{OP}$ using the earlier result for $\overrightarrow{AP}$.

From part (b)(i),
$\overrightarrow{AP}=\frac{2}{5}(b-a).$
So
$\overrightarrow{OP}=\overrightarrow{OA}+\overrightarrow{AP}$
$=a+\frac{2}{5}(b-a)$
$=a+\frac{2}{5}b-\frac{2}{5}a$
$=\frac{3}{5}a+\frac{2}{5}b.$

2. Use that $M$ is the midpoint of $OP$ to find $\overrightarrow{OM}$.

Since $M$ is the midpoint of $OP$,
$\overrightarrow{OM}=\frac{1}{2}\overrightarrow{OP}.$
Hence
$\overrightarrow{OM}=\frac{1}{2}\left(\frac{3}{5}a+\frac{2}{5}b\right)$
$=\frac{3}{10}a+\frac{1}{5}b.$

1. Find $\overrightarrow{PB}$ using the given ratio on $AB$.

Since $AP:PB=2:3$,
$\overrightarrow{PB}=\frac{3}{5}\overrightarrow{AB}.$
Also,
$\overrightarrow{AB}=b-a.$
Hence
$\overrightarrow{PB}=\frac{3}{5}(b-a).$

2. Find $\overrightarrow{MN}$ from the position vectors of $M$ and $N$.

From part (ii),
$\overrightarrow{OM}=\frac{3}{10}a+\frac{1}{5}b.$
Since $N$ is the midpoint of $OB$,
$\overrightarrow{ON}=\frac{1}{2}b.$
Therefore
$\overrightarrow{MN}=\overrightarrow{ON}-\overrightarrow{OM}$
$=\frac{1}{2}b-\left(\frac{3}{10}a+\frac{1}{5}b\right)$
$=-\frac{3}{10}a+\frac{3}{10}b$
$=\frac{3}{10}(b-a).$

3. Compare the relevant sides and identify the quadrilateral.

Since
$\overrightarrow{PB}=\frac{3}{5}(b-a)$
and
$\overrightarrow{MN}=\frac{3}{10}(b-a),$
$\overrightarrow{PB}$ and $\overrightarrow{MN}$ are parallel, but not equal.

Also,
$\overrightarrow{BN}=\overrightarrow{ON}-\overrightarrow{OB}=\frac{1}{2}b-b=-\frac{1}{2}b,$
and
$\overrightarrow{PM}=\overrightarrow{OM}-\overrightarrow{OP}.$
From part (ii), since $M$ is the midpoint of $OP$,
$\overrightarrow{OP}=2\overrightarrow{OM}=\frac{3}{5}a+\frac{2}{5}b.$
So
$\overrightarrow{PM}=\left(\frac{3}{10}a+\frac{1}{5}b\right)-\left(\frac{3}{5}a+\frac{2}{5}b\right)$
$=-\frac{3}{10}a-\frac{1}{5}b.$
Thus $BN$ is not parallel to $PM$ in general. Therefore only one pair of opposite sides, $PB$ and $MN$, is parallel.

Hence $PBNM$ is a trapezium.