The position vector of A is $\begin{pmatrix}3\\4\end{pmatrix}$ and the position vector of B is $\begin{pmatrix}7\\10\end{pmatrix}$. Find the column vector $\overrightarrow{AB}$. Answer ... Find $\lvert\overrightarrow{AB}\rvert$. Answer ... Given that $\overrightarrow{AB} = 3\overrightarrow{BC}$, find the coordinates of point C. Answer ... Given $\overrightarrow{PQ}=\begin{pmatrix}2\\1\end{pmatrix}$, $\overrightarrow{PR}=\begin{pmatrix}-1\\-\tfrac{1}{2}\end{pmatrix}$ and the point Q has coordinates (-2, 4). Deduce the gradient of the line PQ and hence, find the equation of the line PQ. Answer ... Write down 2 facts about P, Q and R.

1. Find the displacement from A to B by subtracting position vectors.

$\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=\begin{pmatrix}7\\10\end{pmatrix}-\begin{pmatrix}3\\4\end{pmatrix}=\begin{pmatrix}4\\6\end{pmatrix}$

1. Find the magnitude of the vector using $\sqrt{x^2+y^2}$.

Since $\overrightarrow{AB}=\begin{pmatrix}4\\6\end{pmatrix}$,

$|\overrightarrow{AB}|=\sqrt{4^2+6^2}=\sqrt{16+36}=\sqrt{52}=2\sqrt{13}.$

1. Find the vector from B to C using $\overrightarrow{AB}=3\overrightarrow{BC}$.

From 7(a)(i),
$\overrightarrow{AB}=\begin{pmatrix}4\\6\end{pmatrix}.$
Given $\overrightarrow{AB}=3\overrightarrow{BC}$,
$\overrightarrow{BC}=\frac{1}{3}\overrightarrow{AB}=\frac{1}{3}\begin{pmatrix}4\\6\end{pmatrix}=\begin{pmatrix}\frac{4}{3}\\2\end{pmatrix}.$

2. Add the displacement from B to C to the coordinates of B.

The position vector of B is $\begin{pmatrix}7\\10\end{pmatrix}$, so
$\vec{OC}=\vec{OB}+\overrightarrow{BC}=\begin{pmatrix}7\\10\end{pmatrix}+\begin{pmatrix}\frac{4}{3}\\2\end{pmatrix}=\begin{pmatrix}\frac{25}{3}\\12\end{pmatrix}.$
Hence, $C=(\frac{25}{3},12)$.

1. Deduce the gradient of line $PQ$ from the vector $\overrightarrow{PQ}$.

Since $\overrightarrow{PQ}=\begin{pmatrix}2\\1\end{pmatrix}$,

$\text{gradient of }PQ=\frac{\text{vertical change}}{\text{horizontal change}}=\frac{1}{2}.$

2. Use the gradient and point $Q(-2,4)$ to find the equation of the line.

Let the equation be $y=\dfrac{1}{2}x+c$.

Substitute $Q(-2,4)$:

$4=\frac{1}{2}(-2)+c$

$4=-1+c$

$c=5$

So the equation is

$y=\frac{1}{2}x+5.$

1. Compare the two given vectors using scalar multiplication.

$\overrightarrow{PQ}=\begin{pmatrix}2\\1\end{pmatrix}, \qquad \overrightarrow{PR}=\begin{pmatrix}-1\\-\tfrac12\end{pmatrix}$

Since
$\overrightarrow{PR}=-\frac12\overrightarrow{PQ},$
the vectors are parallel but in opposite directions.

2. State two geometric facts about the points.

Because $\overrightarrow{PR}$ and $\overrightarrow{PQ}$ are in opposite directions from the same point $P$, the points $R$, $P$ and $Q$ lie on the same straight line, with $P$ between $R$ and $Q$.

Also, since the scale factor is $\frac12$, $PR$ is half of $PQ$.