Review question

# Can we find the position vector of the intersection from the given ratios? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6007

## Solution

The position vectors of three points $O, A$ and $B$ are $\begin{pmatrix}0\\0\end{pmatrix}, \begin{pmatrix}3\\3.5\end{pmatrix}$ and $\begin{pmatrix}6\\-1.5\end{pmatrix}$ respectively. Given that $\mathbf{OD} = \dfrac{1}{3}\mathbf{OB}$ and $\mathbf{AE} = \dfrac{1}{4}\mathbf{AB}$ write down the position vectors of $D$ and $E$.

For convenience, we define $\mathbf{a}=\mathbf{OA}$ and similarly $\mathbf{b}$, $\mathbf{d}$, $\mathbf{e}$ and $\mathbf{x}$.

We are told that $\mathbf{d}=\dfrac{1}{3}\mathbf{b}=\begin{pmatrix}2\\-\frac{1}{2}\end{pmatrix}.$

We also have that \begin{align*} \mathbf{AE} &= \frac{1}{4}\mathbf{AB} \\ \implies\quad (\mathbf{e}-\mathbf{a}) &= \frac{1}{4}(\mathbf{b}-\mathbf{a}) \\ \implies\quad \mathbf{e} &= \frac{1}{4}\mathbf{b} + \frac{3}{4}\mathbf{a} = \cmarraystretch{1.4}\begin{pmatrix}\frac{15}{4}\\\frac{9}{4}\end{pmatrix}. \end{align*}

Given also that $OE$ and $AD$ intersect at $X$, and that $\mathbf{OX} = p\:\mathbf{OE}$, and that $\mathbf{XD} = q\:\mathbf{AD}$, find the position vector of $X$ in terms of (i) $p$ (ii) $q$.

We have $\mathbf{x} = p\:\mathbf{e} = \cmarraystretch{1.4}\begin{pmatrix}\frac{15}{4}p\\\frac{9}{4}p\end{pmatrix}.$

We also know that \begin{align*} \mathbf{x} &= \mathbf{d} + \mathbf{DX}\\ &=\mathbf{d} + q(\mathbf{a}-\mathbf{d})\\ &=\begin{pmatrix}2\\-\frac{1}{2}\end{pmatrix}+q\begin{pmatrix}1\\4\end{pmatrix}\\ &=\cmarraystretch{1.4}\begin{pmatrix}2+q\\-\frac{1}{2}+4q\end{pmatrix}. \end{align*}

Hence calculate $p$ and $q$.

Our two expressions for $\mathbf{x}$ must be the same, so $\cmarraystretch{1.4}\begin{pmatrix}\frac{15}{4}p\\\frac{9}{4}p\end{pmatrix} = \begin{pmatrix}2+q\\-\frac{1}{2}+4q\end{pmatrix}.$

Equating the $x$ and $y$ components separately gives us two simultaneous equations which we can solve to find $p = \dfrac{2}{3} \quad\text{and}\quad q = \dfrac{1}{2}.$