Can we find the position vector of the intersection from the given ratios?

Question

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 $E$ .

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) $q$ .

Hence calculate $p$ and $q$ .

Suggestion

How can we express the position vector of $E$ in terms of things we already know? (A diagram might help.)

What other ways can we express $\mathbf{XD}$ ? Which of these use the position vector of $X$ ? Can we rearrange this in a helpful way?

How can we derive two simultaneous equations in $p$ and $q$ from our work above?

Solution

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} = \begin{pmatrix}\frac{15}{4}\\\frac{9}{4}\end{pmatrix}. \end{align*}

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

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}\\ &=\begin{pmatrix}2+q\\-\frac{1}{2}+4q\end{pmatrix}. \end{align*}

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

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