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 \(\mathbf{x}\).

the points O, A, B, D, E and 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 \[p = \dfrac{2}{3} \quad\text{and}\quad q = \dfrac{1}{2}.\]