\(O\) is the origin, \(\overrightarrow{OA} = \mathbf{a}\), and \(\overrightarrow{OB} = \mathbf{b}\). The point \(X\) is on \(AB\) such that \(AX = 2XB\) and \(Y\) is the midpoint of \(OX\).
Find \(\overrightarrow{OX}\) and hence show that \(\overrightarrow{BY} = \dfrac{1}{6}\mathbf{a}-\dfrac{2}{3}\mathbf{b}\).
The vector \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\). Thus \[\overrightarrow{OX} = \overrightarrow{OA} + \overrightarrow{AX} = \mathbf{a} + \dfrac{2}{3}(\mathbf{b}-\mathbf{a}) = \dfrac{1}{3}\mathbf{a}+ \dfrac{2}{3}\mathbf{b}.\]
This means that \[\overrightarrow{OY} = \dfrac{1}{2}\overrightarrow{OX} = \dfrac{1}{6}\mathbf{a} + \dfrac{1}{3}\mathbf{b}.\]
Now we can say that \[\overrightarrow{BY} = \overrightarrow{OY} - \overrightarrow{OB} = \left(\dfrac{1}{6}\mathbf{a} + \dfrac{1}{3}\mathbf{b}\right) - \mathbf{b} = \dfrac{1}{6}\mathbf{a} - \dfrac{2}{3}\mathbf{b},\] as required.
\(BY\) produced meets \(OA\) at \(Z\). Using the facts that \[\overrightarrow{OZ} = \mathbf{b} + k\,\overrightarrow{BY} \quad \text{for some value of } k\] and \[\overrightarrow{OZ} = h\,\mathbf{a} \quad \text{for some value of } h,\] find the position vector of \(Z\).
We can see from the diagram why the two given facts are true. Firstly, \(Z\) is on \(BY\), and secondly, \(Z\) is on \(OA\).
The two facts together give us that \[h\,\mathbf{a}=\mathbf{b}+k\,\overrightarrow{BY} = \mathbf{b} + k\left(\frac{1}{6}\mathbf{a} - \frac{2}{3}\mathbf{b}\right) \implies h\,\mathbf{a}= \frac{k}{6}\mathbf{a}+\left(1-\dfrac{2k}{3}\right)\mathbf{b}.\]
Assuming \(\mathbf{a}\) and \(\mathbf{b}\) are linearly independent vectors (i.e. they are not parallel), we can equate the coefficients of \(\mathbf{a}\) and \(\mathbf{b}\) on either side of this equation.
Thus \[h = \frac{k}{6} \;\text{ and }\; 0 = 1-\frac{2k}{3} \quad\implies\quad k = \frac{3}{2} \;\text{ and }\; h = \frac{1}{4}.\]
So \(\overrightarrow{OZ} = \dfrac{1}{4}\mathbf{a}\), and we have the position vector of \(Z\).
What would happen if \(\mathbf{a}\) and \(\mathbf{b}\) were parallel or anti-parallel?
