If $Z$ is on $OA$ and also on $BY$, what is its position vector?

If $Z$ is on $OA$ and also on $BY$, what is its position vector?

Question

$O$ 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 $\overrightarrow{BY} = \dfrac{1}{6}\mathbf{a}-\dfrac{2}{3}\mathbf{b}$.

$BY$ 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 $Z$.

Solution

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.

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?