I found this question on vector geometry in an IGCSE text. I got these solutions for the following vectors: $\vec {BP}=\mathbf a-\mathbf b$, $\vec {AB}=\mathbf b-3\mathbf a$, $\vec {MB}=\frac 12(\mathbf b-3\mathbf a)$. Then for $\vec {MX}$ I got this really long and complicated expression in terms of $\mathbf a$, $\mathbf b$ and $\mathit k$: $$\vec {MX}=\frac 12(\mathbf b-3\mathbf a)+\mathit k(\mathbf a-\mathbf b)=\left(\mathit k-\frac 32\right)\mathbf a+\left(\frac 12-\mathit k\right)\mathbf b$$
As I understand, if $\vec {MX}$ is parallel to $\vec {BO}$, $\vec {MX}$ would be something times $\vec {BO}$. But how do you use that result to find the value of $\mathit k$?
Two vectors $\mathbf u$ and $\mathbf v$ are parallel if there is a scalar $c$, so $\mathbf u =c\mathbf v$. So if $\vec{MX}$ is parallel to $\vec{OB}$, then there should be $c$: $$ \left(k-\frac32\right)\mathbf a+\left(\frac12-k\right)\mathbf b=c\mathbf b $$
But since $\mathbf a$ isn't parallel to $\mathbf b$, it can't be true until $k-3/2=0$.