The points $A, B, P$ have position vectors $a, b, p$ respectively relative to an origin $O$.
Write down in vector terms an expression for the cosine of $\angle{AOP}$ and hence obtain the condition that $\angle{AOP} = \angle{BOP}$.
Given that $\overrightarrow{AP} = k\overrightarrow{AB}$, that $a, b$ are $\begin{pmatrix}3\\4\end{pmatrix}$, $\begin{pmatrix}12\\5\end{pmatrix}$ respectively and that $\angle{AOP} = \angle{BOP}$,
Use your condition in order to evaluate $k$.
I got the first and second part of question using the dot product property.
$cos \angle{AOP} = \dfrac{a.p}{\|a\| \|p\|}$
Also,
$cos \angle{BOP} = \dfrac{b.p}{\|b\| \|p\|}$
When, $\angle{AOP} = \angle{BOP}$ $ \begin{align} \dfrac{a.p}{\|a\| \|p\|} &= \dfrac{b.p}{\|b\| \|p\|} \\ \dfrac{a.p}{\|a\|} &= \dfrac{b.p}{\|b\|} \\ \hat a \cdot p &= \hat b.p \\ \end{align} $
I am having difficulty with the next part. I am not able to connect the expression I got with evaluating $k$. Also I am not sure that the expression I obtained is correct, can you guys please verify it's accuracy?
Can you guys give me a hint on how to evaluate $k$?
Thanks for all your help!