Find expression for when angle between 2 vectors and common vector is equal

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!

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1 Answer

Yes, your equation for the angles is correct (if by $\hat a$ you mean the unit vector along $a$).

To use it to evaluate $k$, note that $\overrightarrow{AP} = p-a$ and $\overrightarrow{AB}=b-a$, so the condition given allows you to express $p$ in terms of $a$, $b$ and $k$. Since $a$ and $b$ are given, substituting that into your equation for the angles yields an equation with the single unknown $k$.

(It will also be of use to know the Pythagorean triples $(3,4,5)$ and $(5,12,13)$.)

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Thanks! I got $k = \dfrac{5}{18}$. My brain seems to forget all algebra when confronted with vectors instead of variables. All in all, that was actually a fun problem to solve. After you helped me figure it out that is! :) –  mathguy80 Jul 29 '11 at 15:03
You're welcome :-) –  joriki Jul 29 '11 at 15:15