The four points $a$, $c$, $b$ and $d$ are collinear (lie on line $l$) and in that sequence.
The origin $o$ is some point not on the line $l$.
Given: $\vec{ac}=2\vec{cb}$, $\ \vec{ad}=2\vec{bd}\ $ and $\ |\vec{a}|=2|\vec{b}|$
Using scalar product or otherwise show $\angle{cod}=\frac{\pi}{2}$
[Hint: ($2\vec{a}+\vec{b})(\vec{a}-\vec{b})=2|\vec{a}|^2-2\vec{a}.\vec{b}+\vec{a}.\vec{b}-|\vec{b}|^2$]
My solution:
I first let, $\vec{a}=a_1\vec{i}+a_2\vec{j}$ and $\vec{b}=b_1\vec{i}+b_2\vec{j}$
I then found both $\vec{c}$ and $\vec{d}$ in terms of $\vec{a}$ and $\vec{b}$.
Then by carrying out the operation $\vec{a}.\vec{b}$, I found that $\vec{a}.\vec{b}=0$ hence proving that $\angle{cod}=\frac{\pi}{2}$
$\color{blue}{\text{My question is:}}$
what is the meaning of the hint? I can't see how that fits in. It's probably exactly what I did or something obvious but I just can't see it.