Let $S$ be a circle with it s center at point $O$ and a radius of $1$. Let $\triangle$ABC be a triangle such that all its vertices are on $S$ and $AB:AC=3:2$. As shown in the figure, let $D$ be a point on the extension of side $BC$ and $k$ be the number where
$BC:CD=2:k$.
Moreover, set
$\vec{OA}=\vec{a},$ $\vec{OB}=\vec{b},$ $\vec{OC}=\vec{c},$
Since the equality
$|\vec{b}-\vec{a}|=\frac{3}{2}|\vec{c}-\vec{a}|$
holds, by expressing the inner product $\vec{a}\cdot\vec{b}$ in terms of the inner product $\vec{a}\cdot\vec{c}$, we have
$\vec{a}\cdot\vec{b}=\frac{F}{G}\vec{a}\cdot\vec{c}-\frac{H}{I}$
Find F, G, H and I.
This is my scholarship exam practice assuming high school math knowledge.
The answer key provided is 9, 4, 5 and 4. I do not know how to begin here, could you please give me a hint to start on this question?
Hint: Expand $\lvert\vec{b}-\vec{a}\rvert^2=\frac94\lvert\vec{c}-\vec{a}\rvert^2$ and remember $\vec{a}^2=\vec{b}^2=\vec{c}^2=1$. The point $D$ plays no part here.