# Expressing the inner product $\vec{a}\cdot\vec{b}$ in terms of the inner product $\vec{a}\cdot\vec{c}$

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.
• I think I would like some hint on my last question as well saying that: It follows that when the tangent to $S$ at the point $A$ passes through the point $D$, then $k=?$ – Trey Anupong Jun 19 at 15:09
• Oops, yes, it should be $\vec{a}$ perpendicular to $\overrightarrow{AD}$ so with $\overrightarrow{AD}=\vec{b}-\vec{a}+\frac{2+k}2(\vec{c}-\vec{b})$ we get $k$ by dotting with $\vec{a}$. – user10354138 Jun 19 at 16:19
• So we dot $\vec{a}$ with $\vec{AD}$ which is equal to $\vec{OD}$, am I correct? – Trey Anupong Jun 19 at 16:24
• $\overrightarrow{AD}=\overrightarrow{OD}-\vec{a}$. We know $\overrightarrow{AD}\cdot\vec{a}=0$ which gives us an equation for $k$ in terms of $\vec{a}\cdot\vec{b}$ and $\vec{a}\cdot\vec{c}$. – user10354138 Jun 19 at 16:26