Question regarding vectors within a circle in an $x$- and $y$-plane.

Question

In an $x$ and $y$ coordinate plane, with respect to the points $A, B$, and $C$ on a circle of radius $1$, find the minimum value of $\vec {AB} \cdot \vec {AC}$

So far, taking $O$ as the origin I've gotten

$\vec {AB} \cdot \vec {AC} = (\vec {OB} - \vec {OA}) \cdot (\vec {OA} - \vec {OC}) $. Let's call this $1$. And then the answer book says that this is equal to

$1+ \vec {OB} \cdot \vec {OC} - (\vec {OB} + \vec {OC}) \cdot \vec {OA}$. Let's call this $2$.

Could you please explain how this jump from $1$ to $2$ is justified? I can't seem to figure it out. Thank you!

Answer 1

I think you did a small mistake. $\vec{AC}=\vec{OC}-\vec{OA}$

When you open the dot product $$(\vec {OB}-\vec{OA}).(\vec{OC}-\vec{OA})$$ we get $$\vec{OB}.\vec{OC}+|\vec{OA}|^2-\vec{OB}.\vec{OA}-\vec{OA}.\vec{OC}$$ $$=1+\vec{OB}.\vec{OC}-(\vec{OB}+\vec{OC}).\vec{OA}$$ Since $\vec{OA}$ is radius vector, and its modulus is $1$

Answer 2

Fist of all your expression 1 is not correct. It should be:

$$\vec{AB}\cdot\vec{AC} = ( \vec{OB}-\vec{OA})\cdot(\vec{OC}-\vec{OA})$$

and you have only to expand the product:

$$ ( \vec{OB}-\vec{OA})\cdot(\vec{OC}-\vec{OA}) = \vec{OB}\cdot\vec{OC} - \vec{OB}\cdot\vec{OA} - \vec{OA}\cdot\vec{OC} + \vec{OA}\cdot\vec{OA}$$

and as:

$$\vec{OA}\cdot\vec{OA} = 1$$

you can get the expression 2