If I define vectors $\displaystyle \vec{CA}=\vec{V}+\vec{U}$ and $\displaystyle \vec{CB}=\vec{V}-\vec{U}$ is the following true?
$\displaystyle \vec{CA}\cdot\vec{CB}=\vec{V}^{2}-\vec{U}^{2} $
thanks for the help!
You are correct. The dot product is a scalar.
I think your mistake is in notation.
$\displaystyle \vec{CA}\cdot\vec{CB}=\vec{V}\cdot\vec{V}-\vec{U}\cdot\vec{U}$. Do see why that is the case?
Now $\displaystyle \vec{V}\cdot\vec{V}=\|\vec{V}\|^2$, the length squared.
So we have $\displaystyle \|\vec{V}\|^2-\|\vec{U}\|^2$, a number.
Suppose that $\displaystyle AB$ is the diameter of a circle with center $\displaystyle O$, and that $\displaystyle C$ is a point on one of the two arcs joining $\displaystyle A$ and $\displaystyle B$. Show that $\displaystyle \vec{CA}$ and $\displaystyle \vec{CB}$ are orthogonal. $\displaystyle \vec{V}$ is from $\displaystyle O$ to $\displaystyle C$. $\displaystyle \vec{U}$ is from $\displaystyle O$ to $\displaystyle B$. $\displaystyle \vec{-U}$ is from $\displaystyle O$ to $\displaystyle A$.
I know if their dot product is zero, they are orthogonal. If the dot product is $\displaystyle \vec{U}^{2}-\vec{V}^{2}$ it would have to be zero because their magnitudes are the same.
Is this enough proof? I think it is.