Question: Could someone explain how to prove any angle inscribed in a semicircle is a right angle using vectors. I understand that the dot product of two vectors is 0 is they are perpendicular but I don't know how to show this in a semicircle.

Answer: 
Wlog. we can assume the semicircle to be the upper unit semicircle with centre in the plane origin $(0, 0)$. Let $(x, y)$ be a fixed point on the semicircle with $y>0$.

Consider the vector $v\_1$, which is the difference between the positional vectors $(x, y)$ and $(-1, 0)$, and the vector $v\_2$, which is the difference between the positional vectors $(x, y)$ and $(1, 0)$.

Now note that the angle inscribed in the semicircle is a right angle if and only if the two vectors are perpendicular. Using the scalar product, this happens precisely when $v\_1\cdot v\_2=0$.

So just compute the product $v\_1\cdot v\_2$, using that $x^2+y^2=1$ since $(x, y)$ lies on the unit circle.
If the semicirce has radius $a$ you can represents the two vectors as the difference between the coordinates of the points $(-a,0)$ and $(a,0)$ with respect to a generic point $(a \cos \theta, a \sin \theta)$ :
$$
\vec v\_1=(a\cos \theta -a; a \sin \theta)^T \quad and \quad \vec v\_2=(a\cos \theta +a, a \sin \theta)^T 
$$
so you have:
$$
(\vec v\_2,\vec v\_2)=a^2(\cos \theta -1)(\cos \theta +1)+a^2 \sin^2 \theta=a^2(\cos^2 \theta -1 + \sin^2 \theta)=0
$$