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.
Let's consider a semicircle of radius $1$. The semicircle is in the right side of cartesian axes. So all points save for the extremes have positive $X$.

Let $P=(x,y)$ be a point in the semicircle which is not an extreme. 

$x^2+y^2=1 \implies x=+{\sqrt{1-y^2}} \implies P=(+{\sqrt{1-y^2}},y)$ 

The vectors to the extremes are :  

Vector to top extreme : $v\_t = (0,1) - (+{\sqrt{1-y^2}},y) = ( -{\sqrt{1-y^2}},1-y)$  

Vector to bottom extreme : $v\_b = (0,-1) - (+{\sqrt{1-y^2}},y) = ( -{\sqrt{1-y^2}},-1-y)$ 

As you said if the dot product is 0 the angle is right :  

$v\_t.v\_b = ( -{\sqrt{1-y^2}} )^2 + (-y+1)(-y-1) = (1-y^2) + (-y)^2 - 1^2 = 0$

Rotating the semicircle and vectors construction does not change the angle, thus the property of being a right angle is general for all semicircles of radius 1.  

Scaling the semicircle and vectors construction does not change the angle, thus the property of being a right angle is general for all semicircles.