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: 
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:
$$
(\vec v\_2,\vec v\_2)=a^2(\cos \theta -1)(\cos \theta +1)+a^2 \sin^2 \theta=a^2(\cos^2 \theta -1 + \sin^2 \theta)=0
$$
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.