I'm trying to learn more about geometry and vectors and I'm lost with an proposed problem.
I have the following triangle coordinates:
$A = (0,1/2,3/2)$
$B = (1,1,0)$
$C = (1,0,1)$
How can I calculate the perimeter of the triangle ABC ?
Should I add the vertices of the triangle like this?
$AB: \sqrt{(0-1)^2+(1/2-1)^2+(3/2-0)^2}$
$BC: \sqrt{(1-1)^2+(1-0)^2+(0-1)^2}$
$CA: \sqrt{(1-0)^2+(0-1/2)^2+(1-3/2)^2}$
And how can I calculate the value of the internal angle of $C$?
If you want to use vectors, find vectors $\vec CB$ and $\vec CA$:
The components of $\vec CB$ are $ (0,1,-1)$, $\vec CA$ are $ (-1,0.5,0.5)$. Then you can find length (magnitude) of the vectors, i.e. : $||CB||=\sqrt{0^2+1^2+(-1)^2}$.
To find cosine of the angle C, use the law of cosines: $$\cos C=\frac{\vec CA \cdot \vec CB}{||CA||\cdot ||CB||}$$ where $\vec CA \cdot \vec CB=0\cdot(-1)+1\cdot0.5+(-1)\cdot0.5=0$ is the "dot" product.