# 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 ?

I have the following triangle coordinates:
$A=\left(0,1/2,3/2\right)$
$B=\left(1,1,0\right)$
$C=\left(1,0,1\right)$
How can I calculate the perimeter of the triangle ABC ?
Should I add the vertices of the triangle like this?
$AB:\sqrt{\left(0-1{\right)}^{2}+\left(1/2-1{\right)}^{2}+\left(3/2-0{\right)}^{2}}$
$BC:\sqrt{\left(1-1{\right)}^{2}+\left(1-0{\right)}^{2}+\left(0-1{\right)}^{2}}$
$CA:\sqrt{\left(1-0{\right)}^{2}+\left(0-1/2{\right)}^{2}+\left(1-3/2{\right)}^{2}}$
And how can I calculate the value of the internal angle of C?
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ahem37
If you want to use vectors, find vectors $\stackrel{\to }{C}B$ and $\stackrel{\to }{C}A$:
The components of $\stackrel{\to }{C}B$ are (0,1,−1), $\stackrel{\to }{C}A$ are (−1,0.5,0.5). Then you can find length (magnitude) of the vectors, i.e. : $||CB||=\sqrt{{0}^{2}+{1}^{2}+\left(-1{\right)}^{2}}$.
To find cosine of the angle C, use the law of cosines:
$\mathrm{cos}C=\frac{\stackrel{\to }{C}A\cdot \stackrel{\to }{C}B}{||CA||\cdot ||CB||}$
where $\stackrel{\to }{C}A\cdot \stackrel{\to }{C}B=0\cdot \left(-1\right)+1\cdot 0.5+\left(-1\right)\cdot 0.5=0$ is the "dot" product.