# Calculate surface normal of each equilateral triangle in a tetrahedron

How can I calculate the surface normal of each equilateral triangle that makes up a tetrahedron?

These are the xyz coordinates of the tetrahedron

(+1, +1, +1) (−1, −1, +1) (−1, +1, −1) (+1, −1, −1)

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There are slicker ways of doing this, but...

Each side is determined by 3 points. Ignore one of the points and you're left with points which determine one of the four possible triangular sides. Then by taking differences of points you'll obtain vectors parallel to that side. Finally cross product two such vectors and you'll get a normal (then normalize).

For example: (Ignore point number 4) $(1,1,1) - (-1,-1,1) = (2,2,0)$ and $(1,1,1) - (-1,1,-1) = (2,0,2)$. Take the cross product and get $(4,-4,-4)$ finally normalize and get $(1/\sqrt{3})(1,-1,-1)$

Now repeat for other sides.

(if they need to be outward pointing, a quick sketch will help determine if you need to multiply by -1)

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1. If those are the vertices, say $a,b,c,d$, then how should you find the edge vectors? That is, the displacement vectors between vertices that represent the edges' directions?