If you have an equilateral triangle in 3D space, where all the sides are of length 1, there are two points that you could use to form a tetrahedron. One floating out in front of the triangle, and one behind it. Given the coordinates of the three known vertices, how would you calculate either of the possible fourth vertices?
I would really appreciate it if you can show how to do it with the Processing vector class definition
Average your three points to get the center of the triangle:
center = (a + b + c) / 3
Calculate the normal vector by taking the cross product of two of the sides:
normal = (c - a) x (b - a)
Normalize the normal vector (make it of unit length):
unit_normal = normal / |normal|
Scale the normal by the height of regular tetrahedron:
scaled_normal = unit_normal * sqrt(2/3)
Now, your two points are:
top = center + scaled_normal
bottom = center - scaled_normal
(a + b + c)/3 (centre of the triangle)
+/- ((a-b) x (b-c) (cross product of two sides of the triangle, hence perpendicular to both)
* some constant or other) (the height of a regular tetrahedron divided by the length of that cross product, the length being 1 * 1 * sin(60 degrees) = sqrt(3)/2)
This can probably be simplified.
[Edit: height is sqrt(2/3), so the constant is 2*sqrt(2)]
[Second edit: any fourth point not in the plane of the first three forms a tetrahedron. ITYM a regular tetrahedron ;-)]
