I want to calculate the fourth vertex of a regular tetrahedron. I have the coordinates
{0, 0, Sqrt[2/3] - 1/(2 Sqrt[6])}, {-(1/(2 Sqrt[3])), -(1/2), -(1/(2 Sqrt[6]))} and {-(1/(2 Sqrt[3])), 1/2, -(1/(2 Sqrt[6]))}
Can anybody please help?
Find the center of face
cx = (x1 + x2 + x3)/3 and similar for y,z
Get two edge vectors
e2x = x2 - x1
e2y = y2 - y1
e2z = z2 - z1
e3x = x3 - x1
e3y = y3 - y1
e3z = z3 - z1
Calculate edge len
elen = sqrt(e2x*e2x+e2y*e2y+e2z*e2z)
Calculate vector product to get normal to this face
nx = e2y*e3z - e2z*e3y
ny = e2z*e3x - e2x*e3z
nz = e2x*e3y - e2y*e3x
Make unit normal
nlen = sqrt(nx*nx+ny*ny+nz*nz)
nx = nx / nlen
...
Make normal of needed length (tetrahedron height)
lnx = nx * sqrt(2/3) * elen
...
Add this normal to the face center
x4 = cx +/- lnx
y4 = cy +/- lny
z4 = cz +/- lnz
+/- signs correspond to two possible positions of the fourth vertex