How to determine the normal of the triangles faces of tetrahedral?

Question

Tetrahedral is given by the points: p0 = (2, −1,0), p1 = (2,1,0), p2 = (−1,1,0), p3 = (1,0,4)

How can I determine the unit normal (facing outside) of the triangles faces of the tetrahedral?

And how can I calculate the implicit equation of the planes that contain the tetrahedral’s faces?

Answer 1

You can use vector cross product.

The cross product of two vectors is another vector that is at right angles to both.

Say the perpendicular vector we are calculating is called C, and we are calculating it using vectors A and B, cross product is calculated as follows:

C.x = A.y * B.z - A.z * B.y

C.y = A.z * B.x − A.x * B.z

C.z = A.x * B.y − A.y * B.x

Now, the only thing missing are the vectors A and B, which you can calculate from the points you have given. Say A would be the vector between points p0 and p1, in that case, you just do:

A = (p1.x - p0.x, p1.y - p0.y, p1.z - p0.z)

Find whichever B vector you want to and you can calculate the perpendicular vector to both with cross product.