Given three points on a tetrahedron, Find the 4th

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 Pr开发者_开发知识库ocessing vector class definition

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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

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(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 ;-)]

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Since, 3D has never been my interest, I guess I can only provide a way to do this, rather than exact coordinates.

A point which lies at a distance of sqrt(2/3) from the centroid of the triangle and on a line perpendicular to the plane formed by the triangle and containing the centroid.