# Tetrahedron Cartesian Coords [duplicate]

I was wondering if anyone could provide a 'clear' way of establishing the vertices of a tetrahedron. This is a regular/standard/all faces are congruent tetrahedron.

To make it more generalized, let's say that the center of the object sits at $(0,0,0)$ and the first known vertex sits at $(0,-1, 0)$ giving the sphere that contains the tetrahedron a radius of $1$ unit. A second vertex must sit at $(0, y > 0, z < 0)$.

Furthermore, $(x < 0, y > 0, z > 0)$ and $(x > 0, y > 0, z > 0)$.

The desired end result is that I could provide any initial coordinate for the vertex and receive the other three vertices in relation to the first.

I figured this was the best route get a solution... --

Correction, in my grasping for straws here, I mentioned 3-Simplex which appears to not be conducive to what I'm trying to learn here. I simply want a clear cut way to represent a REGULAR tetrahedron when one of the vertices is very specific to the sphere that encompasses said tetrahedron. By clear cut I mean... break the tetrahedron into it's separate triangles and multiply by the sin of 180/pi.

This is a personal problem that I'm sure is beneath the initial design of this Stack but it is keeping me up at night and scouring wikipedia and an old Trig book is not helping

## marked as duplicate by Steven Stadnicki, Jyrki Lahtonen, t.b., J. M. is a poor mathematician, Henning MakholmJul 20 '12 at 12:38

IMHO the simplest description of a regular tetrahedron has its vertices at four corners of a cube. So start with the vertices $(0,0,0),(1,1,0),(1,0,1),(0,1,1)$. If you want the center of the tetrahedron to be at the origin, translate by $(-1/2,-1/2,-1/2)$. If you want a specific vertex at some other point, do another translation. If you want the tetrahedron in some other orientation, apply a 3x3 orthogonal matrix. If you want a tetrahedron of a different size, apply a scaling factor.