Surface Area of General Tetrahedron Using Vertex Coordinates I'm trying to figure out what the formula is to find the surface area of a non-regular tetrahedron when given the vertices coordinates.
Example:
Vertex $A$: $[0, 0, 5]$
Vertex $B$: $[-1, -1, 0]$
Vertex $C$: $[1, 0, 0]$
Vertex $D$: $[0, 2, 0]$
Basically I want to turn this user input into surface area.
Thank you
 A: Hint: As a numerical analyst, the most common way to compute the face are of a bunch of tetrahedra is using cross product, since it is the most easily vectorized/paralleled algorithm. The face opposite to $A$ denoted as $F_A$ is spanned by the vector $\overrightarrow{BC}$ and $\overrightarrow{BD}$, then the area is
$$
|F_A| = \frac{1}{2} |\overrightarrow{BC}\times \overrightarrow{BD}|
$$
More likely since you mentioned "input", I am guessing you are writing some subroutine, if using cyclic notation that a tetrahedron has vertices $V_i$, $i=1,\cdots,4$, then above formula can be vectorized using the following MATLAB code snippets assuming you have your $i$-th vertex of the $n$-th tetrahedron stored in a 3d-array V(n,:,i):
face_normal = cross(V(:,:,i+1) - V(:,:,i-1), ...
V(:,:,i+1) - V(:,:,i+2), 2);
face_area = 0.5*sqrt(sum(face_normal.^2,2));

which could be easily ported to Python, C or Fortran.
A: You have four triangles for which you can compute the sides from the Pythagorean theorem.  For example, $CD=\sqrt{1^2+2^2+0^2}=\sqrt 5$.  Then use Heron's formula
