Given an irregular hexahedron (six faces) having 8 vertices, is there a simple general formula to compute its volume knowing the coordinates of its vertices?
I know that the hexahedron can be subdivided in 5 tetrahedra, and we can compute the volume for each tetrahedron using a simple matrix determinant. More precisely, let us number the vertices from 1 to 8, one face being 1234, its opposite face 5678 (with the same orientation), and vertices 1 and 5 being on the same segment. The hexahedron can be divided in the following tetrahedra:
- A: 1245
- B: 2347
- C: 2567
- D: 4578
- E: 2457
This gives us five 4x4 determinants to compute. For instance, for the tetrahedron A:
$$V_A = \frac 16 \left|\begin{matrix} x_1 & y_1 & z_1 & 1\\ x_2 & y_2 & z_2 & 1\\ x_4 & y_4 & z_4 & 1\\ x_5 & y_5 & z_5 & 1 \end{matrix}\right|$$
Now, by definition, a single 5x5 determinant is decomposed precisely into five 4x4 determinants! For this reason, I have been looking for a 5x5 matrix whose determinant would be equal to the sum of the five 4x4 determinants above, unsuccesfully.
Is there such a 5x5 matrix? Otherwise, is there a simpler calculation than the one above?