Tetrahedron area not using triangle area Is there any formula for the area of a general tetrahedron that does not determine the area by summing up its 4 triangle areas but considers the tetrahedron as a unit? The tetrahedron is given by 4 points in $\mathbb{R}^3$.
Sometimes, you can embed the tetrahedron inside a rectangular prism, with the edges of the tetrahedron crossing the diagonals of the sides of the prisim.  The volume of the tetrahedron will be 1/3 the volume of the prism.

e.g. the tetrahedron with vertexes at:

$(1,1,1), (-1,-1,1), (-1,1,-1), (1,-1,-1)$

Can be embedded in a cube, with additional vertices at $(-1,1,1),(1,-1,1),(1,1,-1), (-1,-1,-1)$

The volume of this cube is 8.  So the volume of the tetrahedron is $\frac {8}{3}$

If you can place one vertex at the origin.  Arrange the other vertices as row vectors in a matrix.  The determinant of this matrix will be $6 \times$ the volume of the tetrahedron.

If we had a tetrahedron with vertices at:
$(0,0,0),(0,2,2),(2,0,2),(2,2,0)$

Set it up as a matrix like this:

$\begin{bmatrix} 0&amp;2&amp;2\\ 2&amp;0&amp;2\\2&amp;2&amp;0\end{bmatrix}$

Calculate the determinant (16)

$V = \frac {8}{3}$  Which shouldn't be a surprise as these two tetrahedra are congruent.
Sometimes, you can embed the tetrahedron inside a rectangular prism, with the edges of the tetrahedron crossing the diagonals of the sides of the prisim.  The volume of the tetrahedron will be 1/3 the volume of the prism.

e.g. the tetrahedron with vertexes at:

$(1,1,1), (-1,-1,1), (-1,1,-1), (1,-1,-1)$

Can be embedded in a cube, with additional vertices at $(-1,1,1),(1,-1,1),(1,1,-1), (-1,-1,-1)$

The volume of this cube is 8.  So the volume of the tetrahedron is $\frac {8}{3}$

If you can place one vertex at the origin.  Arrange the other vertices as row vectors in a matrix.  The determinant of this matrix will be $6 \times$ the volume of the tetrahedron.

If we had a tetrahedron with vertices at:
$(0,0,0),(0,2,2),(2,0,2),(2,2,0)$

Set it up as a matrix like this:

$\begin{bmatrix} 0&amp;2&amp;2\\ 2&amp;0&amp;2\\2&amp;2&amp;0\end{bmatrix}$

Calculate the determinant (16)

$V = \frac {8}{3}$  Which shouldn't be a surprise as these two tetrahedra are congruent.