Proof that the volume of a tetrahedron is given by a $4\times 4$ determinant

Question

I have seen that the volume of a tetrahedron $ABCD$ where $$A = (x_1,y_1,z_1), B = (x_2,y_2,z_2), C = (x_3,y_3,z_3), D = (x_4,y_4,z_4)$$ and $A$ being the top can be described as $$V = \dfrac{1}{6}|\det M| \qquad\text{where}\qquad M = \begin{bmatrix}x_1&y_1&z_1&1\\x_2&y_2&z_2&1\\x_3&y_3&z_3&1\\x_4&y_4&z_4&1\\\end{bmatrix}$$ but I cannot find the proof anywhere. How is the following true? $$\vec{BC}\cdot(\vec{BD} \times \vec{BA})=\det M$$

Any kind of explanation is welcome.

Answer 1

Hint,

Consider $A(x_1,y_1,z_1)$ as the base of the tetrahedron, and note that its volume = the volume of the pyramid with that base.

That is, $V=\frac{1}{3}$ area of $A \times$ the height.

Answer 2

Translate all points so $B$ is at the origin; $A,C,D$ now represent position vectors relative to $B$ and the determinant does not change. By cofactor expansion, $\det M$ is $$\begin{vmatrix}x_1'&y_1'&z_1'&1\\0&0&0&1\\x_3'&y_3'&z_3'&1\\x_4'&y_4'&z_4'&1\end{vmatrix}=\begin{vmatrix}x_1'&y_1'&z_1'\\x_3'&y_3'&z_3'\\x_4'&y_4'&z_4'\end{vmatrix}$$ This smaller $3×3$ determinant can be computed using the given triple product, which proves the last equation $\vec{BC}\cdot(\vec{BD} \times \vec{BA})=\det M$.