How to find the volume of a tetrahedron?

Question

For this question, how do we find the volume of a tetrahedron in $\Bbb R^4$ if we have a 4 by 3 matrix? The cross product and determinant only work for square matrices. I'm not sure what to do after we subtract the points from the emanating point. Here is what I have so far. Can anyone please help me out?

Find the volume of the tetrahedron in $\Bbb R^4$ with vertices $(1,0,0,1),(-1,2,0,1), (3,0,1,1), and (-1,4,0,1)$.

$(-1,2,0,1)-(1,0,0,1) = [-2,2,0,0]$

$(3,0,1,1)-(1,0,0,1) = [2,0,1,0]$

$(-1,4,0,1)-(1,0,0,1) = [-2,4,0,0]$

Answer 1

You are almost there. You just need to notice that all points are on the same hyperplane (last element is 1 for all points). Then the problem reduces to calculating the volume of the tetrahedron with three vertices given by the first three components of your three differences, and the fourth vertex is the origin. So your vertices are $[0,0,0], [-2,2,0], [2,0,1],$ and $[-2,4,0]$

Answer 2

In $\mathbb{R}^3$, up to a sign, the volume of a tetrahedron with vertices at $\vec{v}_0 = \vec{0}$, $\vec{v}_1$, $\vec{v}_2$ and $\vec{v}_3$ is given by a scalar triple product: $$\verb/Volume/ = \frac16 \left| \vec{v}_1 \cdot ( \vec{v}_2 \times \vec{v}_3 ) \right|$$ If one construct a $3 \times 3$ matrix $\Delta$ whose $i^{th}$ column equals to $\vec{v}_i$, above formula becomes

$$\verb/Volume/ = \frac16 | \det\Delta |$$

Let $G(\vec{v}_1,\vec{v}_2,\vec{v}_3) = \Delta^T\Delta$ be the Gram matrix associated with the vectors $\vec{v}_1$, $\vec{v}_2$ and $\vec{v}_3$. For the purpose of this question, it is simply a $3 \times 3$ matrix whose entry at row $i$, column $j$ equals to $\vec{v}_i\cdot\vec{v}_j$. Since these entries depend only on inner products, the expression continues to work even when the points belong to some higher dimension space.

In terms of the Gram matrix, the 3-volume of a tetrahedron with vertices $\vec{u}_0$, $\vec{u}_1$, $\vec{u}_2$, $\vec{u}_3 \in \mathbb{R}^n$ for any $n \ge 3$ equals to

$$\verb/Volume/ = \frac16\sqrt{\det G(\vec{v}_1,\vec{v}_2,\vec{v}_3)} \quad\text{ where }\quad \vec{v}_i = \vec{u}_i - \vec{u}_0$$

For the problem at hand

$$\begin{cases} \vec{v}_1 = (-2,2,0,0)\\ \vec{v}_2 = (\phantom{+}2,0,1,0)\\ \vec{v}_3 = (-2,4,0,0) \end{cases} \implies G(\vec{v}_1,\vec{v}_2,\vec{v}_3) = \begin{bmatrix} \vec{v}_1\cdot\vec{v}_1 & \vec{v}_1\cdot\vec{v}_2 & \vec{v}_1\cdot\vec{v}_3\\ \vec{v}_2\cdot\vec{v}_1 & \vec{v}_2\cdot\vec{v}_2 & \vec{v}_2\cdot\vec{v}_3\\ \vec{v}_3\cdot\vec{v}_1 & \vec{v}_3\cdot\vec{v}_2 & \vec{v}_3\cdot\vec{v}_3\\ \end{bmatrix} = \begin{bmatrix} 8 & -4 & 12 \\ -4 & 5 & -4 \\ 12 & -4 & 20 \end{bmatrix} $$

This leads to $$\verb/Volume/ = \frac16 \sqrt{\left|\begin{matrix} 8 & -4 & 12 \\ -4 & 5 & -4 \\ 12 & -4 & 20 \end{matrix}\right|} = \frac16\sqrt{16} = \frac{2}{3}$$

Update

For an alternative formula for the volume, we can apply Cauchy-Binet formula to $\det(\Delta^T\Delta)$ and decompose it as a sum of squares of determinants of $3 \times 3$ sub-matrices of $\Delta$. More precisely, let $\vec{r}_1, \vec{r}_2, \ldots, \vec{r}_n \in \mathbb{R}^3$ be the row vectors of $\Delta$, we have $$\det(G) = \det(\Delta^T\Delta) = \sum_{1 \le i < j < k \le n} | \vec{r}_i \cdot ( \vec{r}_j \times \vec{r}_k )|^2$$

When $n = 4$, the volume of a tetrahedron becomes

$$\verb/Volume/ = \frac16\sqrt{ | \vec{r}_1 \cdot ( \vec{r}_2 \times \vec{r}_3 )|^2 +| \vec{r}_1 \cdot ( \vec{r}_2 \times \vec{r}_4 )|^2 +| \vec{r}_1 \cdot ( \vec{r}_3 \times \vec{r}_4 )|^2 +| \vec{r}_2 \cdot ( \vec{r}_3 \times \vec{r}_4 )|^2}$$

For the problem at hand, $\begin{cases} \vec{r}_1 = (-2,2,-2)\\ \vec{r}_2 = (2,0,4)\\ \vec{r}_3 = (0,1,0)\\ \vec{r}_4 = (0,0,0) \end{cases}$. Since $\vec{r}_4 = \vec{0}$, only one triple product survives and $$\verb/Volume/ = \frac16 | \vec{r}_1 \cdot (\vec{r}_2 \times \vec{r}_3)| = \frac16 \left|\begin{bmatrix} -2 & 2 & -2\\ 2 & 0 & 4\\ 0 & 1 & 0 \end{bmatrix}\right| = \frac{|(-2)4 - (-2)(2)|}{6} = \frac23 $$ Same answer as before. In general, this illustrate if one component of $u_k$ is the same, then some row vectors $\vec{r}_k = \vec{0}$. Up to a constant, the formula of volume reduce to a scalar triple product again.

Answer 3

A general approach is to use the Cayley-Menger determinant. You may note that this was mentioned in a comment by Blue, from whom I first learned this same fact in a comment on MSE five years earlier - when I posed a heretofore unresolved question about tetrahedra in MSE 351913, which I eventually migrated (still without success) to MathOverflow as MO 142983.