# How to find the volume of a tetrahedron?

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 $(1,0,0,1),(-1,2,0,1), (3,0,1,1), and (-1,4,0,1)$.

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

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

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

• Having moved one vertex to the origin, you could find a rigid transformation (that is, a transformation corresponding to orthogonal matrix) that rotates the moved tetrahedron into $\mathbb{R}^3$, and then use the cross product formula you know. Alternatively, you could calculate the distances between pairs of points, and then use the Cayley-Menger determinant.
– Blue
Jul 4, 2018 at 23:09

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.

• What is Gram Matrix? Jul 5, 2018 at 0:18
• @dg123 follows the link in answer and refs there. It is a tool to study higher dimension simplex and stuff. For example, one use of it is to prove the formula of volume in terms of Cayler Menger determinant (the generalization of Heron's formula for area of triangle). For your purposes, it is a matrix with entries $\vec{v}_i\cdot\vec{v}_j$ and you don't need to know anything more. Jul 5, 2018 at 0:28
• The scalar triple product is mistyped, should be $(v_1×v_2)\cdot v_3$. Jul 5, 2018 at 1:00
• @OscarLanzi Or $v_1 \cdot (v_2 \times v_3)$. Jul 5, 2018 at 1:22

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 $[-2,4,0]$

• But how do we calculate the determinant if its a 3 by 4 matrix? Jul 4, 2018 at 23:15
• Use the scalar triple product. Jul 4, 2018 at 23:25
• Ok, I got the answer now, but can you explain a bit further on how you converted from $\Bbb R^4$ to $\Bbb R^3$? Jul 4, 2018 at 23:48
• And when you say the last element is 1 for all points, do you mean 0? Jul 4, 2018 at 23:49
• Look at your original vectors. They all end in 1. The differences end in 0. Jul 5, 2018 at 0:09

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.