Tetrahedron - Volume

Volume

The volume of a tetrahedron is given by the pyramid volume formula:

where A0 is the area of the base and h the height from the base to the apex. This applies for each of the four choices of the base, so the distances from the apexes to the opposite faces are inversely proportional to the areas of these faces.

For a tetrahedron with vertices a = (a1, a2, a3), b = (b1, b2, b3), c = (c1, c2, c3), and d = (d1, d2, d3), the volume is (1/6)·|det(a − b, b − c, c − d)|, or any other combination of pairs of vertices that form a simply connected graph. This can be rewritten using a dot product and a cross product, yielding

If the origin of the coordinate system is chosen to coincide with vertex d, then d = 0, so

where a, b, and c represent three edges that meet at one vertex, and a · (b × c) is a scalar triple product. Comparing this formula with that used to compute the volume of a parallelepiped, we conclude that the volume of a tetrahedron is equal to 1/6 of the volume of any parallelepiped that shares three converging edges with it.

The triple scalar can be represented by the following determinants:

6 \cdot V =\begin{vmatrix} \mathbf{a} & \mathbf{b} & \mathbf{c} \end{vmatrix} or 6 \cdot V =\begin{vmatrix} \mathbf{a} \\ \mathbf{b} \\ \mathbf{c} \end{vmatrix} where is expressed as a row or column vector etc.

Hence

36 \cdot V^2 =\begin{vmatrix} \mathbf{a^2} & \mathbf{a} \cdot \mathbf{b} & \mathbf{a} \cdot \mathbf{c} \\ \mathbf{a} \cdot \mathbf{b} & \mathbf{b^2} & \mathbf{b} \cdot \mathbf{c} \\ \mathbf{a} \cdot \mathbf{c} & \mathbf{b} \cdot \mathbf{c} & \mathbf{c^2} \end{vmatrix} where etc.

which gives

where α, β, γ are the plane angles occurring in vertex d. The angle α, is the angle between the two edges connecting the vertex d to the vertices b and c. The angle β, does so for the vertices a and c, while γ, is defined by the position of the vertices a and b.

Given the distances between the vertices of a tetrahedron the volume can be computed using the Cayley–Menger determinant:

288 \cdot V^2 = \begin{vmatrix} 0 & 1 & 1 & 1 & 1 \\ 1 & 0 & d_{12}^2 & d_{13}^2 & d_{14}^2 \\ 1 & d_{12}^2 & 0 & d_{23}^2 & d_{24}^2 \\ 1 & d_{13}^2 & d_{23}^2 & 0 & d_{34}^2 \\ 1 & d_{14}^2 & d_{24}^2 & d_{34}^2 & 0 \end{vmatrix}

where the subscripts represent the vertices {a, b, c, d} and is the pairwise distance between them—i.e., the length of the edge connecting the two vertices. A negative value of the determinant means that a tetrahedron cannot be constructed with the given distances. This formula, sometimes called Tartaglia's formula, is essentially due to the painter Piero della Francesca in the 15th century, as a three dimensional analogue of the 1st century Heron's formula for the area of a triangle.

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