Editing Tetrahedron (section)

===Volume===
The volume of a tetrahedron can be obtained in many ways. It can be given by using the formula of the pyramid's volume: 
<math display="block"> V = \frac{1}{3}Ah. </math>
where <math> A </math> is the [[base (geometry)|base]]' area and <math> h </math> is the height from the base to the apex. This applies for each of the four choices of the base, so the distances from the apices to the opposite faces are inversely proportional to the areas of these faces. Another way is by dissecting a triangular prism into three pieces.{{sfn|Alsina|Nelsen|2015|p=[https://books.google.com/books?id=FEl2CgAAQBAJ&pg=PA67 67]}}

Given the vertices of a tetrahedron in the following:
<math display="block"> \begin{align}
 \mathbf{a} &= (a_1, a_2, a_3), \\
 \mathbf{b} &= (b_1, b_2, b_3), \\
 \mathbf{c} &= (c_1, c_2, c_3), \\
 \mathbf{d} &= (d_1, d_2, d_3).
\end{align} </math>
The volume of a tetrahedron can be ascertained in terms of a [[determinant]] <math display="inline"> \frac{1}{6} \det(\mathbf{a} - \mathbf{d}, \mathbf{b} - \mathbf{d}, \mathbf{c} - \mathbf{d}) </math>,{{sfn|Fekete|1985|p=[https://books.google.com/books?id=3_AIXBO11bEC&pg=PA68 68]}} or any other combination of pairs of vertices that form a simply connected [[Graph (discrete mathematics)|graph]]. Comparing this formula with that used to compute the volume of a [[parallelepiped]], we conclude that the volume of a tetrahedron is equal to {{sfrac|1|6}} of the volume of any parallelepiped that shares three converging edges with it.

The absolute value of the scalar triple product can be represented as the following absolute values of determinants:
:<math>6 \cdot V =\begin{Vmatrix}
\mathbf{a} & \mathbf{b} & \mathbf{c}
\end{Vmatrix}</math>{{pad|2em}}or{{pad|1em}}<math>6 \cdot V =\begin{Vmatrix}
\mathbf{a} \\ \mathbf{b} \\ \mathbf{c}
\end{Vmatrix}</math>{{pad|2em}}where{{pad|1em}}<math>\begin{cases}\mathbf{a} = (a_1,a_2,a_3), \\
 \mathbf{b} = (b_1,b_2,b_3), \\
 \mathbf{c} = (c_1,c_2,c_3), \end{cases}</math>{{pad|1em}}are expressed as row or column vectors.

Hence
:<math>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}</math>{{pad|1em}}where{{pad|1em}}<math>\begin{cases}\mathbf{a} \cdot \mathbf{b} = ab\cos{\gamma}, \\ \mathbf{b} \cdot \mathbf{c} = bc\cos{\alpha}, \\ \mathbf{a} \cdot \mathbf{c} = ac\cos{\beta}. \end{cases}</math>

where <math>a = \begin{Vmatrix} \mathbf{a} \end{Vmatrix} </math>, <math>b = \begin{Vmatrix} \mathbf{b} \end{Vmatrix} </math>, and <math>c = \begin{Vmatrix} \mathbf{c} \end{Vmatrix} </math>, which gives
:<math>V = \frac {abc} {6} \sqrt{1 + 2\cos{\alpha}\cos{\beta}\cos{\gamma}-\cos^2{\alpha}-\cos^2{\beta}-\cos^2{\gamma}}, \,</math>
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'''.

If we do not require that '''d''' = 0 then
:<math>6 \cdot V = \left| \det \left( \begin{matrix}
a_1 & b_1 & c_1 & d_1 \\
a_2 & b_2 & c_2 & d_2 \\
a_3 & b_3 & c_3 & d_3 \\
 1  &  1  &  1  &  1
\end{matrix} \right) \right|\,.</math>

Given the distances between the vertices of a tetrahedron the volume can be computed using the [[Cayley–Menger determinant]]:
:<math>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}</math>
where the subscripts {{nowrap|''i'', ''j'' ∈ {1, 2, 3, 4}|}} represent the vertices {{nowrap|{'''a''', '''b''', '''c''', '''d'''}|}} and ''d{{sub|ij}}'' 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.<ref>[http://www.mathpages.com/home/kmath664/kmath664.htm "Simplex Volumes and the Cayley-Menger Determinant"], MathPages.com</ref>

Let <math> a </math>, <math> b </math>, and <math> c </math> be the lengths of three edges that meet at a point, and <math> x </math>, <math> y </math>, and <math> z </math> be those of the opposite edges. The volume of the tetrahedron <math> V </math> is:{{sfn|Kahan|2012|p=11}}
<math display="block"> V = \frac{\sqrt{4 a^2 b^2 c^2-a^2 X^2-b^2 Y^2-c^2 Z^2+X Y Z}}{12}</math>
where
<math display="block"> \begin{align}
 X &= b^2+c^2-x^2, \\
 Y &= a^2+c^2-y^2, \\
 Z &= a^2+b^2-z^2.
\end{align}</math>
The above formula uses six lengths of edges, and the following formula uses three lengths of edges and three angles.{{sfn|Kahan|2012|p=11}}
<math display="block"> V = \frac{abc}{6} \sqrt{1 + 2\cos{\alpha}\cos{\beta}\cos{\gamma}-\cos^2{\alpha}-\cos^2{\beta}-\cos^2{\gamma}}</math>

[[Image:Six edge-lengths of Tetrahedron.png|class=skin-invert-image|right|thumb|240px|Six edge-lengths of Tetrahedron]]
The volume of a tetrahedron can be ascertained by using the Heron formula. Suppose <math> U </math>, <math> V </math>, <math> W </math>, <math> u </math>. <math> v </math>, and <math> w </math> are the lengths of the tetrahedron's edges as in the following image. Here, the first three form a triangle, with <math> u </math> opposite <math> U </math>, <math> v </math> opposite <math> V </math>, and <math> w </math> opposite <math> W </math>. Then,
<math display="block"> V = \frac{\sqrt {\,( - p + q + r + s)\,(p - q + r + s)\,(p + q - r + s)\,(p + q + r - s)}}{192\,u\,v\,w}</math>
where
<math display="block"> \begin{align}
 p = \sqrt{xYZ}, &\quad q = \sqrt{yZX}, \\
 r = \sqrt{zXY}, &\quad s = \sqrt {xyz},
\end{align} </math>
and
<math display="block"> \begin{align}
 X = (w - U + v)(U + v + w), &\quad x = (U - v + w)(v - w + U), \\
 Y = (u - V + w)(V + w + u), &\quad y = (V - w + u)(w - u + V), \\
 Z = (v - W + u)(W + u + v), &\quad z = (W - u + v)\,(u - v + W).
\end{align}
</math>

Any plane containing a bimedian (connector of opposite edges' midpoints) of a tetrahedron [[bisection|bisects]] the volume of the tetrahedron.{{sfn|Bottema|1969}}

For tetrahedra in [[hyperbolic space]] or in three-dimensional [[elliptic geometry]], the [[dihedral angle]]s of the tetrahedron determine its shape and hence its volume. In these cases, the volume is given by the [[Murakami–Yano formula]], after Jun Murakami and Masakazu Yano.{{sfn|Murakami|Yano|2005}} However, in Euclidean space, scaling a tetrahedron changes its volume but not its dihedral angles, so no such formula can exist.

Any two opposite edges of a tetrahedron lie on two [[skew lines]], and the distance between the edges is defined as the distance between the two skew lines. Let <math> d </math> be the distance between the skew lines formed by opposite edges <math> a </math> and <math> \mathbf{b} - \mathbf{c} </math> as calculated [[Skew lines#Distance between two skew lines|here]]. Then another formula for the volume of a tetrahedron <math> V </math> is given by
<math display="block"> V = \frac {d |(\mathbf{a} \times \mathbf{(b-c)})|}{6}. </math>