Triple product

CC Image
Parallelepiped spanned by vectors A, B, and C (shown in red).

In analytic geometry, a triple product is a common term for a product of three vectors A, B, and C leading to a scalar (a number). The absolute value of this scalar is the volume V of the parallelepiped spanned by the three vectors:

$V={\big |}\mathbf {A} \cdot (\mathbf {B} \times \mathbf {C} ){\big |},$

where B × C is the cross product of two vectors (resulting into a vector) and the dot indicates the inner product between two vectors (a scalar).

The triple product is sometimes called the scalar triple product to distinguish it from the vector triple product A×(B×C). The scalar triple product is often written as [A B C]. The vector triple product can be expanded by the aid of the baccab formula.

Explanation

Let n be a unit normal to the parallelogram spanned by B and C (see figure). Let h be the height of the terminal point of the vector A above the base of the parallelepiped. Recall:

Volume V of parallelepiped is height h times area S of the base.

Note that h is the projection of A on n and that the area S is the length of the cross product of the vectors spanning the base,

$h=\mathbf {A} \cdot \mathbf {n} \quad {\hbox{and}}\quad S=|\mathbf {B} \times \mathbf {C} |.$

Use

$V=(\mathbf {A} \cdot \mathbf {n} )\;(|\mathbf {B} \times \mathbf {C} |)=\mathbf {A} \cdot (\mathbf {n} \,|\mathbf {B} \times \mathbf {C} |)=\mathbf {A} \cdot (\mathbf {B} \times \mathbf {C} ),$

where it is used that

$\mathbf {n} \;|\mathbf {B} \times \mathbf {C} |=\mathbf {B} \times \mathbf {C} .$

(The unit normal n has the direction of the cross product B × C).

If A, B, and C do not form a right-handed system, A•n < 0 and we must take the absolute value: | A• (B×C)|.

Triple product as determinant

Take three orthogonal unit vectors i , j, and k and write

$\mathbf {A} =A_{1}\mathbf {i} +A_{2}\mathbf {j} +A_{3}\mathbf {k} ,\quad \mathbf {B} =B_{1}\mathbf {i} +B_{2}\mathbf {j} +B_{3}\mathbf {k} ,\quad \mathbf {C} =C_{1}\mathbf {i} +C_{2}\mathbf {j} +C_{3}\mathbf {k} .$

The triple product is equal to a 3 × 3 determinant

$\mathbf {A} \cdot (\mathbf {B} \times \mathbf {C} )={\begin{vmatrix}A_{1}&A_{2}&A_{3}\\B_{1}&B_{2}&B_{3}\\C_{1}&C_{2}&C_{3}\\\end{vmatrix}}.$

Indeed, writing the cross product as a determinant we find

${\begin{aligned}\mathbf {A} \cdot (\mathbf {B} \times \mathbf {C} )&=\mathbf {A} \cdot {\begin{vmatrix}\mathbf {i} &\mathbf {j} &\mathbf {k} \\B_{1}&B_{2}&B_{3}\\C_{1}&C_{2}&C_{3}\\\end{vmatrix}}\\&={\big (}A_{1}\mathbf {i} +A_{2}\mathbf {j} +A_{3}\mathbf {k} {\big )}\cdot {\big [}(B_{2}\,C_{3}-B_{3}\,C_{2})\;\mathbf {i} +(B_{3}\,C_{1}-B_{1}\,C_{3})\;\mathbf {j} +(B_{1}\,C_{2}-B_{2}\,C_{1})\;\mathbf {k} {\big ]}\\&=A_{1}\;(B_{2}\,C_{3}-B_{3}\,C_{2})+A_{2}\;(B_{3}\,C_{1}-B_{1}\,C_{3})+A_{3}\;(B_{1}\,C_{2}-B_{2}\,C_{1})\\&={\begin{vmatrix}A_{1}&A_{2}&A_{3}\\B_{1}&B_{2}&B_{3}\\C_{1}&C_{2}&C_{3}\\\end{vmatrix}}.\end{aligned}}$

Since a determinant is invariant under cyclic permutation of its rows, it follows

$\mathbf {A} \cdot (\mathbf {B} \times \mathbf {C} )=\mathbf {B} \cdot (\mathbf {C} \times \mathbf {A} )=\mathbf {C} \cdot (\mathbf {A} \times \mathbf {B} ).$

Reference

M. R. Spiegel, Theory and Problems of Vector Analysis, Schaum Publishing, New York (1959) p. 26