Triple product

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 &times; C is the cross product of two vectors (resulting into a vector) and the dot indicates the inner product between two vectors (a scalar).

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 &times; 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&times;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 &times; 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{align} \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{align} $$

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