The cross product between two 3-D vectors produces
a new vector that is perpendicular to both.

Consider the two vectors

$$\begin{array}{l}A={a}_{1}\widehat{i}+{a}_{2}\widehat{j}+{a}_{3}\widehat{k}\text{\hspace{0.17em}}\text{\hspace{0.17em}},\\ B={b}_{1}\widehat{i}+{b}_{2}\widehat{j}+{b}_{3}\widehat{k}\text{\hspace{0.17em}}\text{\hspace{0.17em}}.\end{array}$$

In terms of a matrix determinant involving the basis vectors $$\widehat{i}$$, $$\widehat{j}$$,
and $$\widehat{k}$$,
the cross product of *A* and *B* is

$$\begin{array}{l}C=A\times B=\left|\begin{array}{cc}\begin{array}{cc}{\widehat{i}}_{}& {\widehat{j}}_{}\end{array}& {\widehat{k}}_{}\\ \begin{array}{cc}\begin{array}{c}{a}_{1}\\ {b}_{1}\end{array}& \begin{array}{c}{a}_{2}\\ {b}_{2}\end{array}\end{array}& \begin{array}{c}{a}_{3}\\ {b}_{3}\end{array}\end{array}\right|\\ \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=({a}_{2}{b}_{3}-{a}_{3}{b}_{2})\widehat{i}+({a}_{3}{b}_{1}-{a}_{1}{b}_{3})\widehat{j}+({a}_{1}{b}_{2}-{a}_{2}{b}_{1})\widehat{k}\text{\hspace{0.17em}}\text{\hspace{0.17em}}.\end{array}$$

Geometrically, $$A\times B$$ is
perpendicular to both *A* and *B*.
The magnitude of the cross product, $$\Vert A\times B\Vert $$,
is equal to the area of the parallelogram formed using *A* and *B* as
sides. This area is related to the magnitudes of *A* and *B* as
well as the angle between the vectors by

$$\Vert A\times B\Vert =\Vert A\Vert \text{\hspace{0.17em}}\Vert B\Vert \mathrm{sin}\alpha \text{\hspace{0.17em}}\text{\hspace{0.17em}}.$$

Thus, if *A* and *B* are parallel,
then the cross product is zero.