The cross product of two vectors $\boldsymbol{\mathrm{a} }$ and $\boldsymbol{\mathrm{b} }$, separated by an angle of $\theta$ radians, is defined as follows:
\[\boldsymbol{\mathrm{a} }\times\boldsymbol{\mathrm{b} }=\lvert\boldsymbol{\mathrm{a} }\rvert\lvert\boldsymbol{\mathrm{b} }\rvert \sin{(\theta)}\;\boldsymbol{\mathrm{\hat{n} } }\]
$\boldsymbol{\mathrm{\hat{n} } }$ is a unit vector perpendicular to $\boldsymbol{\mathrm{a} }$ and $\boldsymbol{\mathrm{b} }$ such that $\boldsymbol{\mathrm{a} }$, $\boldsymbol{\mathrm{b} }$ and $\boldsymbol{\mathrm{\hat{n} } }$ form a right-handed system.
The cross product gives a vector which is perpendicular to both $\boldsymbol{\mathrm{a}}$ and $\boldsymbol{\mathrm{b}}$
Expressing the cross product in the form of a matrix determinant allows us to calculate each component explicitly:
\begin{align} \boldsymbol{\mathrm{a} }\times\boldsymbol{\mathrm{b} } &= \begin{vmatrix} \boldsymbol{\mathrm{e} }_1 & \boldsymbol{\mathrm{e} }_2 & \boldsymbol{\mathrm{e} }_3 \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} \\ &=(a_2b_3-a_3b_2)\boldsymbol{\mathrm{e} }_1 + (a_3b_1-a_1b_3)\boldsymbol{\mathrm{e} }_2 + (a_1b_2-a_2b_2)\boldsymbol{\mathrm{e} }_3 \end{align}
Recall that $\boldsymbol{\mathrm{e} }_1$, $\boldsymbol{\mathrm{e} }_2$ and $\boldsymbol{\mathrm{e} }_3$ form the unit basis vectors.
The cross product is anticommutative:
The cross product is distributive over addition:
Since the angle between parallel vectors is $0$, and $\sin{0}=0$, the cross product of a vector with itself is $0$:
By the same argument, if $\boldsymbol{\mathrm{a} }$ and $\boldsymbol{\mathrm{b} }$ are parallel vectors, then their cross product is $0$:
Given vectors $\boldsymbol{\mathrm{a} }=(4,\,7,\,-3)$ and $\boldsymbol{\mathrm{b} }=(5,-8,11)$, calculate $\boldsymbol{\mathrm{a} }\times\boldsymbol{\mathrm{b} }$.
Recall that the cross product can be calculated by evaluating a determinant.
\begin{align} \boldsymbol{\mathrm{a} }\times\boldsymbol{\mathrm{b} } &= \begin{vmatrix} \boldsymbol{\mathrm{e} }_1 & \boldsymbol{\mathrm{e} }_2 & \boldsymbol{\mathrm{e} }_3 \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} \\ &=(a_2b_3-a_3b_2)\boldsymbol{\mathrm{e} }_1+(a_3b_1-a_1b_3)\boldsymbol{\mathrm{e} }_2+(a_1b_2-a_2b_1). \end{align}
For the given vectors the cross product is
\begin{align} \boldsymbol{\mathrm{a} }\times\boldsymbol{\mathrm{b} } &= \begin{vmatrix} \boldsymbol{\mathrm{e} }_1 & \boldsymbol{\mathrm{e} }_2 & \boldsymbol{\mathrm{e} }_3 \\ 4 & 7 & -3 \\ 5 & -8 & 11 \end{vmatrix} \\ &=(7\cdot11-(-3)\cdot(-8))\boldsymbol{\mathrm{e} }_1+((-3)\cdot5-4\cdot11)\boldsymbol{\mathrm{e} }_2+(4\cdot(-8)-7\cdot5)\boldsymbol{\mathrm{e} }_3 \\ &=(77-24)\boldsymbol{\mathrm{e} }_1+(-15-44)\boldsymbol{\mathrm{e} }_2+(-32-35)\boldsymbol{\mathrm{e} }_3 \\ &=53\boldsymbol{\mathrm{e} }_1-59\boldsymbol{\mathrm{e} }_2-67\boldsymbol{\mathrm{e} }_3 \end{align}
Hence,
\[\boldsymbol{\mathrm{a} }\times\boldsymbol{\mathrm{b} }= 53\boldsymbol{\mathrm{e} }_1-59\boldsymbol{\mathrm{e} }_2-67\boldsymbol{\mathrm{e} }_3.\]
Or equivalently,
\[\boldsymbol{\mathrm{a} }\times\boldsymbol{\mathrm{b} }=(53,\,-59,\,-67).\]
This workbook produced by HELM is a good revision aid, containing key points for revision and many worked examples.
Test yourself: Numbas test on vectors including dot and cross product