Question

How do you take the cross product of a $3\mathrm{d}$ vector $?$

Answer 1

Cross product of $3\mathrm{d}$ vector :

The cross product of two vectors $\stackrel{\rightharpoonup }{\alpha }$ and $\stackrel{\rightharpoonup }{\beta }$ is another vector given by:

$\stackrel{\rightharpoonup }{\mathrm{\alpha }}\times \stackrel{\rightharpoonup }{\mathrm{\beta }}=||\mathrm{\alpha }||||\mathrm{\beta }|||\sin \mathrm{\theta }|\stackrel{\rightharpoonup }{\mathrm{u}}$

Where, $\theta$ is the angle between $\stackrel{\rightharpoonup }{\alpha }$ and $\stackrel{\rightharpoonup }{\beta }$

$\stackrel{\rightharpoonup }{u}$ is a unit vector perpendicular to the plane that contains $\stackrel{\rightharpoonup }{\alpha }$ and $\stackrel{\rightharpoonup }{\beta }$

The direction of $\stackrel{\rightharpoonup }{u}$ is given by the so called right-hand rule.

Hence, we can take the cross product of a 3d vector by the formula $\stackrel{\mathbf{\rightharpoonup }}{\mathbf{\alpha }}\mathbf{}\mathbf{\times }\mathbf{}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{\beta }}\mathbf{}\mathbf{=}\mathbf{}||\alpha ||||\beta ||\mathbf{|}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{}\mathbf{\theta }\mathbf{|}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{u}}$