# Question #8e005

Jan 22, 2017

$2 {m}^{2} {n}^{2} {\sin}^{2} \theta$, where $\theta = \angle$ between $\vec{m} \mathmr{and} \vec{n}$. If $\vec{m} \mathmr{and} \vec{n}$ are unit vectors, m and n are 1 and this becomes $2 {\sin}^{2} \theta$

#### Explanation:

Assuming that the expression is

$| \vec{m} X \vec{n} {|}^{2} + | \vec{n} X \vec{m} {|}^{2}$, it is

$2 {m}^{2} {n}^{2} {\sin}^{2} \theta$,

where $\theta = \angle$ between $\vec{m} \mathmr{and} \vec{n}$

$= 2 {\sin}^{2} \theta$, for unit vectors, having m and n as 1,

I have used

$\vec{m} X \vec{n} = - \vec{n} X \vec{m}$ and

$\vec{m} X \vec{n} = m n \sin \theta \vec{z}$,

where $\vec{z}$ is the unit vector in the direction of $\vec{m} X \vec{n}$.