# What is the angle between A×B andB×A?

Jun 8, 2018

See below

#### Explanation:

From the definition:

$\boldsymbol{A} \times \boldsymbol{B} = \left\mid \boldsymbol{A} \right\mid \left\mid \boldsymbol{B} \right\mid \sin {\alpha}_{\text{AB}} \setminus \boldsymbol{\hat{n}}$

So $\boldsymbol{A} \times \boldsymbol{B}$ lies in the direction that is normal to both $\boldsymbol{A}$ and $\boldsymbol{B}$, labelled $\boldsymbol{\hat{n}}$, and has magnitude $\left\mid \boldsymbol{A} \right\mid \left\mid \boldsymbol{B} \right\mid \sin {\alpha}_{\text{AB}}$, where ${\alpha}_{\text{AB}}$ is the angle between $\boldsymbol{A}$ and $\boldsymbol{B}$

On the other hand:

$\boldsymbol{B} \times \boldsymbol{A} = \left\mid \boldsymbol{B} \right\mid \left\mid \boldsymbol{A} \right\mid \sin {\alpha}_{\textcolor{red}{\boldsymbol{\text{BA}}}} \setminus \boldsymbol{\hat{n}}$

$= \left\mid \boldsymbol{A} \right\mid \left\mid \boldsymbol{B} \right\mid \sin \left(- {\alpha}_{\textcolor{g r e e n}{\boldsymbol{\text{AB}}}}\right) \setminus \boldsymbol{\hat{n}}$

$= \boldsymbol{\textcolor{red}{-}} \left\mid \boldsymbol{A} \right\mid \left\mid \boldsymbol{B} \right\mid \sin {\alpha}_{\text{AB}} \setminus \boldsymbol{\hat{n}}$

IOW, these vectors have same magnitude, but are anti-parallel, so the angle between them is 180 degrees

It is therefore true to say that:

$\boldsymbol{A} \times \boldsymbol{B} = - \boldsymbol{B} \times \boldsymbol{A}$

And vice versa.