If $\overrightarrow{\mathrm{A}}\times \overrightarrow{\mathrm{B}}=\overrightarrow{\mathrm{B}}\times \overrightarrow{\mathrm{A}}$, then the angle between $\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{B}}$ is 

$\overrightarrow{\mathrm{A}}\times \overrightarrow{\mathrm{B}}=\overrightarrow{\mathrm{B}}\times \overrightarrow{\mathrm{A}}$

$\Rightarrow (\overrightarrow{\mathrm{A}}\times \overrightarrow{\mathrm{B}})-(\overrightarrow{\mathrm{B}}\times \overrightarrow{\mathrm{A}})=0$ 

$(\overrightarrow{\mathrm{A}}\times \overrightarrow{\mathrm{B}})+(\overrightarrow{\mathrm{A}}\times \overrightarrow{\mathrm{B}})=0$$\Rightarrow 2(\overrightarrow{\mathrm{A}}\times \overrightarrow{\mathrm{B}})=0$ 

So, $2\mathrm{ABsin}\mathrm{\theta }\widehat{\mathrm{n}}=0$ As $\mathrm{A}\ne 0 \mathrm{nor} \mathrm{B}\ne 0, \mathrm{so} \sin \mathrm{\theta }=0.$

Hence,$\mathrm{\theta }=0^0\text{ or }\mathrm{\pi }$