Mathematics - Vector Question with Solution | TestHub

$\overrightarrow{a}=3\widehat{i}+2\widehat{j}+2\widehat{k},\overrightarrow{b}=i+2\widehat{j}-2\widehat{k}$
$\because \overrightarrow{\alpha }=\overrightarrow{a}+\overrightarrow{b}=4\widehat{i}+4\widehat{j}$
$\overrightarrow{\beta }=\overrightarrow{a}-\overrightarrow{b}=2\widehat{i}+4\widehat{k}$
$\therefore \overrightarrow{\alpha }\times \overrightarrow{\beta }=\left|\begin{array}{ccc}\widehat{i}& \widehat{j}& \widehat{k}\\ 4& 4& 0\\ 2& 0& 4\end{array}\right|=\widehat{i}\left(16\right)-\widehat{j}\left(16\right)+\widehat{k}\left(-8\right)=16\widehat{i}-16\widehat{j}-8\widehat{k}$
$\therefore$Unit vector perpendicular to$\overrightarrow{\alpha }\&\overrightarrow{\beta }$is
$\widehat{n}=\frac{\overrightarrow{\alpha }\times \overrightarrow{\beta }}{\left|\overrightarrow{\alpha }\times \overrightarrow{\beta }\right|}=\frac{2\widehat{i}-2\widehat{j}-\widehat{k}}{3}$
$\therefore$Required vector$=12\frac{\left(2\widehat{i}-2\widehat{j}-\widehat{k}\right)}{3}=4\left(2\widehat{i}-2\widehat{j}-\widehat{k}\right)$