Mathematics - Vector Question with Solution | TestHub

As,$\overrightarrow{a}$lies in the plane of the vectors$\overrightarrow{b}$and$\overrightarrow{c}$, So,$\overrightarrow{a}=\lambda \left(\frac{\widehat{i}+\widehat{j}}{\sqrt{2}}+\frac{\widehat{\mathrm{i}}+\widehat{\mathrm{j}}+4\widehat{\mathrm{k}}}{3\sqrt{2}}\right)=\frac{\lambda }{3\sqrt{2}}\left[3\widehat{i}+3\widehat{j}+\widehat{i}-\widehat{j}+4\widehat{k}\right]=\frac{\lambda }{3\sqrt{2}}\left[4\widehat{i}+2\widehat{j}+4\widehat{k}\right]$
Compare with$\overrightarrow{a}=a\widehat{i}+2\widehat{j}+\beta \widehat{k}$
$\frac{2\lambda }{3\sqrt{2}}=2\Rightarrow \lambda =3\sqrt{2}$
$\overrightarrow{a}=4\widehat{i}+2\widehat{j}+4\widehat{k}$
Not in option
So, now consider$\overrightarrow{a}=\mu \left(\frac{\widehat{i}+\widehat{j}}{\sqrt{2}}-\frac{\widehat{i}-\widehat{j}+4\widehat{k}}{3\sqrt{2}}\right)$
$\overrightarrow{a}=\frac{\mu }{3\sqrt{2}}\left(3\widehat{i}+3\widehat{j}-\widehat{i}+\widehat{j}-4\widehat{k}\right)$
$=\frac{\mu }{3\sqrt{2}}\left(2\widehat{i}+4\widehat{j}-4\widehat{k}\right)$
Compare with$\overrightarrow{a}=\alpha \widehat{i}+2\widehat{j}+\beta \widehat{k}$
$\frac{4\mu }{3\sqrt{2}}=2\Rightarrow \mu =\frac{3\sqrt{2}}{2}$
$\overrightarrow{a}=\widehat{i}+2\widehat{j}-2\widehat{k}$
$\overrightarrow{a}.\widehat{k}+2=0$