Mathematics - Vector Question with Solution | TestHub

Given,

The vector $\overrightarrow{a}=-\hat{i}+2\hat{j}+\hat{k}$ is rotated through a right angle, passing through the $y$-axis in its way and the resulting vector is $\overrightarrow{b}$

Now taking, $\overrightarrow{b}=\lambda \overrightarrow{a}\times \left(\overrightarrow{a}\times \hat{j}\right)$ {note here we have taken cross vector with $\hat{j}$ as it is rotated along $y-$axis}

$\Rightarrow \overrightarrow{b}=\lambda \left(-\hat{i}+2\hat{j}+\hat{k}\right)\times \left(\left(-\hat{i}+2\hat{j}+\hat{k}\right)\times \hat{j}\right)$

$\Rightarrow \overrightarrow{b}=\lambda \left(-\hat{i}+2\hat{j}+\hat{k}\right)\times \left(-\hat{i}-\hat{k}\right)$

$\Rightarrow \overrightarrow{b}=\lambda \left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -1& 2& 1\\ -1& 0& -1\end{array}\right|$

$\Rightarrow \overrightarrow{b}=\lambda \left(-2\hat{i}-2\hat{j}+2\hat{k}\right)$

Also magnitude of both vector will be equal,

So,$\left|\overrightarrow{b}\right|=\left|\overrightarrow{a}\right|\Rightarrow \sqrt{6}=\sqrt{12} \left|\lambda \right|$

$\Rightarrow \lambda =\pm \frac{1}{\sqrt{2}}$

(Now $\lambda =\frac{1}{\sqrt{2}}$ rejected $\because \overrightarrow{\mathit{b}}$ makes acute angle with y axis )

So, $\overrightarrow{b}=-\sqrt{2}\left(-\hat{i}-\hat{j}+\hat{k}\right)$

Now projection of  $3\overrightarrow{a}+\sqrt{2}\overrightarrow{b}$ on $\overrightarrow{c}$ is given by,

$\frac{\left(3\overrightarrow{a}+\sqrt{2b}\right)·\overrightarrow{c}}{\left|\overrightarrow{c}\right|}=3\sqrt{2}$