Question

The vector $\vec{a}=-\hat{i}+2\hat{j}+\hat{k}$ is rotated through a right angle, passing through the y-axis in its way and the resulting vector is $\vec{b}$. Then the projection of $3\vec{a}+\sqrt{2}\vec{b}$ on $\vec{c}=5\hat{i}+4\hat{j}+3\hat{k}$ is :

• A. $3\sqrt{2}$
• B. 1
• C. $2\sqrt{3}$
• D. $\sqrt{6}$

Answer 1

Answer: (A)

$\vec{b}=\lambda \vec{a}\times (\vec{a}\times \hat{j})$
$\Rightarrow \vec{b}=\lambda (-2\hat{i}-2\hat{j}+2\hat{k})$
$|\vec{b}|=|\vec{a}|\therefore \sqrt{6}=\sqrt{12}|\lambda |\Rightarrow \lambda =\pm \frac{1}{\sqrt{2}}$
$\left(\lambda =\frac{1}{\sqrt{2}}\text{ rejected }\because \vec{b}\text{ makes acute angle with y axis }\right)$
$\vec{b}=-\sqrt{2}(-\hat{i}-\hat{j}+\hat{k})$
$\frac{(3\vec{a}+\sqrt{2}\vec{b})\cdot \vec{c}}{|\vec{c}|}=3\sqrt{2}$