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}$