The vector a=-i+2 j+k is rotated through a right angle, passing t…

MCQ

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

✓
$3 \sqrt{2}$
B
$1$
C
$\sqrt{6}$
D
$2 \sqrt{3}$

✓

Answer

Correct option: A.

$3 \sqrt{2}$

a
$\overrightarrow{ b }=\lambda \overrightarrow{ a } \times(\overrightarrow{ a } \times \hat{ j })$

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

$|\overrightarrow{ b }|=|\overrightarrow{ a }| \quad \therefore \sqrt{6}=\sqrt{12}|\lambda| \Rightarrow \lambda=\pm \frac{1}{\sqrt{2}}$

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

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

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

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