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

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$\sqrt{6}$

$2 \sqrt{3}$

$3 \sqrt{2}$

answer is A.

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Detailed Solution

$\begin{array}{l} \vec{b} = λ \vec{a} + μ \hat{j} \\ = (λ (- \hat{i} + 2 \hat{j} + \hat{k}) + μ \hat{j}) \\ \vec{b} . \vec{a} = 0 \\ (λ \vec{a} + μ \hat{j}) \cdot \vec{a} = 0 \\ 6 λ + 2 μ = 0 \\ μ = - 3 λ \\ \vec{b} = λ (\vec{a} - 3 \hat{j}) = λ (- \hat{i} - \hat{j} + \hat{k}) \Rightarrow | \vec{b} | = | \vec{a} | \Rightarrow λ = \pm \sqrt{2} \\ ∴ \sqrt{2} \vec{b} = (2, 2, - 2) \\ ∴ 2 \vec{a} + \sqrt{2} \vec{b} = (- 1, 8, 1) \\ ∴ \frac{(3 \bar{a} + \sqrt{2} \bar{b}) . \vec{c}}{| \bar{c} |} = 3 \sqrt{2} \end{array}$