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

<p>First, we write $\overrightarrow{b}$ as a linear combination of $\overrightarrow{a}$ and $\overrightarrow{j}$ since $\overrightarrow{b}$ is a rotation of $\overrightarrow{a}$ about the y-axis.</p> <p>$\vec{b}=\lambda \vec{a}+\mu \hat{j}=\lambda(-\hat{i}+2 \hat{j}+\hat{k})+\mu \hat{j}=-\lambda \hat{i}+(2 \lambda+\mu \hat{j})+\lambda \hat{k}$</p> <p>$\overrightarrow{b}$ is orthogonal to $\overrightarrow{a}$ due to the right angle rotation, so $\overrightarrow{b} \cdot \overrightarrow{a} = 0$. </p> <p>This implies :</p> <p>$\begin{aligned} & (-\hat{i}+2 \hat{j}+\hat{k}) \cdot(-\lambda \hat{i}+(2 \lambda+\mu) \hat{j}+\lambda \hat{k})=0 \\\\ & \lambda+2(2 \lambda+\mu)+\lambda=0 \Rightarrow 6 \lambda+2 \mu=0 \Rightarrow \mu+3 \lambda=0\end{aligned}$</p> <p>$\therefore$ $\vec{b}=\lambda \vec{a}-3 \lambda \hat{j}=\lambda(-\hat{i}+2 \hat{j}+\hat{k})-3 \lambda \hat{j}=\lambda(-\hat{i}-\hat{j}+\hat{k})$</p> <p>The magnitude of $\overrightarrow{b}$ is the same as the magnitude of $\overrightarrow{a}$ because a rotation doesn&#39;t change the magnitude of a vector. This gives us :</p> <p>$$ |\vec{b}|=\sqrt{3}|\lambda|=\sqrt{6}[\because|a|=\sqrt{6}] \Rightarrow|\lambda|=\sqrt{2} \Rightarrow \lambda \neq \sqrt{2} $$ <br/><br/>as for this value of $\lambda$ angle between $b$ and $y$-axis is not acute. <br/><br/>Therefore $\lambda=-\sqrt{2}$</p> <p>Thus, we have :</p> <p>$\overrightarrow{b} = -\sqrt{2}(-\hat{i} - \hat{j} + \hat{k}) = \sqrt{2}\hat{i} + \sqrt{2}\hat{j} - \sqrt{2}\hat{k}$</p> <p>Then, we find the vector $3\overrightarrow{a} + \sqrt{2}\overrightarrow{b}$ :</p> <p>$3\overrightarrow{a} + \sqrt{2}\overrightarrow{b} = 3(-\hat{i} + 2\hat{j} + \hat{k}) + \sqrt{2}(\sqrt{2}\hat{i} + \sqrt{2}\hat{j} - \sqrt{2}\hat{k}) $ <br/><br/>$= -3\hat{i} + 6\hat{j} + 3\hat{k} + 2\hat{i} + 2\hat{j} - 2\hat{k} = -\hat{i} + 8\hat{j} + \hat{k}$</p> <p>The projection of this vector onto $\overrightarrow{c}$ is given by the dot product divided by the magnitude of $\overrightarrow{c}$ :</p> <p>$\frac{(-\hat{i} + 8\hat{j} + \hat{k}) \cdot (5\hat{i} + 4\hat{j} + 3\hat{k})}{\sqrt{5^2 + 4^2 + 3^2}} = \frac{-5 + 32 + 3}{\sqrt{50}} = \frac{30}{5\sqrt{2}} = 3\sqrt{2}$</p> <p>So the projection of $3\overrightarrow{a} + \sqrt{2}\overrightarrow{b}$ onto $\overrightarrow{c}$ is $3\sqrt{2}$ which is option D.</p>