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Step-by-Step Solution
Step 1: Understand the Problem
We have an initial vector
$ \overrightarrow{a} = -\hat{i} + 2 \hat{j} + \hat{k} $.
It is rotated in three-dimensional space through a right angle about the y-axis, leading to a new vector
$ \overrightarrow{b}. $
We need to find the projection of
$ 3 \overrightarrow{a} + \sqrt{2} \,\overrightarrow{b} $
onto the vector
$ \overrightarrow{c} = 5 \hat{i} + 4 \hat{j} + 3 \hat{k}. $
Finally, we must evaluate this projection and match it with one of the given options.
Step 2: Express $\overrightarrow{b}$ in Terms of $\overrightarrow{a}$
Since $ \overrightarrow{b} $ is obtained by rotating $ \overrightarrow{a} $ by a right angle (90°) in a plane passing through the y-axis, $ \overrightarrow{b} $ must be orthogonal to $ \overrightarrow{a}. $ Thus, we set:
$ \overrightarrow{b} \cdot \overrightarrow{a} = 0. $
Let us write
$ \overrightarrow{b} = \lambda \, \overrightarrow{a} + \mu \, \hat{j}. $
On substituting
$ \overrightarrow{a} = -\hat{i} + 2 \hat{j} + \hat{k} ,$
we get
$ \overrightarrow{b} = \lambda(-\hat{i} + 2\hat{j} + \hat{k}) + \mu \hat{j}.$
Step 3: Impose Orthogonality Condition
Since $ \overrightarrow{b} \cdot \overrightarrow{a} = 0:$
$$
(-\hat{i} + 2\hat{j} + \hat{k}) \,\cdot\, \bigl[-\lambda \hat{i} + (2\lambda + \mu) \hat{j} + \lambda \hat{k}\bigr] = 0.
$$
This dot product simplifies to:
$$
\lambda + 2(2\lambda + \mu) + \lambda = 0
\quad \Longrightarrow \quad
6\lambda + 2\mu = 0
\quad \Longrightarrow \quad
\mu + 3\lambda = 0.
$$
Thus,
$$
\mu = -3\lambda.
$$
Hence,
$$
\overrightarrow{b} = \lambda\, (-\hat{i} + 2\hat{j} + \hat{k}) - 3\lambda\, \hat{j}
= \lambda (-\hat{i} - \hat{j} + \hat{k}).
$$
Step 4: Apply Equal Magnitude Condition
A pure rotation does not change the magnitude of a vector. The magnitude of
$ \overrightarrow{a} $
is
$ |\overrightarrow{a}| = \sqrt{(-1)^2 + 2^2 + 1^2} = \sqrt{6}. $
Since
$ \overrightarrow{b} $
is obtained by a 90° rotation of the same magnitude:
$$
|\overrightarrow{b}| = |\lambda| \, \sqrt{(-1)^2 + (-1)^2 + 1^2} = |\lambda| \sqrt{3} = \sqrt{6}.
$$
So,
$$
|\lambda| \sqrt{3} = \sqrt{6}
\quad \Longrightarrow \quad
|\lambda| = \sqrt{2}.
$$
Given additional constraints (the direction with respect to the y-axis), we take
$$
\lambda = -\sqrt{2}.
$$
Therefore,
$$
\overrightarrow{b}
= -\sqrt{2}\bigl(-\hat{i} - \hat{j} + \hat{k}\bigr)
= \sqrt{2}\,\hat{i} + \sqrt{2}\,\hat{j} - \sqrt{2}\,\hat{k}.
$$
Step 5: Compute $3\overrightarrow{a} + \sqrt{2}\,\overrightarrow{b}$
Now we add:
$$
3\overrightarrow{a}
= 3\bigl(-\hat{i} + 2\hat{j} + \hat{k}\bigr)
= -3\hat{i} + 6\hat{j} + 3\hat{k},
$$
and
$$
\sqrt{2}\, \overrightarrow{b}
= \sqrt{2}\bigl(\sqrt{2}\,\hat{i} + \sqrt{2}\,\hat{j} - \sqrt{2}\,\hat{k}\bigr)
= 2\hat{i} + 2\hat{j} - 2\hat{k}.
$$
Thus,
$$
3\overrightarrow{a} + \sqrt{2}\,\overrightarrow{b}
= (-3\hat{i} + 6\hat{j} + 3\hat{k}) + (2\hat{i} + 2\hat{j} - 2\hat{k})
= -\hat{i} + 8\hat{j} + \hat{k}.
$$
Step 6: Project onto $\overrightarrow{c}$
The projection of a vector $ \overrightarrow{v} $ onto another vector $ \overrightarrow{w} $ is given by
$$
\text{Proj}_{\overrightarrow{w}}(\overrightarrow{v})
= \frac{\overrightarrow{v} \cdot \overrightarrow{w}}{|\overrightarrow{w}|}.
$$
Here,
$ \overrightarrow{v} = 3\overrightarrow{a} + \sqrt{2}\,\overrightarrow{b} = -\hat{i} + 8\hat{j} + \hat{k}, $
and
$ \overrightarrow{w} = \overrightarrow{c} = 5\hat{i} + 4\hat{j} + 3\hat{k}. $
First, compute the dot product:
$$
(-\hat{i} + 8\hat{j} + \hat{k}) \,\cdot\, (5\hat{i} + 4\hat{j} + 3\hat{k})
= (-1)\times 5 + 8\times 4 + 1\times 3
= -5 + 32 + 3
= 30.
$$
The magnitude of $ \overrightarrow{c} $ is
$$
|\overrightarrow{c}| = \sqrt{5^2 + 4^2 + 3^2}
= \sqrt{25 + 16 + 9}
= \sqrt{50}
= 5\sqrt{2}.
$$
Hence, the projection is:
$$
\frac{30}{5\sqrt{2}} = \frac{30}{5\sqrt{2}} = 3\sqrt{2}.
$$
Step 7: Final Answer
The projection of $ 3\overrightarrow{a} + \sqrt{2}\,\overrightarrow{b} $ on $ \overrightarrow{c} $ is
$ 3\sqrt{2}, $
which corresponds to the correct option.
