JEE Mathematics 2023 Question with Solution

Given: $\vec{a} = -\hat{i} + 2\hat{j} + \hat{k}$ and $\vec{c} = 5\hat{i} + 4\hat{j} + 3\hat{k}$.

Find: The projection of $3\vec{a} + \sqrt{2}\vec{b}$ on $\vec{c}$, where $\vec{b}$ is obtained by rotating $\vec{a}$ through a right angle and it passes through the $y$-axis in its way.

Since the rotation is through a right angle, $\vec{b}$ is perpendicular to $\vec{a}$.

$$\vec{a} \cdot \vec{b} = 0$$

the solution indicates the valid direction is chosen using the condition that $\vec{b}$ makes an acute angle with the $y$-axis. Using that accepted orientation, the working concludes with:

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

Hence the scalar projection of $3\vec{a} + \sqrt{2}\vec{b}$ on $\vec{c}$ is

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

Therefore, the correct option is A.