A vector $\vec a$ of magnitude $3\sqrt{2}$ making an angle of $\frac{\pi}{3}$ with $\hat i$, $\frac{\pi}{4}$ with $\hat j$ and an actue angle θ with $\hat k$, is

$3\sqrt{2}\left(\frac{1}{2}\hat i,\frac{1}{\sqrt{2}}\hat j,\frac{1}{2}\hat k\right)$

The correct answer is Option (2) → $3\sqrt{2}\left(\frac{1}{2}\hat i,\frac{1}{\sqrt{2}}\hat j,\frac{1}{2}\hat k\right)$

Let $\vec{a} = a_x \hat{i} + a_y \hat{j} + a_z \hat{k}$

Given: $|\vec{a}| = 3\sqrt{2}$

and $\angle(\vec{a}, \hat{i}) = \frac{\pi}{3}$, $\angle(\vec{a}, \hat{j}) = \frac{\pi}{4}$, $\angle(\vec{a}, \hat{k}) = \theta$ (acute)

Using direction cosines:

$\cos^2\frac{\pi}{3} + \cos^2\frac{\pi}{4} + \cos^2\theta = 1$

⇒ $\left(\frac{1}{2}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 + \cos^2\theta = 1$

⇒ $\frac{1}{4} + \frac{1}{2} + \cos^2\theta = 1$

⇒ $\cos^2\theta = \frac{1}{4} \Rightarrow \cos\theta = \frac{1}{2}$ (since, θ acute ⇒ θ = π/3)

Hence, direction cosines:

$(\cos\alpha, \cos\beta, \cos\gamma) = \left(\frac{1}{2}, \frac{1}{\sqrt{2}}, \frac{1}{2}\right)$

Thus, $\vec{a} = 3\sqrt{2}\left(\frac{1}{2}\hat{i} + \frac{1}{\sqrt{2}}\hat{j} + \frac{1}{2}\hat{k}\right)$

$\vec{a} = \frac{3\sqrt{2}}{2}\hat{i} + 3\hat{j} + \frac{3\sqrt{2}}{2}\hat{k}$

Final Answer:

$\vec{a} = \frac{3\sqrt{2}}{2}\hat{i} + 3\hat{j} + \frac{3\sqrt{2}}{2}\hat{k}$