In sub-parts (i) and (ii) choose the correct option and in sub-parts (iii) to (v), answer the questions as instructed.
If a unit vector `hat(a)` makes angle `pi/3` with `hat(i), pi/3" with " hat(j)` and an acute angle `theta` with k, then find `theta`.

Answer

To solve the problem, we need to find the acute angle \( \theta \) that a unit vector \( \hat{a} \) makes with the \( z \)-axis (represented by \( \hat{k} \)). The vector makes angles of \( \frac{\pi}{3} \) (or 60 degrees) with both the \( x \)-axis (represented by \( \hat{i} \)) and the \( y \)-axis (represented by \( \hat{j} \)).

### Step-by-Step Solution:

1. **Understanding the Angles**:
- The unit vector \( \hat{a} \) makes an angle of \( \frac{\pi}{3} \) with \( \hat{i} \) and \( \hat{j} \).
- We denote the angles with respect to the axes as follows:
- \( \alpha = \frac{\pi}{3} \) (angle with \( \hat{i} \))
- \( \beta = \frac{\pi}{3} \) (angle with \( \hat{j} \))
- \( \gamma = \theta \) (angle with \( \hat{k} \))

2. **Direction Cosines**:
- The direction cosines \( l, m, n \) of the vector \( \hat{a} \) are given by:
- \( l = \cos(\alpha) = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \)
- \( m = \cos(\beta) = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \)
- \( n = \cos(\gamma) = \cos(\theta) \)

3. **Using the Relation of Direction Cosines**:
- The relation between the direction cosines is given by:
\[ l^2 + m^2 + n^2 = 1 \]
- Substituting the values we have:
\[ \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 + \cos^2(\theta) = 1 \]
- This simplifies to:
\[ \frac{1}{4} + \frac{1}{4} + \cos^2(\theta) = 1 \]
- Combining the fractions:
\[ \frac{2}{4} + \cos^2(\theta) = 1 \]
- Which simplifies to:
\[ \frac{1}{2} + \cos^2(\theta) = 1 \]

4. **Solving for \( \cos^2(\theta) \)**:
- Rearranging gives:
\[ \cos^2(\theta) = 1 - \frac{1}{2} = \frac{1}{2} \]

5. **Finding \( \theta \)**:
- Taking the square root:
\[ \cos(\theta) = \pm \frac{1}{\sqrt{2}} \]
- Since \( \theta \) is acute, we take the positive value:
\[ \cos(\theta) = \frac{1}{\sqrt{2}} \]
- Thus,
\[ \theta = \cos^{-1}\left(\frac{1}{\sqrt{2}}\right) \]
- We know that:
\[ \cos^{-1}\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4} \]

### Final Answer:
\[ \theta = \frac{\pi}{4} \text{ radians} \quad \text{or} \quad 45^\circ \]