A vector `vecr` is equally inclined with the vectors
`veca=cos thetahati+sin theta hatj, vecb=-sin theta hati+cos theta hatj` and `vecc=hatk`, then the angle between `vecr` and `veca` is

Options

A. `cos^(-1)((1)/(sqrt2))`
B. `cos^(-1)((1)/(3))`
C. `cos^(-1)((1)/(sqrt3))`
D. `cos^(-1).(1)/(2)`

Answer

To solve the problem, we need to find the angle between the vector \( \vec{R} \) and the vector \( \vec{A} \), given that \( \vec{R} \) is equally inclined to the vectors \( \vec{A} \), \( \vec{B} \), and \( \vec{C} \).

### Step-by-Step Solution:

1. **Define the vectors**:
\[ \vec{A} = \cos(\theta) \hat{i} + \sin(\theta) \hat{j} \]
\[ \vec{B} = -\sin(\theta) \hat{i} + \cos(\theta) \hat{j} \]
\[ \vec{C} = \hat{k} \]

2. **Assume a specific value for \( \theta \)**:
To simplify calculations, we can assume \( \theta = 0 \):
\[ \vec{A} = \cos(0) \hat{i} + \sin(0) \hat{j} = 1 \hat{i} + 0 \hat{j} = \hat{i} \]
\[ \vec{B} = -\sin(0) \hat{i} + \cos(0) \hat{j} = 0 \hat{i} + 1 \hat{j} = \hat{j} \]
\[ \vec{C} = \hat{k} \]

3. **Determine vector \( \vec{R} \)**:
Since \( \vec{R} \) is equally inclined to \( \vec{A} \), \( \vec{B} \), and \( \vec{C} \), we can express \( \vec{R} \) as:
\[ \vec{R} = R_1 \hat{i} + R_2 \hat{j} + R_3 \hat{k} \]
For equal inclination, we can assume \( R_1 = R_2 = R_3 = R \). Thus:
\[ \vec{R} = R(\hat{i} + \hat{j} + \hat{k}) \]

4. **Calculate the angle between \( \vec{R} \) and \( \vec{A} \)**:
The angle \( \phi \) between two vectors \( \vec{R} \) and \( \vec{A} \) can be found using the dot product:
\[ \vec{R} \cdot \vec{A} = |\vec{R}| |\vec{A}| \cos(\phi) \]

5. **Calculate the magnitudes**:
\[ |\vec{A}| = \sqrt{1^2 + 0^2} = 1 \]
\[ |\vec{R}| = \sqrt{R^2 + R^2 + R^2} = R\sqrt{3} \]

6. **Calculate the dot product**:
\[ \vec{R} \cdot \vec{A} = (R \hat{i} + R \hat{j} + R \hat{k}) \cdot \hat{i} = R \]

7. **Set up the equation**:
\[ R = |\vec{R}| |\vec{A}| \cos(\phi) \implies R = (R\sqrt{3})(1) \cos(\phi) \]
\[ \cos(\phi) = \frac{R}{R\sqrt{3}} = \frac{1}{\sqrt{3}} \]

8. **Find the angle**:
\[ \phi = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right) \]

### Final Answer:
The angle between \( \vec{R} \) and \( \vec{A} \) is \( \cos^{-1}\left(\frac{1}{\sqrt{3}}\right) \).