Let `bara` be a unit vector perpendicular to unit vectors `barb and barc` inclined at an angle a. then `barb times barc` is

Options

A. `(cosa)bara`
B. `(coseca)bara`
C. `(sina)bara`
D. None of these

Answer

To solve the problem, we need to find the value of the cross product of two unit vectors \( \mathbf{b} \) and \( \mathbf{c} \) given that they are inclined at an angle \( a \) and that there is a unit vector \( \mathbf{a} \) which is perpendicular to both \( \mathbf{b} \) and \( \mathbf{c} \).

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- We have two unit vectors \( \mathbf{b} \) and \( \mathbf{c} \).
- The angle between these two vectors is \( a \).
- There is a unit vector \( \mathbf{a} \) which is perpendicular to both \( \mathbf{b} \) and \( \mathbf{c} \).

2. **Using the Cross Product Formula**:
- The magnitude of the cross product \( \mathbf{b} \times \mathbf{c} \) can be calculated using the formula:
\[ |\mathbf{b} \times \mathbf{c}| = |\mathbf{b}| |\mathbf{c}| \sin(a) \]
- Since both \( \mathbf{b} \) and \( \mathbf{c} \) are unit vectors, their magnitudes are 1:
\[ |\mathbf{b}| = 1 \quad \text{and} \quad |\mathbf{c}| = 1 \]
- Therefore, the formula simplifies to:
\[ |\mathbf{b} \times \mathbf{c}| = 1 \cdot 1 \cdot \sin(a) = \sin(a) \]

3. **Direction of the Cross Product**:
- The direction of \( \mathbf{b} \times \mathbf{c} \) is given by the right-hand rule and is along the direction of the unit vector \( \mathbf{a} \) (since \( \mathbf{a} \) is perpendicular to both \( \mathbf{b} \) and \( \mathbf{c} \)).
- Thus, we can express the cross product as:
\[ \mathbf{b} \times \mathbf{c} = |\mathbf{b} \times \mathbf{c}| \cdot \hat{\mathbf{n}} = \sin(a) \cdot \hat{\mathbf{a}} \]
- Here, \( \hat{\mathbf{a}} \) is the unit vector in the direction of \( \mathbf{a} \).

4. **Final Expression**:
- Therefore, we conclude:
\[ \mathbf{b} \times \mathbf{c} = \sin(a) \cdot \hat{\mathbf{a}} \]

### Conclusion:
The value of \( \mathbf{b} \times \mathbf{c} \) is \( \sin(a) \cdot \hat{\mathbf{a}} \).