If vector A and B have an angle `theta` between them, then value of `|hatA -hatB|` will be ,

Options

A. `2 cos""(theta)/(2)`
B. `2 tan""(theta)/(2)`
C. `2 Sin""(theta)/(2)`
D. None of these

Explanation

To solve the problem of finding the value of \(|\hat{A} - \hat{B}|\) where \(\hat{A}\) and \(\hat{B}\) are unit vectors with an angle \(\theta\) between them, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Unit Vectors**:
Since \(\hat{A}\) and \(\hat{B}\) are unit vectors, we have:
\[ |\hat{A}| = 1 \quad \text{and} \quad |\hat{B}| = 1 \]

2. **Use the Formula for the Magnitude of the Difference of Two Vectors**:
The magnitude of the difference of two vectors can be expressed using the formula:
\[ |\hat{A} - \hat{B}| = \sqrt{|\hat{A}|^2 + |\hat{B}|^2 - 2 |\hat{A}| |\hat{B}| \cos(\theta)} \]

3. **Substitute the Values**:
Since both \(\hat{A}\) and \(\hat{B}\) are unit vectors, we substitute \( |\hat{A}| = 1 \) and \( |\hat{B}| = 1 \):
\[ |\hat{A} - \hat{B}| = \sqrt{1^2 + 1^2 - 2 \cdot 1 \cdot 1 \cdot \cos(\theta)} \]

4. **Simplify the Expression**:
This simplifies to:
\[ |\hat{A} - \hat{B}| = \sqrt{1 + 1 - 2\cos(\theta)} = \sqrt{2 - 2\cos(\theta)} \]

5. **Factor Out the Common Term**:
We can factor out the 2 from the square root:
\[ |\hat{A} - \hat{B}| = \sqrt{2(1 - \cos(\theta))} \]

6. **Use the Identity for Sine**:
We know from trigonometric identities that:
\[ 1 - \cos(\theta) = 2\sin^2\left(\frac{\theta}{2}\right) \]
Substituting this into our expression gives:
\[ |\hat{A} - \hat{B}| = \sqrt{2 \cdot 2\sin^2\left(\frac{\theta}{2}\right)} = \sqrt{4\sin^2\left(\frac{\theta}{2}\right)} \]

7. **Final Simplification**:
Taking the square root results in:
\[ |\hat{A} - \hat{B}| = 2\sin\left(\frac{\theta}{2}\right) \]

### Conclusion:
Thus, the final answer is:
\[ |\hat{A} - \hat{B}| = 2\sin\left(\frac{\theta}{2}\right) \]