If the magnitude of difference of two unit vectors is `sqrt(3)`, then the magnitude of sum of the two vectors is

Options

A. `1/2`unit
B. 1 unit
C. 2 unit
D. 3 unit

Explanation

To solve the problem, we need to find the magnitude of the sum of two unit vectors \( \mathbf{A} \) and \( \mathbf{B} \) given that the magnitude of their difference \( |\mathbf{A} - \mathbf{B}| = \sqrt{3} \).

### Step-by-Step Solution:

1. **Understanding the Vectors**:
Let \( \mathbf{A} \) and \( \mathbf{B} \) be two unit vectors. This means:
\[ |\mathbf{A}| = 1 \quad \text{and} \quad |\mathbf{B}| = 1 \]

2. **Using the Magnitude of the Difference**:
We are given that:
\[ |\mathbf{A} - \mathbf{B}| = \sqrt{3} \]
Squaring both sides gives:
\[ |\mathbf{A} - \mathbf{B}|^2 = 3 \]

3. **Expanding the Squared Magnitude**:
The squared magnitude can be expanded as follows:
\[ |\mathbf{A} - \mathbf{B}|^2 = |\mathbf{A}|^2 + |\mathbf{B}|^2 - 2 \mathbf{A} \cdot \mathbf{B} \]
Substituting the magnitudes:
\[ 1^2 + 1^2 - 2 \mathbf{A} \cdot \mathbf{B} = 3 \]
This simplifies to:
\[ 1 + 1 - 2 \mathbf{A} \cdot \mathbf{B} = 3 \]
\[ 2 - 2 \mathbf{A} \cdot \mathbf{B} = 3 \]

4. **Solving for the Dot Product**:
Rearranging the equation:
\[ -2 \mathbf{A} \cdot \mathbf{B} = 3 - 2 \]
\[ -2 \mathbf{A} \cdot \mathbf{B} = 1 \]
Dividing by -2:
\[ \mathbf{A} \cdot \mathbf{B} = -\frac{1}{2} \]

5. **Finding the Magnitude of the Sum**:
Now we need to find the magnitude of the sum \( |\mathbf{A} + \mathbf{B}| \):
\[ |\mathbf{A} + \mathbf{B}|^2 = |\mathbf{A}|^2 + |\mathbf{B}|^2 + 2 \mathbf{A} \cdot \mathbf{B} \]
Substituting the known values:
\[ |\mathbf{A} + \mathbf{B}|^2 = 1 + 1 + 2 \left(-\frac{1}{2}\right) \]
This simplifies to:
\[ |\mathbf{A} + \mathbf{B}|^2 = 2 - 1 = 1 \]

6. **Taking the Square Root**:
Finally, taking the square root gives:
\[ |\mathbf{A} + \mathbf{B}| = \sqrt{1} = 1 \]

### Conclusion:
The magnitude of the sum of the two unit vectors \( \mathbf{A} \) and \( \mathbf{B} \) is:
\[ \boxed{1} \]