If the magnitude of difference of two unit vectors is `sqrt3`, then the magnitude of sum of the two vectors is

Options

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

Answer

To solve the problem, we need to find the magnitude of the sum of two unit vectors \( \vec{a} \) and \( \vec{b} \) given that the magnitude of their difference is \( \sqrt{3} \).

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- We know that both \( \vec{a} \) and \( \vec{b} \) are unit vectors. Therefore, we have:
\[ |\vec{a}| = 1 \quad \text{and} \quad |\vec{b}| = 1 \]
- The magnitude of the difference of the two vectors is given as:
\[ |\vec{a} - \vec{b}| = \sqrt{3} \]

2. **Using the Magnitude of the Difference**:
- We can use the formula for the magnitude of the difference of two vectors:
\[ |\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2 \vec{a} \cdot \vec{b} \]
- Substituting the known magnitudes:
\[ (\sqrt{3})^2 = 1^2 + 1^2 - 2 \vec{a} \cdot \vec{b} \]
- This simplifies to:
\[ 3 = 1 + 1 - 2 \vec{a} \cdot \vec{b} \]
- Which further simplifies to:
\[ 3 = 2 - 2 \vec{a} \cdot \vec{b} \]

3. **Solving for \( \vec{a} \cdot \vec{b} \)**:
- Rearranging the equation gives:
\[ 2 \vec{a} \cdot \vec{b} = 2 - 3 \]
- Thus:
\[ 2 \vec{a} \cdot \vec{b} = -1 \]
- Dividing both sides by 2, we find:
\[ \vec{a} \cdot \vec{b} = -\frac{1}{2} \]

4. **Finding the Magnitude of the Sum**:
- Now we can find the magnitude of the sum of the two vectors using the formula:
\[ |\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2 \vec{a} \cdot \vec{b} \]
- Substituting the known values:
\[ |\vec{a} + \vec{b}|^2 = 1^2 + 1^2 + 2 \left(-\frac{1}{2}\right) \]
- This simplifies to:
\[ |\vec{a} + \vec{b}|^2 = 1 + 1 - 1 = 1 \]
- Taking the square root gives:
\[ |\vec{a} + \vec{b}| = \sqrt{1} = 1 \]

### Final Answer:
The magnitude of the sum of the two vectors is \( 1 \).

---