If the sum of two unit vectors is a unit vector, then magnitude of difference is-

Options

A. `sqrt(2)`
B. `sqrt(3)`
C. `1//sqrt(2)`
D. `sqrt(5)`

Explanation

To solve the problem, we need to find the magnitude of the difference between two unit vectors \( \mathbf{n_1} \) and \( \mathbf{n_2} \) given that their sum is also a unit vector.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- Let \( \mathbf{n_1} \) and \( \mathbf{n_2} \) be two unit vectors.
- According to the problem, \( \mathbf{n_1} + \mathbf{n_2} = \mathbf{n} \) where \( \mathbf{n} \) is also a unit vector.

2. **Magnitude of the Sum**:
- Since \( \mathbf{n} \) is a unit vector, we have:
\[ |\mathbf{n}| = 1 \]
- The magnitude of the sum of two vectors can be expressed as:
\[ |\mathbf{n_1} + \mathbf{n_2}|^2 = |\mathbf{n}|^2 \]
- Expanding the left-hand side:
\[ |\mathbf{n_1} + \mathbf{n_2}|^2 = |\mathbf{n_1}|^2 + |\mathbf{n_2}|^2 + 2 \mathbf{n_1} \cdot \mathbf{n_2} \]
- Since \( \mathbf{n_1} \) and \( \mathbf{n_2} \) are unit vectors:
\[ |\mathbf{n_1}|^2 = 1 \quad \text{and} \quad |\mathbf{n_2}|^2 = 1 \]
- Therefore:
\[ |\mathbf{n_1} + \mathbf{n_2}|^2 = 1 + 1 + 2 \mathbf{n_1} \cdot \mathbf{n_2} = 1 \]
- Simplifying gives:
\[ 2 + 2 \mathbf{n_1} \cdot \mathbf{n_2} = 1 \]
- Rearranging leads to:
\[ 2 \mathbf{n_1} \cdot \mathbf{n_2} = 1 - 2 \quad \Rightarrow \quad \mathbf{n_1} \cdot \mathbf{n_2} = -\frac{1}{2} \]

3. **Finding the Angle Between the Vectors**:
- The dot product \( \mathbf{n_1} \cdot \mathbf{n_2} \) can also be expressed in terms of the angle \( \theta \) between them:
\[ \mathbf{n_1} \cdot \mathbf{n_2} = |\mathbf{n_1}| |\mathbf{n_2}| \cos(\theta) = 1 \cdot 1 \cdot \cos(\theta) = \cos(\theta) \]
- Thus:
\[ \cos(\theta) = -\frac{1}{2} \]
- This implies:
\[ \theta = 120^\circ \]

4. **Calculating the Magnitude of the Difference**:
- The magnitude of the difference of the two unit vectors is given by:
\[ |\mathbf{n_1} - \mathbf{n_2}|^2 = |\mathbf{n_1}|^2 + |\mathbf{n_2}|^2 - 2 \mathbf{n_1} \cdot \mathbf{n_2} \]
- Substituting the known values:
\[ |\mathbf{n_1} - \mathbf{n_2}|^2 = 1 + 1 - 2(-\frac{1}{2}) \]
- Simplifying this:
\[ |\mathbf{n_1} - \mathbf{n_2}|^2 = 2 + 1 = 3 \]
- Taking the square root gives:
\[ |\mathbf{n_1} - \mathbf{n_2}| = \sqrt{3} \]

### Final Answer:
The magnitude of the difference between the two unit vectors is \( \sqrt{3} \).