If the sum of two unit vectors is also a unit vector. Then magnituce of their difference and angle between the two given unit vectors is

Options

A. `sqrt(3),60^(@)`
B. `sqrt(3), 120^(@)`
C. `sqrt(2), 60^(@)`
D. `sqrt(2),120^(@)`

Answer

To solve the problem, we need to find the angle between two unit vectors A and B, given that their sum is also a unit vector. We also need to find the magnitude of their difference.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- Let A and B be two unit vectors. Therefore, |A| = 1 and |B| = 1.
- The sum of the two unit vectors is given as another unit vector C, i.e., |A + B| = 1.

2. **Using the Magnitude of the Sum of Vectors**:
- The magnitude of the sum of two vectors can be expressed as:
\[ |A + B|^2 = |A|^2 + |B|^2 + 2|A||B|\cos\theta \]
- Since A and B are unit vectors, we have:
\[ |A + B|^2 = 1^2 + 1^2 + 2(1)(1)\cos\theta \]
- This simplifies to:
\[ |A + B|^2 = 1 + 1 + 2\cos\theta = 2 + 2\cos\theta \]

3. **Setting Up the Equation**:
- Since |A + B| = 1, we can write:
\[ 1^2 = 2 + 2\cos\theta \]
- This leads to:
\[ 1 = 2 + 2\cos\theta \]

4. **Solving for Cosine**:
- Rearranging the equation gives:
\[ 2\cos\theta = 1 - 2 \]
\[ 2\cos\theta = -1 \]
\[ \cos\theta = -\frac{1}{2} \]

5. **Finding the Angle**:
- The angle θ for which cosθ = -1/2 is:
\[ \theta = 120^\circ \]

6. **Finding the Magnitude of the Difference**:
- The magnitude of the difference of the two vectors A and B is given by:
\[ |A - B|^2 = |A|^2 + |B|^2 - 2|A||B|\cos\theta \]
- Substituting the values, we get:
\[ |A - B|^2 = 1^2 + 1^2 - 2(1)(1)\cos(120^\circ) \]
- Since cos(120°) = -1/2, we have:
\[ |A - B|^2 = 1 + 1 - 2(-\frac{1}{2}) = 2 + 1 = 3 \]
- Therefore, the magnitude is:
\[ |A - B| = \sqrt{3} \]

### Final Answers:
- The angle between the two unit vectors A and B is **120 degrees**.
- The magnitude of their difference is **√3**.