If magnitude of sum of two unit vectors is `sqrt2` then find the magnitude of subtraction of these unit vectors.

Explanation

To solve the problem, we need to find the magnitude of the subtraction of two unit vectors \( \mathbf{A} \) and \( \mathbf{B} \) given that the magnitude of their sum is \( \sqrt{2} \).

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- We know that both \( \mathbf{A} \) and \( \mathbf{B} \) are unit vectors. This means:
\[ |\mathbf{A}| = 1 \quad \text{and} \quad |\mathbf{B}| = 1 \]
- The magnitude of the sum of these vectors is given as:
\[ |\mathbf{A} + \mathbf{B}| = \sqrt{2} \]

2. **Using the Formula for the Magnitude of the Sum**:
- The magnitude of the sum of two vectors can be expressed as:
\[ |\mathbf{A} + \mathbf{B}| = \sqrt{|\mathbf{A}|^2 + |\mathbf{B}|^2 + 2 |\mathbf{A}||\mathbf{B}| \cos \theta} \]
- Substituting the values we have:
\[ |\mathbf{A} + \mathbf{B}| = \sqrt{1^2 + 1^2 + 2 \cdot 1 \cdot 1 \cdot \cos \theta} \]
- This simplifies to:
\[ |\mathbf{A} + \mathbf{B}| = \sqrt{2 + 2 \cos \theta} \]

3. **Setting Up the Equation**:
- Since we know that \( |\mathbf{A} + \mathbf{B}| = \sqrt{2} \), we can set up the equation:
\[ \sqrt{2 + 2 \cos \theta} = \sqrt{2} \]

4. **Squaring Both Sides**:
- Squaring both sides of the equation gives:
\[ 2 + 2 \cos \theta = 2 \]

5. **Solving for \( \cos \theta \)**:
- Rearranging the equation:
\[ 2 \cos \theta = 0 \]
- This implies:
\[ \cos \theta = 0 \]
- Therefore, the angle \( \theta \) between the two vectors is:
\[ \theta = 90^\circ \]

6. **Finding the Magnitude of the Subtraction**:
- Now we need to find the magnitude of the subtraction \( |\mathbf{A} - \mathbf{B}| \):
\[ |\mathbf{A} - \mathbf{B}| = \sqrt{|\mathbf{A}|^2 + |\mathbf{B}|^2 - 2 |\mathbf{A}||\mathbf{B}| \cos \theta} \]
- Substituting the known values:
\[ |\mathbf{A} - \mathbf{B}| = \sqrt{1^2 + 1^2 - 2 \cdot 1 \cdot 1 \cdot \cos(90^\circ)} \]
- Since \( \cos(90^\circ) = 0 \):
\[ |\mathbf{A} - \mathbf{B}| = \sqrt{1 + 1 - 0} = \sqrt{2} \]

7. **Final Answer**:
- Therefore, the magnitude of the subtraction of the two unit vectors is:
\[ |\mathbf{A} - \mathbf{B}| = \sqrt{2} \]