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

Options

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

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 their difference \( \vec{a} - \vec{b} \) is also a unit vector.

### Step-by-step Solution:

1. **Define the unit vectors**: Let \( \vec{a} \) and \( \vec{b} \) be two unit vectors. This means:
\[ |\vec{a}| = 1 \quad \text{and} \quad |\vec{b}| = 1 \]

2. **Express the difference**: We know that the difference of the two vectors is also a unit vector:
\[ |\vec{a} - \vec{b}| = 1 \]

3. **Square the difference**: Squaring both sides gives:
\[ |\vec{a} - \vec{b}|^2 = 1^2 \]
Using the property of magnitudes, we have:
\[ |\vec{a}|^2 + |\vec{b}|^2 - 2 \vec{a} \cdot \vec{b} = 1 \]
Substituting \( |\vec{a}|^2 = 1 \) and \( |\vec{b}|^2 = 1 \):
\[ 1 + 1 - 2 \vec{a} \cdot \vec{b} = 1 \]

4. **Simplify the equation**: This simplifies to:
\[ 2 - 2 \vec{a} \cdot \vec{b} = 1 \]
Rearranging gives:
\[ 2 \vec{a} \cdot \vec{b} = 1 \quad \Rightarrow \quad \vec{a} \cdot \vec{b} = \frac{1}{2} \]

5. **Find the magnitude of the sum**: Now, we need to find the magnitude of the sum \( |\vec{a} + \vec{b}| \):
\[ |\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2 \vec{a} \cdot \vec{b} \]
Substituting the known values:
\[ |\vec{a} + \vec{b}|^2 = 1 + 1 + 2 \left(\frac{1}{2}\right) \]
This simplifies to:
\[ |\vec{a} + \vec{b}|^2 = 2 + 1 = 3 \]

6. **Take the square root**: Therefore, the magnitude of the sum is:
\[ |\vec{a} + \vec{b}| = \sqrt{3} \]

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