If `vecm_a,vecm_band vecm_c` are 3 units vectors such that `vecm_a.vecm_b= vecm_a.vecm_c=0` and the angle between `vecm_b.vecm_c " is " pi/3,` then then value of `|vecm_axxvecm_b -vecm_axxvecm_c |` is equal to

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we need to find the value of \( |\vec{m_a} \times \vec{m_b} - \vec{m_a} \times \vec{m_c}| \) given the conditions about the vectors \( \vec{m_a}, \vec{m_b}, \) and \( \vec{m_c} \).

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- \( \vec{m_a}, \vec{m_b}, \vec{m_c} \) are unit vectors, meaning \( |\vec{m_a}| = |\vec{m_b}| = |\vec{m_c}| = 1 \).
- The dot products \( \vec{m_a} \cdot \vec{m_b} = 0 \) and \( \vec{m_a} \cdot \vec{m_c} = 0 \) imply that \( \vec{m_a} \) is perpendicular to both \( \vec{m_b} \) and \( \vec{m_c} \).
- The angle between \( \vec{m_b} \) and \( \vec{m_c} \) is \( \frac{\pi}{3} \).

2. **Using the Cross Product**:
- We want to find \( |\vec{m_a} \times \vec{m_b} - \vec{m_a} \times \vec{m_c}| \).
- We can factor out \( \vec{m_a} \):
\[ \vec{m_a} \times \vec{m_b} - \vec{m_a} \times \vec{m_c} = \vec{m_a} \times (\vec{m_b} - \vec{m_c}). \]

3. **Magnitude of the Cross Product**:
- The magnitude of the cross product can be expressed as:
\[ |\vec{m_a} \times (\vec{m_b} - \vec{m_c})| = |\vec{m_a}| \cdot |\vec{m_b} - \vec{m_c}| \cdot \sin(\theta), \]
where \( \theta \) is the angle between \( \vec{m_a} \) and \( \vec{m_b} - \vec{m_c} \).

4. **Finding the Magnitude of \( \vec{m_b} - \vec{m_c} \)**:
- We can use the formula for the magnitude of the difference of two vectors:
\[ |\vec{m_b} - \vec{m_c}|^2 = |\vec{m_b}|^2 + |\vec{m_c}|^2 - 2(\vec{m_b} \cdot \vec{m_c}). \]
- Since both \( \vec{m_b} \) and \( \vec{m_c} \) are unit vectors, \( |\vec{m_b}|^2 = 1 \) and \( |\vec{m_c}|^2 = 1 \).
- The dot product \( \vec{m_b} \cdot \vec{m_c} = |\vec{m_b}| |\vec{m_c}| \cos(\frac{\pi}{3}) = 1 \cdot 1 \cdot \frac{1}{2} = \frac{1}{2} \).
- Therefore:
\[ |\vec{m_b} - \vec{m_c}|^2 = 1 + 1 - 2 \cdot \frac{1}{2} = 2 - 1 = 1. \]
- Thus, \( |\vec{m_b} - \vec{m_c}| = 1 \).

5. **Finding the Angle \( \theta \)**:
- Since \( \vec{m_a} \) is perpendicular to both \( \vec{m_b} \) and \( \vec{m_c} \), the angle \( \theta \) between \( \vec{m_a} \) and \( \vec{m_b} - \vec{m_c} \) is \( 90^\circ \) or \( \frac{\pi}{2} \).
- Therefore, \( \sin(\theta) = \sin\left(\frac{\pi}{2}\right) = 1 \).

6. **Final Calculation**:
- Now substituting back into the magnitude:
\[ |\vec{m_a} \times (\vec{m_b} - \vec{m_c})| = 1 \cdot 1 \cdot 1 = 1. \]

### Conclusion:
The value of \( |\vec{m_a} \times \vec{m_b} - \vec{m_a} \times \vec{m_c}| \) is \( 1 \).