If `veca, vecb,vecc` are unit vectors such that `veca` is perpendicular to the plane of `vecb, vecc` and the angle between `vecb,vecc` is `pi/3`, then `|veca+vecb+vecc|=`

Options

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

Explanation

To solve the problem, we need to find the magnitude of the vector sum \( \vec{a} + \vec{b} + \vec{c} \) given the conditions about the unit vectors \( \vec{a}, \vec{b}, \) and \( \vec{c} \).

### Step-by-Step Solution:

1. **Understanding the Given Information:**
- \( \vec{a}, \vec{b}, \vec{c} \) are unit vectors, which means:
\[ |\vec{a}| = 1, \quad |\vec{b}| = 1, \quad |\vec{c}| = 1 \]
- \( \vec{a} \) is perpendicular to the plane formed by \( \vec{b} \) and \( \vec{c} \). Therefore:
\[ \vec{a} \cdot \vec{b} = 0 \quad \text{and} \quad \vec{a} \cdot \vec{c} = 0 \]
- The angle between \( \vec{b} \) and \( \vec{c} \) is \( \frac{\pi}{3} \).

2. **Finding the Dot Product of \( \vec{b} \) and \( \vec{c} \):**
- Using the formula for the dot product:
\[ \vec{b} \cdot \vec{c} = |\vec{b}| |\vec{c}| \cos\left(\frac{\pi}{3}\right) \]
- Since both \( \vec{b} \) and \( \vec{c} \) are unit vectors:
\[ \vec{b} \cdot \vec{c} = 1 \cdot 1 \cdot \frac{1}{2} = \frac{1}{2} \]

3. **Calculating the Magnitude of \( \vec{a} + \vec{b} + \vec{c} \):**
- We need to find \( |\vec{a} + \vec{b} + \vec{c}|^2 \):
\[ |\vec{a} + \vec{b} + \vec{c}|^2 = (\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c}) \]
- Expanding this using the distributive property:
\[ |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b}) + 2(\vec{b} \cdot \vec{c}) + 2(\vec{c} \cdot \vec{a}) \]
- Substituting the known values:
\[ = 1^2 + 1^2 + 1^2 + 2(0) + 2\left(\frac{1}{2}\right) + 2(0) \]
- This simplifies to:
\[ = 1 + 1 + 1 + 0 + 1 + 0 = 4 \]

4. **Finding the Magnitude:**
- Now, taking the square root to find \( |\vec{a} + \vec{b} + \vec{c}| \):
\[ |\vec{a} + \vec{b} + \vec{c}| = \sqrt{4} = 2 \]

### Final Answer:
\[ |\vec{a} + \vec{b} + \vec{c}| = 2 \]