If `vec(a) , vec(b), vec(c)` are unit vectors such that `vec(a)` is perpendicular to the plane of `vec(b), vec(c)`, and the angle between `vec(b) and vec(c)` is `(pi)/(3)`
Then, what is `|vec(a)+vec(b)+vec(c)|`?

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 vectors. Here’s a step-by-step solution:

### Step 1: Understand the properties of the vectors
We know that:
- \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) are unit vectors.
- \(\vec{a}\) is perpendicular to both \(\vec{b}\) and \(\vec{c}\).
- The angle between \(\vec{b}\) and \(\vec{c}\) is \(\frac{\pi}{3}\).

### Step 2: Use the dot product
Since \(\vec{a}\) is perpendicular to \(\vec{b}\) and \(\vec{c}\), we have:
\[ \vec{a} \cdot \vec{b} = 0 \quad \text{and} \quad \vec{a} \cdot \vec{c} = 0 \]

The dot product of \(\vec{b}\) and \(\vec{c}\) can be calculated using the cosine of the angle between them:
\[ \vec{b} \cdot \vec{c} = |\vec{b}| |\vec{c}| \cos\left(\frac{\pi}{3}\right) \]
Since both \(\vec{b}\) and \(\vec{c}\) are unit vectors, their magnitudes are 1:
\[ \vec{b} \cdot \vec{c} = 1 \cdot 1 \cdot \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \]

### Step 3: Calculate the magnitude of \(\vec{a} + \vec{b} + \vec{c}\)
To find the magnitude \(|\vec{a} + \vec{b} + \vec{c}|\), we first compute the square of the magnitude:
\[ |\vec{a} + \vec{b} + \vec{c}|^2 = (\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c}) \]
Expanding this using the distributive property:
\[ |\vec{a} + \vec{b} + \vec{c}|^2 = \vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{b} + \vec{c} \cdot \vec{c} + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \]
Since \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) are unit vectors:
\[ |\vec{a}|^2 = 1, \quad |\vec{b}|^2 = 1, \quad |\vec{c}|^2 = 1 \]
Thus:
\[ |\vec{a} + \vec{b} + \vec{c}|^2 = 1 + 1 + 1 + 2(0 + \frac{1}{2} + 0) \]
This simplifies to:
\[ |\vec{a} + \vec{b} + \vec{c}|^2 = 3 + 2 \cdot \frac{1}{2} = 3 + 1 = 4 \]

### Step 4: Take the square root
Finally, we take the square root to find the magnitude:
\[ |\vec{a} + \vec{b} + \vec{c}| = \sqrt{4} = 2 \]

### Final Answer
Thus, the magnitude of \(\vec{a} + \vec{b} + \vec{c}\) is:
\[ \boxed{2} \]