Let ` veca ,vecb , vecc` be three unit vectors such that angle between ` veca and vecb is alpha , vecb and vecc " is " beta and vecc and veca " is " gamma. " if " | veca. + vecb + vecc| =2` , then ` cos alpha + cos beta + cos gamma ` =

Options

A. 1
B. ` -1/ 2`
C. ` 3/2`
D. ` 1/2`

Explanation

To solve the problem, we need to find the value of \( \cos \alpha + \cos \beta + \cos \gamma \) given that \( |\vec{a} + \vec{b} + \vec{c}| = 2 \) and that \( \vec{a}, \vec{b}, \vec{c} \) are unit vectors with angles \( \alpha, \beta, \gamma \) between them.

### Step-by-Step Solution:

1. **Understanding the Magnitude of the Sum of Vectors**:
We know that \( |\vec{a}| = |\vec{b}| = |\vec{c}| = 1 \) (since they are unit vectors). We can express the magnitude of the sum of the vectors:
\[ |\vec{a} + \vec{b} + \vec{c}|^2 = (\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c}) \]

2. **Expanding the Dot Product**:
Expanding the dot product gives us:
\[ |\vec{a} + \vec{b} + \vec{c}|^2 = \vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{b} + \vec{c} \cdot \vec{c} + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \]
Since \( |\vec{a}|^2 = |\vec{b}|^2 = |\vec{c}|^2 = 1 \), we have:
\[ |\vec{a} + \vec{b} + \vec{c}|^2 = 1 + 1 + 1 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 3 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \]

3. **Setting Up the Equation**:
We know from the problem statement that:
\[ |\vec{a} + \vec{b} + \vec{c}| = 2 \]
Therefore:
\[ |\vec{a} + \vec{b} + \vec{c}|^2 = 4 \]
Equating the two expressions gives us:
\[ 3 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 4 \]

4. **Solving for the Dot Products**:
Rearranging the equation:
\[ 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 4 - 3 = 1 \]
Dividing by 2:
\[ \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = \frac{1}{2} \]

5. **Substituting the Dot Products**:
Recall that:
\[ \vec{a} \cdot \vec{b} = \cos \alpha, \quad \vec{b} \cdot \vec{c} = \cos \beta, \quad \vec{c} \cdot \vec{a} = \cos \gamma \]
Therefore:
\[ \cos \alpha + \cos \beta + \cos \gamma = \frac{1}{2} \]

### Final Answer:
Thus, the value of \( \cos \alpha + \cos \beta + \cos \gamma \) is:
\[ \boxed{\frac{1}{2}} \]