Let `vec(a), vec(b), vec(c)` are three unit vector such that `vec(a)+vec(b)+vec(c)` os also a unit vector. If pairwise angles between `vec(a), vec(b), vec(c)` are `theta_(1), theta_(2)` and `theta_(3)` respectively then `cos theta_(1)+cos theta_(2)+ cos theta_(3)` equals

Options

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

Explanation

To solve the problem, we need to find the sum of the cosines of the angles between three unit vectors \( \vec{a}, \vec{b}, \vec{c} \) given that their sum is also a unit vector. Let's break this down step by step.

### Step 1: Understand the Given Information
We know:
- \( |\vec{a}| = 1 \)
- \( |\vec{b}| = 1 \)
- \( |\vec{c}| = 1 \)
- \( |\vec{a} + \vec{b} + \vec{c}| = 1 \)

### Step 2: Use the Magnitude Formula
The magnitude of the sum of the vectors can be expressed using the formula:
\[ |\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \]
Substituting the values we know:
\[ 1^2 = 1 + 1 + 1 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \]
This simplifies to:
\[ 1 = 3 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \]

### Step 3: Rearranging the Equation
Rearranging the equation gives us:
\[ 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 1 - 3 \]
\[ 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = -2 \]
Dividing both sides by 2:
\[ \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -1 \]

### Step 4: Relate Dot Products to Cosines
Since \( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta_1 \) and similarly for the other pairs, we have:
\[ \cos \theta_1 + \cos \theta_2 + \cos \theta_3 = -1 \]

### Conclusion
Thus, the final result is:
\[ \cos \theta_1 + \cos \theta_2 + \cos \theta_3 = -1 \]