If `vec(a) , vec( b)` and `vec( c )` are unit vectors such that` vec(a) + vec(b) + vec( c) = 0`, then the values of `vec(a). vec(b)+ vec(b) . vec( c )+ vec( c) .vec(a)` is

Options

A. 1
B. `(3)/(2)`
C. `- (3)/(2)`
D. none of these

Answer

To solve the problem, we need to find the value of \( \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} \) given that \( \vec{a} + \vec{b} + \vec{c} = 0 \) and that \( \vec{a}, \vec{b}, \vec{c} \) are unit vectors.

### Step-by-step Solution:

1. **Start with the given equation:**
\[ \vec{a} + \vec{b} + \vec{c} = 0 \]
This implies that \( \vec{c} = -(\vec{a} + \vec{b}) \).

2. **Square both sides:**
\[ (\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c}) = 0 \]
Expanding this using the dot product:
\[ \vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{b} + \vec{c} \cdot \vec{c} + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0 \]

3. **Substitute the values of the dot products:**
Since \( \vec{a}, \vec{b}, \vec{c} \) are unit vectors:
\[ \vec{a} \cdot \vec{a} = 1, \quad \vec{b} \cdot \vec{b} = 1, \quad \vec{c} \cdot \vec{c} = 1 \]
Thus, we have:
\[ 1 + 1 + 1 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0 \]
Simplifying this gives:
\[ 3 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0 \]

4. **Rearranging the equation:**
\[ 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = -3 \]
Dividing both sides by 2:
\[ \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -\frac{3}{2} \]

### Final Answer:
Thus, the value of \( \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} \) is:
\[ \boxed{-\frac{3}{2}} \]