If `vec(a), vec(b), vec(c)` are three unit vectors such that `vec(a)+ vec(b)+vec(c)= 0`, then `vec(a).vec(b) + vec(b).vec(c) + vec(c).vec(a)` is equal to

Explanation

To solve the problem, we start with the information given:

1. **Given**: Three unit vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) such that:
\[ \vec{a} + \vec{b} + \vec{c} = 0 \]

2. **Objective**: Find the value of:
\[ \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} \]

### Step-by-Step Solution:

**Step 1**: Start from the equation \(\vec{a} + \vec{b} + \vec{c} = 0\).

Since \(\vec{c} = -(\vec{a} + \vec{b})\), we can substitute \(\vec{c}\) in our dot product expression.

**Step 2**: Take the dot product of both sides of the equation \(\vec{a} + \vec{b} + \vec{c} = 0\) with itself:
\[ (\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c}) = 0 \]

**Step 3**: Expand the left-hand side:
\[ \vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{b} + \vec{c} \cdot \vec{c} + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0 \]

**Step 4**: Since \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) are unit vectors, we know:
\[ \vec{a} \cdot \vec{a} = 1, \quad \vec{b} \cdot \vec{b} = 1, \quad \vec{c} \cdot \vec{c} = 1 \]

Thus, we can substitute these values:
\[ 1 + 1 + 1 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0 \]

**Step 5**: Simplify the equation:
\[ 3 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0 \]

**Step 6**: Rearranging gives:
\[ 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = -3 \]

**Step 7**: Divide by 2 to isolate the dot product sum:
\[ \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -\frac{3}{2} \]

### Final Answer:
\[ \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -1.5 \]