If `veca,vecb and vecc` are unit vectors such that `veca+vecb+vecc=vec0` then the value of `veca.vecb+vecb.vecc+vecc.veca` is a) 1 b) 0 c) 3 d) `-3/2`

Options

A. 1
B. 0
C. 3
D. `-3/2`

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 \( \vec{a} + \vec{b} + \vec{c} = \vec{0} \) and that \( \vec{a}, \vec{b}, \vec{c} \) are unit vectors.

### Step-by-Step Solution:

1. **Understanding the Given Condition**:
Since \( \vec{a} + \vec{b} + \vec{c} = \vec{0} \), we can rearrange this to express one vector in terms of the others:
\[ \vec{c} = -(\vec{a} + \vec{b}) \]

2. **Taking the Dot Product**:
We will take the dot product of both sides of the equation \( \vec{a} + \vec{b} + \vec{c} = \vec{0} \) with itself:
\[ (\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c}) = 0 \]

3. **Expanding the Dot Product**:
Expanding the left-hand side using the distributive property of the dot product gives:
\[ \vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{b} + \vec{c} \cdot \vec{c} + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0 \]

4. **Substituting the Magnitudes**:
Since \( \vec{a}, \vec{b}, \vec{c} \) are unit vectors, we know that:
\[ \vec{a} \cdot \vec{a} = 1, \quad \vec{b} \cdot \vec{b} = 1, \quad \vec{c} \cdot \vec{c} = 1 \]
Therefore, substituting these values gives:
\[ 1 + 1 + 1 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0 \]
Simplifying this results in:
\[ 3 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0 \]

5. **Isolating the Dot Product Sum**:
Now, we can isolate the sum \( \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} \):
\[ 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = -3 \]
Dividing both sides by 2 gives:
\[ \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -\frac{3}{2} \]

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