If a, b, c are three unit vectors such that `a+b+c=0`. Where 0 is null vector, then `a.b+a.c+c.a` is

Options

A. `-3`
B. `-2`
C. `-(3)/(2)`
D. 0

Explanation

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

### Step-by-Step Solution:

1. **Understanding the Given Condition**:
We know that \( a + b + c = 0 \). This implies that \( c = - (a + b) \).

2. **Taking the Magnitude**:
Since \( a, b, c \) are unit vectors, we have:
\[ |a| = |b| = |c| = 1 \]
Therefore, we can take the magnitude of both sides of the equation \( a + b + c = 0 \):
\[ |a + b + c| = |0| = 0 \]

3. **Expanding the Magnitude**:
Now, we square the magnitude:
\[ |a + b + c|^2 = 0 \]
Expanding this using the dot product:
\[ (a + b + c) \cdot (a + b + c) = a \cdot a + b \cdot b + c \cdot c + 2(a \cdot b + b \cdot c + c \cdot a) = 0 \]

4. **Substituting the Values**:
Since \( |a|^2 = 1 \), \( |b|^2 = 1 \), and \( |c|^2 = 1 \), we have:
\[ 1 + 1 + 1 + 2(a \cdot b + b \cdot c + c \cdot a) = 0 \]
Simplifying this gives:
\[ 3 + 2(a \cdot b + b \cdot c + c \cdot a) = 0 \]

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

6. **Final Result**:
Thus, the value of \( a \cdot b + a \cdot c + c \cdot a \) is:
\[ -\frac{3}{2} \]