If $\mathbf{a, b}$ and $\mathbf{c}$ are unit vectors such that $\mathbf{a + b + c = 0}$, then the value of $\mathbf{a \cdot b + b \cdot c + c \cdot a}$ is

$-\frac{3}{2}$

The correct answer is Option (3) → $-\frac{3}{2}$ ##

We have, $\mathbf{a + b + c = 0}$

and $a^2 = 1, b^2 = 1, c^2 = 1$

On multiplying by $(\mathbf{a + b + c})$ both sides in Eq. (i), we get

$∵(\mathbf{a + b + c}) \cdot (\mathbf{a + b + c}) = 0$

$\Rightarrow \mathbf{a \cdot a + a \cdot b + a \cdot c + b \cdot a + b \cdot b + b \cdot c + c \cdot a + c \cdot b + c \cdot c = 0}$

$\Rightarrow a^2 + b^2 + c^2 + 2(\mathbf{a \cdot b + b \cdot c + c \cdot a}) = 0$

$[∵\mathbf{a \cdot b = b \cdot a, b \cdot c = c \cdot b} \text{ and } \mathbf{c \cdot a = a \cdot c}]$

$\Rightarrow 1 + 1 + 1 + 2(\mathbf{a \cdot b + b \cdot c + c \cdot a}) = 0 \quad [∵a^2 = b^2 = c^2 = 1]$

$\Rightarrow \mathbf{a \cdot b + b \cdot c + c \cdot a} = -\frac{3}{2}$