Vector Algebra — Class 12 Maths Solution

We have, $|\vec a| = |\vec b| = |\vec c| = 1$
…(i)
As we know, $\vec a + \vec b + \vec c = 0$
…(ii)
Squaring (ii),

we get ${(\vec a + \vec b + \vec c)^2} = 0$
$\Rightarrow |\vec a{|^2} + |\vec b{|^2} + |\vec c{|^2} + 2(\vec a \cdot \vec b + \vec b \cdot \vec c + \vec c \cdot \vec a) = 0$

$\Rightarrow {(1)^2} + {(1)^2} + {(1)^2} + 2(\vec a \cdot \vec b + \vec b \cdot \vec c + \vec c \cdot \vec a) = 0$

Hence, $\vec a \cdot \vec b + \vec b \cdot \vec c + \vec c \cdot \vec a = - \cfrac{3}{2}$