If $\vec{a}, \vec{b},$ and $\vec{c}$ are unit vectors such that $\vec{a}+\vec{b}+\vec{c}=\vec 0$, then value of $\vec{a}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a}$ is :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $-\frac{3}{2}$
$\vec{a}+\vec{b}+\vec{c}=\vec 0$
$(\vec{a}+\vec{b}+\vec{c}).(\vec{a}+\vec{b}+\vec{c})=0$
so $|\vec a|^2+|\vec b|^2+|\vec c|^2+2(\vec a.\vec b+\vec b.\vec c+\vec c.\vec a)=0$
so $\vec a.\vec b+\vec b.\vec c+\vec c.\vec a=-\frac{3}{2}$