# If $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are unit vectors such that$\mid\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\mid=0$,then the value of $\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a}$ is

$(A)\;1\quad(B)\;3\quad(C)\;-\frac{3}{2}\quad(D)\;None\;of\;these$

Toolbox:
• If the magnitude of the three vectors in a triangle equal, then it is an equilateral triangle
• $\overrightarrow {a}\overrightarrow {b}=|\overrightarrow a||\overrightarrow b| \cos \theta$
$|\overrightarrow a|=1,|\overrightarrow b|=1,|\overrightarrow c|=1$
Given $|\overrightarrow a|+|\overrightarrow b|+|\overrightarrow c|=0$
Since the magnitudes of the three vectors are equal and also $\overrightarrow a+\overrightarrow b+\overrightarrow c=0$
The three vectors should form an equilateral triangle whose angles are $60^{\circ}$
$\overrightarrow a.\overrightarrow b=|\overrightarrow a||\overrightarrow b|\cos (\pi-60 ^{\circ})$
But $(\pi-60^{\circ})$
$\qquad =1.1.\bigg(\large\frac{-1}{2}\bigg)$
$\qquad=\large\frac{-1}{2}$
$\overrightarrow b.\overrightarrow c=|\overrightarrow b||\overrightarrow c|\cos (\pi-60 ^{\circ})$
$\qquad =1.1.\bigg(\large\frac{-1}{2}\bigg)$
Hence on substituting the values we get,
Therefore $\overrightarrow a.\overrightarrow b+\overrightarrow b.\overrightarrow c+\overrightarrow c.\overrightarrow a=\bigg(\large\frac{-1}{2}\bigg)+\bigg(\frac{-1}{2}\bigg)+\bigg(\frac{-1}{2}\bigg)$
$=\large\frac{-3}{2}$
Hence the correct option is $C$