If $\overrightarrow{a}$ , $\overrightarrow{b}$ and $\overrightarrow{c}$ are three unit vectors such that $\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} = \overrightarrow{0}$ find the value of $\overrightarrow{a} . \overrightarrow{b}$ + $\overrightarrow{b} . \overrightarrow{c}$ + $\overrightarrow{c} . \overrightarrow{a}$

$\dfrac{3}{2}$

$-\dfrac{3}{2}$

$-\dfrac{2}{3}$

$\dfrac{3}{2}$

Karnataka PU MQP 2023

Solution

The correct answer is

$-\dfrac{3}{2}$

Explanation

$|\overrightarrow{a}|$ =

$|\overrightarrow{b}|$ =

$|\overrightarrow{c}|$ = 1

∴

$(\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c})^2 = 0$

⇒

$|\overrightarrow{a}|^2 + |\overrightarrow{b}|^2 + |\overrightarrow{c}|^2$ +

$2 \overrightarrow{a} . \overrightarrow{b}$ +

$2 \overrightarrow{b} . \overrightarrow{c}$ +

$2 \overrightarrow{c} . \overrightarrow{a}$ =

$0$

∴

$1^2 + 1^2 + 1^2$ +

$2 [\overrightarrow{a} . \overrightarrow{b} + \overrightarrow{b} . \overrightarrow{c} + \overrightarrow{c} . \overrightarrow{a}]$ =

$0$
⇒

$[\overrightarrow{a} . \overrightarrow{b} + \overrightarrow{b} . \overrightarrow{c} + \overrightarrow{c} . \overrightarrow{a}]$ =

$-\dfrac{3}{2}$