Question

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

• A. 1
• B. $\frac{3}{2}$
• C. 3
• D. $-\frac{3}{2}$

Answer 1

The correct option is D $-\frac{3}{2}$

$\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are unit vectors

$\therefore \left|\overrightarrow{a}\right|=\left|\overrightarrow{b}\right|=\left|\overrightarrow{c}\right|=1$
$\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0$ (given)
$\therefore {(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})}^2=0$

${\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2+{\left|\overrightarrow{c}\right|}^2+2(\overrightarrow{a}\cdot \overrightarrow{b}+\overrightarrow{b}\cdot \overrightarrow{c}+\overrightarrow{c}\cdot \overrightarrow{a})=0$

$1+1+1+2(\overrightarrow{a}\cdot \overrightarrow{b}+\overrightarrow{b}\cdot \overrightarrow{c}+\overrightarrow{c}\cdot \overrightarrow{a})=0$

$\therefore \overrightarrow{a}\cdot \overrightarrow{b}+\overrightarrow{b}\cdot \overrightarrow{c}+\overrightarrow{c}\cdot \overrightarrow{a}=\frac{-3}{2}$