Question

If $\overline{a},\overline{b},\overline{c}$ are unit vectors such that $\overline{a}+\overline{b}+\overline{c}=\overline{0}$, then $\overline{a}.\overline{b}+\overline{b}.\overline{c}+\overline{c}.\overline{a}=$

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

Answer 1

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

Given,

$\overline{a}+\overline{b}+\overline{c}=0$

Squaring both sides,

${\left(\overline{a}\right)}^2+{\left(\overline{b}\right)}^2+{\left(\overline{c}\right)}^2+2.\overline{a}.\overline{b}+2.\overline{b}.\overline{c}+2.\overline{c}.\overline{a}=0$

$1+1+1+2(\overline{a}.\overline{b}+\overline{b}.\overline{c}+\overline{c}.\overline{a})=0$

$\Rightarrow (\overline{a}.\overline{b}+\overline{b}.\overline{c}+\overline{c}.\overline{a})=-\frac{3}{2}$ is the correct answer.