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 $-\frac{3}{2}$

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

$\Rightarrow |\overline{a}{|}^2+|\overline{b}{|}^2+|\overline{c}{|}^2+2\times (\overline{a}.\overline{b}+\overline{b}.\overline{c}+\overline{c}.\overline{a})=\overline{0}$

then $\overline{a}.\overline{b}+\overline{b}.\overline{c}+\overline{c}.\overline{a}=\frac{-3}{2}$