Question

If $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are three vectors such that $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}$ and $\left|\overrightarrow{a}\right|=2,\left|\overrightarrow{b}\right|=3,\left|\overrightarrow{c}\right|=5,$ then value of $\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a}$ is

• A. $0$
• B. $1$
• C. $-19$
• D. $38$

Answer 1

The correct option is C $-19$

In identity :

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

$\overline{a}+\overline{b}+\overline{c}=0,|\overline{a}|=2,|\overline{b}|=3$ and $|\overline{c}|=5$

$\Rightarrow 0^2=4+9+25+2(\overline{a}.\overline{b}+\overline{b}.\overline{c}+\overline{c}.\overline{a})$

$\Rightarrow 2(\overline{a}.\overline{b}+\overline{b}.\overline{c}+\overline{c}.\overline{a})=-38$

$\Rightarrow \overline{a}.\overline{b}+\overline{b}.\overline{c}+\overline{c}.\overline{a}=-19$