Question

Let $\overrightarrow{a},\overrightarrow{b}$ and $\overrightarrow{c}$ be three vectors such that $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}$, and $|\overrightarrow{a}|=10,|\overrightarrow{b}|=6$ and $|\overrightarrow{c}|=14$.

Fund the value of $\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a}.$

• A. $-332$
• B. $-166$
• C. $0$
• D. $166$

Answer 1

The correct option is B $-166$

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

$\therefore (\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}).(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})=0$

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

$\therefore 100+36+196+2(\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a})=0$

$\therefore \overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a}=-166$

So, the answer is option (B)