Question

If $\overline{a},\overline{b},\overline{c}$ are three vectors such that $\overline{a}+\overline{b}+\overline{c}=0,\left|\begin{array}{c}\overline{a}\end{array}\right|=1,\left|\begin{array}{c}\overline{b}\end{array}\right|=2,\left|\begin{array}{c}\overline{c}\end{array}\right|=3$, then $\overline{a}\cdot \overline{b}+\overline{b}\cdot \overline{c}+\overline{c}\cdot \overline{a}$ is equal to

• A. $-7$
• B. $7$
• C. $1$
• D. $0$

Answer 1

The correct option is A $-7$

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

Consider
$(\overline{a}+\overline{b}+\overline{c}{)}^2={\overline{a}}^2+{\overline{b}}^2+{\overline{c}}^2+2(\overline{a}\cdot \overline{b}+\overline{b}\cdot \overline{c}+\overline{c}\cdot \overline{a})=0$

$\Rightarrow (\overline{a}\cdot \overline{b}+\overline{b}\cdot \overline{c}+\overline{c}\cdot \overline{a})=-\frac{{\overline{a}}^2+{\overline{b}}^2+{\overline{c}}^2}{2}$

$\Rightarrow (\overline{a}\cdot \overline{b}+\overline{b}\cdot \overline{c}+\overline{c}\cdot \overline{a})=\frac{(1^2+2^2+3^2)}{2}=-7$