Question

If $\overrightarrow{a}$,$\overrightarrow{b}$ and $\overrightarrow{c}$ are three vectors of equal magnitude. The angle between each pair of vectors is $\frac{\pi }{3}$ such that $|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}|=\sqrt{6}$. Then |$\overrightarrow{a}$| is equal to

• A. $2$
• B. $-1$
• C. $1$
• D. $\sqrt{6}$

Answer 1

The correct option is C $1$

$|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}{|}^2=6$
$\Rightarrow |\overrightarrow{a}{|}^2+|\overrightarrow{b}{|}^2+|\overrightarrow{c}{|}^2+2(\overrightarrow{a}\cdot \overrightarrow{b}+\overrightarrow{b}\cdot \overrightarrow{c}+\overrightarrow{c}\cdot \overrightarrow{a})=6$
Given $|\overrightarrow{a}|=$$|\overrightarrow{b}|=$$|\overrightarrow{c}|$ and $\overrightarrow{a}\cdot \overrightarrow{b}=|\overrightarrow{a}||\overrightarrow{b}|\cos \frac{\pi }{3}$
i.e.,

$\overrightarrow{a}\cdot \overrightarrow{b}=\overrightarrow{b}\cdot \overrightarrow{c}=\overrightarrow{c}\cdot \overrightarrow{a}=\frac{1}{2}{\left|\overrightarrow{a}\right|}^2$ $\therefore 3{\left|\overrightarrow{a}\right|}^2+3{\left|\overrightarrow{a}\right|}^2=6$
$\Rightarrow {\left|\overrightarrow{a}\right|}^2=1\Rightarrow \left|\overrightarrow{a}\right|=1$