Question

If $\bar{a},\bar{b},\bar{c}$ are three vectors such that each is inclined an angle of $60^{\circ }$ with the other two and $|\bar{a}|=2,|\bar{b}|=3,|\bar{a}+\bar{b}+\bar{c}|=5$, then find the value of $|\bar{c}|$.

Answer 1

Answer: 1

$|\bar{a}+\bar{b}+\bar{c}|=5$
$\Rightarrow |\bar{a}+\bar{b}+\bar{c}{|}^2=25$
$\iff (\bar{a}+\bar{b}+\bar{c})\cdot (\bar{a}+\bar{b}+\bar{c})=25$
$\iff |\bar{a}{|}^2+|\bar{b}{|}^2+|\bar{c}{|}^2+2\bar{a}\cdot \bar{b}+2\bar{b}\cdot \bar{c}+2\bar{c}\cdot \bar{a}=25$
$\iff 4+9+|\bar{c}{|}^2+2\cos 60^{\circ }$
$(|\bar{a}\Vert \bar{b}|+|\bar{b}\Vert \bar{c}|+|\bar{c}\Vert \bar{a}|)=25$
$\iff 13+|\bar{c}{|}^2+2\times \frac{1}{2}(2\times 3+3|\bar{c}|+|\bar{c}|\times 2)=25$
$\iff 13+|\bar{c}{|}^2+6+3|\bar{c}|+2|\bar{c}|=25$
$\iff |\bar{c}{|}^2+5|\bar{c}|-6=0$
$\iff (|\bar{c}|+6)(|\bar{c}|-1)=0$
$\iff |\bar{c}|=-6$
or $|\bar{c}|=1$
$\Rightarrow |\bar{c}|=1$