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

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