Question

If $\overline{a},\overline{b},\overline{c}$ are vectors of equal magnitudes and each of them inclined of ${60}^o$ to others. If $|\overline{a}+\overline{b}+\overline{c}|=\sqrt{6}$ then find $|\overline{a}|$

Answer 1

$\begin{array}{c}a,b,carethreevectorofequalmagnitudes.\\ \left|a\right|=\left|b\right|=\left|c\right|\to \left(1\right)\\ now\\ \left|a+b+c\right|=\sqrt{\left(a+b+c\right).\left(a+b+c\right)}\\ =\sqrt{a.a+a.b+a.c+b.a+b.b+b.c+c.a+c.b+c.c}\\ =\sqrt{{\left|a\right|}^2+{\left|b\right|}^2+{\left|c\right|}^2+2\left(a.b+b.c+c.a\right)}\\ =\sqrt{{\left|a\right|}^2+{\left|b\right|}^2+{\left|c\right|}^2+\left(2ab\cos 60+2bc\cos 60+2ca\cos 60\right)}\\ =\sqrt{{\left|a\right|}^2+{\left|b\right|}^2+{\left|c\right|}^2+\frac{2}{2}\left|\left(ab+bc+ca\right)\right|}\\ fromequation\left(1\right)\\ ={\left|a\right|}^2+{\left|a\right|}^2+{\left|a\right|}^2+\left|a\right|.\left|a\right|+\left|a\right|.\left|a\right|+\left|a\right|.\left|a\right|\\ =\sqrt{3{\left|a\right|}^2+3{\left|a\right|}^2}\\ =a\sqrt{6}\\ \left|a+b+c\right|=\left|a\right|\sqrt{6}\\ \sqrt{6}=\left|a\right|\sqrt{6}\\ \left|a\right|=1Ans.\end{array}$