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

Solution :
$$\eqalign{ & {\text{Let }}\left| {\overrightarrow a } \right| = \left| {\overrightarrow b } \right| = \left| {\overrightarrow c } \right| = k.{\text{ Then}} \cr & {\text{cos}}\frac{\pi }{3} = \frac{{\overrightarrow a .\overrightarrow b }}{{\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|}} = \frac{{\overrightarrow a .\overrightarrow b }}{{{k^2}}}{\text{ or }}\overrightarrow a .\overrightarrow b = \frac{{{k^2}}}{2} \cr & {\text{Similarly, }}\overrightarrow b .\overrightarrow c = \overrightarrow c .\overrightarrow a = \frac{{{k^2}}}{2}.{\text{ So, }}\overrightarrow a .\overrightarrow b + \overrightarrow b .\overrightarrow c + \overrightarrow c .\overrightarrow a = \frac{{3{k^2}}}{2} \cr & {\text{Now, }}6 = {\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right|^2} = {\left| {\overrightarrow a } \right|^2} + {\left| {\overrightarrow b } \right|^2} + {\left| {\overrightarrow c } \right|^2} + 2\left( {\overrightarrow a .\overrightarrow b + \overrightarrow b .\overrightarrow c + \overrightarrow c .\overrightarrow a } \right) \cr & \therefore \,6 = 3{k^2} + 2.\frac{{3{k^2}}}{2} \cr & \therefore \,k = \pm 1.\,{\text{But }}\left| {\overrightarrow a } \right|\,{\text{is positive}}{\text{. So, }}\left| {\overrightarrow a } \right| = 1. \cr} $$