Let $$\overrightarrow a ,\,\overrightarrow b $$  and $$\overrightarrow c $$ be three vectors having magnitudes $$1,\,1$$  and $$2$$ respectively. If $$\overrightarrow a \times \left( {\overrightarrow a \times \overrightarrow c } \right) + \overrightarrow b = \overrightarrow 0 ,$$      the acute angle between $$\overrightarrow a $$ and $$\overrightarrow c $$ is :

Solution :
$$\eqalign{ & {\text{Here }}\left| {\overrightarrow a } \right| = 1,\,\,\left| {\overrightarrow b } \right| = 1,\,\,\left| {\overrightarrow c } \right| = 2 \cr & {\text{Now, }}\overrightarrow a \times \left( {\overrightarrow a \times \overrightarrow c } \right) + \overrightarrow b = \overrightarrow 0 \cr & {\text{or }}\left( {\overrightarrow a .\overrightarrow c } \right)\overrightarrow a - \left( {\overrightarrow a .\overrightarrow a } \right)\overrightarrow c + \overrightarrow b = \overrightarrow 0 \cr & {\text{or }}2\cos \,\theta \,\overrightarrow a - \overrightarrow c + \overrightarrow b = 0 \cr & {\text{or }}{\left( {\overrightarrow c - 2\cos \,\theta \,\overrightarrow a } \right)^2} = {\overrightarrow b ^2} \cr & {\text{or }}{\left| {\overrightarrow c } \right|^2} + 4{\cos ^2}\theta \,{\left| {\overrightarrow a } \right|^2} - 4\cos \,\theta \,\overrightarrow c .\overrightarrow a = {\left| {\overrightarrow b } \right|^2} \cr & {\text{or }}4 + 4{\cos ^2}\theta - 4\cos \,\theta .2\cos \,\theta = 1 \cr & {\text{or }}4{\cos ^2}\theta = 3 \cr & \therefore \,\,\theta = \frac{\pi }{6}. \cr} $$