What is the interior acute angle of the parallelogram whose sides are represented by the vectors $$\frac{1}{{\sqrt 2 }}\hat i + \frac{1}{{\sqrt 2 }}\hat j + \hat k$$     and $$\frac{1}{{\sqrt 2 }}\hat i - \frac{1}{{\sqrt 2 }}\hat j + \hat k\,?$$

Solution :
$$\eqalign{ & {\text{Let }}a = \frac{1}{{\sqrt 2 }}\hat i + \frac{1}{{\sqrt 2 }}\hat j + \hat k \cr & {\text{and }}b = \frac{1}{{\sqrt 2 }}\hat i - \frac{1}{{\sqrt 2 }}\hat j + \hat k \cr & \therefore \,\cos \,\theta = \frac{{a.b}}{{\left| a \right|\left| b \right|}} \cr & \Rightarrow \cos \,\theta = \frac{{\left( {\frac{1}{{\sqrt 2 }}\hat i + \frac{1}{{\sqrt 2 }}\hat j + \hat k} \right).\left( {\frac{1}{{\sqrt 2 }}\hat i - \frac{1}{{\sqrt 2 }}\hat j + \hat k} \right)}}{{\sqrt {\frac{1}{2} + \frac{1}{2} + 1} .\sqrt {\frac{1}{2} + \frac{1}{2} + 1} }} \cr & \Rightarrow \cos \,\theta = \frac{1}{2}\left[ {\frac{1}{2} - \frac{1}{2} + 1} \right] \cr & \Rightarrow \cos \,\theta = \frac{1}{2} \cr & \Rightarrow \cos \,\theta = \cos \,{60^ \circ } \cr & \therefore \,\theta = {60^ \circ } \cr} $$