Let two vectors $$\vec A = 3\hat i + \hat j + 2\hat k$$    and $$\vec B = 2\hat i - 2\hat j + 4\hat k.$$    Consider the unit vector perpendicular to both $${\vec A}$$ and $${\vec B}$$ is

Solution :
Angle between $${\vec A}$$ and $${\vec B}$$ is given by
$$\cos \theta = \frac{{\vec A \cdot \vec B}}{{AB}} = \frac{3}{{\sqrt {21} }}$$
The unit vector perpendicular to $${\vec A}$$ and $${\vec B}$$ is given by
$$\eqalign{ & \hat n = \frac{{\vec A \times \vec B}}{{\left| {\vec A \times \vec B} \right|}} = \frac{{\left( {3\hat i + \hat j + 2\hat k} \right) \times \left( {2\hat i - 2\hat j + 4\hat k} \right)}}{{\left| {\left( {3\hat i + \hat j + 2\hat k} \right) \times \left( {2\hat i - 2\hat j + 4\hat k} \right)} \right|}} \cr & = \frac{{\hat i - \hat j - \hat k}}{{\sqrt 3 }} \cr} $$