Let $$\vec V = 2\vec i + \vec j - \vec k$$    and $$\vec W = \vec i + 3\vec k.$$   If $${\vec U}$$ is a unit vector, then the maximum value of the scalar triple product $$\left| {\vec U\vec V\vec W} \right|$$   is :

Solution :
Given that $$\vec v = 2\hat i + \hat j - \hat k$$    and $$\vec w = \hat i + 3\hat k$$   and $$u$$ is a unit vector $$\therefore \,\,\left| {\vec u} \right| = 1$$
$$\eqalign{ & {\text{Now, }}\left[ {\vec u\,\vec v\,\vec w} \right] = \vec u.\left( {\vec v \times \vec w} \right) \cr & = \vec u.\left( {2\hat i + \hat j - \hat k} \right) \times \left( {\hat i + 3\hat k} \right) \cr & = \vec u.\left( {3\hat i - 7\hat j - \hat k} \right) \cr & = \sqrt {{3^2} + {7^2} + {1^2}} \,\cos \,\theta \cr} $$
which is maximum when $$\cos \,\theta = 1$$
$$\therefore $$ Maximum value of $$\left[ {\vec u\,\vec v\,\vec w} \right] = \sqrt {59} $$