The angle between the two vectors $$A = 3\hat i + 4\hat j + 5\hat k$$     and $$B = 3\hat i + 4\hat j - 5\hat k$$     will be

Solution :
Angle between two vectors is given as from dot product $$A \cdot B = \left| A \right|\left| B \right|\cos \theta $$
$$\eqalign{ & \cos \theta = \frac{{A \cdot B}}{{AB}} \cr & {\text{Here,}}\,\,A = 3\hat i + 4\hat j + 5\hat k \cr & B = 3\hat i + 4\hat j - 5\hat k \cr & \therefore A = \sqrt {{{\left( 3 \right)}^2} + {{\left( 4 \right)}^2} + {{\left( 5 \right)}^2}} = \sqrt {50} \cr & B = \sqrt {{{\left( 3 \right)}^2} + {{\left( 4 \right)}^2} + {{\left( { - 5} \right)}^2}} = \sqrt {50} \cr & {\text{and}}\,A \cdot B = \left( {3\hat i + 4\hat j + 5\hat k} \right) \cdot \left( {3\hat i + 4\hat j - 5\hat k} \right) \cr & = 9 + 16 - 25 = 0 \cr & \therefore \cos \theta = \frac{0}{{\sqrt {50} \cdot \sqrt {50} }} = 0 \Rightarrow \theta = {90^ \circ } \cr} $$