The angle between vectors A=3i^+4j^+5k^  and B=6i^+8j^+10k^  is

We know that A . B=ABcosθ                                (i)                                 A . B=(3i^+4j+5k^). (6i^+8j^+10k^)                   (ii)                                =3×6+4×8+5×10=100 The magnitudes of A and B are                A=(3)2+(4)2+(5)2=50                                                (iii)And        B=(6)2+(8)2+(10)2=200                                           (iv)Using (ii), (iii) and (iv)  in (i), we have                100=50×200cosθ=100cosθ Or           cosθ=1  or θ=zero .