Question

Find the angle between the vectors 9i+3j+6k and i-2j+3k

Answer 1

Given vectors are

A=2i+3j+k  and  B=-3i+6k.

Cosine of angle θ between vectors A and B is given by

cosθ=(A.B)/(|A||B|)

Calculate A.B as

A.B=(2i+3j+k).(-3i+6k)

     =-6(i.i)+12(i.k)-9(j.i)+18(j.k)-3(k.i)+

         6(k.k)

     =-6+0-0+0-0+6

     =-6+6=0

Thus cosθ=0=>θ=90°.