Question

Find the angle between vector a and vector b,if



a.b=axb

Answer 1

|A |*|B|* cos θ = |A||B| * sin θ

Tan θ = 1

Hence

θ = 45°

Answer 2

Given condition
A.B= |AXB|
=> |A||B| cosμ =|A||B|sinμ
[ μ is the acute angle between vectors A and B]
=> cosμ = sin μ
=> μ=45˚
C=A+B can be visualized as the diagonal of the parallelogram with A and B as adjacent sides. Simple trigonometry reveal that the height of this parallelogram to be |A|/√2. C
It may help u.