A×B=ABsinθ
and A.B=ABcosθ.
Given, ∣A×B∣=3A.B… (i)
but ∣A×B∣=∣A∣∣B∣sinθ=ABsinθ
and A⋅B=∣A∣∣B∣cosθ=ABcosθ Putting these values in Eq. (i), we get ABsinθ=3ABcosθ
or tanθ=3 ∴θ=60∘
The addition of vector A and B can be given by the law of parallelogram. ∴∣A+B∣=A2+B2+2ABcos60∘ =A2+B2+2AB×21 =(A2+B2+AB)1/2