The correct option is B 120∘
Let the angle between the vectors 3P and 2P be θ.
Then,
R=√(9P2)+4P2+2×3P×2P×cosθ
Squaring both sides:
R2=13P2+12P2cosθ....(i)
When 3P→6P, then R→2R, i.e resultant doubles.
⇒2R=√(36P2)+4P2+2×6P×2P×cosθ
Squaring both the sides:
4R2=40P2+24P2cosθ.....(ii)
Putting value of R2 from Eq. (i) in Eq. (ii):
52P2+48P2cosθ=40P2+24P2cosθ
⇒12P2+24P2cosθ=0
⇒cosθ=−12
∴θ=120∘