# 3. Let the angle between two nonzero vectors A and B be 120° and its resultant be C.   (a) C must be equal to |A-B|   (b) C must be less than |A-B|   (c) C must be greater than |A-B|   (d) C may be equal to |A-B|

Arun
25750 Points
7 years ago
you can try formula, which is-
R^2= P^2 +Q^2 +2PQ cos$\theta$            where R is Resultant

But if we see that  |A−B| will happen if both the vectors are in opposite direction that means angle is 180 degree.
but here in question is 120 degree hence it will surely be greater than |A-B|.
vishal tarte
13 Points
6 years ago
if c is thehave resultant of A and B,then . C sqrt=A^2+B^2-AB. [cos120=-1\2] similarly, -C=|A-B|=sqrt,A^2+B^2-2ABcos120 =sqrt A^2+B^2+AB so,|A-B|>c
Muneeb
13 Points
5 years ago
They ar
Krish Gupta
4 years ago
Just go with this formula :

R^2= P^2 +Q^2 +2PQ cos$\theta$            where R is Resultant

But if we see that  |A−B| will happen if both the vectors are in the opposite direction that means angle is 180 degree.
but here in question is 120 degree hence it will surely be greater than |A-B|.
4 years ago
Dear Student,

as we know, R2 = A2 + B2 + 2ABcosθ
as cosθ decreases as we go from θ = 0 to θ = 180, so does the magnitude of the vector R,
The minimum magnitude for vector R can be |A – B| at θ = 180.
Hence, the resultant vector at θ = 120 is greater than |A – B|.

Hope it helps.
Thanks and regards,
Kushagra