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Grade 11Vectors

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|

Profile image of Carlyn medona
9 Years agoGrade 11
Answers icon

5 Answers

Profile image of Arun
9 Years ago
you can try formula, which is-
R^2= P^2 +Q^2 +2PQ cos\theta            where R is Resultant
You will surely get answer.
 
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|.
Profile image of vishal tarte
8 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
Profile image of Muneeb
7 Years ago
They ar
Profile image of Krish Gupta
6 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|.
Profile image of Kushagra Madhukar
6 Years ago
Dear Student,
Please find the attached answer to your question.
 
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
 

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