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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|

Carlyn medona , 8 Years ago
Grade 11
anser 5 Answers
Arun

Last Activity: 8 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|.
vishal tarte

Last Activity: 7 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

Last Activity: 6 Years ago

They ar
Krish Gupta

Last Activity: 5 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|.
Kushagra Madhukar

Last Activity: 5 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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