Q.At what angle the two forces A + B and A - B act, so that their resultant is sqrt(3A^2 + B^2).
You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.
Start your 14-day free trial to unlock the full solution →Using R^2 = P^2 + Q^2 + 2PQ cos(theta) with P = A+B and Q = A-B, and setting R^2 = 3A^2+B^2 as given, solving for cos(theta) gives cos(theta) = 1/2, so theta = 60 degrees.
Let the two vectors be P = (A + B) and Q = (A - B), with an angle theta between them (A and B are the magnitudes of two vectors being added/subtracted). Their resultant magnitude R is given by the parallelogram law of vector addition:
R^2 = P^2 + Q^2 + 2PQ cos(theta)
We are given R = sqrt(3A^2 + B^2), so R^2 = 3A^2 + B^2.
Now:
P^2 + Q^2 = (A+B)^2 + (A-B)^2 = (A^2 + 2AB + B^2) + (A^2 - 2AB + B^2) = 2A^2 + 2B^2
2PQ = 2(A+B)(A-B) = 2(A^2 - B^2)
Substituting into the resultant equation:
3A^2 + B^2 = (2A^2 + 2B^2) + 2(A^2 - B^2) cos(theta)
3A^2 + B^2 - 2A^2 - 2B^2 = 2(A^2 - B^2) cos(theta)
A^2 - B^2 = 2(A^2 - B^2) cos(theta)
Dividing both sides by (A^2 - B^2), assuming A is not equal to B:
cos(theta) = 1/2
theta = 60 degrees
…
Unlock everything free for 14 days
- Full step-by-step solutions
- Concept-first explanations
- Methods, shortcuts & mistakes
- PYQ mapping + timed mock tests
Full access for 14 days. No credit card required.