<p>The resultant of forces \(\vec{P}\) and \(\vec{Q}\) is \(\vec{R}\). If \(\vec{Q}\) is doubled then \(\vec{R}\) is doubled. If the direction of \(\vec{Q}\) is reversed, then \(\vec{R}\) is again doubled. Then \(P^2 : Q^2 : R^2\) is</p>
Step-by-Step Solution
Key Concept: Use the vector addition property R² = P² + Q² + 2PQ·cos(θ) for two different configurations, then create a system of equations by substituting the conditions about doubled and reversed vectors.
Step 1: Let angle between P and Q be θ. Initial resultant: R^2 = P^2 + Q^2 + 2PQ·cos(θ) ... (1) Step 2: When Q is doubled, resultant becomes 2 R :
(2R)^2 = P^2 + (2Q)^2 + 2P(2Q)·cos(θ)
4R^2 = P^2 + 4Q^2 + 4PQ·cos(θ) ... (2) Step 3: When Q is reversed, resultant is again 2 R :
(2R)^2 = P^2 + Q^2 + 2PQ·cos(180° - θ)
4R^2 = P^2 + Q^2 - 2PQ·cos(θ) ... (3) Step 4: From (1): 2PQ·cos(θ) = R^2 - P^2 - Q^2 ... (4) Step 5: Substitute (4) into (3):
4R^2 = P^2 + Q^2 - (R^2 - P^2 - Q^2)
4R^2 = 2P^2 + 2Q^2 - R^2
5R^2 = 2P^2 + 2Q^2 ... (5) Step 6: Substitute (4) into (2):
4R^2 = P^2 + 4Q^2 + 2(R^2 - P^2 - Q^2)
4R^2 = P^2 + 4Q^2 + 2R^2 - 2P^2 - 2Q^2
2R^2 = -P^2 + 2Q^2
P^2 = 2Q^2 - 2R^2 ... (6) Step 7: From (5): P^2 = (2Q^2 + 5R^2 - 2Q^2)/2 = 5R^2/2 - Q^2. Substitute into (6):
5R^2/2 - Q^2 = 2Q^2 - 2R^2
5R^2/2 + 2R^2 = 3Q^2
9R^2/2 = 3Q^2, so Q^2 = 3R^2/2
From (6): P^2 = 2(3R^2/2) - 2R^2 = 3R^2 - 2R^2 = R^2 Step 8: Therefore: P^2 : Q^2 : R^2 = R^2 : (3R^2/2) : R^2 = 2 : 3 : 2 ∴ Answer: C
Correct Answer: C