JEE Main 2021 · 25 July · evening · Single correct · +4 / −1
Two vectors X⃗ and Y⃗ have equal magnitude. The magnitude of (X⃗ - Y⃗) is n times the magnitude of (X⃗ + Y⃗). The angle between X⃗ and Y⃗ is :
- (a) cos⁻¹((-n²-1)/(n²-1))
- (b) cos⁻¹((n²-1)/(-n²-1))
- (c) cos⁻¹((n²+1)/(-n²-1))
- (d) cos⁻¹((n²+1)/(n²-1))
Answer and solution
Answer (b) cos⁻¹((n²-1)/(-n²-1))
Solution
Physics sir · 4 steps
Write both magnitudes with the cosine rule. Their ratio gives cos θ directly, because a² cancels.
- Both magnitudes. Let |X⃗| = |Y⃗| = a and let θ be the angle between them. Then |X⃗+Y⃗|² = 2a²(1+cos θ) and |X⃗-Y⃗|² = 2a²(1-cos θ).
- Square the condition. Square |X⃗-Y⃗| = n |X⃗+Y⃗| and divide. The factor 2a² cancels, so only cos θ is left.
(1-cos θ)/(1+cos θ) = n²
- Solve for cosine. Cross-multiply: 1-cos θ = n² + n²cos θ, so cos θ (1+n²) = 1-n². This gives cos θ = (1-n²)/(1+n²).
- So. Multiply the top and bottom by -1: cos θ = (n²-1)/(-n²-1), so θ = cos⁻¹((n²-1)/(-n²-1)). That is option (b).
The idea to keep. For equal vectors, |X⃗±Y⃗|² = 2a²(1±cos θ). Any ratio of the two gives cos θ. Check your answer: a cosine can never be bigger than 1 in size.
Answer from the NTA final answer key. Solution by PrepWiser.