Two vectors A and B have equal magnitudes. The magnitude of ( A+ B) is ' n ' times the magnitude of ( A- B) . The angle between A and B is:
JEE Main 2019 — Physics Mechanics
2019mcqmedium
Two vectors A and B have equal magnitudes. The magnitude of (A+B) is ' n ' times the magnitude of (A−B) . The angle between A and B is:
Official previous-year question
Held on 10 Jan 2019 · Verified 1 Jul 2026.
Options
A
cos−1[n2+1n2−1]
B
sin−1[n+1n−1]
C
cos−1[n+1n−1]
D
sin−1[n2+1n2−1]
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Solution
Given: ∣A+B∣=n∣A−B∣, so, the magnitudes of this will be given as, A2+B2+2ABcosθ=n(A2+B2−2ABcosθ);where θ is the angle between A and B. Also, ∣A∣=∣B∣, 2A2+2A2cosθ=n2A2−2A2cosθ Squaring both sides: 2A2(1+cosθ)=n22A2(1−cosθ) cosθ=n2+1n2−1 i.e., θ=cos−1[n2+1n2−1]
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