Given X=Y
X2+Y2−2×Ycosθ
=nX2+Y2+2×Ycosθ
Square both sides
2X2(1−cosθ)=n2⋅2X2(1+cosθ)
1−cosθ=n2+n2cosθ
cosθ=1+n21−n2
θ=cos−1[−n2−1n2−1]
Two vectors X and Y have equal magnitude. The magnitude of ( X- Y) is n times the magnitude of ( X+ Y). The angle between X and Y is :
JEE Main 2021 — Physics Mechanics
Two vectors X and Y have equal magnitude. The magnitude of (X−Y) is n times the magnitude of (X+Y). The angle between X and Y is :
Held on 25 Jul 2021 · Verified 1 Jul 2026.
Options
- A
cos−1(n2−1−n2−1)
- B
cos−1(−n2−1n2−1)
- C
cos−1(−n2−1n2+1)
- D
cos−1(n2−1n2+1)
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