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JEE Main201910 Jan 2019Evening ShiftPhysicsMathematics in PhysicsActual

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:

Options

  1. Ac o s - 1 n 2 - 1 n 2 + 1
  2. Bs i n - 1 n - 1 n + 1
  3. Cc o s - 1 n - 1 n + 1
  4. Ds i n - 1 n 2 - 1 n 2 + 1

Correct answer

A. c o s - 1 n 2 - 1 n 2 + 1

Step-by-step solution

Given: A → + B → = n A → - B → , so, the magnitudes of this will be given as, A 2 + B 2 + 2 A B c o s θ = n A 2 + B 2 - 2 A B c o s θ ;where θ is the angle between A → and B → . Also, A → = B → , 2 A 2 + 2 A 2 c o s θ = n 2 A 2 - 2 A 2 c o s θ Squaring both sides: 2 A 2 1 + c o s θ = n 2 2 A 2 1 - c o s θ c o s θ = n 2 - 1 n 2 + 1 i.e., θ = c o s - 1 n 2 - 1 n 2 + 1

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