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JEE Main 2019 — Mathematics in Physics Question with Solution

JEE Main 2019 (10 Jan Shift 2)

Question

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:

Choose an option

Show full solutionCorrect option: A
Correct answer
Acos-1n2-1n2+1

Step-by-step explanation

Given: A→+B→=nA→-B→, so, the magnitudes of this will be given as,

A2+B2+2ABcosθ=nA2+B2-2ABcosθ;where θ is the angle between A→ and B→.

Also, A→=B→,

2A2+2A2cosθ=n2A2-2A2cosθ

Squaring both sides:

2A21+cosθ=n22A21-cosθ

cosθ=n2-1n2+1

i.e., θ=cos-1n2-1n2+1

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About this question

This is a previous-year question from JEE Main 2019, covering the Mathematics in Physics chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.