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JEE Main 202126 AugustEvening Shift
Exam
JEE Main
Year
2021
Date
26 August
Shift
Evening Shift
Subject
Physics
Chapter
Mathematics in Physics

JEE Main 2021 · 26 August · Evening Shift · Physics · Mathematics in Physics

The angle between vector ( → A ) and ( → A - → B ) is :

The angle between vector (A→)( \overset{\to}{A} ) and (A→−B→)( \overset{\to}{A} - \overset{\to}{B} ) is :

The angle between vector ( → A ) and ( → A - → B ) is :

Options

  1. Atan⁡−1(BcosθA−Bsinθ)\tan^{- 1}\left(\dfrac{Bcos\theta}{A - Bsin\theta}\right)
  2. Btan⁡−1(3B2A−B)\tan^{- 1}\left(\dfrac{\sqrt{3}B}{2A - B}\right)
  3. Ctan⁡−1(−B2A−B32)\tan^{- 1}\left(\dfrac{- \dfrac{B}{2}}{A - B\dfrac{\sqrt{3}}{2}}\right)
  4. Dtan⁡−1(A0.7B)\tan^{- 1}\left(\dfrac{A}{0 . 7B}\right)

Correct answer

B. tan⁡−1(3B2A−B)\tan^{- 1}\left(\dfrac{\sqrt{3}B}{2A - B}\right)

Step-by-step solution

The angle between vector ( → A ) and ( → A - → B ) is :

tanϕ=Bsin60°A−Bcos60°tanϕ = \dfrac{Bsin60 °}{A - Bcos60 °}

ϕ=tan⁡−1(Bsin60°A−Bcos60°)=tan⁡−1(3B2A−B)ϕ = \tan^{- 1}\left(\dfrac{Bsin60 °}{A - Bcos60 °}\right) = \tan^{- 1}\left(\dfrac{\sqrt{3}B}{2A - B}\right)

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