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

Correct answer: (B) tan^(-1)(√3 B)/(2 A-B). tanϕ=(Bsin60°)/(A-Bcos60°) ϕ=tan^(-1)(Bsin60°)/(A-Bcos60°)=tan^(-1)(√3B)/(2A-B)

Q71 / 90 · JEE Main 2021 (26 Aug Shift 2) · Physics

The angle between vector $(\vec{A})$ and $(\vec{A} - \vec{B})$ is :

$\tan ϕ = \frac{B \sin 60 °}{A - B \cos 60 °}$

$ϕ = \tan^{- 1} (\frac{B \sin 60 °}{A - B \cos 60 °}) = \tan^{- 1} (\frac{\sqrt{3} B}{2 A - B})$

Hence, the correct answer is option: 'B'.