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

1. ${\tan }^{-1}\left(\frac{B\cos \theta }{A-B\sin \theta }\right)$
2. ${\tan }^{-1}\left(\frac{\sqrt{3} B}{2 A-B}\right)$
3. ${\tan }^{-1}\left(\frac{-\frac{B}{2}}{A-B\frac{\sqrt{3}}{2}}\right)$
4. ${\tan }^{-1}\left(\frac{A}{0.7B}\right)$

Solution

Asked in: JEE Main 2021 (26 Aug Shift 2)

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