JEE Mains · Physics · STD 11 - 3.1 vectors
The angle between vector \((\overrightarrow{{A}})\) and \((\overrightarrow{{A}}-\overrightarrow{{B}})\) is :
- A \(\tan ^{-1}\left(\frac{-\frac{{B}}{2}}{{A}-{B} \frac{\sqrt{3}}{2}}\right)\)
- B \(\tan ^{-1}\left(\frac{{A}}{0.7 {B}}\right)\)
- C \(\tan ^{-1}\left(\frac{\sqrt{3} {B}}{2 {A}-{B}}\right)\)
- D \(\tan ^{-1}\left(\frac{{B} \cos \theta}{{A}-{B} \sin \theta}\right)\)
Answer & Solution
Correct Answer
(C) \(\tan ^{-1}\left(\frac{\sqrt{3} {B}}{2 {A}-{B}}\right)\)
Step-by-step Solution
Detailed explanation
Angle between \(\overrightarrow{{A}}\) and \(\overrightarrow{{B}}, \theta=60^{\circ}\) Angle betwenn \(\overrightarrow{{A}}\) and \(\overrightarrow{{A}}-\overrightarrow{{B}}\) \(\tan \alpha=\frac{B \sin \theta}{A-B \cos \theta}\)…
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