JEE Mains · Physics · STD 11 - 3.1 vectors
Two vectors \(\overrightarrow{ A }\) and \(\overrightarrow{ B }\) have equal magnitudes. If magnitude of \(\overrightarrow{ A }+\overrightarrow{ B }\) is equal to two times the magnitude of \(\overrightarrow{ A }-\overrightarrow{ B }\), then the angle between \(\overrightarrow{ A }\) and \(\overrightarrow{ B }\) will be .......................
- A \(\sin ^{-1}\left(\frac{3}{5}\right)\)
- B \(\sin ^{-1}\left(\frac{1}{3}\right)\)
- C \(\cos ^{-1}\left(\frac{3}{5}\right)\)
- D \(\cos ^{-1}\left(\frac{1}{3}\right)\)
Answer & Solution
Correct Answer
(C) \(\cos ^{-1}\left(\frac{3}{5}\right)\)
Step-by-step Solution
Detailed explanation
\(\left(a^{2}+b^{2}+2 a b \cos \theta\right)=4\left(a^{2}+b^{2}-2 a b \cos \theta\right)\) put \(a\) = \(b\) we get \(2 a^{2}+2 a^{2} \cos \theta=8 a^{2}-8 a^{2} \cos \theta\) \(\cos \theta=\frac{3}{5}\)
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