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JEE Mains · Physics · STD 11 - 3.1 vectors

दो सदिश \(\overrightarrow{ A }\) एवं \(\overrightarrow{ B }\) के परिमाण एक समान है। यदि \(\overrightarrow{ A }+\overrightarrow{ B }\) का परिमाण \(\overrightarrow{ A }-\overrightarrow{ B }\) के परिमाण का दो गुना है तो \(\overrightarrow{ A }\) एवं \(\overrightarrow{ B }\) के बीच कोण होगा \(-\)

  1. A \(\sin ^{-1}\left(\frac{3}{5}\right)\)
  2. B \(\sin ^{-1}\left(\frac{1}{3}\right)\)
  3. C \(\cos ^{-1}\left(\frac{3}{5}\right)\)
  4. D \(\cos ^{-1}\left(\frac{1}{3}\right)\)
Verified Solution

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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