Verified Solution • 2025–26 Session moderate

QJEEM19Ph1357: JEE Main 2019 Physics - Class 11 JEEM19Ph1357 Solution: Two vectors and have equal magnitudes. The magnitude of is 'n'…

Two vectors A→\overrightarrow A and B→\overrightarrow B have equal magnitudes. The magnitude of (A→+B→)\left( {\overrightarrow A + \overrightarrow B } \right) is 'n' times the magnitude of (A→−B→)\left( {\overrightarrow A - \overrightarrow B } \right) . The angle between A→{\overrightarrow A } and B→{\overrightarrow B } is -

Multiple Choice Options

  • A sin⁡−1[n−1n+1]{\sin ^{ - 1}}\left[ {{{n - 1} \over {n + 1}}} \right]
  • B sin⁡−1[n2−1n2+1]{\sin ^{ - 1}}\left[ {{{{n^2} - 1} \over {{n^2} + 1}}} \right]
  • C cos⁡−1[n2−1n2+1]{\cos ^{ - 1}}\left[ {{{{n^2} - 1} \over {{n^2} + 1}}} \right]
  • D cos⁡−1[n−1n+1]{\cos ^{ - 1}}\left[ {{{n - 1} \over {n + 1}}} \right]

Step-by-Step Verified Solution

∣A→+B→∣=2acos⁡θ/2\left| {\overrightarrow A + \overrightarrow B } \right| = 2a\cos \theta /2       . . . (1)

∣A→−B→∣=2acos⁡(π−θ)2=2asin⁡θ/2\left| {\overrightarrow A - \overrightarrow B } \right| = 2a\cos {{\left( {\pi - \theta } \right)} \over 2} = 2a\sin \theta /2       . . . (2)

⇒   n(2acos⁡θ2)=2asin⁡θ2\Rightarrow \,\,\,n\left( {2a\cos {\theta \over 2}} \right) = 2a{{\sin \theta } \over 2}

⇒   tan⁡θ2=n\Rightarrow \,\,\,\tan {\theta \over 2} = n

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