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

 

  • Option 1)

    \sin ^{-1}\left [ \frac{n-1}{n+1} \right ]

  • Option 2)

    \cos ^{-1}\left [ \frac{n^{2}-1}{n^{2}+1} \right ]

  • Option 3)

    \sin ^{-1}\left [ \frac{n^{2}-1}{n^{2}+1} \right ]

  • Option 4)

    \cos ^{-1}\left [ \frac{n-1}{n+1} \right ]

Answers (1)

best_answer

 

Triangle law of vector Addition -

If two vector are represented by both magnitude and direction by two sides of triangle taken in same order then their resultant is represented by 3rd side of triangle. 

- wherein

Represents triangle law of vector Addition

A^{2}+A^{2}+2AB\cos \theta = n^{2}\left ( A^{2}+A^{2}-2AB\cos \theta \right )

A^{2}+A^{2}+2A^{2}\cos \theta = n^{2}\left ( A^{2}+A^{2}-2A^{2}\cos \theta \right )

2A^{2}\left ( 1+\cos \theta \right ) = 2A^{2}n^{2}\left ( 1-\cos \theta \right )

\cos \theta \left ( 1+n^{2} \right ) = n^{2}-1

\cos \theta =\frac{n^{2}-1}{n^{2}+1}

\theta = \cos ^{-1}\left ( \frac{n^{2}-1}{n^{2}+1} \right )

 


Option 1)

\sin ^{-1}\left [ \frac{n-1}{n+1} \right ]

Option 2)

\cos ^{-1}\left [ \frac{n^{2}-1}{n^{2}+1} \right ]

Option 3)

\sin ^{-1}\left [ \frac{n^{2}-1}{n^{2}+1} \right ]

Option 4)

\cos ^{-1}\left [ \frac{n-1}{n+1} \right ]

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