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Two vectors , (vector A) and (vector B) have equal magnitudes. If magnitude of (vector A + vector B) is equal to n times the magnitude of (vector A – vector B), then the angle between (vector A) and (vector B) is: Answer : cos^-1(n^2 – 1/n^2 + 1) or negative cosine of n(square) minus 1 over n(square) plus 1. could some one explain the solution. this question is from odisha JEE 2010.

Two vectors ,  (vector A) and (vector B) have equal magnitudes. If magnitude of  (vector A + vector B) is equal to n times the magnitude of (vector A – vector B), then the angle between (vector A) and (vector B) is:
 
Answer :  cos^-1(n^2 – 1/n^2 + 1)
or negative cosine of n(square)  minus 1 over n(square)  plus 1.
 
could some one explain the solution.  this question is from odisha JEE 2010.
 

Grade:11

2 Answers

Piyush
askIITians Faculty 112 Points
8 years ago
\left |A+B \right |^{2}=n^{2}\left | A-B \right |^{2}
A^{2}+B^{2}+2ABcos\theta =n^{2}\left \{ A^{2}+B^{2}-2ABcos\theta \right \}
From here you will get
cos\theta =\frac{n^{^{2}}-1}{n^{2}+1}
RIYAZSHAIK
59 Points
8 years ago
 
its ok are not .please feedback me.cos\theta =\frac{n^{^{2}}-1}{n^{2}+1}From here you will getA^{2}+B^{2}+2ABcos\theta =n^{2}\left \{ A^{2}+B^{2}-2ABcos\theta \right \}\left |A+B \right |^{2}=n^{2}\left | A-B \right |^{2}
if u satisfied with my answer you can approve me

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