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Grade 11Vectors

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.

Pushkar soni

11 Years agoGrade 11

2 Answers

Piyush

11 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

10 Years ago

its ok are not .please feedback me. $cos\theta =\frac{n^{^{2}}-1}{n^{2}+1}$ From here you will get $A^{2}+B^{2}+2ABcos\theta =n^{2}\left \{ A^{2}+B^{2}-2ABcos\theta \right \}$ $\left |A+B \right |^{2}=n^{2}\left | A-B \right |^{2}$

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