Two vector A and B have equal magni...
Question
Two vector A and B have equal magnitude. The magnitude of (A + B) is n times the magnitude of (A − B). The angle between A and B is
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Two vector ( vec{A} ) and ( vec{B} ) have equal magnitude. The
magnitude of ( (vec{A}+vec{B}) ) is ( n ) times the magnitude of
( (vec{A}-vec{B}) . ) The angle between ( vec{A} ) and ( vec{B} ) is:

JEE/Engineering Exams
Physics
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Two vector A and B have equal magnitude. The magnitude of (A + B) is n times the magnitude of (A − B). The angle between A and B is
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where " ( theta ) is angle between vector

( |A-B|=sqrt{A^{2}+B^{2}-2 A B cos theta}={text {  }}{} )
[
|A+B|=sqrt{A^{2}+B^{2}+2 A B cos theta}=
]
as given A, B have same magnitudes
[
begin{array}{c}
(A-B)=sqrt{2 A^{2}-2 A^{2} cos theta}
(A+B)=sqrt{2 A^{2}+2 A^{2} cos theta}
qquad begin{array}{c}
g_{i v e n} cdot nleft|A-B|_{}=right| A+B mid
n sqrt{2 A^{2}-2 A^{2} cos theta}=sqrt{2 A^{2}+2 A^{2} cos theta}
end{array}
end{array}
]

Squaring on both sides
[
begin{array}{c}
n^{2}left(2 A^{2}-2 A^{2} cos thetaright)=left(2 A^{2}+2 A^{2} cos thetaright) ^{2} times 2A^{2}(1 - cos theta) = 2A^{2}(1 +  cos theta)
n^{2}-n^{2} cos theta=1+cos theta
n^{2}-1=cos thetaleft(1+n^{2}right)
cos theta=frac{n^{2}-1}{n^{2}+1}
theta=cos ^{2}left(frac{n^{2}-1}{n^{2}+1}right)
end{array}
]

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