Angle for n-fold Magnitude Increase of A→ +B→

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SUMMARY

To determine the angle θ between two vectors A→ and B→ of equal magnitude such that the magnitude of A→ + B→ is larger than A→ - B→ by a factor of n, one must analyze the resultant vectors in two dimensions. By setting the vectors as (1,0) and (cos(θ), sin(θ)), the resultant for the sum is given by 2a cos(θ/2) and for the difference by 2a sin(θ/2). Solving the equation 2a cos(θ/2) = n * 2a sin(θ/2) leads to the necessary angle θ.

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Two vectors A→ and B→ have precisely equal magnitude .For the magnitude of A→ +B→ to be larger than the magnitude of
A→ - B→ by the factor n, what must be the angle between them?
 
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Clearly, 1) the actual length of the vectors doesn't matter, so let's just make them length 1 and 2) since there are only two vectors involved, we can work in two dimensions. So call one vector (1,0) and the other one (cos(theta),sin(theta)). Now just set up your question and solve for theta.
 
The resultant of both constructions are diagonals of a rhombi.

This means that for the sum the resultant will be

[tex]2a\cos(\frac{\theta}{2})[/tex]

where [tex]\theta[/tex] is the require angle and [tex]a[/tex] is the magnitude.

For the difference the angle between the vectors will be [tex]180^o - \theta[/tex]. In this case the resultant amounts to

[tex]2a\sin(\frac{\theta}{2})[/tex]
 

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