Finding the Angle Between Two Planes: A Different Approach

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Homework Statement



I got everything in the answer, just that my answer was 2∫ d∅ was from [0 to ∏]. Same answer, but different approach.


The Attempt at a Solution



How can the angle between 2 planes be greater than ∏? I took 2∫ d∅ from [0 to ∏] because I considered 2 cases, where y > 0, and y < 0...
 

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This boils down to the angle between their normal vectors. If one vector is fixed in the sense that it is always the reference in relation to which the angle is measured, then the angle with the other one is in [0, 2\pi]. If the reference is not fixed, then either vector may be used as one, and then you have ambiguity in the definition of the angle, it can always be taken to be less and greater than \pi.
 
I think using the standard interpretation of the angle between two planes, that angle is never greater than ##\pi/2##. You calculate it by calculating the angle between the two normals using$$
\theta =\arccos\left(\frac{n_1\cdot n_2}{|n_1||n_2|}\right)$$and taking the supplementary angle if that comes out between ##\pi/2## and ##\pi##.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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