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I have two bounded planes $\pi$ and $\rho$ in three dimensional space. Each plane is bounded by a coplanar rectangle. How can I find the orthogonal projection area of $\pi$ over $\rho$?

Thanks in advance.

Federico

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I'm not sure what you mean, but one can easily find the dihedral angle $\phi$ between two planes: $\cos\,\phi=\frac{a^\prime a+b^\prime b+c^\prime c}{\sqrt{a^2+b^2+c^2}\sqrt{{a^\prime}^2+{b^\prime}^2+{c^\prime}^2}}$, and then use Rodrigues's rotation matrix, using the line formed by the two planes as the axis. –  J. M. Sep 12 '11 at 16:53
    
@J.M.: yes, you're right. The problem is that I'm working with bounded planes, but I wanted to ask for the general case, that's why I wrongly thought I could think of a translation and a rotation, but as you said, all I have between two planes is an angle. –  Federico Sep 12 '11 at 17:17
    
@J.M.: +1 for your clarification; and I have posted the real problem I'm working on. –  Federico Sep 12 '11 at 17:32
    
What is a bounded plane? What are the coplanar rectangles coplanar with? The only way I can make sense of the question is to think that you're calling the plane surface surrounded by a rectangle a "bounded plane" and "coplanar" refers to the fact that the rectangle is coplanar with the plane surface it surrounds. That would be a very unusual use of these words. –  joriki Sep 13 '11 at 20:47
    
@joriki: yes, the interpretation you've made is correct, that's exactly what I was trying to convey; I really don't know of a better way to say it. –  Federico Sep 20 '11 at 18:33

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