Planar geometry, orthognal projections of a piece of a plane

natsu igneel
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Homework Statement



I have a piece of a plane in 3 dimensions (imagine holding an enevelope in the air), and two orthognal projections which form quadrilaterals, one on the xy plane (i.e. looking at the enevelope from above) and one on the xz plane (looking at it from the side). We know the equation for a plane is f=ax+by+cz, and I need to reconstruct this equation given only the two projections. So in essence the problem statement is this:

Given only two orthognal projections of a finite plane, recreate that plane.

Homework Equations



The gradient, angle between vectors etc.

The Attempt at a Solution



The normal, n, to the plane f will be <a,b,c> and the normal to the xy and xz planes will be k <0,0,1> and j <0,1,0> respectively.

The angle between n and k is
cos(theta) = n *dot* k / |n| |k|

which works out to be c/sqrt(a^2 + b^2 + c^2)

The same can be done with the angle between j and k, alpha, except the numerator is b instead of c.

So that's the information I've worked out. I also know a 3D surface needs to be paramaterised with two paramaters, I've called them s and t. Since the cross product of two vectors gives the normal I also know that
∂R/∂s *cross* ∂R/∂t = n = <a,b,c>.

I'm not looking for an answer here just a pointer in the right direction. At the moment I suppose I'm trying to find expressions for a, b and c. But I don't see how to do that form what I have, let alone with only two projections not three.

Any help would be greatly appreciated.
 
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Assuming the boundary of the finite plane can be parameterized using one parameter t as (x(t),y(t),z(t)), it would have a constant projection to (a,b,c), and the projection onto xy and xz planes are easily given by (x(t),y(t),0) and (x(t),0,z(t)). Seems there are lots of equations you can come up with.
 
That's a good idea and for a plane one paramter like that will work, but I've specifically been told to have two paramaters, I think later on we're moving to more general surfaces so learning how to do it with two paramaters now will be helpful
 
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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