It's not solvable. Even with the normal, there are an infinite number of rectangles you can have in that plane given the diagonal there. I mean, just taking your diagram:
---D --/ C --/ /--- ---/ /-- --/ /-- --/ /-- --/ /-- A /-- B---
Same diagonal, same plane, different rectangle.
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It's not solvable. Even with the normal, there are an infinite number of rectangles you can have in that plane given the diagonal there. I mean, just taking your diagram:
---D --/ C --/ /--- ---/ /-- --/ /-- --/ /-- --/ /-- A /-- B---
Same diagonal, same plane, different rectangle.
He does have the dimensions as well, which reduces it to two choices (and it's possible that could be disambiguated further by other contraints, such as axis alignment or winding).
Still, this seems like a pain. So to the OP, I'll ask again: what is this for exactly, and why this representation? There are a number of other representations for a rectangle in 3D that would be much more convenient.
tnx guyz for your replies, but could you please tell me what's the best way of representing 3D rectangle? cos I think by having two points which are the opposite corners and a normal to that surface you should be able to define that rectangle.
Quote:Original post by jeff_rowa cos I think by having two points which are the opposite corners and a normal to that surface you should be able to define that rectangle.
Nope. As noted previously, in the absence of additional information there is an infinite number of solutions. Knowing the dimensions of the rectangle makes the number of solutions finite, but there is still ambiguity.
A better representation would be a center point and two orthonormal basis vectors (a coordinate frame), and the dimensions (width and height). Other variations of this that offered essentially the same information would work as well.
oh, sorry :) I think I found a way for solving it, by having opposite corners and knowing the angle is 90 and the normal: since normal is always one of the i,j or k vectors, it should looks like as if rectangle is drawn in 2D. So for instance if one corner is 0,0,0 and the other is 1,1,0 there should be a way to calculate the two other points which are 1,0,0 and 0,1,0. cos the normal is also 0,0,1
I think it's solveable only if you're saying it's a square, not a rectangle.
Consider the 2d case: In the upper row you clearly see that you can form several rectangles from the same two points. In the lower row you see that you can only build one square from two points.
It works exactly the same in 3d but with the exception that you need a normal to define the plane, otherwise you'll have an infinite number of squares rotated arounf the axis formed by the two points.
Hi eq, tnx for your reply, yeah I meant square, sorry about my english guyz!!, so by saying that I want to draw a square, how can I calculate the other two points based on the other two points that you marked in your diagram and the normal in 3D.
