Interection of a plane and a 3D curve

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Discussion Overview

The discussion revolves around the intersection of a 3D curve (or surface) with a plane, specifically how to determine the points of intersection and describe the resulting cross-section. Participants explore both specific cases, such as the intersection of the surface defined by Sin[x*y]=z with the xy-plane, and more general scenarios involving arbitrary planes.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants question the terminology used, noting that a "curve" is one-dimensional and does not have a cross-section in the traditional sense.
  • One participant suggests that to find the intersection of the surface Sin[x*y]=z with the xy-plane, one can set z=0, leading to the equation sin(xy)=0, which results in a series of hyperbolae.
  • Another participant inquires about defining a plane that is not parallel to any axis and seeks guidance on how to derive the intersection points with a surface.
  • Participants discuss the method of defining a plane using three non-collinear points and the mathematical formulation that allows for determining if a point lies on that plane.
  • There is a suggestion that solving the intersection equations in a specific form could yield the desired cross-section.

Areas of Agreement / Disagreement

Participants express differing views on the terminology and the nature of curves versus surfaces. While some agree on the method to find intersections, there is no consensus on the best approach for arbitrary planes or the implications of the definitions used.

Contextual Notes

Limitations include the need for clarity on the definitions of curves and surfaces, as well as the assumptions made about the planes being discussed. The discussion does not resolve the complexities involved in defining intersections with arbitrary planes.

soandos
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is there a way to find out what a given "cross-section" of a curve looks like?
assuming that the curve itself is known and can be easily plotted?
is there a way to define a plane in a way that the points of intersection form the cross section?
thanks.
 
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Strictly, a "curve" is a one-dimensional object and so does not have any "cross section". Do you mean the cross section where a plane intersects a region of space? If so, how is the region given?
 
so i guess a "curve" is the wrong way to describe it.
lets say that i want to find the points of intersection between Sin[x*y]=z with the xy plane.
i.e. find an equation that can be plotted on two axis that will describe such a 1D curve.
is there a way to do this?
 
Okay, that's the intersection of a surface and a plane. Of course, the xy-plane is defined by z= 0 so just put z= 0 in sin(xy)= z to get sin(xy)= 0. Since sin(u)= 0 is satified for u= n\pi for n any integer, that is xy= n\pi or y= n\pi/x, a hyperbola, in the z= 0 plane: The intersection is a series of hyperbolae (x, n\pi/x, 0) for n any integer.
 
and if i wanted a plane in general, not one that is parallel to any axis?
 
For a general plane (ax + by + cz = d), you could try solving for one of the coordinates in the equation for the plane, and inserting it into the equation for your surface.
 
1) if i wanted to define a plane, can i just pick three points on the plane and then solve for a,b,c?
2) would solving the above equation and finding the points of intersection with the surface in the form y= yield the "cross section" that i want?
 
1) Yes you can. There's a nice formula for that as well: given three non-collinear points (x1, y1, z1), (x2, y2, z2), and (x3, y3, z3) on a plane, a point (x, y, z) lies on the plane iff
\begin{vmatrix}<br /> x_1 &amp; y_1 &amp; z_1 &amp; 1 \\<br /> x_2 &amp; y_2 &amp; z_2 &amp; 1 \\<br /> x_3 &amp; y_3 &amp; z_3 &amp; 1 \\<br /> x &amp; y &amp; z &amp; 1<br /> \end{vmatrix} = 0.
That is, the point (x, y, z) is an affine combination of the three given points.

Obviously, a nice choice of your three points would simplify these calculations significantly; for instance:
\begin{vmatrix}<br /> x_1 &amp; 0 &amp; 0 &amp; 1 \\<br /> 0 &amp; y_2 &amp; 0 &amp; 1 \\<br /> 0 &amp; 0 &amp; z_3 &amp; 1 \\<br /> x &amp; y &amp; z &amp; 1<br /> \end{vmatrix} = 0 \Longleftrightarrow \frac{x}{x_1} + \frac{y}{y_2} + \frac{z}{z_3} = 1,
provided that x_1 y_2 z_3 \ne 0.

2) Yes.
 

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