1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Interection of a plane and a 3D curve

  1. Dec 20, 2008 #1
    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.
     
  2. jcsd
  3. Dec 21, 2008 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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?
     
  4. Dec 21, 2008 #3
    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?
     
  5. Dec 21, 2008 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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 [itex]u= n\pi[/itex] for n any integer, that is [itex]xy= n\pi[/itex] or [itex]y= n\pi/x[/itex], a hyperbola, in the z= 0 plane: The intersection is a series of hyperbolae [itex](x, n\pi/x, 0)[/itex] for n any integer.
     
  6. Dec 22, 2008 #5
    and if i wanted a plane in general, not one that is parallel to any axis?
     
  7. Dec 22, 2008 #6
    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.
     
  8. Dec 22, 2008 #7
    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?
     
  9. Dec 22, 2008 #8
    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
    [tex]\begin{vmatrix}
    x_1 & y_1 & z_1 & 1 \\
    x_2 & y_2 & z_2 & 1 \\
    x_3 & y_3 & z_3 & 1 \\
    x & y & z & 1
    \end{vmatrix} = 0.[/tex]
    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:
    [tex]\begin{vmatrix}
    x_1 & 0 & 0 & 1 \\
    0 & y_2 & 0 & 1 \\
    0 & 0 & z_3 & 1 \\
    x & y & z & 1
    \end{vmatrix} = 0 \Longleftrightarrow \frac{x}{x_1} + \frac{y}{y_2} + \frac{z}{z_3} = 1,[/tex]
    provided that [tex]x_1 y_2 z_3 \ne 0[/tex].

    2) Yes.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Interection of a plane and a 3D curve
  1. 3D Plane question (Replies: 3)

Loading...