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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?
  2. jcsd
  3. Dec 21, 2008 #2


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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?
  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


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    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
    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:
    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.
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