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Perpendicular Planes

  1. Jun 15, 2005 #1

    I suppose my question has to do with planes in general, rather than just tangent planes. Say you have a plane given by the equation

    [tex]z=\frac{\partial f}{\partial x}(x)+\frac{\partial f}{\partial y}(y).[/tex]

    How would you find the equation of the plane perpendicular to this one?

    Thanks for your help.
  2. jcsd
  3. Jun 16, 2005 #2


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    General approach:

    Equation of plane ax+by+cz+d=0, where at least one of (a,b,c) is not =0.
    Assume it is a, then bx-ay+m=0 or cx-az+n=0, where m and n are arbitrary. The general idea is to set up a nonzero vector perpendicular to (a,b,c) and use it as coefficients for x,y,z. The constant term is arbitrary.
  4. Jun 16, 2005 #3
    Where did the [tex]bx-ay+m=0[/tex] and [tex]cx-az+n=0[/tex] come from?
  5. Jun 16, 2005 #4
    Well, I guess I can understand that. If you have one vector on a plane "r-r0 = (x-x0)i + (y-y0)j + (z-z0)k" and the gradient vector "ai + bj +ck" tangent to f(x,y), would you compute the cross product between these two vectors to find that perpendicular vector, then use that as coefficients?

    (ex.) <(x-x0), (y-y0), (z-z0)> x <a, b, c> = <p, q, t>


    So... the tangent plane is px+qy+cz=d.

    Is this correct?
  6. Jun 17, 2005 #5


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    I tried to give an answer for planes in general. I am a little rusty on your special case (tangent planes).
    The plane equations in the previous post just simply reflect the fact that (b,-a,0) and (c,0,-a) are both non-zero and perpendicular to (a,b,c) as long as a is not 0.
  7. Jun 17, 2005 #6
    Alright that makes sense then. By which method did you compute that (b, -a, 0) and (c, 0, -a) are perpendicular to (a, b, c)?

    The reason I ask is that I have the equation [tex]z=\frac{\partial g}{\partial x}(x)+\frac{\partial g}{\partial y}(y)[/tex] and need to find the perpendicular plane (the problem actually has to do with projections).

    Thanks again.
    Last edited: Jun 17, 2005
  8. Jun 18, 2005 #7


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    In your case a=dg/dx, b=dg/dy, c=-1. Since c not=0, you can use (1,0,dg/dx) or
    (0,1,dg/dy) or any linear combination of the two vectors.
  9. Jun 18, 2005 #8
    Alright I understand now. Thanks a lot for your help.
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