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General form of plane equation

  1. Jul 23, 2010 #1
    ...is Ax + By + Cz + D = 0.

    The vector <A, B, C> is a normal vector of the plane. My question is: does the value of D have any geometric significance/interpretation?

    I have an algorithm (that I didn't write myself) to evaluate the best-fit plane for a set of points in R3, and the value of D is coming back extremely large (10^20 or something obviously ridiculous). I wonder if D is just a mathematical artifact and I shouldn't worry, or if there is a problem with the algorithm (or my usage of it).

    I don't think D should be that large based on the equation for distance between a point (x0, y0, z0) and a plane Ax + By + Cz + D, which is

    D = abs(A*x0 + B*y0 + C*z0+ D) / sqrt(A^2 + B^2 + C^2)

    My values of A, B, and C are all -1<value<1, but D is so big that it will completely dominate that expression.

    Maybe this is a junior question and this thread should be re-categorized as such.

  2. jcsd
  3. Jul 23, 2010 #2
    The equation of a plane with normal [tex]\mathbf{n}[/tex] going through the point [tex]\mathbf{a}[/tex] is given by [tex](\mathbf{x} - \mathbf{a})\cdot \mathbf{n}=0[/tex]. Can you see how that helps?
  4. Jul 23, 2010 #3
    sure. Expanding the dot product gives the point-normal form (as opposed to the general form) of the plane equation. If a = (x0, y0, z0), x = <x, y, z> and n = <A, B, C>, we have A(x-x0) + B(y-y0) + C(z-z0) = 0

    That means D = -(x0 + y0 + z0), but I still don't see what the significance of that quantity is...
  5. Jul 24, 2010 #4
    Well, using your notation we have:

    [tex] (\mathbf{x} - \mathbf{a})\cdot \mathbf{n} =0 \quad \textrm{means} \quad Ax+By+Cz = \mathbf{n}\cdot \mathbf{a}[/tex]

    So your D is given by:

    [tex] D = - \mathbf{n}\cdot \mathbf{a} = -|\mathbf{a}| \cos \theta = |\mathbf{a}| \cos (\theta - \pi)[/tex]

    where [tex]\theta[/tex] is the angle made between the vectors [tex]\mathbf{a}[/tex] and [tex]\mathbf{n}[/tex]. So you have a geometrical interpretation of D.
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