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Find the center of the curve in 3D made by 4-points

  1. Nov 6, 2008 #1
    A(Ax,Ay,Az)
    B(Bx,By,Bz)
    C(Cx,Cy,Cz)
    D(Dx,Dy,Dz)

    Example:
    Points X Y Z
    a -2.00 0.00 -1.00
    b -2.00 0.00 -1.59
    c -1.59 0.00 -2.00
    d -1.00 0.00 -2.00
    The Answer is supposed to be ? Center: X:-1.0 Y:0.0 Z:-1.0

    What I am trying to solve is to find the equation of the plane based on two vectors AB , AC; then find the normal vector, and then find the plane equation, which look like this:

    mX+nY+oZ = m(Ax) + n(Ay) + o(Az) where
    m = (By-Ay)(Cz-Az) - (Bz-Az)(Cy-Ay)
    n = (Cx-Ax)(Bz-Az) - (Bx-Ax)(Cz-Az)
    o = (Bx-Ax)(Cy-Ay) - (By-Ay)(Cx-Ax)

    Now that I have the equation of the plane I am trying to find the the center of curve by using the equation of the plane.
    I believe i need to find a perpendicular vector to AB vector in my plane, and another perpendicular vector to CD vector in my plane, and find where the two vector meets, and that point would be the center of curve drawn. But I don't know how to do that??
     

    Attached Files:

  2. jcsd
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