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Math Help - vector/plane proof

  1. #1
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    vector/plane proof

    Prove that if a, b, c are not all zero, then the equation: ax+by+cz+d=0 represents a plane and \langle a,b,c\rangle is a normal vector to the plane.

    Hint if a \neq0, then ax+by+cz+d=0 is equivalent to a(x+\frac{d}{d})+b(y-0)+c(z-0)=0

    My Calc III teacher loves proofs, but I'm not real good with them. How can I prove this? I'm not even sure how to approach it. Can anyone help? Thanks!
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  2. #2
    A Plied Mathematician
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    Well, the usual equation for a plane looks like this:

    \mathbf{n}\cdot(\mathbf{x}-\mathbf{x}_{0})=0,

    where \mathbf{n} is a unit vector normal to the surface, \mathbf{x} is a vector to the arbitrary point on the plane, and \mathbf{x}_{0} is a vector to any particular point on the plane.

    Can you get your equation to look like mine? If so, you're done.
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  3. #3
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    n \cdot (x-x_{0})=0
    n=\langle a, b, c\rangle
    \langle a, b, c\rangle \cdot (x-x_{0})=0
    \langle a,b,c\rangle\langle x,y,z\rangle-\langle a,b,c\rangle\langle x_{0},y_{0},z_{0}\rangle=0
    ax+by+cz-ax_{0}-by_{0}-cz_{0}=0
    ax+by+cz+d=0

    I know that -ax_{0}-by_{0}-cz_{0}=d, but is this enough to prove it? Where did you get the usual equation for a plane from? Thanks!
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  4. #4
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    It is simply using the fact that two vectors are perpendicular if and only if their dot product is 0. Let (x_0, y_0, z_0) be any fixed point and (x, y, z) be any other point. Then the vector connecting them is [tex]<x- x_0, y- y_0, z- z_0>[tex] and the "set of all points (x, y, z) such that the vector [tex]<x- x_0, y- y_0, z- z_0>[tex] is perpendicular to <a, b, c> (the tangent plane to that vector containing (x_0, y_0, z_0)) must satisfy
    <a, b, c>\cdot<x- x_0, y-y_0, z-z_0>= 0.
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