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Math Help - Equation of a plane

  1. #1
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    Equation of a plane

    Determine the equation of the plane that passes through points (1,0,-1), (0,2,3), and (0,0,0).

    Letting the equation of the plane be ax+by+cz+d=0

    Skipping through the steps...

    cx-\frac {3}{2}cy+cz=0

    2x-3y+2z=0

    How do you know c=2?
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  2. #2
    Senior Member Twig's Avatar
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    Hi

    You could use a normal vector to the plane and ANY point on the plane to get your equation.

    Your plane goes through the origin. You have two vectors in the plane:

     (1,0,-1) \mbox{ and } (0,2,3)

    We use the fact that a crossing two vectors (cross-product) produces a vector that is normal to the plane spanned by these two vectors:

    (1,0,-1) \times (0,2,3) = (2,-3,2)

    Now, since these vector is normal to the plane,  \underbrace{(2,-3,2) \cdot (x,y,z)=0}_{\text{Scalarproduct}} \Longleftrightarrow 2x-3y+2z=0
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  3. #3
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    I've never learned cross product, only inner product.
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  4. #4
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by chengbin View Post
    Determine the equation of the plane that passes through points (1,0,-1), (0,2,3), and (0,0,0).

    Letting the equation of the plane be ax+by+cz+d=0

    Skipping through the steps...

    cx-\frac {3}{2}cy+cz=0
    Just divide throughout by c and rearrange to get the equation of the plane ie 2x-3y+2z=0.
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