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Proof, equation of a plan, linear algebra

  1. Mar 14, 2009 #1
    1. The problem statement, all variables and given/known data
    Equation of a plan that contains the point p0=x0,y0,z0 and the normalvector N=A,B,C is N*P*P0=0 where P is(x,y,z)
    =>A(x-xo)+B(y-y0)+c(z-z0)=0

    Is it possible to prove this?
    Can you help me if it is?
     
  2. jcsd
  3. Mar 14, 2009 #2
    Maybe it should be N*(P-P0)=0 instead of N*P*P0?

    If I understand correctly, you are supposed to prove that the plane that contains the point P0 and has normal N is given by the equation N*(P-P0)=0. This is true. Did you attempt to prove it?
     
  4. Mar 14, 2009 #3
    No I didn´t because I have no idea how to or if it´s possible.
    Is it?
    And if it is, can you tell me how to do it.
     
  5. Mar 14, 2009 #4
    P - P0 is a vector that is parallel to the plane, so this must be orhogonal to the normal vector. Therefore:

    N dot (P - P0) = 0
     
  6. Mar 14, 2009 #5

    lanedance

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    Homework Helper

    Hi sciencegirl1

    drawing a picture always seems to help, but in line with yyat's comments, but the geometric reasoning is as follows

    first take a point on the plane
    [tex]\textbf{p}_0 = (x_0,y_0,z_0)[/tex]

    next take any other point on the plan
    [tex]\textbf{p} = (x,y,z)[/tex]

    the difference is a vector parallel to the plane
    [tex]\textbf{p}-\textbf{p}_0 = (x- x_0,y-y_0,z-z_0)[/tex]

    so if we have a vector normal to the plane
    [tex]\textbf{n}= (a,b,c)[/tex]

    then by defintion this is perpindciular to any vector parallel to the plane, so
    [tex]\textbf{n}*(\textbf{p}-\textbf{p}_0)= 0[/tex]

    i'm not too sure what you want to do here, but if you want to prove this try writing a formula for a plane, find the normal vector, then take the dot product with the difference of any 2 points on the plane and see if you can show its 0
     
    Last edited: Mar 14, 2009
  7. Mar 14, 2009 #6
    If you have shown that if P is in the plane, then:

    N dot (P-P0) = 0

    Then what you need to show is that the reverse is also true. This is equivalent to saying that if P is not in the plane, you have

    N dot (P - P0) not equal to zero.

    This follows from the fact that if P - P0 is not in the plane, it must have a component along N.

    So, you then have a necessary and sufficient condition for P to be in the plane: The equation

    N dot (P - P0) = 0

    This is therefore the equation of the plane.
     
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