1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Proof, equation of a plan, linear algebra
  1. Linear Algebra proof (Replies: 21)

Loading...