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Equation of a plane

  1. Mar 28, 2012 #1
    1. The problem statement, all variables and given/known data
    Determine the equation of the plane that passes through (5,-5,5) and is perpendicular to the lines of intersection of the planes 3x-2z+1=0 and 4x+3y+7=0


    2. Relevant equations

    Ax+By+Cz+D=0

    3. The attempt at a solution

    I found the cross product of the normals of the planes given. Then used that as the direction vector of the line of intersection. Then I let z=0 and solved for x and y using the equations of the planes given in order to find a point on the line of intersection. The equation for the line of intersection is r=(-1/3,-17/9,0) + s(6,8,9).

    I'm not sure what to do now. Can someone please explain?
     
  2. jcsd
  3. Mar 28, 2012 #2

    HallsofIvy

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    Great! Now use the fact that if <A, B, C> is the normal to the plane and the plane contains the point [itex](x_0, y_0, z_0)[/itex] then the plane is given by [itex]A(x- x_0)+ B(y- y_0)+ C(z- z_0)= 0[/itex]. Of course, you use [itex](x_0, y_0, z_0)= (5, -5, 5).

    The "equation of the line of intersection" is irrelevant. The point you are given is not near that line.
     
  4. Mar 28, 2012 #3
    Ohhh I see! Thanks for the help!
     
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