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Equation of the plane

  1. Jul 6, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the equation of the plane through the line of intersection of the planes

    x-z=1 and y+2z=3

    AND perpendicular to the plane x+y-2z=1

    2. Relevant equations
    The normal of the first plane is n1=(1,0,-1), second is n2=(0,1,2)

    3. The attempt at a solution

    I started off by finding the cross product of the intersecting planes, which is (1,2,1).
    I think this vector is the direction vector of the line that defines the intersection of the two planes. So I think this vector has to be ON the plane (THE plane who's equation we're trying to find).

    But now I'm not sure where to go from here. Since my plane has to be parallel to the third plane, I guess that the normal for that plane would be the normal for my plane. So I have to normal.....I need points.....
    I'm guessing this is an easy question, but I've just mixed myself up a lot!
  2. jcsd
  3. Jul 7, 2008 #2
    The line of intersection of the planes (1) x-z=1 and (2) y+2z=3 is on our plane.
    To find the line of intersection of these planes, evaluate 2(1)+(2) to obtain (3) 2x+y=5
    Let y=[tex]\lambda[/tex], then
    Sub into (1), to get z=(7-[tex]\lambda[/tex])/2
    Therefore the line of intersection can be represented by (x,y,z) = (5/2,0,7/2) + [tex]\lambda[/tex](-1/2,1,-1/2)
    The direction vector (-1/2,1,-1/2) is therefore parallel to the plane. Simplify this to (-1,2,-1)

    Now consider the fact that the plane we want is perpendicular to the plane x+y-2z=1. Thus it is parallel with the normal to x+y-2z=1, ie parallel to (1,1,-2).

    Now we have two lines parallel to our plane, (1,1,-2) and (-1,2,-1)
    Cross these to give (3,3,3) and simplify to (1,1,1). Thus (1,1,1) is a normal to our plane.
    Thus our plane is in the form x+y+z=D, where D is a number
    Since we already know that (5/2,0,7/2) is on the plane, sub this in to get D=6

    The plane you want is x+y+z=6
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