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Finding the Equation of a plane

  1. Nov 8, 2009 #1
    1. The problem statement, all variables and given/known data
    The plane that passes through the line of intersection of the plane x-z=1 and y+2z=3 and is perpendicular to the plane x+y-2z=1


    2. Relevant equations



    3. The attempt at a solution

    I found the intersection point of the 2 planes by setting z=0 yielding the point (1,3,0). Next I found the normal of the 2 planes that is intersecting, (1,0,-1)x(0,1,2)= (1,-2,1). the normal of the perpendicular plane is (1,1,-2). since the 2 normal are not parallel, i found another normal (1,-2,1) x (1,1,-2) which yields(3,3,3). Then substituting the point (1,3,0) and (3,3,3) into the plane equation. So 3(x-1)+3(y-3)+3z=d

    I am unsure if this is the correct answer or this is how you do it? It would be nice if someone would confirm this with me or lead me to the right direction on how to solve this.

    Thanks in advance
     
  2. jcsd
  3. Nov 8, 2009 #2

    Mark44

    Staff: Mentor

    The equation you found is incorrect. The normal to this plane is <3, 3, 3>. In your other work, you found the normal to the plane in question to be <1, 1, -2>.

    If the normal to a plane if n = <A, B, C> and a point on the plane is (x0, y0, z0), then the equation of the plane is A(x - x0) + B(y - y0) + C(z - z0) = 0.

    The point you found, (1, 3, 0) is on the line of intersection of the two given planes, and the normal to the third plane is <1, 1, -2>. These two things are all you need to find the equation of the plane you want.
     
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