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Parametric Equation of a Line from the intersection of two planes

  1. Feb 14, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the parametric equation for a line of intersection of these two planes

    x+2y+3z=0
    4x+5y+6z=5
    2. Relevant equations
    Normal to plane 1= <1,2,3>
    Normal to plane 2= <4,5,6>


    3. The attempt at a solution

    I know the way to do this problem is to take cross product of two normals etc etc,
    but i want to know if the way i did this is correct also.

    I already turned the work in so theres nothing i can do to change it but the curiosity is killing me.

    First, i set the two planes equal to each other

    4x+5y+6z-5=0
    x+2y+3z=0

    => x+2y+3z=4x+5y+6z-5 (please correct me my thinking process is wrong,im winging it)
    => 3x+3y+3z=5
    => x+y+z=5/3 (now im thinking this is the equation of the intersection of the two planes but this isnt the equation of a line, it looks like a plane, or is it?)

    so i took a point on this set, (5/3,0,0) and two other points (0,5/3,0) and (0,0,5/3)

    i did <0,0,5/3> - <0,5/3,0> = <0,-5/3,5/3> as a directional vector.

    so L(t)= (5/3,0,0) + <0,-5/3,5/3>t

    x=5/3
    y=-5/3t
    z=5/3t

    IS any of this wrong?!
     
  2. jcsd
  3. Feb 14, 2010 #2

    Dick

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    Try it out. Put your x,y and z into the two original plane equations. At t=0 you get (5/3,0,0). That's not on either of the original planes, is it?
     
  4. Feb 14, 2010 #3
    Ah i do notice that, but

    x+2y+3z=4x+5y+6z-5

    if i plug them into that those points solve the equation, which is the x,y,z such that those two planes are equal, or is that me failing at winging a problem?
     
  5. Feb 14, 2010 #4

    Dick

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    If you mix up two planes you get a third plane. A line in that plane may or may not be in either of the two planes you started with. I admire your spirit of winging it, but when you scamble the two planes, you loose information.
     
  6. Feb 14, 2010 #5
    so i just needed to restrict that plane to a set of values that lie in both planes. so im just wondering what exactly did i do by setting the two planes equal to each other? i notice i get the same result if i subtract equation 1 from equation 2, so did i just subtract the two planes by setting them equal? im confused as to what that actually does
     
  7. Feb 14, 2010 #6

    vela

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    Yeah, you effectively just subtracted one equation from the other. This gives you a new plane in which the line of intersection lies, so you're just finding the equation of a plane rotated about that line.
     
  8. Feb 15, 2010 #7

    Dick

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    Try a simpler example. Take the planes x=0 and z=0. They intersect along the line (0,t,0). Subtract them and get x-z=0. That's a different plane. Still contains (0,t,0) though. Adding and subtracting the plane equations isn't getting you any closer to finding the intersection. I think you'd better stick with the normal and cross product method.
     
  9. Feb 15, 2010 #8
    okay thanks guys for all your help. i understand this a lot better now
     
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