1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: 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

    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


    => 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


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


    User Avatar
    Science Advisor
    Homework Helper

    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


    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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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


    User Avatar
    Science Advisor
    Homework Helper

    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook