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: A frustrating plane question

  1. Aug 31, 2009 #1
    1. The problem statement, all variables and given/known data

    Prove that the line of intersection of the planes x+2y-z=2 and 3x+2y+2z=7 is parallel to the line x=1+6t, y=3-5t, z=2-4t. find an equation of the plane determined by the two lines

    3. The attempt at a solution[/
    cross product of n1 by n2 to determine direction of intersecting line, that answer is
    (6,-5,-4) which is parallel to l2 as its direction vector is the same so the scalar c=1, thus parallel. From here, I let x=0 in both plane 1 and 2 to determine a position vector for my intersecting line, thus
    2y-z=2 and 2y+2z=7
    z=2y-2 so y=11/6 and z=10/6, x=0
    thus the intersecting line is determined by r=(0,11/6,10/6)+t(6,-5,-4)
    now is where I get stuck. Intuitively, I want to find one point p on r say the position vector
    (0,11/6,10/6), then find two point p1, p2 on l2, then find two vectors pp1, pp2, cross product them, and find my answer...unfortunately this doesn't work thus after an hour of frustration I beg for help
  2. jcsd
  3. Aug 31, 2009 #2


    User Avatar
    Homework Helper

    this is correct.

    :confused: I am not sure what you are doing here. You already found the line of intersection between the two planes. What exactly are you trying to do?
  4. Aug 31, 2009 #3
    I am trying to find a plane that contains (I presume that is one what determines means) these two lines
  5. Sep 1, 2009 #4


    User Avatar
    Science Advisor

    Choose any point on the first line, say t= 0, which gives (1, 3, 2). Choose any point on the second line, say t= 0, which gives (0, 11/6, 10/6). Then the vector from (0, 11/6, 10/6) to (1, 3, 2), <1, 7/6, 2/6> is a vector in the plane. Now, you don't need another point because you know the direction vector of both lines, <6, -5, -4> is also in the plane. Use the cross product of those two vectors to find a vector perpendicular to the plane.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook