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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

    rock.freak667

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    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

    HallsofIvy

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    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.
     
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