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Equation of a plane containing two lines

  1. Nov 15, 2009 #1
    1. The problem statement, all variables and given/known data

    Find an equation for the plane containing the lines:
    x-3= 4t , y-4 = t, z-1 = 0
    x+1 = 12t, y-7 = 6t, z-5 = 3t

    3. The attempt at a solution

    I found the equations of both lines.
    First line:
    (3, 4, 1) + t(4, 1, 0)
    (-1, 7, 5) + t(12, 6, 3)

    Now I take the cross product of (4, 1, 0) and (12, 6, 3) to get (3, -12, 12)

    I found two answers from reviewing this question and I'm confusing myself, so I was wondering which answer is right. (If any)

    3x - 12y + 12z = 27
    or
    3x - 12y + 12z = 0

    Thanks in advance.
     
  2. jcsd
  3. Nov 15, 2009 #2

    LCKurtz

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    Well, you know several points on the plane. Try a few in your equations.
     
  4. Nov 15, 2009 #3

    Dick

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    I don't think either one is right. You are right that the plane equation is 3x - 12y + 12z=C. Find the correct value for C by putting say t=0 into your line equations and solving for it.
     
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