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Finding an equation for a plane

  1. Dec 18, 2011 #1
    Find an equation for the plane that contains the line x = 1+t, y = 3t, z = 2t and is parallel to the line of intersection of the planes -x+2y+z = 0 and x + z + 1 = 0.

    On the attachment, the answer is there. My question is, (on the first page) right after he wrote, x,y, leading, z free, he wrote down the matrix form of the equation. Where did he get the values of the matrix right after z. The values are -1, -1 and 1. How did he find them, or where did he get them? Please explain, thanks!
     

    Attached Files:

  2. jcsd
  3. Dec 18, 2011 #2

    LCKurtz

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    If you write out the equations that step represents they are:

    x + z = -1
    y + z = -1/2 - z
    z = z

    Solve for (x,y,z) written as a column vector.
     
  4. Dec 18, 2011 #3
    Wait, how exactly do you get that though? I see how you get x + z = -1, but how do you get y + z = -1/2 -z? I also understand z = z. Can you explain the second line? thanks.
     
  5. Dec 18, 2011 #4

    LCKurtz

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    Isn't it obvious to you that it is just a typo? Did you try fixing it to see if it answered your question?
     
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