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Finding Parametric equations for the line of intersection of two plane

  1. Jun 16, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the parametric equations for the line of intersection of two planes

    2. Relevant equations

    Equations for the two planes...
    z=x+y,-------(1)
    2x-5y-z=1 -----(2)

    3. The attempt at a solution

    My answers are not correct so I guess I'm going about it the wrong way. Someone please walk me through it from the top.
     
    Last edited: Jun 16, 2013
  2. jcsd
  3. Jun 16, 2013 #2
    Also, it would be helpful if you solved it with the elimination method because that's the way we were shown in class.
     
  4. Jun 16, 2013 #3

    Dick

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    You really need to show your attempt before anyone can help you, please?
     
  5. Jun 16, 2013 #4
    First thing I did was eliminate x.
    Rearranged the first equation to get, x+y-z=0 -------(3)

    Multiplied (3) through by 2 to get, 2x+2y-2z=0 -------(4)
    Subtracted (2) from (4) to get, 7y-z=-1 or y=(z-1)/7 ----(5)
    Let z=t, so I got y=(t-1)/7.

    Did the same thing for y, multiplying the first equation through by -5. And I got my final answer to be
    x=(1+6t)/(-7) then z=t

    The answers are x=6t, y=(-1/6)+t, z=(-1/6)+7t
    My process could be completely wrong, which is why I need someone to help me out from the top. I'm just so confused.
     
  6. Jun 16, 2013 #5

    HallsofIvy

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    Yes, this is correct.

    This is incorrect. Multiplying x+y- z= 0 by 5 gives 5x+ 5y- 5z= 0. Adding that to 2x- 5y- z= 1 gives 7x- 6z= 1 so that 7x= 1+ 6z, x= (1+ 6z)/7, not "-7".

    You understand that there can be many different parameters for the same line don't you?
    From your "x= (1+ 6t)/7, y= (t- 1)/7, z= t" (that negative sign corrected), we can let x= 6s and have, from the first equation, 6s= (1+6t)/7 so 1+ 6t= 42s, 6t= 42s- 1, t= 7s- 1/6. Then y= (t- 1)/7= (7s- 1/6- 1)/7= (7s- 7/6)/7= s- 1/6. That is x= 6s, y= (-1/6)+ s, z= (-1/6)+ 7s, just what you give, except with "s" instead of "t".

     
  7. Jun 16, 2013 #6
    Keep scrolling...
     
    Last edited: Jun 16, 2013
  8. Jun 16, 2013 #7
    Okay, Why did you add the two equations? instead of subtract like in the first one. Aren't you supposed to do the same thing to both parts?

    Hmmmm....I didn't know that about the final answer. Must have missed it in class. Now let's just clarify the middle part. Thanks and thanks in advance! :-).
     
  9. Jun 17, 2013 #8

    HallsofIvy

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    I added 5x+ 5y- 5z= 0 to 2x- 5y- z= 1 in order to eliminate "y"": 5y- 5y= 0.

    What two parts are you talking about? I did add both parts of the equations: on the left (5x+ 5y- 5z)+ (2x-5y- z)= (5x+ 2x)+ (5y- 5y)+ (-5z- z)= 7x- 6z and on the right 0+ 1= 1
    7x- 6z= 1.

     
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