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, intersection of two planes

  1. Jun 3, 2009 #1
    urgent, intersection of two planes

    1. The problem statement, all variables and given/known data

    Find the equation of the intersection between 2x + 3y - 4z = 12 and 5x + 2y + 3z = 7.



    2. Relevant equations

    none. I don't know how to use cross products, but is there another way?



    3. The attempt at a solution


    I don't know how to do this and it will be on my final tomorrow... I tried various things but right now the closest I have is, set x to 0 and find for y and z. Please help, I have no idea what else to do.

    I know it is in this form (or at least, I think it is):

    [x, y, z] + t(a, b, c)

    but what are those? t is for time. is the x y z a point on the plane, and how would I find that?
     
  2. jcsd
  3. Jun 3, 2009 #2

    rock.freak667

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    Re: urgent, intersection of two planes

    t is not for time. t is a parameter.

    so for 2x + 3y - 4z = 12 and 5x + 2y + 3z = 7, what are the normals for these two planes?
    What will you get when you take the cross-product of these two normals?
     
  4. Jun 3, 2009 #3
    Re: urgent, intersection of two planes

    normals are [2, 3, -4] and [5, 2, 3] right?

    I don't know, I never learned to cross product. Is that the only way? I used to be able to do this I think like a year ago (can't find the notes for it) and I didn't use cross producting...how do you cross product? I tried googling it but didn't understand it.
     
  5. Jun 3, 2009 #4

    rock.freak667

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    Re: urgent, intersection of two planes

    ok, well there is another way to do it.

    Take 2x + 3y - 4z = 12 and 5x + 2y + 3z = 7

    Eliminate either x,y or z. When you get an equation with two of variables left, then let either of the variables be your parameter (t,a,b, whatever you like) and then find all the other variables in terms of that parameter.
     
  6. Jun 3, 2009 #5
    Re: urgent, intersection of two planes

    Ok is this how you do it?

    Let z = 0

    2x + 3y = 12 and 5x + 2y = 7

    x ends up equaling 1, y = 10/3

    I'm not sure what happens next? so is it

    [1, 10/3, 0] + t (a, b, c)

    How do i find the other variables in terms of that parameter?
     
  7. Jun 3, 2009 #6

    rock.freak667

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    Re: urgent, intersection of two planes

    Nope.

    2x + 3y - 4z = 12 is the same as 10x+15y-20z=60 (x5 throughout)

    5x + 2y + 3z = 7 is the same as 10x+4y+6z=14 (x2 throughout)



    10x+15y-20z=60 ........(1)
    10x+4y+6z=14 ...........(2)

    Find (1)-(2)
     
  8. Jun 3, 2009 #7
    Re: urgent, intersection of two planes

    OH OKAY. So then:

    9y - 26z = 46

    y = 46/9 + 26z/9

    And then you plug that into the equation above?

    Okay thanks, now I get how to get x, y, and z, but I'm still fuzzy on how to get it into [x,y,z] + t(a, b, c) form? Does that still apply?
     
  9. Jun 3, 2009 #8

    rock.freak667

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    Re: urgent, intersection of two planes

    it's not best to put the parameter as one of your variables, so we'll use t instead

    so you let z=t, and you found y in terms of t

    y=46/9 +26t/9

    So put that into any of the equations of the planes to find x in terms of t.
    What do you get then?
     
  10. Jun 3, 2009 #9
    Re: urgent, intersection of two planes

    9y - 26t = 46
    9 (46/9 +26t/9) - 26t = 46

    like that?
     
  11. Jun 3, 2009 #10

    rock.freak667

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    Re: urgent, intersection of two planes

    Not into that equation. Into either of two equations of the planes.

    Either put y=46/9 +26t/9 and z=t into 2x + 3y - 4z = 12 or 5x + 2y + 3z = 7.

    EDIT:

    When you get x in terms of t. Put x,y and z in the vector form as such

    [tex]\left( \begin{array}{c}
    x\\
    y \\
    z
    \end{array}\right)[/tex]
     
    Last edited: Jun 3, 2009
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