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Intersection of two planes

  1. Jun 11, 2010 #1
    1. The problem statement, all variables and given/known data

    P1 = 3x+4y=1
    P2 = x +y -z=3

    Find the intersection.

    2. Relevant equations

    x = x0 +at
    y = y0 + bt
    z = z0 + ct

    3. The attempt at a solution
    I used a quicker method. I first did P1 x P2 and got its cross product <-4,3,-1>

    Then I followed what our teacher had shown us in class - we did substitution.
    3x+4y=1
    x+y -z =3
    let x = 0, then 4y = 1 means y = 1/4 and y -z = 3 will give us -11/4

    then i had both direction vector <-4,3,-1> and point (0,3,-11/4) and i could write the equations of the line of intersection using the parametric equations

    x = -4t + 0
    y = 3t + 1/4
    z = -1t -11/4

    the book gives
    x = 4t +11
    y = -3t - 8
    z = t

    it seems like the author made z = 0 (in the parametric equation z0 = 0)
    my vector was <-4,3,-1> and his was <4,-3,1> and they were scalar multiple of -1

    i know there are many possible parametric equations describing the same intersection. but is my result valid, even though my x0, y0, and z0 are different?
     
  2. jcsd
  3. Jun 11, 2010 #2

    rock.freak667

    User Avatar
    Homework Helper

    Yes they are the same essentially.
     
  4. Jun 11, 2010 #3
    thank you rock.
     
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