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Show the line of intersection of both planes by finding the coordinates of a point

  1. Aug 19, 2012 #1
    1. The problem statement, all variables and given/known data

    The planes ∏1 and ∏2 have equations 3x - y - z = 2 and x + 5y + z = 14 respectively. Show that the point (3,1,6) lies on both planes.

    The Question:
    By finding the coordinates of another point lying in both planes, or otherwise, show that the line of intersection of ∏1 and ∏2 has equation r = 3i + j + 6k + t ( i -j + 4k ).


    2. Relevant equations

    Show point (3,1,6) lies in both planes:
    Substitute point (3,1,6) into the r of both plane equations.


    3. The attempt at a solution

    The first part:

    Let point (3,1,6) be a

    Substitute a into r of both plane equations.

    The dot product of a and r = ''d" of the plane equations.
    LHS=RHS [Shown]

    How do I attempt the second part of the question? I do not quite understand the second part of the question.
    Please help me. :uhh:
    Thanks!
     
  2. jcsd
  3. Aug 19, 2012 #2

    LCKurtz

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    Re: Show the line of intersection of both planes by finding the coordinates of a poin

    You just need to find x,y, and z that work in both equations. You have 2 equations in 3 unknowns, so it should be easy with an extra unknown. Try something like letting y = 0 and see if you can find an x and z that work.
     
  4. Aug 20, 2012 #3

    HallsofIvy

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    Re: Show the line of intersection of both planes by finding the coordinates of a poin

    You are given planes 3x - y - z = 2 and x + 5y + z = 14 and want to show that their line of intersection is r = 3i + j + 6k + t ( i -j + 4k ).
    Since you are given a line you don't have to solve for it, just check as you did for the point.

    You have already shown that (3, 1, 6) is in both planes so all you now need to do is show that the vector i -j + 4k lies in both planes and so in the line of intersection. And you can do that by showing it is perpendicular to the normal vectors to both planes: 3i- j- k and i+ 5j+ k.
     
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