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Desperate for love and affection- converting a parametric equation to Cartesian

  1. Sep 14, 2010 #1
    1. The problem statement, all variables and given/known data
    Give a Cartesian equation of the given hyperplane
    The plane passing through (1,2,2) and orthogonal to the line x = (5,1,-1) + t(-1,1,-1)


    2. Relevant equations



    3. The attempt at a solution
    So I've looked at this for a few hours and still can't figure out how to do it. It's supposed to be x(sub1) - x(sub2) + x(sub3) = 1, but I can't understand how one gets that solution. I know there should be a system of equations, with one of the equations being (1,-1,1) (which is orthogonal to the line in the question) but how about the other equations? Is the other hyperplane used for the system (5,1,-1) - (1,2,2) or something? Someone please help?!
     
  2. jcsd
  3. Sep 14, 2010 #2

    gabbagabbahey

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    Hint: If I gave you the normal to a plane and a point in the plane, could you give me the equation of the plane?
     
  4. Sep 14, 2010 #3
    Is the parametric equation the line x = (1 +t, 2-t, 2+t)? I simply plugged in the point and the normal of the vector from the question.
    And I still can't figure out how to convert this to Cartesian form, but I might need to look at it some more.
     
  5. Sep 14, 2010 #4

    gabbagabbahey

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    No, a plane is not a line...
     
  6. Sep 14, 2010 #5
    Uhh, you know what I mean. Help please?
     
  7. Sep 14, 2010 #6

    hunt_mat

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    Do you know what vector describes the diection of the line you have?
     
  8. Sep 14, 2010 #7
    You take the vector and divide it by its distance. So in the case of (-1,1,-1), it would be -1/sqrt3, 1/sqrt3, -1/sqrt3
     
  9. Sep 14, 2010 #8

    hunt_mat

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    It doesn't have to be of unit length. Now what does it mean to be orthogonal? Say if I had a point (x,y,z) which was orthogonal to (-1,1,-1)?
     
  10. Sep 14, 2010 #9
    Orthogonal=perpendicular. So... (1,-1,1) obviously isn't orthogonal to my first line
     
  11. Sep 14, 2010 #10

    hunt_mat

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    Think in terms of dot products of vectors, what does it mean to be orthogonal?
     
  12. Sep 14, 2010 #11
    Oh yes the dot product has to be zero. But couldn't then the normal vector be any number of lines?
     
  13. Sep 14, 2010 #12

    gabbagabbahey

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    No, I don't. I asked if you could determine the equation of a plane, given the normal and a point in the plane. You spat out the equation of a line.

    Don't worry about the line you are given yet. If I tell you that the normal to a plane is (a,b,c) and that the point [itex](x_0,y_0,z_0)[/itex] lies in that plane, what is the equation of that plane?
     
  14. Sep 14, 2010 #13
    Lets see, (x,y,z)+t(a,b,c). So...
     
  15. Sep 14, 2010 #14

    gabbagabbahey

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    No, that's the equation of a line (Actually, it isn't even an equation! But assuming you meant [itex](x_0,y_0,z_0)+(a,b,c)t = 0[/itex], its a line). Open up your textbook/notes and read the section on determining the equation of a plane (or Google it), and then try again.
     
  16. Sep 14, 2010 #15

    hunt_mat

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    You had the direction of te line correct, it was (-1,1,-1). There are an infitite number of lines which make up the plane which you're interested in.
     
  17. Sep 14, 2010 #16
    It's a(sub1)x(sub1)+a(sub2)x(sub2)+a(sub3)x(sub3) = c
     
  18. Sep 14, 2010 #17

    gabbagabbahey

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    Oh, and what are a(sub1), a(sub2), a(sub3) and "c" in terms of the vector and point I gave you?
     
  19. Sep 14, 2010 #18

    hunt_mat

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    If
    [tex]
    (x_{1},x_{2},x_{3})
    [/tex]
    if perpendicular to (-1,1,-1) then?
     
  20. Sep 14, 2010 #19
    [itex]
    ax_0+by_0+cz_0 = 0
    [/itex]
     
  21. Sep 14, 2010 #20

    gabbagabbahey

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    No.

    Straight from wikipedia:

    THIS is the equation you should have looked up.

    Anyways, if I give you a line that is orthoganl to a plane, what can you say about the direction of the given line and the normal of the given plane?
     
    Last edited by a moderator: May 4, 2017
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