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Math Help - Vector Parametric Equation

  1. #1
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    Vector Parametric Equation

    How do I find the equation of a plane with this information:
    Its through the point (-3, 9, 10)
    and x = 5+t y = 4t and z = 3-2t
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    Quote Originally Posted by acg716 View Post
    How do I find the equation of a plane with this information:
    Its through the point (-3, 9, 10)
    and x = 5+t y = 4t and z = 3-2t
    I'm guessing that the plane contains the line you mention. If so, then the vector

    \bold{u} = < 1, 4, -2>

    is on the plane. We can also find another vector on the plane, a point on the line to the point given

    \bold{v} = <5-(-3),0-9 , 3 -10> = <8,-9,-7>

    The normal to the plane is then obatin from

    \bold{n} = \bold{u} \times \bold{v} (*)

    If \bold{n} = <n_1,n_2,n_3>

    then the plane is given by

    n_1(x+3) + n_2(x-9) + n_3(x-10) = 0

    So all you need to do is find n *
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    When I did that I got -46(x+3)-9(y-9)-41(z-10)=0
    but that isn't the correct answer...
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    Quote Originally Posted by acg716 View Post
    When I did that I got -46(x+3)-9(y-9)-41(z-10)=0
    but that isn't the correct answer...
    Well, when you substitute the point (-3,9,10) into this plane we get 0, right? And when we substitute x = 5+t,\;\;y = 4t,\;\;z = 3-2t we get

     -46(5+t+3)-9(4t-9)-41(3-2t-10)=
     -46(8+t)-9(4t-9)-41(-7-2t)=
     -368 - 46 t +81-36t + 287 + 82t = 0

    So both the point and line lie on the plane. Did the question say that the line is perpendicular to the plane?
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    yes it is perpendicular, so sorry that i forgot to include that!
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    Quote Originally Posted by acg716 View Post
    yes it is perpendicular, so sorry that i forgot to include that!
    Makes a huge difference (and easier). The normal to the plane will be parallel so the line and so

    \bold{n} = <1,4,-2>

    so the line is

    (x+3) + 4(x-9) - 2(x-10)=0

    See, way easier.
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  7. #7
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    ahh it still says it wrong so maybe I'm leaving something out from the directions so here is what the directions say exactly:

    Find an equation of the plane through the point and perpendicular to the given line.

    (-3, 9, 10)
    x= 5 + t
    y= 4t
    z= 3-2t
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    Quote Originally Posted by acg716 View Post
    ahh it still says it wrong so maybe I'm leaving something out from the directions so here is what the directions say exactly:

    Find an equation of the plane through the point and perpendicular to the given line.

    (-3, 9, 10)
    x= 5 + t
    y= 4t
    z= 3-2t
    What answer are you given? Are you given x + 4y - 2z = 13?
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  9. #9
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    I'm not given an answer, its on my webassign homework so when I submit it it tells me if I'm right or not.
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  10. #10
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    Quote Originally Posted by danny arrigo View Post
    Makes a huge difference (and easier). The normal to the plane will be parallel so the line and so

    \bold{n} = <1,4,-2>

    so the line is

    (x+3) + 4(x-9) - 2(x-10)=0

    See, way easier.
    Quote Originally Posted by danny arrigo View Post
    What answer are you given? Are you given x + 4y - 2z = 13?
    Did you try both answers? If so, I don't know what to suggest. Can some of the other regulars chime in here.
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  11. #11
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    The second one was correct, thank you so much!!!
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