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Convert into vector form

  1. Jul 11, 2010 #1

    Mentallic

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    1. The problem statement, all variables and given/known data
    How do I convert [tex]ax_1+bx_2+cx_3+d=0[/tex] into vector form?


    3. The attempt at a solution
    I am completely at a loss here, mainly because I don't quite understand vector geometry.
     
  2. jcsd
  3. Jul 11, 2010 #2

    HallsofIvy

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    Re: Vectors

    If we are to assume that "[itex]x_1[/itex]", "[itex]x_2[/itex]", and "[itex]x_3[/itex]" are components of a vector then that equation would be written [itex]<a, b, c> \cdot <x_1, x_2, x_3>+ d= 0[/itex] where the first term is a "dot product".
     
  4. Jul 11, 2010 #3

    Mentallic

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    Re: Vectors

    I better go read up on dot products then. Thanks.
     
  5. Jul 11, 2010 #4

    Mentallic

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    Re: Vectors

    Before I go on, x1, x2 and x3 are just variables in 3 dimensions such as x,y,z. Not exactly sure if that is what you were assuming.

    Ok so given the formula for a dot product of two vector a and b is [tex]|a||b|cos\theta[/tex] then we have [tex]\sqrt{(a^2+b^2+c^2)(x_1^2+x_2^2+x_3^2)}cos\theta+d=0[/tex]

    This doesn't seem right... I don't know how to find the angle between each vector and this isn't anywhere near the kind of answer I'm looking for, it should be of a form similar to this:

    [tex]<x_1,x_2,x_3>=<0,0,d>+\lamda<a,0,0>[/tex]

    Although I'm possibly just using the dot product all wrong.
     
  6. Jul 11, 2010 #5
    Re: Vectors

    That's true; however, there's a much simpler definition of the dot product in this case:

    [tex]
    <a,b,c> \cdot <x_1,x_2,x_3> = ax_1+bx_2+cx_3.
    [/tex]

    As an additional remark, note that, for a plane in [tex] R^3 [/tex], we have the following:

    [tex] \vec{\textbf{n}} \cdot \vec{\textbf{x}} = 0, [/tex] where [tex] \vec{\textbf{n}} = <a,b,c> [/tex] is a normal vector to the plane and [tex] \vec{\textbf{x}} = <x_1,x_2,x_3> [/tex] is any point on the plane. This is intuitive when we consider the definition of the dot product that you provided. The angle between any point on the plane and a corresponding normal vector is 90 degrees. Thus, [tex] \cos(\theta) = \cos(90) = 0. [/tex]

    I hope this helps.
     
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