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Find cartesian equation of hyperplane spanned by a set of vectors

  1. Mar 10, 2012 #1
    Let W be a hyperplane in R4 spanned by the column vectors v1 , v2, and v3, where

    Note that these are suppose to be COLUMN vectors:

    v1 = [3,1, -2 , -1], v2 = [0, -1, 0 , 1], v3= [1,2 ,6, -2]

    Find the Cartesian (i.e., linear) equation for W.

    I'm not quite sure where to start or how to interpret this problem. I was thinking about first finding the span or column space?... But after that I would not know how to convert into a cartesian equation.

    Any guidance or tips would be appreciated.

    Thank you.
     
  2. jcsd
  3. Mar 10, 2012 #2

    tiny-tim

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    Good morning, Mr. Johnson, how are you! :smile:

    Hint: how would you find the Cartesian equation of the plane in R3 spanned by two column vectors v1 and v2 ? :wink:
     
  4. Mar 10, 2012 #3
    Ok so I found an example but I feel like it isn't very intuitive and that there are better methods and approaches to this problem.

    So here is what I did:

    The hyperplane W is of the form Ax1 + Bx2 + Cx3 + Dx4 = 0 since it must pass through the origin.

    rref ( 3 1 -2 -1) = [1 0 0 0]
    (0 -1 0 1 ) [ 0 1 0 -1]
    (1 2 6 -2 ) [0 0 1 0]

    Thus A = C =0
    B = D

    => 0x1 + Bx2 +0x3 + Bx4 =0

    Dividing by B => x2 + x4 =0 which is the final answer for the cartesian equation for W.

    Can anyone verify? I know there is a better approach to this problem.
     
  5. Mar 10, 2012 #4

    tiny-tim

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    not actually following what you've done there :redface:,

    but your result x2 + x4 = 0 is obviously correct! :smile:

    the more general method i was thinking of (for two vectors in ℝ3) was to find their cross product using a determinant …

    can you see a 4D equivalent of that? :wink:
     
  6. Mar 10, 2012 #5

    HallsofIvy

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    A little more direct way: Any vector in the span of those three vectors can be written
    <x, y, z, t>= a<3, 1, -2, -1>+ b<0, -1, 0, 1>+ c<1, 2, 6, -2>= <3a+ c, a- b+ 2c, -2a+ 6c, -a+ b- 2c>

    so that x= 3a+ c, y= a- b+ 2c, z= -2a+ 6c, t= -a+ b- 2c. Those are parametric equations for the plane, also there are three equations in the three "unknowns", a, b, and c. Can you solve the equations for a, b, and c, to get just a single equation in x, y, z.
     
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