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Math Help - Finding a vector perpendicular to three vectors in R^4

  1. #1
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    Finding a vector perpendicular to three vectors in R^4

    I'm trying to find all vectors in R^4 that are perpendicular to the following three vectors:

    [ 1 1 1 1 ]
    [ 1 2 3 4 ]
    [ 1 9 9 7 ]


    I'm not sure exactly how to set up a matrix that would help me to find these vectors. Any help setting this up would be appreciated. Thanks!
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  2. #2
    A Plied Mathematician
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    Here's what I would do: set up a system of three equations in four unknowns. You've got your three known vectors, call them \mathbf{r}_{1}, \mathbf{r}_{2}, \mathbf{r}_{3}. The vector you're trying to produce that's perpendicular to all of them, we'll call \mathbf{s}. Then you can get three equations by setting

    \mathbf{s}\cdot\mathbf{r}_{1}=0,

    \mathbf{s}\cdot\mathbf{r}_{2}=0, and

    \mathbf{s}\cdot\mathbf{r}_{3}=0.

    Use normal Gaussian elimination to proceed. Does that make sense?
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  3. #3
    MHF Contributor

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    Another way of saying the same thing: call the vector <w, x, y, z>. Saying that is orthogonal to each of the given vectors means that
    w+ x+ y+ z= 0
    w+ 2x+ 3y+ 4z= 0
    w+ 9x+ 9y+ 7z= 0

    That is three equations in 4 unknow so you should be able to solve for three of them in terms of the fourth. That is, you should be able to find, say, w= az, x= bz, y= cz, for specific numbers a, b, and c, so that <w, x, y, z>= <az, bz, cz, z>= z<a, b, c, 1> and so the set of all such vectors is the one dimensional subspace, the set of all multiples of <a, b, c, z>.
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  4. #4
    MHF Contributor Bruno J.'s Avatar
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    Or you can just do a 4-dimensional cross-product!

    (Note that while you can do the cross product of two vectors in three dimensional space, you need three vectors in four dimensional space.)
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