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Finding all vectors normal to a given set of vectors

  1. Sep 24, 2008 #1
    How do you find all vectors perpendicular to a set of three vectors in R^4? I know that the dot product of a normal vector and each given vector will be equal to 0. How could I set up the system of equations in matrix form?
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  3. Sep 25, 2008 #2


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    Hmm, you just have to write out the equations for each dot product. For example, suppose the three vectors are [itex]v_1, v_2, v_3[/itex]. Then let the perpendicular vector be [itex]u[/itex]. So we know that [itex]u \cdot v_i = 0[/itex]. Let [itex]u = (a_1, a_2, a_3, a_4)^T[/itex] for concreteness.

    So if [tex]v_1 = (x_1, x_2, x_3, x_4)^T[/tex], then we have, after applying the dot product of u and v1:
    [tex]a_1 x_1 + a_2 x_2 +a_3 x_3 + a_4 x_4 = 0[/tex]. Do the same for the other two vectors and you have a system of equations which you can then express in matrix form.
  4. Sep 25, 2008 #3
    could you not cross product??
  5. Sep 25, 2008 #4


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    I believe the cross product is defined only in R^3, well as far as i know :P
  6. Sep 25, 2008 #5


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  7. Sep 25, 2008 #6


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    As the Wiki article says, there's more than one way to extend the cross product to higher dimensions. One very straightforward way extends it to all dimensions of Rn, but it's a little peculiar: In dimension n it becomes a function of n-1 vectors rather than a simple product of two vectors. This is mentioned in Spivak's classic monograph, Calculus on Manifolds, chapter 4, p83 in the old Benjamin edition. I'll quote the relevant snippet here:

  8. Sep 25, 2008 #7


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    Welcome to PF!

    Hi mds9668! Welcome to PF! :smile:

    You use the wedge product, ⋀, of all three vectors (with x⋀x = 0).

    That will give you a 3-form, which is a linear combination a.x⋀y⋀z + b.y⋀z⋀t + c.z⋀x⋀t + d.x⋀y⋀t.

    The vectors perpendicular to all three will be the multiples of a.t + b.x + c.y + d.z

    (because (a.t + b.x + c.y + d.z)(a.x⋀y⋀z + b.y⋀z⋀t + c.z⋀x⋀t + d.x⋀y⋀t) = a² + b² + c² + d²) :smile:
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