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Subspace / basis problem

  1. Dec 12, 2014 #1
    1. The problem statement, all variables and given/known data
    Let S, a subspace of ℝ3 be the set of vectors orthogonal to vector (1,2,3)
    a)describe Set S
    b) find a basis for Set S

    2. Relevant Equations

    That a basis has to be linearly independent and span R^3


    3. The attempt at a solution

    I would do this:
    I know that vector (1,2,3) is the cross product of 2 vectors v1xv2
    so I could put it in a matrix (where v1=a,b,c and v2=d,e,f)

    a b c
    d e f

    But Im lost as to describe set S... Wouldn't I need to row reduce to see which variable is free, and then I could say whether or not it is a line or a plane ( well the dimension)
     
  2. jcsd
  3. Dec 12, 2014 #2

    Mark44

    Staff: Mentor

    What's another way to show that an arbitrary vector <x, y, z> is orthogonal to a given vector <1, 2, 3>?
    If you take constant multiples of the given vector <1, 2, 3>, what sort of geometric object do you get? What's the dimension of this subspace?
     
  4. Dec 12, 2014 #3

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    What you attempt is to find a,b,c,d,e,f such that (a,b,c) x (d,e,f) = (1,2,3)
    Might seem reasonable, but it's a bit tedious: you want to find six variables from three equations. And you don't really need the lengths (that comes in part b).

    Orthogonal to a vector ##\vec v## is anything that has ##\vec a \cdot \vec v = 0 ## with ##|\vec a| > 0\;##. That's only one equation with three unknowns.
    Pick a vector v1 that satisfies that equation.
    Pick a different one and call that v2.
    I think then you have a basis already according to this link (exercise doesn't ask for orthogonal or orthonormal basis !)

    But if you do want them orthonormal you can do v1 x (1,2,3) to get a v2 that is perpendicular to both.
    Then normalize v1 and v2.

    Re describing S: Why do you think there is the possibility that S is a line ?
     
  5. Dec 12, 2014 #4
    Ah I see what you're getting at, once I've found two indepedent solutions i can just put it in a matrix and solve it. I see what you mean by not spanning R^3 too. it will span R^2 if I understand correctly.

    I could also find the dot product, apart from that I am not too sure what I could use to prove the orthogonality. Geometrically speaking it is a plane, so the vector would be the normal.

    As for the post above me, I meant in any case ( wasn't solved and I wasn't entirely sure). However, I expected it to be a plane.
     
  6. Dec 12, 2014 #5

    Mark44

    Staff: Mentor

    The dot product is the primary tool to show orthogonality.

    Yes, set S is a plane.
     
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