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Linear Algebra - Find unit vector orthogonal to 2, 4-space vectors?

  1. Jul 15, 2009 #1
    1. The problem statement, all variables and given/known data

    Given the vectors
    u = (2, 0, 1, -4)
    v = (2, 3, 0, 1)
    Find any unit vector orthogonal to both of them

    2. Relevant equations

    I know that two vectors are orthogonal if their dot product is zero...

    3. The attempt at a solution

    I don't even know how to begin! I know the unit vector for one vector is the vector over its magnitude, but what about two of them? How do I find something that's orthogonal to both at the same time?

    Please help ASAP! :) Thanks.
  2. jcsd
  3. Jul 15, 2009 #2


    Staff: Mentor

    Hint: Gram-Schmidt
  4. Jul 15, 2009 #3
    I haven't learned that in class, but I did wikipedia it..
    according to that, the answer wouldnt change since <u,v> = 0 and the GS method just uses u - projvu which would be <u,v>v/<v,v> with a 0 on top... giving only (2, 0, 1, -4) which would not work.

    any other suggestions please? or advice? Thank you :)

    edit: please consider helping me with this, even if just a tiny tiny bit of guidance. I have been trying to figure this out for hours.. it's over 2 am my time.. and I'm really tired :/

    I will work hard and along with you! I need to understand this problem to do other problems in the HW too...

    we haven't learned the method you mentioned in class... we have been talking about euclidean inner product and vectors in R^n space
    Last edited: Jul 15, 2009
  5. Jul 15, 2009 #4
    So you know you are looking for a unit vector, call it [itex]\vec{a}=<a_1,a_2,a_3,a_4> [/itex]. Since [itex]\vec{a}[/itex] is a unit vector we must have [itex]|\vec{a}|=a_1^2+a_2^2+a_3^2+a_4^2=1[/itex].

    What else do you know?
  6. Jul 15, 2009 #5
    that 2a_1 + a_3 - 4a_4 = 0
    and 2a_1 + 3a_2 + a_4 = 0
  7. Jul 15, 2009 #6
    True, what can you do with these three equations?
  8. Jul 15, 2009 #7
    put them in matrix form and try to eliminate one of the variables?
  9. Jul 15, 2009 #8
    See what happens
  10. Jul 15, 2009 #9
    I did that before.. but whenever I would eliminate one of the variables, another would "return"

    like if I put it in the form

    [2 0 1 -4 | 0]
    [2 3 0 1 | 0]

    lets say I subtracted row one (R1) from row two (R2) ... R2 - R1
    then I get

    [2 0 1 -4 | 0]
    [0 3 -1 5 | 0]
    .. which makes me lose a variable but gain another in the process.

    this is where I got stuck before
    Last edited: Jul 15, 2009
  11. Jul 15, 2009 #10
    So we either need one more equation or to get 'rid' of one variable
  12. Jul 15, 2009 #11
    Suppose, [itex]\vec{a}=<b,c,d,0>[/itex]. How many equations and unkowns do we have now?
  13. Jul 15, 2009 #12
    I don't understand what I could eliminate or add though...

    unless you mean solving for something like a_1 to get it in terms of the other three? but that seems kinda like we're I'm stuck at now...
  14. Jul 15, 2009 #13
    i mean, set a_4=0
  15. Jul 15, 2009 #14
    If that is the case, then we have:

    2b + d = 0
    2b + 3c = 0
    b + c + d = 1


    so three equations and 3 unknowns?
    which I think I can actually solve...

    is this allowable? and why so?
  16. Jul 15, 2009 #15
    Yes thats right. Well, the problem states to find any unit vector orthogonal to u and v
  17. Jul 15, 2009 #16
    And I guessed that we needed 4 equations and 4 unkowns or 3 eqns and 3 unknowns
  18. Jul 15, 2009 #17
    that's true.. so hypothetically, if I wanted to, I could make another variable 0?

    I'm not going to in this case because I can solve the problem fine with the method you gave me.. but in the future I am allowed to put as many 0's as I'd like? Or is there some kind of limit?

    thank you so much for your help and patience!
  19. Jul 15, 2009 #18
    Try it, and see if you get a suitable vector (unit length orthogonal to both)
  20. Jul 15, 2009 #19
    To tell you the truth, I dont know... I think you might lose too much information if you use too many zeros
  21. Jul 15, 2009 #20
    I usually just eliminate enough variables so that it fits with the number of equations i have (and pray it works)
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