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Linear algebra

  1. Dec 5, 2013 #1
    1. The problem statement, all variables and given/known data

    Find two vectors in R4 of norm 1 that are orthogonal to the vectors u = (2, 1, −4, 0),
    v = (−1, −1, 2, 2) and w = (3, 2, 5, 4).

    2. Relevant equations



    3. The attempt at a solution

    What i did was, i let a vector x = (x1, x2, x3, x4) that has a norm of 1 and orthogonal to u,v and w
    i.e. x.u = 0; x.v = 0 and x.w = 0.
    This where i'm stuck. Some help would be appreciated. Thanks.
     
  2. jcsd
  3. Dec 5, 2013 #2

    tiny-tim

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    hi sevag00! :smile:

    (try using the X2 button just above the Reply box :wink:)

    hint: how would you find a vector (x1,x2,x3) orthogonal to (a,b,c) and (d,e,f) ? :wink:
     
  4. Dec 5, 2013 #3
    Using dot product. x1.a + x2.b + x3.c = 0. Same for (d, e, f).
     
  5. Dec 5, 2013 #4

    tiny-tim

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  6. Dec 5, 2013 #5
    Yeah, but the chapter only explains the dot product. So, i think the problem should be solved using dot product.
     
    Last edited: Dec 5, 2013
  7. Dec 5, 2013 #6

    tiny-tim

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    ok, in that case you have three simultaneous equations (from the dot products) …

    solve them the usual way :smile:
     
  8. Dec 5, 2013 #7
    3 equations with 4 unknowns. Will this yield a solution?
     
  9. Dec 5, 2013 #8

    tiny-tim

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    3 equations with 3 unknown ratios :wink:

    so yes, you can find the ratios :smile:
     
  10. Dec 5, 2013 #9
    Ah yes. 4 equations with 4 unknowns.
     
  11. Dec 5, 2013 #10

    HallsofIvy

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    Further, the cross product is defined only in three dimensions while this problem is in four dimensions.
     
  12. Dec 5, 2013 #11
    Wait a minute. You got me confused. There are 3 equations with 4 unknown. The equation of the norm cannot be included with the other three.
     
  13. Dec 5, 2013 #12

    tiny-tim

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    i was talking about three dimensions
    there are only three unknown ratios

    so you can find the ratios
     
  14. Dec 5, 2013 #13
    What ratios? I want to find the components of x.
     
  15. Dec 5, 2013 #14

    tiny-tim

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    x2/x1, x3/x1, x4/x1
     
  16. Dec 5, 2013 #15
    Not getting your approach.
     
  17. Dec 5, 2013 #16

    tiny-tim

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    divide your three dot-product=0 equations by x1

    that gives you three equations in x2/x1, x3/x1, and x4/x1

    then solve them :smile:
     
  18. Dec 5, 2013 #17
    I think i did it right. I replaced x2/x1 with X x3/x1 with Y and x4/x1 with Z. Then solved 3 equations with 3 unknowns.
     
  19. Dec 5, 2013 #18

    Ray Vickson

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    You can write the three equations in four unknowns as three equations in three unknowns but with the fourth unknown on the right-hand-side. In other words, you can express three of the x_i as functions of the fourth. For example, you can write the second equation ##-1 x_1 - 1 x_2 + 2 x_3 + 2 x_4 = 0## as ##-1 x_1 - 1 x_2 + 2 x_3 = -2 x_4##, or ##x_1 + x_2 - 2x_3 = 2x_4##. Do the same thing for the third equation; the first already has the right form. Now you can solve for ##x_1, x_2, x_3##; this will give them as formulas involving ##x_4##.

    BTW: the method above is a standard way of dealing with such problems; it is used all the time in such fields as linear optimization (linear programming).
     
  20. Dec 5, 2013 #19
    I know that method. But then those equations will become parameters. They will depend on the value of x4 which has infinite solutions.
     
  21. Dec 5, 2013 #20

    Ray Vickson

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    Of course; that is the whole point. So, if you want to find two different solutions, just give two different values for ##x_4##---any two values you like, except 0. (Putting ##x_4 = 0## gives the zero vector, which cannot be normalized to length = 1!)
     
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