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Vectors coursework question

  1. Feb 2, 2009 #1
    if the vector x and the scalar m satisfy the eqs.
    a x x = m + b and a.x = 1

    where a=i+2j, b=2i+j-2k

    Find the value of m and the expression ofr the vector x in terms of i,j and k
     
    Last edited: Feb 2, 2009
  2. jcsd
  3. Feb 2, 2009 #2

    tiny-tim

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    Hi gtfitzpatrick! :wink:

    Show us what you've tried, and where you're stuck, and then we'll know how to help. :smile:
     
  4. Feb 2, 2009 #3
    Re: Vectors

    i got the cross product of the leftside which gave me 2(x3)i - (x3)j + ((x2)-2(x1))k and cleaned up the right side which gave (m+2)i + (2m+1)j - 2k and let the i, j ,k parts equal each other
    2(x3) =(m+2)
    -(x3) = (2m+1)
    (x2)-2(x1) = -2

    and tried to solve but i'm getting 0
     
    Last edited: Feb 2, 2009
  5. Feb 2, 2009 #4
    Re: Vectors

    wait a sec i actually got m= -4/5 and vector x = -3/5k

    i got both the i and j parts = 0
    from a.x = 1 gives (x1)+2(x2)=1
    and eq 1 i got 2(x2)+ (x1) =1 and (x2)-2(x1) = -2 which gives (x1)=(x2)=0
     
    Last edited: Feb 2, 2009
  6. Feb 3, 2009 #5
    Re: Vectors

    i think i'm doing this right, But any hints would be greatly appreciated
    Thanks
     
  7. Feb 3, 2009 #6

    tiny-tim

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    Hi gtfitzpatrick! :smile:

    I was a little confused at first, but I think you meant

    a x x = ma + b :wink:

    Hint: whenever you see a cross-product, try dotting it with something

    in this case, try dot-producting this equation with a, and you should get an easy equation for m. :smile:

    (and then dot-product it with … ?)
     
  8. Feb 5, 2009 #7
    Re: Vectors

    Thanks Tiny-Tim,
    That is what i meant well spoted. Dot producting both side is a good tip, It was a quicker way of finding M, I'm still not sure of the second part though...
     
  9. Feb 5, 2009 #8

    tiny-tim

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    show us what you've got so far :smile:
     
  10. Feb 6, 2009 #9
    Re: Vectors

    Hello,
    I crossed the left side giving me 2X[tex]_{3}[/tex]i - X[tex]_{3}[/tex]j + (X[tex]_{2}[/tex] - 2X[tex]_{1}[/tex])i

    (1,2,3 are subsets but i cant get them to work)

    and the right side cleans up to give (m+2)i + (2m+1)j - 2k

    and then let the i's, j's and k's equal each other an i ended up with 3 eqs
    2X[tex]_{3}[/tex] = m +2
    -X[tex]_{3}[/tex] = 2m +1
    (X[tex]_{2}[/tex]-2X[tex]_{1}[/tex]) = -2
    and from a.x = 1 i got X[tex]_{1}[/tex] + 2X[tex]_{2}[/tex] 1

    from the first 2 eqs i get m = -4/5 and x3 = 3/5

    and from eqs 3&4 i get x2 = 0 and x1 = 1
     
    Last edited: Feb 6, 2009
  11. Feb 9, 2009 #10

    tiny-tim

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    Hello gtfitzpatrick! :smile:

    I'm sorry I've taken so long to reply. :redface:
    Yes, that looks fine! …

    x = (1,0,3/5), and so x x a = (6/5,-3/5,-2) = (-4/5,8/5,0) + (2,1,-2) = -4/5 a + b. :smile:
     
  12. Feb 9, 2009 #11
    Re: Vectors

    cheers tiny tim,
    all the help is much appreciated
     
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