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Vector Component Problem

  1. Sep 17, 2004 #1
    Guys, I am stuck on this problem. I have the answer from the back of the book, but I don't know how to solve it. I tried to solve it by isolating Ux, Vy, and trying to solve it by simultaneous equations, but I always end up getting something like Vy = Vy , or something like Vy - Vy = 0. It keeps going in circles, maybe this isn't the way to go. But anyways, this is the problem:



    (note: the x, y , and z are written as small sub next to the U, V, and W, I just can't type it like it is in the book)

    The three vectors

    U = Uxi + 3j + 2k
    V = -3i + Vyj + 3k
    W = -2i + 4j + Wzk

    are mutually perpendicular. Use the dot product to determine the compnents of Ux, Vy, and Wz.


    The answers in the back of the book are Ux = 2.857, Vy = 0.857, Wz = -3.143.


    I am using the dot product definition: U dot V = UxVx+UyVy+UzVz


    so I get (Ux)(-3)+(3)(Vy)+(2)(3) = 0

    I set it all to zero because they are perpendicular to each other.

    So simplifying for Ux, i get Ux = Vy + 2. THen I plug Vy +2 for Ux, but then I go in circles from there.
     
  2. jcsd
  3. Sep 17, 2004 #2
    Using the dot product definition: U dot V = UxVx+UyVy+UzVz = |U||V|cos(a)
    you have a system of 3 equation:

    -3U + 3V + 6 = 0
    6 + 4V +3W = 0
    -2U + 12 + 2W = 0

    You can solve it by Cramer or by substitution
    The solution are: U =20/7 V=6/7 W=-22/7
     
    Last edited: Sep 17, 2004
  4. Sep 17, 2004 #3
    what is cramer, what do i substitute?
     
  5. Sep 17, 2004 #4
    What do I put for the |U||V|, since those are the magnitudes, I don't know how to find them because we are given only the components, and some are (Ux,Vy,Wz) are not actuallly given, so I cant compute it in the radical to get a value.
     
    Last edited: Sep 17, 2004
  6. Sep 17, 2004 #5

    Pyrrhus

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    Cinder are you familiar with System fo equations?, because i see no problem with Final's answer...
     
  7. Sep 17, 2004 #6
    how did u get the -3, 4, and the 12, they are not given in the problem? They are only variables Ux, Vy, and Wz
     
  8. Sep 17, 2004 #7

    Pyrrhus

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    In Final's post U is Ux, V is Vy, and W is Wz.
     
  9. Sep 17, 2004 #8
    Yes, but in his 3 equations, he has -3U in the first one, and 6, in the second, and , man sorry i just dont see it
     
  10. Sep 17, 2004 #9
    can someone show me the steps to get those values, cause when i tried to isolate Ux, or Vy, or Wz, in a system, then i would get like Vy - Vy =0 or something useless.
     
  11. Sep 17, 2004 #10

    Pyrrhus

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    The first equation is U dot V
    The second equation is V dot W
    The third equation is U dot W
     
  12. Sep 17, 2004 #11

    Pyrrhus

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    There are 3 unknowns and 3 equations, this is a solvable system, check what you are doing.
     
  13. Sep 17, 2004 #12

    HallsofIvy

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    No offense intended, CinderBlockFist, but this a lot like a guy asking a question about calculus and not being able to do arithmetic. If you are studying linear algebra, you certainly would be expect to be able to solve 3 linear equations in 3 unknowns. You may have had that so long ago you have forgotten.

    The three vectors are
    U = Uxi + 3j + 2k
    V = -3i + Vyj + 3k
    W = -2i + 4j + Wzk

    U.V= -3Ux+ 3Vy+ 6= 0
    U.W= -2Ux+ 12+ 2Wz= 0
    W.V= 6+ 4Vy+ 3Wz= 0. Three linear equations in 3 unknowns.

    There are many different ways to solve these but the simplest is probably:
    Multiply the first equation by 2 to get -6Ux+ 6Vy+ 12= 0
    Multiply the second equation by 3 to get -6Ux+ 36 +6Wz=0
    Of course, the purpose of the multiplications was to get Ux with the same coefficient, -6, in both. If we subtract the second of those equations from the first the two "-6Ux" terms cancel and we get 6Vy+ 12- 36- 6Wz= 0 or 6Vy- 6Wz- 24= 0.

    The last equation did not have Ux in it to begin with so now I have two equations in 2 unknowns: 6Vy- 6Wz- 24= 0 and 6+ 4Vy+ 3Wz= 0. If we divide the first of those by 2 we get 3Vy- 3Wz- 12=0 and adding that two the second will cause the "3Wz" terms to cancel: 3Vy- 3Wz- 12+ 6+ 4Vy+ 3Wz= 7Vy- 6= 0. That one equation in one unknown we can solve to get Vy= 6/7.

    Putting that back into the original equation 6+ 4Vy+ 3Wz= 0, we get 6+ 24/7+ 3Wz= 0 so 3Wz= -6- 24/7= -66/7 so Wz= -22/7. Finally, we can put Vy= 6/7 into the original equation -3Ux+ 3Vy+ 6= 0 which is the same as Ux= Vy+ 2= 6/7+ 2= 20/7.

    Putting all of those together, Ux= 20/7, Vy= 6/7, and Wz= -22/7 just as Final said originally.
     
  14. Sep 17, 2004 #13
    Ivy, THank you for the great response, yes I get it! I forgot, I was trying to solve using only 2 of the equations, no wonder i got no where. And I forgot about that technique to match of the coeff. Thanks again =)
     
  15. Sep 17, 2004 #14

    HallsofIvy

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    Ah, yes! The ever-popular "fight with one hand tied behind my back method"!
     
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