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How do you express a vector in terms of I,J,&K?

  1. Sep 3, 2013 #1
    Express the vector <1,0,0> in terms of I,J,&k.

    (a)*I + (b)*J + (c)*K

    You have to find a,b, and c. I know that I,J,&K are unit vectors and so I thought that a = 1, b = 0, c = 0. But the answer has radicals and fractions in it. I don't understand how to arrive at the solution.
     
  2. jcsd
  3. Sep 3, 2013 #2

    mfb

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    If you know the supposed answer, why don't you post it here? This could help to understand where the difference comes from.

    Which background (which course) does the question have?
     
  4. Sep 3, 2013 #3

    Mark44

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    In terms of the unit vectors i, j, and k, <1, 0, 0> is 1i + 0j + 0k, or more simply, i. There shouldn't be any radicals or fractions.
     
  5. Sep 3, 2013 #4
    Its part part of my homework for calc 3. I already tried 1,0,0 and it was wrong.

    The problem gave a hint to do the dot product of the given vector with all the unit vectors. The previous problem was to express the vector <0,0,1> in terms of I,J&,K. The answer to that one was -sqrt(3)/2, 0,1/2. This one doesn't have answer in the back of the book and nothing in the chapter specifically talks about this problem.
     
  6. Sep 3, 2013 #5

    HallsofIvy

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    "i", "j", and "k" (small letters) are the unit vectors in the x, y, and z directions. Perhaps your "I", "J", and "K" (capital letters) mean something else?
     
  7. Sep 4, 2013 #6

    mfb

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    Hmm, I thought about quaternions, but that does not make sense here.
    Unless I,J,K have some special definition given somewhere, I don't see how the result could be different from the one that you calculated in post 1.
     
  8. Sep 4, 2013 #7

    D H

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    That was my thought as well, that these represent some other set of unit vectors other than the canonical i, j, and k, and that nando94 did not specify what these I, J, and K are.


    nando94, what is the problem as written in the textbook, verbatim and in its entirety?
     
  9. Sep 4, 2013 #8

    CompuChip

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    Hmm, ##-\sqrt{3}/2## and 1/2 are typically sines and cosines of "nice" angles such as ##\pi / 6## and ##\pi / 3##.
    But as said: we can keep guessing until you post more information, then we might be able to give you a conclusive answer.
     
  10. Sep 4, 2013 #9
    The I,J,&K were vectors from a previous page that were supposed to be added together to result in <0,0,1>. I got the answer correct. Sorry for the confusion. This textbook is very unclear.
     
  11. Sep 4, 2013 #10

    Mark44

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    This is why we ask you to use the homework template, and post the complete problem statement. We were not able to read your mind that I, J, and K were given in the problem.
    That may be, but it doesn't help if you don't provide all of the given information.
     
  12. Sep 5, 2013 #11

    vanhees71

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    That teaches you the important fact that giving just three components of a vector doesn't tell you anything, if you don't tell wrt. which basis these components are taken. Further you also need to tell us the new basis vectors I, J, K.
     
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