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Homework Help: Two axes in a problem, going from IJK to ijk

  1. Oct 16, 2011 #1
    1. The problem statement, all variables and given/known data

    *see attachment for diagram*
    I'm having trouble understanding a step in an example problem.

    "The Z-axis is vertical and carries the unit vector
    [tex] \vec{K}=\vec{j}cos(\gamma)+\vec{k}sin(\gamma) "[/tex]
    is what I cannot understand.

    3. The attempt at a solution

    I want to split the K vector into components of j and k, I'm pretty sure.

    If I try and do this (see attachment for my triangle), I get
    [tex] \vec{K}=\vec{j}sin(\gamma)+\vec{k}cos(\gamma) [/tex]
    which is the opposite of what it should be.

    Any help would be appreciated.

    Attached Files:

  2. jcsd
  3. Oct 16, 2011 #2


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    Science Advisor

    Your formula can't right. To see this, try looking at the limiting cases. If gamma = 0, then K=j, if gamma = pi/2, then K=k.
  4. Oct 16, 2011 #3
    Just thinking about this,

    how do I know which way I should have the triangle going? For example, in the diagram I've attached, I have the 90 degree angle on the Y-axis. To get the answer it appears I should just do the opposite (always have 90 degree angle NOT on the IJK axes or parallel to them). That method seems to work, but I doubt that the correct way of thinking about it.

    edit: ignore this. It doesn't work. If I have gamma between y and Z axes (the angle here would be gamma) with right angle along y, then I can get the textbook's solution. But then why doesn't this work if I have gamma between Y and z, with right angle along z ?
    they seem to be saying the same thing.
    Last edited: Oct 16, 2011
  5. Oct 16, 2011 #4
    Are you referring to the correct one given by my textbook?
    [tex] \vec{K}=\vec{j}cos(\gamma)+\vec{k}sin(\gamma) [/tex]

    if gamma = 0 , then the axis xyz would be lined up so y corresponded to Z. That's exactly what the diagram is saying, so I don't see a problem with that. It's just how the textbook has the axis set up.
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