# Two axes in a problem, going from IJK to ijk

1. Oct 16, 2011

### NEGATIVE_40

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
$$\vec{K}=\vec{j}cos(\gamma)+\vec{k}sin(\gamma) "$$
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
$$\vec{K}=\vec{j}sin(\gamma)+\vec{k}cos(\gamma)$$
which is the opposite of what it should be.

Any help would be appreciated.

#### Attached Files:

• ###### IJK to ijk conversion.JPG
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2. Oct 16, 2011

### phyzguy

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.

3. Oct 16, 2011

### NEGATIVE_40

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
4. Oct 16, 2011

### NEGATIVE_40

Are you referring to the correct one given by my textbook?
$$\vec{K}=\vec{j}cos(\gamma)+\vec{k}sin(\gamma)$$

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.