I Rotation of a vector along two axes (of which one is angle-dependent)

Andrea94
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I have been trying to determine an expression for a unit vector in the direction of F for hours now.
I think the expression is supposed to look something kind of like this,

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But I don't understand at all how to arrive at this expression.
Any help?
 
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If you relocate the real vector F to point O, keeping its direction, locating the tail exactly at O, what its projections on the three axes would be?
 
Lnewqban said:
If you relocate the real vector F to point O, keeping its direction, locating the tail exactly at O, what its projections on the three axes would be?

On the z-axis it is clearly cos(beta) since that part of the rotation is not influenced by alpha. For the x-axis, I visualize that if alpha=0 then it is -sin(beta) and if alpha != 0 then this is the same as rotating -sin(beta) by cos(alpha). But I cannot figure out the y-axis.
 
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Look down perpendicularly to the x-y plane.
 
Lnewqban said:
Look down perpendicularly to the x-y plane.

So the only way we have a y-component is if beta != 0 AND alpha != 0, in which case the component along y from the beta part is sin(beta) (because this will be a diagonal vector contributing both to the y component and negatively to the x-component). So I can see the sin(beta) part, but I don't understand why I must rotate it by sin(alpha) (and not eg cos(alpha)) to get the correct answer.
 
I think I get it based on spherical coordinate transformation,

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The first column corresponds to my problem, but I have to add a negative sign because of the way my directions are set-up.
 
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Andrea94 said:
So I can see the sin(beta) part, but I don't understand why I must rotate it by sin(alpha) (and not eg cos(alpha)) to get the correct answer.
May it be because it is a projection of one projection of the actual vector F?
 
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Lnewqban said:
May it be because it is a projection of one projection of the actual vector F?
Yes specifically the components I am looking for 😌. Thanks for the help!
 
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