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I'm rotating an object in $3D$ space with respect to a relative base, or reference frame. I'm using a normal vector to represent the rotation angles. Suppose you have an object parallel to the reference frame; in this case the object has the following normal vector: $v=[0, 0, 1]$. So, the normal vector makes $90^\circ$ with x-axis and y-axis and $0^\circ$ with z-axis. (I suppose z-axis is the principle axis).

My question is: If we rotate the object $30^\circ$ around x-axis; what is the direction normal and the angles? is the change only on the angle around x-axis or z-axis also. I noted when I rotate $30^\circ$ around x-axis the z-axis also rotated with the same amount. This confused me. Any Help!.

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  • $\begingroup$ What matrix are you using and what's your final vector? $\endgroup$ – MasterOfBinary Nov 11 '13 at 22:50
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I'll assume you mean the object's position when you say $v = (0,0,1)$. Since the object lies on the $z$-axis, rotation of $30$ degrees about $x$-axis, then the direction normal changes with $\theta$ as it goes from $0$ to $30$, and $x$ gets fixed:

In a right-hand-rule system, rotation around $x$-axis analogous to standard rotation in the $xy$-plane (around $z$-axis), we have that the $y$-axis takes the place of $x$-axis, and $z$-axis takes the place of $y$-axis.

$$ R_{\theta}(x,y,z) = (x, e^{i\theta}\cdot (y + zi)) $$

i.e. treat $(y,z)$ like a complex number and multiply to get the $\cos\theta, \sin\theta$ formulas.

The unit vector tangent to the direction of motion starts out as the $-\hat{y}$ (the unit vector parallel to $y$-axis), and progresses with $\theta$ as $$ R_{\theta}(0, 1, 0) = (0, e^{i\theta} (1 + 0i)) = (0, \cos\theta, \sin\theta) $$

Expanding the first one also gives you a vector normal to the direction of rotation, but clearly there are infinitely many normal vectors to the circular path to choose from.

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