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!.
Best Answer
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.