Why $\frac{\partial \vec{m}} {\partial \omega} = \vec{n}$

Question

Consider a 3 dimensional orthogonal basis $\vec{m}, \vec{n}, \vec{t}$. Consider rotating the $(\vec{m}, \vec{n})$ plane about $\vec{t}$. For further analysis, consider $\omega$ as the anticlockwise angle from some reference datum to $\vec{m}$. Please help me explain why then following derivatives hold: $\begin{equation} \frac{\partial \vec{m}} {\partial \omega} = \vec{n}; \ \ \ \frac{\partial \vec{n}} {\partial \omega} = - \vec{m} \end{equation} $ enter image description here

Answer 1

This is same as working of derivatives of sin and cos functions. If you are bothering about the direction of it then ,when you consider a vector(x,y) and start rotating it about z axis in anticlock wise x directional magnitude start decresing where as y direction magnitude starts increasing with decreasing angle.The angle is decreasing because we are rotating the points in anti clockwise direction.Now to prove the magnitude you can use taylors series of expansion for sine and cos and work on it because every point on periphery of a circle of radius R,can be epressed in terms of sine and cos based on angle. Hope this helps you.

Answer 2

Consider a plane $(\vec{u},\vec{v})$, consider other two vectors $\vec{m}$ and $\vec{n}$ in the $(\vec{u},\vec{v})$ such that $\vec{m} \cdot \vec{n}=0$. $\vec{m}$ is at angle $\omega$ from the vector $\vec{u}$. Then:

$\vec{m}=u\ cos \omega \textbf{i} + v\ sin \omega \textbf{j} \\ \vec{n}= -u\ sin \omega \textbf{i} + v\ cos \omega \textbf{j}\\ u = |\vec{u}|\ , v = |\vec{v}|$

From there, both the derivatives follow easily.