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Let $R(\phi, \boldsymbol{n})$ be a member of Lie Group SO(3). According to Wikipedia

If $R(\phi, \boldsymbol{n})$ denotes a counter-clockwise 3D rotation through an angle $\phi$ about the axis specified by the unit vector $\boldsymbol{n}$ , then

$\left.{\operatorname{d}\over\operatorname{d}\phi} \right|_{\phi=0} R(\phi,\boldsymbol{n}) \boldsymbol{x} = \boldsymbol{n} \times \boldsymbol{x}$

for every vector $\boldsymbol{x}$ in $\textsf{R}^3$.

Can someone provide a little bit of intuition for this result?

As I understand it, the operation $\boldsymbol{n} \times \boldsymbol{x}$ in 3 dimensions produces a vector normal to the plane containing $\boldsymbol{n}$ and $\boldsymbol{x}$. So what does it mean to say that evaluating the expression on the left produces this particular vector?

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  • $\begingroup$ Fixed $\boldsymbol x$ then $R(\phi,\boldsymbol n)\boldsymbol x$ is a curve with parameter $\phi$. You are looking for the velocity for $\phi=0$. $\endgroup$ – Hideyuki Kabayakawa May 3 '16 at 14:42
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Consider the expression of the rotation matrix as angle and axis $$\frac{\delta}{\delta \phi} R(\phi,\mathbf{n}) = \frac{\delta}{\delta \phi} (\cos\phi \mathbf{I} + (1-cos\phi)\mathbf{n}\mathbf{n}^T - sin\phi \mathbf{n}^\times )= -\sin(\phi)\mathbf{I} + \sin\phi \mathbf{n}\mathbf{n}^T - cos\phi \mathbf{n}^\times = - \boldsymbol{x}^\times (\cos\phi \mathbf{I} + (1-cos\phi)\mathbf{n}\mathbf{n}^T - sin\phi \mathbf{n}^\times) = - \boldsymbol{x}^\times R(\phi,\mathbf{n}) $$

now if we consider the rotation of a vector

$$\left.{\operatorname{d}\over\operatorname{d}\phi} \right|_{\phi=0} R(\phi,\boldsymbol{n}) \boldsymbol{x} = - \boldsymbol{x}^\times \left.R(\phi,\mathbf{n})\right|_{\phi=0} \boldsymbol{x} = -\boldsymbol{x}^\times \boldsymbol{I} \boldsymbol{x} = -\boldsymbol{n} \times \boldsymbol{x}$$

That is the intuition behind.

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Fixed $\textbf n$ and $\textbf x$, let $\gamma (\phi)=R(\phi,\textbf n)\textbf x$, the trajectory curve of the extreme of the vector $\textbf x$. As $\displaystyle\frac{\partial R}{\partial\phi}(\phi,\textbf n)|_{\phi=0}\textbf x=\gamma'(0)$ is the velocity in $\gamma(0)=\textbf x$, and $\gamma$ is a circle, then $\gamma'(0)$ is normal to $\textbf n$ and $\textbf x$. Furthermore the rotation is counter-clockwise, so $\gamma'(0)$ have the same direction as $\textbf n\times\textbf x$.

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On the other hand, the radius of the trajectory is $|\textbf x|\cdot \sin\alpha$, with $\alpha$ the angle between $\textbf x$ and $\textbf n$. Hence $\Delta s=|\textbf x|\cdot \sin\alpha\cdot \Delta \phi$, and $$\displaystyle|\frac{\partial R}{\partial\phi}(\phi,\textbf n)|_{\phi=0}\textbf x|=|\gamma'(0)|=\frac{ds}{d\phi}=|\textbf x|\cdot \sin\alpha=|\textbf n\times\textbf x|.$$

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