Let $[ \bx, \by, \bz]$ be a frame of right-handed orthonormal vectors in $\real^3$ , and let $\bv = a\bx + b\by + c\bz$ (with $a, b, c \in \real$ ) be any vector to be rotated on the $\bz$ axis, by an angle $\theta$ counter-clockwise.

The image vector $\bv'$ is the vector $\bv$ with its component in the $\bx,\by$ plane rotated, so we can write $$ \bv' = a\bx' + b\by' + c\bz\,,\\ $$ where $\bx'$ and $\by'$ are the rotations by angle $\theta$ of the $\bx$ and $\by$ vectors in the $\bx,\by$ plane. By the rotation formula in two dimensions, we have

We attempt to simplify further: $$ \bv' = \bv + \sin\theta (\bz \times \bv) + (\cos \theta - 1)(\bv - (\bv \cdot \bz)\bz )\,. $$ Since $\bz \times \bv$ is linear in $\bv$ , this transformation is represented by a linear operator $A$ . Under a right-handed orthonormal basis, the matrix representation of $A$ is directly computed to be $$ A\bv = \bz \times \bv = \begin{bmatrix} 0 & -z_3 & z_2 \\ z_3 & 0 & -z_1 \\ -z_2 & z_1 & 0 \end{bmatrix} \, \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} \,. $$ We also have

		(rotate by )
		(rotate by )

Relation with the matrix exponential

Second proof: If we regard $\theta$ as time, and differentiate the equation $\bv' = a\bx' + b\by' + c\bz$ with respect to $\theta$ , we obtain $d\bv'/d\theta = a\by' - b\bx' = \bz \times \bv' = A \bv'$ , whence the solution (to this linear ODE) is $\bv' = e^{\theta A} \bv$ .

Remark: The operator $e^{\theta A}$ , as $\theta$ ranges over $\real$ , is a one-parameter subgroup of $\mathrm{SO}(3)$ . In higher dimensions $n$ , every rotation in $\mathrm{SO}(n)$ is of the form $e^A$ for a skew-symmetric $A$ , and the second proof above can be modified to prove this more general fact.

Footnotes

...http://planetmath.org/encyclopedia/Order7.html ¹

If we want to use coordinate-free language, then in this section, ``matrix'' should be replaced by ``linear operator'' and transposes should be replaced by the adjoint operation.