Rotation matrices vs quaternions? It seems we can describe every rotation in $SO(3)$ by at least one unit vector axis $u$ and angle $\theta$ pair.
Each of these pairs can also be described by a rotation matrix:

I've heard quarternions can also be used to achieve the same goal.
How do you translate from the $(u, \theta)$ representation of a rotation to a quaternion representation?
How do you use the quaternion representation of a rotation to rotate some point $P \in R^3$ about the origin?
 A: The relation is as follows: Given the rotation angle $\theta$ and the unit vector (axis) $\mathbf{u}$, you have to form the quaternion
$$
\mathbf{q}=\cos\frac{\theta}{2}+\sin\frac{\theta}{2}\mathbf{u}.
$$
Then the double-sided action
$$
R(\mathbf{v})=\mathbf{q}\mathbf{v}\mathbf{q^*}
$$
(where $\mathbf{q^*}$ is the conjugate quaternion and the operation is quaternion multiplication) gives you the rotated vector.
A: I'll expand on the existing answer a little. Write $c:=\cos\frac{\theta}{2},\,s:=\sin\frac{\theta}{2},\,q=c+su$ with $u^2=-1$. The most general form of $v$ is $A+Bu+Cw$ with $w^2=-1,\,[u,\,w]=0$ (if we identify imaginary quaternions with $3$-dimensional vectors in the obvious way, the last condition is $\vec{u}$'s orthogonality to $\vec{w}$). The $A+Bu$ part belongs to the same representation in $\Bbb H$ of $\Bbb C$ as do $q,\,q^\ast$, so commutes with them, viz. $$q(A+Bu)q^\ast=A+Bu.$$On the other hand, $$(c+su)Cw(c-su)=(c+su)(cCw+sCuw)=(c^2-s^2)Cw+2csCuw.$$In terms of $\Bbb R^3$, this is $$(\cos\theta)C\vec{w}+\sin\theta\vec{u}\times C\vec{w}.$$Note that $R_{ij}-\delta_{ij}=(1-u_i u_j)(1-\cos\theta)-\epsilon_{ijk}u_k\sin\theta$.
