rotations in 3d space

Question

When looking at rotations in 3d space, does specifying two points (say point A is rotated to point B) determine the whole rotation, or is there a degree of freedom left?

Answer 1

Think about rotating the sphere by sending the North Pole to the South Pole and the South Pole to the North Pole and send $(1,0,0)$ to $(0,1,0)$.

Answer 2

Even for rotations in 2d space there more to specify. How much to rotate, the angle. In your words take the circle centred at B and radius equalling the length of AB. Then A could have moved to any other point A$'$ on that circumference. (

In 3d there is more degree of freedom: namely the choice of plane that contains the line segment AB, so our circle of radius AB could be in any of those planes.

Answer 3

Suppose, to take an easy example, that $A=(0,0,1)$ (the north pole in the usual way of drawing coordinate axes) and $B=(1,0,0)$ (on the equator). Then you could send $A$ to $B$ by a $90^\circ$ rotation about the $y$-axis (i.e., the line through $(0,1,0)$). But you could also send $A$ to $B$ by a $180^\circ$ rotation about the point midway between $A$ and $B$ $\left(\text{namely} \space \left(\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}}\right)\right)$. Indeed, you could choose any point $X$ on the sphere that is equidistant from $A$ and $B$ (there's a whole great circle of such $X$'s) and send $A$ to $B$ by a rotation about $X$. That is, there's a $1$-parameter family of rotations, all sending $A$ to $B$.