# Is the speed of recession equal to the speed of approach for any elastic collision?

I am comfortable with the idea that in a one-dimensional elastic collision, the speed of approach of the two colliding bodies is equal to the speed of recession following the collision, which can be expressed as $$v_{1\text{i}}-v_{2\text{i}}=v_{2\text{f}}-v_{1\text{f}}$$ Does $$v_\text{app}=v_\text{rec}$$ hold true in two- and three-dimensional elastic collisions? I know that in these cases, the corresponding equations would be more complex vector equations, but I'm wondering if the underlying principle is constant regardless of the spatial dimension of the system.

Furthermore, is the coefficient of restitution of a collision $$e=\frac{v_\text{rec}}{v_\text{app}}$$ defined in two- and three-dimensional collisions? If so, is it always true that $$e=1$$ when the glancing collision is perfectly elastic?

Finally, does anyone know if the impact parameter $$b$$ of a collision (the distance between the centers of the colliding objects measured perpendicular to the direction of the line of impact) is defined only for elastic glancing collisions, or is this a relevant quantity in inelastic collisions as well? (My hunch would be that it is, since this parameter should be measurable for any collision and no real-world collision is ever actually perfectly elastic.)

If $$\boldsymbol{n}$$ is a vector normal to the contacting surfaces, then the law of contact states
$$\boldsymbol{n} \cdot ( \boldsymbol{v}_{2f} - \boldsymbol{v}_{1f} ) = - \boldsymbol{n} \cdot ( \boldsymbol{v}_{2i} -\boldsymbol{v}_{1i} )$$
where $$\cdot$$ is the vector dot product. The above projects the relative velocity along the contact normal to extract the relative speed of approach or retreat.