Decomposition of the acceleration vector

A basis for any 3-dimensional vector space must have 3 vectors in it.

So the acceleration of any object in [itex]ℝ^{3}[/itex] can be decomposed into the standard basis vectors for [itex]ℝ^{3}[/itex].

However, I have seen another decomposition, namely, into the tangential and normal (centripetal) acceleration vectors. This set contains only two vectors. This seems to contradict the fact that at least three vectors are necessary to span [itex]ℝ^{3}[/itex].

Either I am missing a vector, or something is terribly wrong. All insight is appreciated.

EDIT: The decomposition I am referring to is [tex] a = \frac{dv}{dt}T + κ||v||^{2}N[/tex]

where [itex] κ [/itex] is the curvature of the path.

Thanks!

BiP