What does it mean to take the cross product of velocity and acceleration? This is from a practice question I am working on.
The osculating plane to the curve given by the vector valued function 
$r(t) =\langle\cos(t), (t-1)^2, -\sin(t)\rangle$
at the point corresponding to $t = 0$ is _____.
Plane $B$ is $2x + y - 2z - 3 = 0$
Plane $C$ is $x - y - 2z - 0$
By plugging $r(0)$ in you get $\langle 1,\ 1,\ 0\rangle $ which satisfies $B$ and $C$.
However the solution states taking $r'(0) \times r''(0)$.
Source: 
http://www.math.lsa.umich.edu/courses/215/17exampractice/pdf/exam1w12sol.pdf
 A: Don't think of them as velocity and acceleration here. We are only interested in their direction, not their magnitude, so it's better to think of them as the tangent vector and the vector that points in the direction of the centre of curvature (perhaps this second vector has a name too; I don't know it).
The osculating plane contains the tangent of the curve, given by $\mathbf{r}'(0)$, and the centre of curvature, which is in the direction $\mathbf{r}''(0)$. So the normal to the plane is perpendicular to both of these vectors: it is therefore (some multiple of)
$$\mathbf{r}'(0) \times \mathbf{r}''(0)$$
A: Very old question. Typing this as a note for others.
In short, it provides the binormal direction that is perpendicular to your osculating plane.
To understand better, see the derivation:
Call $\vec{v}(t) = r'(t)$, $\vec{a}(t) = \vec{r}''(t)$, then
$$\vec{v}(t) \times \vec{a}(t)$$
is a frequently appearing term in various formulas, e.g. curvature.
Split the acceleration vector into orthogonal complements:
$$\vec{a} = a_T \hat{T} + a_N \hat{N}$$
and remember velocity is speed times unit tangent vector:
$$\vec{v}(t) = v(t) \hat{T}$$
Since cross products of vectors with itself gives zero, we see that
$$\vec{v}(t) \times \vec{a}(t) = (v(t) \hat{T}) \times (a_N(t) \hat{N}) = v(t) a_N(t) \hat{B}(t) $$
So the interpretation of the vector $\vec{v}(t) \times \vec{a}(t)$ is that it's the binormal direction with magnitude being speed times lateral acceleration.
