A homework problem I have been assigned is telling me to find the tangential and normal components of the acceleration vector corresponding to $r=\langle3t-t^3,3t^2\rangle.$ While the tangential component can be found with the formula $a_T=\dfrac{r'\cdot r''}{|r'|}$ yielding $a_T=6t,$ the formula given for the normal component involves taking the cross product of $r'$ and $r''.$
Since $r'$ and $r''$ are two-dimensional vectors, you can't take the cross product of them. Is there another way to obtain the normal component of a two-dimensional vector $r$?
\begin{align*} \mathbf{r} &= (3t-t^3,3t^2) \\ \mathbf{v} &= (3-3t^2,6t) \\ \mathbf{a} &= (-6t,6) \\ \mathbf{a_{\parallel}} &= \frac{\mathbf{a\cdot v}}{v^{2}} \mathbf{v} \\ &= \frac{(3-3t^2)(-6t)+(6t)(6)}{(3-3t^2)^2+(6t)^2} (3-3t^2,6t) \\ &= \frac{6t}{1+t^2} (1-t^2,2t) \\ \mathbf{a_{\perp}} &= \mathbf{a}-\mathbf{a_{\parallel}} \\ &= \frac{6}{1+t^2}(-2t,1-t^2) \end{align*}