Schaum's Outline of Vector Calculus Problem 3.54 states:
Prove that the curvature of the space curve $\mathbf{r}=r(t)$ is given numerically by $\kappa=\dfrac{|\dot{r}\times \ddot{r}|}{|\dot{r}|^3}$ where dots denote differentiation with respect to $t$.
In 2D, the formula is terrible, but barely comprehensible. At least from there it's evident how you'd argue the existence of the cross product term in the original expression.
The question however, seems to suggest probably a formula consistent in 3D. Here's my reasoning:
\begin{align*} \dfrac{ds}{dt} &= |\dot{r}| \\ \hat{T} &= \dfrac{\dfrac{d\vec{r}}{dt}}{\dfrac{ds}{dt}} = \dfrac{\dot{r}}{|\dot{r}|} \\ \\ \dfrac{d}{dt}\dfrac{1}{|\dot{r}|} &= \dfrac{d}{dt}(\dot{r}\cdot\dot{r})^{-\frac12}=-\dfrac{\dot{r}\cdot\ddot{r}}{|\dot{r}|^3} \\ \\ \dfrac{d\hat{T}}{ds} &=\dfrac{\ddot{r}}{|\dot{r}|^2}-\dfrac{\dot{r}(\dot{r} \cdot \ddot{r})}{|\dot{r}|^4} \tag*{$\left(\dfrac{d\hat{T}}{ds}=\dfrac{\frac{d\hat{T}}{dt}}{\frac{ds}{dt}}\right)$}\\ &= \dfrac{\ddot{r}(\dot{r}\cdot\dot{r})-\dot{r}(\dot{r} \cdot \ddot{r})}{|\dot{r}|^4} \\ \dfrac{d\hat{T}}{ds} \cdot \dfrac{d\hat{T}}{ds} &= \dfrac{(\ddot{r} \cdot \ddot{r})(\dot{r}\cdot\dot{r})^2+(\dot{r}\cdot\dot{r})(\dot{r} \cdot \ddot{r})^2-2(\dot{r}\cdot\ddot{r})(\dot{r}\cdot\dot{r})(\dot{r}\cdot\ddot{r})}{|\dot{r}|^8} \\ &= \dfrac{(\dot{r}\cdot\dot{r})\Bigl((\ddot{r} \cdot \ddot{r})(\dot{r}\cdot\dot{r})-(\dot{r}\cdot\ddot{r})^2\Bigr)}{|\dot{r}|^8} \\ &= \dfrac{(\ddot{r} \cdot \ddot{r})(\dot{r}\cdot\dot{r})-(\dot{r}\cdot\ddot{r})^2}{|\dot{r}|^6} \\ \left|\dfrac{d\hat{T}}{ds}\right| &= \dfrac{\sqrt{(\ddot{r} \cdot \ddot{r})(\dot{r}\cdot\dot{r})-(\dot{r}\cdot\ddot{r})^2}}{|\dot{r}|^3} \end{align*}
That's as much as I could do short of conjuring matrices from thin air. Most of it looks done, but I need help where I'm stuck. This specifically:
$\sqrt{(\ddot{r} \cdot \ddot{r})(\dot{r}\cdot\dot{r})-(\dot{r}\cdot\ddot{r})^2} = |\dot{r} \times \ddot{r}|$
My amateur guess is that maybe $(\ddot{r} \cdot \ddot{r})(\dot{r}\cdot\dot{r})-(\dot{r}\cdot\ddot{r})^2=(\dot{r}\times\ddot{r})\cdot(\dot{r}\times\ddot{r})$ (strategy worked well for me the whole way anyway) but I have no idea what to do with that.
Your guess $$ (\ddot{r} \cdot \ddot{r})(\dot{r}\cdot\dot{r}) - (\dot{r}\cdot\ddot{r})^{2} = (\dot{r}\times\ddot{r})\cdot(\dot{r}\times\ddot{r}), $$ which may be rewritten $$ \|\ddot{r}\|^{2}\, \|\dot{r}\|^{2} - (\dot{r}\cdot\ddot{r})^{2} = \|\dot{r}\times\ddot{r}\|^{2}, $$ is an example of Lagrange's identity for the cross product.
If $\theta$ denotes the angle between $\dot{r}$ and $\ddot{r}$, and we recall \begin{align*} \dot{r}\cdot\ddot{r} &= \|\dot{r}\|\, \|\ddot{r}\| \cos\theta, \\ \|\dot{r}\times\ddot{r}\| &= \|\dot{r}\|\, \|\ddot{r}\| \sin\theta, \end{align*} the rewritten identity may further be seen as an instance of $1 - \cos^{2}\theta = \sin^{2}\theta$.