A parametric equation is defined by $r\left(t\right)=cos\left(-7t\right)i+sin\left(-7t\right)j+6tk$. Compute the normal component of the acceleration vector.
So I got that $r'(t)$ was $\left(-7sin\left(7t\right),\:-7cos\left(7t\right),\:6\right)$ and $r''(t)$ was $\left(-49cos\left(7t\right),\:49sin\left(7t\right),\:0\right)$.
The cross product was $\left(-294sin\left(7t\right),\:-294cos\left(7t\right),\:-343sin^2\left(7t\right)-343cos^2\left(7t\right)\right)$ and its magnitude was $\sqrt{-343cos^4\left(7t\right)+86093}$.
The magnitude for $r'(t)$ was $\sqrt{85}$ so I through the normal component was $\frac{\sqrt{-343cos^4\left(7t\right)+86093}}{\sqrt{85}}$, but this doesn't seem to be the right answer.
Does anyone know where I may have gone wrong?
You've obtained the cross product correctly (It would have been easier to do the exercise if you would have used the identity $\sin^2(u)+\cos^2(u)=1$, an identity one should never forget!): $$\mathbf{r}'(t)\times \mathbf{r}''(t)=\begin{bmatrix} -294\sin(7t) \\ -294\cos(7t) \\ -343\sin^2(7t)-343\cos^2(7t) \end{bmatrix}=\begin{bmatrix} -294\sin(7t) \\ -294\cos(7t) \\ -343 \end{bmatrix}$$ The magnitude, however is wrong: $$\begin{align}\|\mathbf{r}'(t)\times \mathbf{r}''(t)\|&=\sqrt{(-294\sin(7t))^2+(-294\cos(7t))^2+(-343)^2}\\&=\sqrt{294^2\sin^2(7t)+294^2\cos^2(7t)+343^2}\\&=\sqrt{294^2+343^2}\\&=49\sqrt{85} \end{align}$$ The rest of your work seems correct.