Show that the tangent lines to the regular parameterized curve $\alpha(t)=(3t,2t^2,2t^3)$ make a constant angle with the line $y=0,z=x$.
1) The tangent line at each point is given, I believe, by $\alpha'(t)=(3,4t,6t^2)$
2) So the angle this makes with $(x,0,x)$ is given I believe by $\frac{\arccos(u\cdot v)}{|u||v|}$?
$$\frac{\arccos(3tx+6t^2x)}{\sqrt{(2x^2)(9+16t^2+36t^4)}}$$
Two concerns, should $x$ in my line $y=0,z=x$ be parameterized by $t$? Clearly my dot product is wrong, since it won't give a constant angle?
HINT: the direction of the line is $\left(\begin{matrix}1\\0\\1\end{matrix}\right)$
therefore you need to show that $\theta$ is constant, where $$\theta=\cos^{-1}\frac{(\underline{u}\cdot\underline{v})}{|\underline{u}||\underline{v}|}$$ and $\underline{u}=\left(\begin{matrix}1\\0\\1\end{matrix}\right)$ and $\underline{v}=\left(\begin{matrix}3\\4t\\6t^2\end{matrix}\right)$
Can you finish it?