How to show that the curve $\displaystyle \alpha(t)=(3, 4t, 6t^2)$ makes a constant angle with the line $\displaystyle x=z, y=0$?
The angle between the two curves would be the angle between their tangent vectors, right? If so, is the statement of the problem really true?
The line can be parametrized by $\displaystyle (x,y,z) = t(1,0,1)$, so you see that its tangent will point in the direction of $\displaystyle (1,0,1)$. The derivative of the curve is non-constant. What am I misunderstanding?
As Tonio said, the curve and line do not intersect so it really makes no sense to talk about the "angle they make". Even if we imagine one or the other being "shifted", without changing orientation, so they do intersect, the angle between them is NOT constant:
At any t, the tangent vector to the first curve is < 0, 4, 12t> and the tangent vector to the second curve is <1, 0, 1>. Their dot product is $\displaystyle <0, 4, 12t>\cdot<1, 0, 1>= 12t$ while their lengths are $\displaystyle \sqrt{16+ 144t^2}= 2\sqrt{1+ 9t^2}$ and $\displaystyle \sqrt{2}$. If $\displaystyle \theta$ is the angle between the vectors, $\displaystyle cos(\theta)= \frac{12t}{2\sqrt{2+ 18t^2}}$ which is NOT a constant so the angle is not a constant.