I want to find the point(s) where the curve defined by $\mathbf{r}(t) = \langle 2t^2-t,3t+1,1-2t \rangle$ and the $yz$-plane intersect.
The $yz$-plane is defined by the equation $x=0$, so if I set the $x$ component of $\mathbf{r}$ to $0$ and solve for $t$, I should be able to find my points of intersection.
The parametric equation for the $x$ component of $\mathbf{r}$ is
$$x = t(2t-1)$$
Letting $x=0$, we see that $t = 0$ or $t=\tfrac{1}{2}$. Therefore, the points of intersection are $(0,1,1)$ and $(0,\tfrac{5}{2},0)$.
If this is correct, then I also have seemingly extra information that I did not provide on this post, like deriving the normal vector $\mathbf{n}$ of the $yz$-plane and constructing all three parametric equations for $x,y,z$ components of $\mathbf{r}$, but it appears that what I have done above should suffice. Am I wrong? Am I not taking other factors into account?
the normal vector of the yz-plane is given by $$\vec{n}=(1;0;0)$$