Tangent line parallel to line of equations $x= -8z$ and $x= 2y$

Question

Problem:

If $c(t)=(2t,t^2,-\frac{t^3}{3})$, then "find the instant $t_0$ where the tangent line to $c(t)$ at $c(t_0)$ is parallel to the line of equations $x=-8z$ and $x=2y$."

I started by choosing two distinct vectors belonging to the latter, $A=(8,4,-1)$ and $B=(-8,-4,1)$, but I don't know how to proceed next.

What should I do?

Answer 1

The line can be written in this vector form:

$$x=t, y=\frac{1}{2}t, z=-\frac{1}{8}t$$

The direction vector of this line is then $(1, \frac{1}{2}, -\frac{1}{8})$. The tangent direction of the curve $c(t)$ is $(x'(t), y'(t), z'(t))$. Since you want the tangent of the curve parallel to the line, $(x'(t), y'(t), z'(t))=D(1, \frac{1}{2}, -\frac{1}{8})$ for some constant $D$. You can then solve for $t_0$.