Question: Suppose you head toward the $xy$-plane from the surface $x^2+y^2-z^2=-1$ at the point $(1, 1, \sqrt{3})$ by following the normal line to the surface at that point. What are the $x$ and $y$ coordinates at which you will hit the $xy$-plane?$
My attempted Solution: I know that the normal line has the parametric equation
$$r(t) = <1+2t, 1+2t, \sqrt{3}+2\sqrt{3}t>$$ but I'm unsure of how to find the $xy$ coordinates at which this line hits the $xy$ plane. I tried setting $z=0$ and having $\sqrt{3}+2\sqrt{3}t = 0$. In this case, $t=-1/2$. Solving for $x$ and $y$, they are somehow also both zero.
Could someone please show me how this is done? Thanks!
Normal line has equation $$\vec{r}(t)=(1,1,\sqrt 3)+t(2,2,-2\sqrt 3)$$ Setting $z=0$ we get $\sqrt 3-2t\sqrt 3=0\to t=\frac12$
Thus the intersection point with $xy$-plane has coordinates $(2,2,0)$.