Direction of t (Vector Space) 
Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. $$x = e^{-t}\cos t, y = e^{-t} \sin t, z = e^{-t}; (1, 0, 1). $$

The solution states I form $r(t)$ then get the $r'(t)$. After, it says: The point $(1, 0, 1)$ correspond to $t = 0$. Then they solve $r'(0)$ and thus, the tangent line is parallel to the vector $r'(0)$. 
I'm pretty confused by the whole process.
 A: The curve you mention is given by $$r(t) = (e^{-t}\cos t, e^{-t}\sin t, e^{-t}).$$ (You may have $\textbf{r}(t)$ instead) More precisely, the image of this function is the curve under consideration, so every point on the curve is of the form $r(t)$ for some $t$. When they say the point $(1, 0, 1)$ corresponds to $t = 0$, they mean that the point $(1, 0, 1)$ is on the curve and $r(0) = (1, 0, 1)$. 
How did they determine this? By solving the equation $r(t) = (1, 0, 1)$ which reduces to simultaneously solving the three equations \begin{align} e^{-t}\cos t &= 1\\ \\e^{-t}\sin t &= 0\\ \\e^{-t} &= 1.\end{align} The third equation is the easiest to solve as the other two involve products. The equation $e^{-t} = 1$ has unique solution $t = 0$. However, we need to solve these equations simultaneously. That is, we need to find the value(s) of $t$ such that all three equations hold, not just one of them. So we know that $e^t = 1$ holds if $t = 0$. As $e^0\cos 0 = 1\times 1 = 1$ and $e^0\sin 0 = 1\times 0 = 0$, $t = 0$ also satisfies the first and second equations as well. We can then conclude that the point $(1, 0, 1)$ is on the curve, and it corresponds to $t = 0$; that is, $r(0) = (1, 0, 1)$.
If you really want to make sure you have understood this process, I encourage you to try the following for yourself. I have hidden the answers below the questions. You can see them by placing your mouse over the box but you should have a go at them yourself first.
Is the point $(e, 0, e)$ on the curve? If so, which $t$ does it correspond to?

 Looking at the three equations as above, but for the point $(e, 0, e)$, the third equation becomes $e^{-t} = e$ which has unique solution $t = -1$. However, $e^{-(-1)}\cos(-1) = e\cos(-1) \neq e$ as $\cos(-1) \neq 1$. Therefore there is no $t$ such that $r(t) = (e, 0, e)$, so the point $(e, 0, e)$ is not on the curve.  

Is the point $(\frac{1}{\sqrt{2}}\sqrt[4]{e^{\pi}}, -\frac{1}{\sqrt{2}}\sqrt[4]{e^{\pi}}, \sqrt[4]{e^{\pi}})$ on the curve? If so, which $t$ does it correspond to?

 Again, looking at the three equations, the third becomes $e^{-t} = \sqrt[4]{e^{\pi}} = e^{\pi/4}$ so $t = -\frac{\pi}{4}$ is the unique solution. Now we need to check the other two equations. As $e^{-(-\pi/4)}\cos\left(-\frac{\pi}{4}\right) = \sqrt[4]{e^{\pi}}\frac{1}{\sqrt{2}}$, $t = -\frac{\pi}{4}$ satisfies the first equation, and as $e^{-(-\pi/4)}\sin\left(-\frac{\pi}{4}\right) = \sqrt[4]{e^{\pi}}\frac{1}{\sqrt{2}}$, $t = -\frac{\pi}{4}$ also satisfies the second equation. Therefore the point $(\frac{1}{\sqrt{2}}\sqrt[4]{e^{\pi}}, -\frac{1}{\sqrt{2}}\sqrt[4]{e^{\pi}}, \sqrt[4]{e^{\pi}})$ is on the curve and corresponds to $t = -\frac{\pi}{4}$; that is, $r\left(-\frac{\pi}{4}\right) = (\frac{1}{\sqrt{2}}\sqrt[4]{e^{\pi}}, -\frac{1}{\sqrt{2}}\sqrt[4]{e^{\pi}}, \sqrt[4]{e^{\pi}})$.  

Does the curve intersect the $xy$-plane?

 The $xy$-plane is given by the equation $z = 0$, so if the curve intersects the $xy$-plane, it does so at a point of the form $(a, b, 0)$. That is, if the curve does intersect the $xy$-plane, there is some $t$ corresponding to that point such that $r(t) = (a, b, 0)$. But then the third equation reads $e^{-t} = 0$ which has no solutions. Therefore the curve does not intersect the $xy$-plane.

