Line tangent to a curve (Vector style)

[Whitishcube]

Homework Statement

Let L be the line tangent to the curve $\vec{G}(t)=10\text{Cos}(t)\mathbf{i} + 10\text{Sin}(t)\mathbf{j} + 16t \mathbf{k}$ at the point $(\frac{10}{\sqrt{2}}, \frac {10}{\sqrt{2}}, 4 \pi)$
Find the point at which L intersects the x-y plane.

Homework Equations

The Attempt at a Solution

So taking the derivative of G gives

$$\vec{G}'(t)= -10\sin (t)\mathbf{i} + 10\cos (t)\mathbf{j} +16\mathbf{k}$$

So L should be defined as follows (with $t_0=\frac{\pi }{4}$):

$$L=\vec{G}\left(t_0\right)+t\left(\vec{G}'\left(t_0\right)\right)$$

$$L=\left\{\frac{10}{\sqrt{2}},\frac{10}{\sqrt{2}},4\pi \right\}+t\left\{\frac{-10}{\sqrt{2}},\frac{10}{\sqrt{2}},16\right\}$$
$$L=\frac{10}{\sqrt{2}}(1-t)\mathbf{i} + \frac{10}{\sqrt{2}}(1+t)\mathbf{j}\text{ }+(16t+4\pi )\mathbf{k}$$

So, solving for where the x and y components are both 0:

$$\frac{10}{\sqrt{2}}(1-t) =0$$
$$\frac{10}{\sqrt{2}}(1+t)=0$$

So the line L passes through the x-y plane at the point $(1,-1)$
does everything I've done here look correct? I'm not sure if I'm going about this the easiest (or correct) way. Thanks for the input!

 

[SEngstrom]

the x-y plane is defined by z=0 - other than that it looks good to me.