Tangent line to level curve of function at point I understand that the gradient is perpendicular to the surface curve, or surface plane, depending on the dimensional space. So if I were trying to find the equation of the tangent plane, I know that I could use the gradient as the normal because they're parallel.
Yet what if I want to find a tangent line? 
I know that the gradient and an aribitrary vector tangent to the surface, say $<a,b>$, will be perpendicular, thus their cross-product would be $0$. Then I could have the equations $x = x_0 + at, y = y_0 + bt$, and thus a line. 
Except, since $a$ will be in terms of $b$ and vice-versa, it's hard for me to just put it in just terms of $x$ and $y$ sometimes. So I wanted to know if there's a faster method.
Another way seems to be to find the gradient at a specific point, and set $z=0$. i.e. $ ax +by = d$. 
This way seems easier, but it makes less sense geometrically than the first approach. All it does is set $z=0$, and doesn't really seem to explain why this would be a line tangent to the surface. 
So could someone explain the second method to me, and why by simply setting $z=0$, an equation of a line is formed.
 A: Given an $n\geq2$, a function $f: \>{\Bbb R}^n\to{\Bbb R}$, and a point $p$ define $q:=f(p)$. The set
$$S:=f^{-1}(q)=\{x \>|\>f(x)=q\}$$
is called the level line (or level surface) of $f$ through $p$. If $\nabla f(p)\ne0$ then there is a  neighborhood $U$ of $p$ such that $S\cap U$ is  an $(n-1)$-dimensional hypersurface in the sense of differential geometry. $S$ then has an $(n-1)$-dimensional tangent plane $T_pS$ at $p$, and this tangent plane has a well defined orthogonal complement which is the one-dimensional space $\langle\nabla f(p)\rangle\subset T_p{\Bbb R}^n$.
Any nonzero vector $v\in T_pS$ can serve as tangent vector to a curve $\gamma\subset S$, and these vectors are all of "equal right". There is no justification to set one variable to $0$ to single out one of them. If $n=2$ then $n-1=1$, and all such $v$ are equal up to a nonzero scalar. 
A: Use any functional form of the curve f=f(t) (f values are n-dimensional, t is a real value).
And for the point of interest P that corresponds to t=x take f'(t) at x.  Divide by modulus of f'(t) computed at x (is going to be a n-dimensional quantity).
And voila! the unit tangent vector to your curve.
A: I would refer to $\mathbb R^2$ cut surfaces in particular.
If the surface is given in implicit form $F(x,y,z)=0,$ then at any level $ z_1= c $ where $c$ is a constant level parameter it produces various level curves by horizontal slicings/intersections. And if we can express  $ z_1 = f(x,y) $ it comes out into this Monge form. For example all spiric sections obtained by cutting a torus at $y=y_0$.
At any section $z=c$ we have curvature $ k_{n(\psi)}$ in the plane of the section inclined at angle $\psi$ to the principal directions characterized by $ F=0, f=0 $ coefficients of first and second fundamental forms in classical surface theory. We need to pick the particular $\psi$ from the tangent bundle of filaments $S^1$ on the surface $S^2$.
By Euler's theorem we have for normal curvature of a filament $\kappa_n$ in the bundle rotating around a common normal to surface.
$$ k_{n(\psi)} =  ( \frac{k_1+ k_2}{2}   + \frac{k_1 - k_2}{2} \cos 2\psi ) =H \pm D \cos 2\psi   = k_1 \cos ^2 \psi + k_2  \sin^2 \psi $$
$$ \psi = \frac12\cos^{-1} \frac {( k_{n(\psi)}-H)}{D} $$
The mean curvature and angle between level line and principal curvature line $\psi$ marked (2,3) in the figure are represented on the Mohr's circle.We just need to orient / match normal curvature scalar at any point of level curve and we have the full picture with  tangent line in planes parallel to $(x,y)$ as
$$ \frac{y-y_1}{x-x_1}= \frac{\partial y/\partial s}{\partial x/ \partial s} $$ where the $s$ is arc along intersection boundary line.

