Tangent line to a parameterized curve implicitly defined

Question

Let

$F=\sin(xe^y)+e^y\cos z-1$

and

$G=x^2-e^{xz}-z+1$.

Consider the system

$$\left\{\begin{matrix}
F(x,y,z)=0& & \\
G(x,y,z)=0& &
\end{matrix}\right.$$

(i) Prove that the system implicitly defines a curve in space, in a neighbourhood of $(0,0,0)$. (ii) Calculate the tangent to the curve at the origin.

Applying two times the implicit function theorem, we end up writing the solutions to the system near $(0,0,0)$ as $(x,α(x),β(x))$, which is a curve with parameter $x$, and the first point (i) is done. We also have $α'$ and $β'$.

But how can we approach the second? I found that the direction of the tangent line in such a case should be parallel to the vector given by the cross product between the gradients of the two functions. Is this true? Why?

If we find the direction of the tangent, can we also find its equation in this problem?

Thanks in advance!

Answer 1

For the second part, you may approximate $F=G=0$ near $(0,0,0)$ as follows,

$$F(x, y,z) \approx x+(1+y)-1 = x+y=0$$

$$G(x,y,z)\approx -z =0$$

Their normal vectors at origin are $(1,1,0)$ and $(0,0,-1)$ respectively. Thus, the tangent is $(1,1,0)\times (0,0,-1)=(-1,1,0)$ at the origin.