Differential geometry problem about curves Please, help me to solve the following differential geometry problem.
Equations $F(x, y, z) = 0$ and $G(x, y, z) = 0$ denote space curve $L$.
Gradients of $F$ and $G$ are not collinear in some point $M(x_0, y_0, z_0)$ which belongs to the curve $L$.
Find:
*

*The tangent line (its equation) in the point $M$

*The osculating plane in the point $M$

*The curvature in the point $M$

The 1st part is quite evident for me:
one can simply differentiate $F(x(t), y(t), z(t))$ and $G(x(t), y(t), z(t))$ obtaining two equations on $x'$, $y'$ and $z'$. Such simultaneous equations have a solution due to the condition of non-collinearity. Then, assuming $x' = 1$, it's easy to find $y'$ and $z'$ which gives us a vector $(x', y', z')$. This vector together with the point $M$ denote the desired tangent line.
But what to do with the other two parts?
Thanks in advance.
 A: The curvature vector is defined to be the derivative of the unit tangent vector. Let us say that $\gamma(s)$ is the arc length parametrization of your curve. That is to say: $$F(\gamma(s))=0$$ $$G(\gamma(s))=0.$$ To find the tangent vector you derived and obtained: $$\nabla F(\gamma)\cdot\gamma'=0$$ $$\nabla G (\gamma)\cdot \gamma'=0.$$
To get the answer to the second question you need the direction of $\gamma''$ (having supposed $\|\gamma'\|=1$). So you derive and get: $$\mathcal{H}F(\gamma)[\gamma',\gamma']+ \nabla F(\gamma)\cdot\gamma''=0$$ $$\mathcal{H}G(\gamma)[\gamma',\gamma']+ \nabla G(\gamma)\cdot\gamma''=0.$$
Where $\mathcal{H}F$ is the hessian matrix and the square parenthesis represent the linear product.
Once you solve it you will get the answer. This will answer to your third question too.
A: Better, perhaps, than assuming $x' = 1$, is to get the tangent line by letting 
$$
\mathbf u = \nabla F(x_0, y_0, z_0) \times \nabla G(x_0, y_0, z_0).
$$
(Why is $u$ nonzero?)
Then the tangent line is given by 
$$
\{ X + t \mathbf u \mid t \in \mathbb R\}.
$$
