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.

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.
