Finding a plane is tangent to a curve - $H=\left\{\left(6-s^2-t^2,\:s,\:t\right):\:s,t\in \mathbb{R}\:\right\}$

Question

$H=\left\{\left(6-s^2-t^2,\:s,\:t\right):\:s,t\in \mathbb{R}\:\right\}$ At the point $P=(1,1,2)$
so in order to get to solution, my try:
I defined:
$F\left(s,t\right):=\left(6-s^2-t^2,\:s,\:t\right)$
got:
$\text{grad}F\left(s,t\right)=\begin{pmatrix}-2s&-2t\\ 1&0\\ 0&1\end{pmatrix}$
Then I found out in order for the point to be correct, I must get $s=1$ and $t=2$ Thus, I did:
$\text{grad}F\left(1,2\right)=\begin{pmatrix}-2&-4\\ 1&0\\ 0&1\end{pmatrix}$
And from here my problem, usually In order to find a plane is tangent to a curve I used linearity with the equality:
$L(t)=(x_0,y_0,z_0)t + (x_1,y_1,z_1)$
but here it wont work since the points are with different dimensions.
How can I continue? please help.

Answer 1

If the implicit equation of a curve is $F(x,y,z)=0$ and the position vector of the curve is $\mathbf r=[x,y,z] $ where $x,y \text{ and }z $ are given parametrically as functions of $s$ and $t$ then the chain rule gives $$\nabla F\bullet \frac {\partial \mathbf r}{\partial s}=\mathbf 0,\nabla F\bullet \frac {\partial \mathbf r}{\partial t}=\mathbf 0$$ Then $\nabla F$ is in the direction of $$\frac{\partial \mathbf r}{\partial s}\times \frac{\partial \mathbf r}{\partial t}$$. Indeed, note that by the implicit function theorem, the cndition for $F$ to exist locallly in a neighbourhood of a given $.s=s_0,t=t_0$ is for the rank of the Jacobean of $$(s,t) \mapsto (x(s,t),y(s,t),z(s,t))$$ at$ s-s_0, t=t_0 $ to be 2, which is exactly the condition for the cross-product $\frac{\partial \mathbf r}{\partial s}\times \frac{\partial \mathbf r}{\partial t}$ not to be $\mathbf 0.$Thus a normal vector to the tangent plane is $\frac{\partial \mathbf r}{\partial s}\times \frac{\partial \mathbf r}{\partial t}$. The equation of the tangent plane is$$(\frac{\partial \mathbf r}{\partial s}\times \frac{\partial \mathbf r}{\partial t}) \bullet [x-1,y-1,z-2]=0.$$