Checking if there's a tangent plane to a surface (without the surface equation, but with $3$ given curves on the surface).

Question

Here's the question:

Given that the three curves:
$l_1: x=4t+1, y=4t^2-1, z=4t^2+2t$

$l_2: x=s^2, y=s, z=-s-1$

$l_3: x=3u+2, y=3u, z=9u^2-1$

are all on a surface S, can we pass a tangent plane to surface S in the point $(1,-1,0)$?

I know that we can pass a tangent plane if the function is differentiable, but here I don't have the equation of the surface, I've found out that the give point $(1,-1,0)$ is a point where all the three curves meet in xy plane, I'm not sure if that helps me get toward the answer and I'm confused on what next steps I have to do, I guess I have to somewhat get the equation of the surface using the three curves given, but I can't see how three curves can decide how my surface looks like, so I decided to come here and ask for some help.

Any help is appreciated, thanks in advance to everyone.

Answer 1

What you have said is correct but given we do not have equation of the plane. Here is another way we can approach it using the below definition of tangent plane -

If P be a point on a surface S and let C be any curve passing through P, lying entirely in surface S. If the tangent lines to all such curves C at P lie in the same plane, then this plane is called the tangent plane to S at P.

Below three curves are in a surface S.

$C_1: x=4t+1, y=4t^2-1, z=4t^2+2t$
$C_2: x=s^2, y=s, z=-s-1$
$C_3: x=3u+2, y=3u, z=9u^2-1$

We need to check whether a tangent plane can pass through the surface at point P $(1, -1, 0)$.

As you pointed out, all three curves pass through point P $(1, -1, 0)$. So we can first find tangent vectors to the curves at point P.

At point P, $t = 0$ for $C_1$ , $s = -1$ for $C_2$ and $u = -\frac{1}{3}$ for $C_3$.

Taking derivative, we find tangent vectors as $ \ C'_1(0) = (4, 0, 2); \ C'_2(-1) = (-2, 1, -1); \ C'_3(-\frac{1}{3}) = (3, 3, -6)$

Now if we have a tangent plane passing through P, all the $3$ vectors must lie in the same plane. So we can first evaluate cross product of first two vectors and then check if the dot product of the resulting vector with the third vector is zero.

In this case, it turns out that all three vectors are not in the same plane. So there cannot be a tangent plane to the surface at P.

Answer 2

In this case, the tangent vectors to your curves at the point have to be pointing along the surface of the tangent plane. Calculate all three tangent vectors and see if they are coplanar.

Does the problem say anything about S? Technically S could be very badly behaved outside of the curves, in which case there wouldn't be a tangent plane. I don't think this is what the problem is asking though.