Tangent plane to a parametric curve

Question

Find the tangent plane to a parametric curve $$\vec r(u) = x(u)\mathbf i + y(u)\mathbf j + z(u)\mathbf k$$

I know that the tangent line at a point $u = u_0$ is given by $$\vec \ell(t) = \vec r(u_0) + \vec r'(u_0)(t - u_0)$$ but what about the plane, though?

Answer 1

There is no tangent plane, just a tangent line. The reason is that the your curve is a 1-dimensional manifold and the tangent space has the same dimension as the manifold.

Intuitively, a tangent vector $v$ represents the speed and direction of a body moving on the manifold while remaining "on" that manifold. For example, if you're walking on earth while remaining with your feet on the ground, you can only move tangent to the ground. In your case you have a curve, which is like a path, so you can only move forward or backward with different speeds, so the tangent space is a line.

Answer 2

You have one input, so you can imagine a whole number line of all the inputs, each of which has one corresponding (3-dimensional) output. Now imagine all of the possible inputs going to their outputs, i.e. imagine that number line moving to become some other line in 3-dimensions. This is your parametric curve, the set of all possible outputs as you range the input.

Notice, your parametric curve is only a line (though it likely curves and bends in 3-dimensions), and not a plane, thus you can find a tangent line, but it doesn't really make sense to find a tangent plane. All possible planes that contain that line will be just touching the curve at that point (and thus tangent), and since their is no curving of a surface in other directions, there is no way to choose if any one of them is better (with a parametric surface, most of them will intersect the surface somewhere next to the line, and won't be tangent, so it makes sense to talk about a tangent plane).

I hope this helped!