Plane curves from interval I to a three dimensional space

Question

I'm studying differential geometry and for the definition of curve we have said that it's the image of a fucntion $r(t)$ , $t\in I$, where $I$ is an interval of the real number. Suppose we have that $r: I -> \Bbb R^3$ .

Now what I know, is that a curve is called "plane curve" when $r(t) , \forall t \in I$, are inside a plane. If $r : I -> \Bbb R^2$, it's prety obvious that is is plane since it's codomain is the Euclidean plane. But in the case I wrote above, where we have $\Bbb R^3$, I can't understand how a curve can be plane.

Can someone explain with an example or maybe with a graph? I already looked online but I can't find anything. Thank you in advance.

Answer 1

For example you can choose $r(t)=t(1,1,0)$ that it is a line that lying on the plane $\Pi=\{(x,y,z): z=0\}$

In general you can associate a map $r: I\to \mathbb{R}^2$ a plane curve $r^\sim$ of $\mathbb{R}^3$ in this way:

for any $t\in I$ $r^\sim (t)=(r_1(t),r_2(t),0)$

Or, for example

for any $t\in I$ $r^\sim (t)=(r_1(t),0,r_2(t))$ (In this case the curve lying on the plane $y=0$)