How to verify if a vector in three space lies on another vector?

Question

I can find the intersection points of graphs when asked, but I am unsure what to do when I must "verify if the curve lies on the surface".

Verify that the curve $r(t)$ lies on the given surface. Give the name of the surface:

$r(t) = (0,\;2\text{cos}(t),\;3\text{sin}(t))$

$x^2 + y^2/4 + z^2/9 = 1$

Appreciate the help

Answer 1

The points $(x,y,z)$ of the curve are such that there exist $t$ satisfying

$$x=0, y=2\cos(t), y=3\sin(t)$$

the points $ (x,y,z)$ of the surface must satisfy $$x^2+y^2/4+z^2/9=\color{red}{1}$$

the curve lies in the surface is each point of the curve is in the surface.

but if we replace, we find that

$$0^2+(2\cos(t))^2/4+(3\sin(t))^2/9=\color{red}1$$

done.

Answer 2

The surface is called an ellipsoid.

You can check that $x=0, y=2\cos t,z=3\sin t$ satisfies $x^2+\dfrac{y^2}4+\dfrac {z^2}9=0^2+\dfrac{4\cos^2t}4+\dfrac {9\sin^2t}9=\cos^2t+\sin^2t=1$.

Thus the curve lies on the surface.