1

Is the vector $(0, 0, -1)$ tangent to the surface $\mathbf{r} (u, v) = (2 \cos{v}, 3 \sin{v}, u)?$

What I've done:

The normal vector to the surface is $\mathbf{N} = (-3 \cos{v}, 2 \sin{v}, 0),$ so I want to see if $\mathbf{N} \cdot (0, 0, -1) = 0.$ In fact it is, but I don't know if that's enough to conclude that the vector is tangent to the surface.

Another thing I noticed is that $\partial \mathbf{r}_{u} (u, v) = (0, 0, 1),$ so $- \partial \mathbf{r}_{u} (u, v) = (0, 0, -1).$ Since $\partial \mathbf{r}_{u} (u, v)$ is tangent to the surface, can I conclude that $-\partial \mathbf{r}_{u} (u, v)$ is also tangent to the surface? And if so, why?

Chirs Erickson
  • 131
  • A tangent vector is orthogonal to a normal vector by definition of the two. So, yes, you check that the dot product is zero, which you've done. – Sammy Black May 19 '14 at 02:20
  • The tangent vectors form a vector space, spanned by by the directional derivatives for the parameters $u$ and $v$. So, any linear combination of those directional derivatives are also tangent. – Sammy Black May 19 '14 at 02:23
  • @SammyBlack Okay, I understand. Thanks. – Chirs Erickson May 19 '14 at 02:24

0 Answers0