# Thread: Showing that a curve is normal to a surface

1. ## Showing that a curve is normal to a surface

Show that the curve

$\displaystyle r(t)=t^2\mathbf{i}+t\mathbf{j}+(5t-4)\mathbf{k}$ is normal to the surface $\displaystyle 2x^2+y^2+5z^2=8$ at the point $\displaystyle (1,1,1)$

Would I start off by finding the tangent to the surface, and then showing that the vector given by the curve is normal to the tangent plane?

2. Would I start off by finding the tangent to the surface, and then showing that the vector given by the curve is normal to the tangent plane?
Sounds good to me. What do you get?

3. Originally Posted by Ackbeet
Sounds good to me. What do you get?
For the tangent plane to the surface, I get, $\displaystyle 2x+y+5z=8$

Not too sure what to do from here?

4. Looks good so far. Now you need to find the direction of the vector r at the point (1,1,1) (which is on the curve, as you can easily see). How can you do that?

5. 1) Determine if the curve and surface do intersect at (1, 1, 1)!
2) Determine the tangent vector to the curve at that point.
3) Determine the normal vector to the plane at that point.

The curves intersects the surface at right angles at that point if and only if the tangent vector to the curve is a multiple of the normal vector to the plane.