On 26 Oct 2006 20:59:14 -0700, firstname.lastname@example.org wrote:
>Got it. Now I have the main tool to answer a question like this: > >" Find all points on the surface z = x^2 + y^2 -1 at which the normal >line to the surface coincides with the line joining the origin to the >point. " [From a previous year's Multivariable Calculus midterm] > >The actual curve equation is not important, I just think I lack another >tool to finish the question off (which can be applied to any surface). > >For this example (my effort): > >define f(x,y,z) = x^2 + y^2 -1 - z > >Therefore f(x,y,z) = 0 is a level surface of f that coincides with >surface in question. > >gradient of f at (x,y,z) is normal to the level surface through (x,y,z) > > >therefore: > >del f(x,y,z) = (2x, 2y, -1) is normal to surface in question at (x,y,z) > >So the line defined in direction of this vector is (2x, 2y, -1) t for >all t in R > >Now the line from origin through the point is obviously (x,y,z) s for >all s in R > >So to get all points where normal coincides with this line from origin: > >(2x, 2y, -1) t = (x, y, z) s > >which gives: > >(1) 2xt = xs => x(2t-s) = 0 >(2) 2yt = ys => y(2t-s) = 0 >(3) -t = zs >
You are on the right track, but you don't need two parameters here. The vectors are parallel if one is a multiple of the other
2x = xs so x(2-s)=0 2y = ys so y(2-s)=0 sz = -1
So either s = 2 (check the ramifications of that) or x = 0 and y = 0 (check the ramifications of that).