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Topic: how to find normal to a surface in R^3
Replies: 3   Last Post: Oct 27, 2006 12:23 AM

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Lynn Kurtz

Posts: 1,278
Registered: 12/6/04
Re: how to find normal to a surface in R^3
Posted: Oct 27, 2006 12:23 AM
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On 26 Oct 2006 20:59:14 -0700, lite.on.beta@gmail.com 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).

--Lynn





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