Finding Normal Unit Vectors

From: Will Oram (spamguy_at_ihatespam.com)
Date: 09/10/04 

Date: Fri, 10 Sep 2004 17:49:04 -0400

I understand how to find a tangential unit vector T of a parametric 3D
function. What I "don't" understand, however, is how to find a normal
unit vector N from T.

I am told (from books and MathWorld) that

      dT/dt
N = _________
     |dT/dt|

...does that mean that

      a
N = _____
     |a|

since T is based off velocity? If so, that doesn't seem to correspond
with N being perpendicular to T.

Let r(t) equal, say, <t, t^2, t^3>, and I want to find the tangential
and normal at t = -1 (I vow this does not correspond to homework
problems or the like, etc etc.; this problem is one I generated for this
purpose).

Then v(t) = <1, 2t, 3t^2>, and a(t) = <0, 2, 6t>. I find T by dividing v
by its scalar, getting <1, -2, 3>/sqrt(14).

If I find N by the method above, I get <0, 2, -6>/sqrt(40). It doesn't
take much effort to see N and T are not perpendicular as they should be
(dot product != 0), therefore I'm not doing something right.

But what?

Best,

Will

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