PDA

View Full Version : calcuating the magnitude of a binormal vector (B)?


mr_coffee
Oct6-05, 02:12 AM
Hello everyone, I was suppose to show B(t) = T(t) x N(t) is perpendicular to both T(t) and N(t). I'm also to find |B(t)| which is the magnitude of B(t). B stands for a Binormal vector. T is the tagnent unit vector and N is the unit normal. The vector b(t) = T(t) x N(t) is perpendicualr to both T and N and is also a unit vector. Well i got the first part i think, but I don't know how i'm suppose to find the magnitude of B(t), any ideas? Here is my proof and work for the first part, maybe that will help figure out what he wants me to find for part (b).
http://img133.imageshack.us/img133/8076/88eb1.jpg
if that link is slow, this one also has my work:
http://show.imagehosting.us/show/764387/0/nouser_764/T0_-1_764387.jpg
Thanks!

Also, i just thought...because it said the binormal vector is also a unit vector, doens't that mean that |B(t)| = 1? Because I know |T(t)| = 1, and T is also a unit vector.

Galileo
Oct6-05, 04:26 AM
The question is rather trivial if you look at the geometrical definition of a cross product. If A,B are vectors, then A X B is a vector perpendicular to A and B. It's in the definition. If A and B are perpendicular, then |A X B|=|A||B|. This also follows directly from the definition.

mr_coffee
Oct6-05, 09:39 AM
Well i was wondering what |B(t)| is equal too, that does make sense though but this will make him happy thats why I did it that way

mr_coffee
Oct6-05, 07:47 PM
My idea was right, since B is a unit vector its magnitude is 1.