# Calcuating the magnitude of a binormal vector (B)?

1. Oct 6, 2005

### mr_coffee

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 [Broken]
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.

Last edited by a moderator: May 2, 2017
2. Oct 6, 2005

### Galileo

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.

3. Oct 6, 2005

### mr_coffee

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

4. Oct 6, 2005

### mr_coffee

My idea was right, since B is a unit vector its magnitude is 1.