Normalizing a cross product

1. Dec 17, 2007

issisoccer10

[SOLVED] normalizing a cross product

1. The problem statement, all variables and given/known data
How does normalizing (T(t) x T'(t)) equal ||T'(t)||?

2. Relevant equations
r'(t) x r''(t) = ||r'(t)||^2·(T(t) x T'(t))

||r'(t) x r''(t)|| = ||r'(t)||^2·||T'(t)||

3. The attempt at a solution
This doesn't bode well with me. Simply don't get it. Please shed some light on this for me

2. Dec 17, 2007

HallsofIvy

Staff Emeritus
What do you mean by "normalizing"? At first, I thought you meant dividing by its length to get a unit vector. But of course T'(t) itself is not necessarily a unit vector.
Do you mean simply show that ||T(x) x T'||= ||T'(t)||? That's easy:

T(t) is, by definition, the unit vector in the direction of of the curve, r(t). Since it has constant length, its derivative, T'(t), is perpendicular to T(t). In general the cross product of two vectors, u and v, has length $||u||||v||cos(\theta)$ where $\theta$ is the angle between u and v. Here, ||T(t)||= 1 and $\theta= \pi/2$ . Put those into the formula for the length of T(t) x T'(t).

3. Dec 17, 2007

issisoccer10

thanks soo much...I think you mean ||u||||v||sin(theta) but i get it.

4. Dec 17, 2007

HallsofIvy

Staff Emeritus
Oh dear! Yes, of course it was supposed to be $sin(\theta)$.