Why does normalizing T(t) × T'(t) equal ||T'(t)||?

[issisoccer10]

[SOLVED] normalizing a cross product

Homework Statement

How does normalizing (T(t) x T'(t)) equal ||T'(t)||?

Homework Equations

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

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

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

 

[HallsofIvy]

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).

 

[issisoccer10]

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

 

[HallsofIvy]

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