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Normalizing a cross product

  1. Dec 17, 2007 #1
    [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. jcsd
  3. Dec 17, 2007 #2


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    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 [itex]||u||||v||cos(\theta)[/itex] where [itex]\theta[/itex] is the angle between u and v. Here, ||T(t)||= 1 and [itex]\theta= \pi/2[/itex] . Put those into the formula for the length of T(t) x T'(t).
  4. Dec 17, 2007 #3
    thanks soo much...I think you mean ||u||||v||sin(theta) but i get it.
  5. Dec 17, 2007 #4


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    Oh dear! Yes, of course it was supposed to be [itex]sin(\theta)[/itex].
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