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Confused about unit vector equivalence

  1. Feb 19, 2008 #1
    x is a unit vector [tex]\in \Re^{2}[/tex]. My textbook states that [tex]\frac{x}{||x||}=\frac{1}{||x||}x[/tex]. What is the point of including [tex]\frac{1}{||x||}[/tex]; why do they divide the vector by its length?

    Edit: I just looked at a book in Google's database, and from what I understand:

    e.g. [tex]\sqrt{{2^2+2^2+1^2}}=3[/tex] so that becomes [tex](\frac{2}{3}) ,(\frac{2}{3}),(\frac{1}{3}) = 1[/tex] due to the the vector rule (add by component). Basically, the answer to my question lies in the proof.
    Last edited: Feb 19, 2008
  2. jcsd
  3. Feb 19, 2008 #2
    We shrink or stretch the vector such that it points the same direction with the original vector but its length is 1. That shrinking or stretching factor is the reciprocal of its length. So, basically we don't do anything different than the scalar version [tex]5\frac{1}{5} = 1[/tex], only difference is that the length notion becomes [tex]\sqrt{a_1^2 + a_2^2 + \ldots + a_n^2}[/tex].
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