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Dot product arithmetic

  1. Jan 14, 2013 #1
    If you square the magnitude of a vector you get the dot product, correct?

    ||v||^2 = v . v

    Can you also say that

    ||v|| = sqrt(v . v)?
     
    Last edited: Jan 14, 2013
  2. jcsd
  3. Jan 14, 2013 #2
    Of course, basic algebra.
     
  4. Jan 14, 2013 #3
    Thanks. I didn't know if some weird cosine rule existed in there
     
  5. Jan 14, 2013 #4
    Okay. Just to cement this:

    If ##\vec{v} = <a,b>## and ##\vec{w} = <c,d>## then ##\vec{v} \cdot \vec{w} = ac+bd## and ##\vec{v} \cdot \vec{v} = a^2+b^2##

    So if ##|| \vec{v} || ^2 = \vec{v} \cdot \vec{v} = a^2+b^2## then ##\sqrt{|| \vec{v} || ^2} = || \vec{v} || = \sqrt{a^2+b^2}##
     
  6. Jan 14, 2013 #5
    Hey cytochrome.

    If the inner product is valid then all of your statements are true.
     
  7. Jan 15, 2013 #6
    cosine rule cannot bother you here because the angle between "vectors" is zero.
     
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