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Dot Product

  1. Jul 13, 2008 #1


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    If [tex]\vec{u},\vec{v} \in R^n[/tex], find [tex]\vec{u}.\vec{v}[/tex] given that [tex]|\vec{u}+\vec{v}|=1[/tex] and that [tex]|\vec{u}-\vec{v}|=5[/tex].

    When i first looked at this i thought i knew how to do it, but i got a bit stuck. I started by finding the dot product:

    (\overrightarrow u - \overrightarrow v ) \cdot (\overrightarrow u - \overrightarrow v ) = \left| {\overrightarrow u } \right|^2 + \left| {\overrightarrow v } \right|^2 - 2\overrightarrow u \cdot \overrightarrow v \\
    \therefore \left| {\overrightarrow u - \overrightarrow v } \right|^2 = \left| {\overrightarrow u } \right|^2 + \left| {\overrightarrow v } \right|^2 - 2\overrightarrow u \cdot \overrightarrow v \\
    \therefore \overrightarrow u \cdot \overrightarrow v = \frac{{\left| {\overrightarrow u } \right|^2 + \left| {\overrightarrow v } \right|^2 - \left| {\overrightarrow u - \overrightarrow v } \right|^2 }}{2} \\


    And thats where im stuck. Im sure its something simple, but i haven't managed to see it :smile:
  2. jcsd
  3. Jul 13, 2008 #2


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    You are given both [itex]|\vec{u}-\vec{v}|[/itex] and [itex]|\vec{u}+\vec{v}|[/itex] but don't seem to have used the value of [itex]|\vec{u}+ \vec{v}|[/itex]. Did you consider looking at [itex](\vec{u}-\vec{v})\cdot(\vec{u}+\vec{v})= |u|^2- |v|^2[/itex]?
  4. Jul 13, 2008 #3
    I couldn't find relationship between |u|^2-|v|^2

    but, continuing your way
    (u-v).(u-v) = u^2+v^2-2u.v
    (u+v).(u+v) = u^2+v^2+2u.v

    now, it's just one step since you know both left side values
  5. Jul 15, 2008 #4


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    Oh so now i can just solve by eliminating u^2+v^2?

    Thanks for the help guys :smile:
  6. Jul 15, 2008 #5
    if that answers the question ...
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