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Thread: Vector Help

  1. #1
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    Vector Help

    Given:

    $\displaystyle \vec{u}\bullet\vec{v}$ = $\displaystyle |\vec{u}||\vec{v}|\cos{\theta}$

    Prove that $\displaystyle |\vec{u}\bullet\vec{v}| \leq |\vec{u}||\vec{v}|$

    I have:

    $\displaystyle \vec{u}\bullet\vec{v}$ = $\displaystyle |\vec{u}||\vec{v}|\cos{\theta}\leq|\vec{u}||\vec{v }| (1)$ since $\displaystyle |\cos{\theta}|\leq1$

    How do I jump from that to $\displaystyle |\vec{u}\bullet\vec{v}| \leq |\vec{u}||\vec{v}|$. I realize it's probably very simple. If someone could just help me take the leap I would appreciate it!

    I guess I know that the dot product is a scalar so I'm wondering..I don't know...geez louise.

    P.S. It's like 1 in the morning, and this is my last quesiton
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  2. #2
    o_O
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    You already got it don't you?

    If $\displaystyle \vec{u} \cdot \vec{v} = \left| \vec{u}\right| \left|\vec{v}\right| \cos \theta$, then:

    $\displaystyle \left|\vec{u} \cdot \vec{v}\right| = \left| \vec{u}\right| \left|\vec{v}\right| |\cos \theta| \leq \left| \vec{u}\right| \left|\vec{v}\right|(1)$

    since $\displaystyle \left| \vec{u}\right| \geq 0$ and $\displaystyle \left| \vec{v}\right| \geq 0$ already. Just a matter of dealing with $\displaystyle \cos \theta$ which you've promptly handled.
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