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Math Help - vectors

  1. #1
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    vectors

    Vectors \vec{a} and \vec{b} both have the same length, which is 1. The angle between them is 60^0.

    Find t so that  2\vec{a} + \vec{b} and  t\vec{a} + 5\vec{b} are
    perpendicular

    I got that far:

    cos \phi = \frac{\vec{a} \cdot \vec{b}}{\mid \vec{a} \mid \cdot \mid \vec{b} \mid} = \frac{\vec{a} \cdot \vec{b}} {1 \cdot 1} \Rightarrow  \phi = \frac{1}{2} \cdot 1 \cdot 1 = \vec{a} \cdot \vec{b}

    then i put  2\vec{a} = \frac{1}{\vec{b}} into

    (2\vec{a} + \vec{b}) \cdot (t\vec{a} + 5\vec{b}) = 0 and got that:

    (\frac{1}{\vec{b}} + \vec{b}) \cdot (t \cdot \frac{1}{2\vec{b}} + 5\vec{b}) = 0

    and continued for a couple more lines hoping it would become easier but i got nothing useful..how do you solve this?
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  2. #2
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    Quote Originally Posted by metlx View Post
    Vectors \vec{a} and \vec{b} both have the same length, which is 1. The angle between them is 60^0.

    Find t so that  2\vec{a} + \vec{b} and  t\vec{a} + 5\vec{b} are
    perpendicular

    I got that far:

    cos \phi = \frac{\vec{a} \cdot \vec{b}}{\mid \vec{a} \mid \cdot \mid \vec{b} \mid} = \frac{\vec{a} \cdot \vec{b}} {1 \cdot 1} \Rightarrow  \phi = \frac{1}{2} \cdot 1 \cdot 1 = \vec{a} \cdot \vec{b}

    then i put  2\vec{a} = \frac{1}{\vec{b}} into

    (2\vec{a} + \vec{b}) \cdot (t\vec{a} + 5\vec{b}) = 0 and got that:

    (\frac{1}{\vec{b}} + \vec{b}) \cdot (t \cdot \frac{1}{2\vec{b}} + 5\vec{b}) = 0

    and continued for a couple more lines hoping it would become easier but i got nothing useful..how do you solve this?
    \left( {2\overrightarrow a  + \overrightarrow b } \right) \cdot \left( {t\overrightarrow a  + 5\overrightarrow b } \right) = 2t\overrightarrow a \overrightarrow { \cdot a}  + 2\overrightarrow a  \cdot \overrightarrow b  + t\overrightarrow a  \cdot \overrightarrow b  + 5\overrightarrow b  \cdot \overrightarrow b
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