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Math Help - Linear combination and a proof for the area.

  1. #1
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    Linear combination and a proof for the area.

    http://dl.dropbox.com/u/25711584/algebra.jpg

    |DF|=3|AF|

    |BE|=\frac{1}{2}|BC|

    * express \bar{AG} as a linear combination of the vectors u= \bar{AB} and v= \bar{AD}.

    * Show that the triangle AFG always constitutes the same fraction of the area of the Parallelogram ABCD.


    i need help here. On the first question i have tryed to express the vector AG with other vectors to finally come to an linear combination of u and v. But i get problems whit any vector that ends at G. I suppose i should use the symmetry of the triangles somehow but dont really know how?


    the other question i need some tips to get started.

    Thanks.
    Last edited by mechaniac; April 8th 2011 at 11:35 AM.
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  2. #2
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    \Delta AGF ~ \Delta EGB (You know why?) \Rightarrow \frac{AG}{EG}=\frac{AF}{EB}=\frac{GF}{BG}
    AF=\frac{AD}{3}, EB=\frac{BC}{2} and AD=BC \Rightarrow \frac{AF}{EB}=\frac{GF}{BG}=\frac{2}{3}

    Now this should be easy.
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  3. #3
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    Quote Originally Posted by mechaniac View Post
    http://dl.dropbox.com/u/25711584/algebra.jpg

    |DF|=3|AF|

    |BE|=\frac{1}{2}|BC|

    * express \bar{AG} as a linear combination of the vectors u= \bar{AB} and v= \bar{AD}.

    * Show that the triangle AFG always constitutes the same fraction of the area of the Parallelogram ABCD.


    i need help here. On the first question i have tryed to express the vector AG with other vectors to finally come to an linear combination of u and v. But i get problems whit any vector that ends at G. I suppose i should use the symmetry of the triangels somehow but dont really know how?


    the other question i need some tips to get started.

    Thanks.
    1. According to your sketch:

    \overrightarrow{AD} = \overrightarrow{BC}

    \overrightarrow{AF}=\frac14 \overrightarrow{AD}

    2. \overrightarrow{AG} = t \cdot \overrightarrow{AE} = t \cdot (\overrightarrow{AB} + \frac12 \overrightarrow{BC})

    \overrightarrow{AG}= \overrightarrow{AB} + k \cdot \overrightarrow{BF} = \overrightarrow{AB} + k \cdot (-\overrightarrow{AB}+\frac14 \overrightarrow{AD})

    3. Since \overrightarrow{BC} = \overrightarrow{AD} you'll get:

    t \cdot (\overrightarrow{AB} + \frac12 \overrightarrow{BC}) = \overrightarrow{AB} + k \cdot (-\overrightarrow{AB}+\frac14 \overrightarrow{BC})

    Expand both sides of the equation and compare the co-efficient of equal vectors:

    \left|\begin{array}{rcl}t&=&1-k \\ \frac12 t&=&\frac14 k\end{array}\right.

    Solve this system of equations for t and k.

    4. Plug in the value for t or k into the appropriate equation to get the vector \overrightarrow{AG}
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  4. #4
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    Thanks for the help! i used determinants to show the area problem so dont need help on that
    Last edited by mechaniac; April 8th 2011 at 01:38 PM.
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