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Math Help - [SOLVED] Triangle & vectors

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    Member courteous's Avatar
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    Question [SOLVED] Triangle & vectors

    In triangle ABC there is point D on line segment AB, so that |AD|:|DB|=2:1, and there is point E on BC, so that |BE|:|EC|=1:4. Point F is the point where line segments AE and DC intersect. Express the relationship (ratio) between |AF|:|FE| ( \vec{AB}=\vec{a}, \vec{BC}=\vec{b}).



    Exoneration: I'm reposting my thread, as it seems less precarious than bumping the old one. I also can't come up with any new idea.
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    Quote Originally Posted by courteous View Post
    In triangle ABC there is point D on line segment AB, so that |AD|:|DB|=2:1, and there is point E on BC, so that |BE|:|EC|=1:4. Point F is the point where line segments AE and DC intersect. Express the relationship (ratio) between |AF|:|FE| ( \vec{AB}=\vec{a}, \vec{BC}=\vec{b}).



    Exoneration: I'm reposting my thread, as it seems less precarious than bumping the old one. I also can't come up with any new idea.
    According to your sketch you get:

    \overrightarrow{AE}=\vec a + \frac15 \cdot \vec b and

    \overrightarrow{CD}=-\vec b - \frac13 \cdot \vec a

    Now let \overrightarrow{AF}= k\cdot \overrightarrow{AE} = k \cdot \left( \vec a + \frac15 \cdot \vec b \right) and

    \overrightarrow{CF}= r\cdot \overrightarrow{CD} = r \cdot \left( -\vec b - \frac13 \cdot \vec a \right) ............. then the vector \overrightarrow{AF} could be expressed by:

    \overrightarrow{AF}=\vec a + \vec b + r\cdot \overrightarrow{CD} =\vec a + \vec b + r \cdot \left( -\vec b - \frac13 \cdot \vec a \right)

    You'll get the equation:

    k \cdot \left( \vec a + \frac15 \cdot \vec b \right) = \vec a + \vec b + r \cdot \left( -\vec b - \frac13 \cdot \vec a \right) ............. Expand the brackets and collect like terms:

    k\cdot \vec a + \frac15 k \vec b = \left(1-\frac13r\right)\vec a + (1-r) \vec b ............. From this equation you can extract a system of simultaneous equations:
    \left| \begin{array}{lcr}k&=&1-\dfrac13 r \\\dfrac15 k&=&1-r\end{array}\right. ............. I've got k=\frac57~\wedge~r=\frac67

    Therefore

    \overrightarrow{AF} = \frac57 \overrightarrow{AE} and

    \overrightarrow{FE} = \frac27 \overrightarrow{AE}

    The ratio \frac{\overrightarrow{AF}}{\overrightarrow{FE}}= \frac{\frac57 \overrightarrow{AE}}{\frac27 \overrightarrow{AE}} = \boxed{\bold{\frac52}}
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