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Math Help - linear combinations

  1. #1
    Junior Member
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    linear combinations

    The question is:

    In a parallelogram ABCD, F is the midpoint of AD, and E divides BC internally in the ratio 3:2. If P is the point of intersection of AE and BF, show that P divides AE in the ratio 5:6.

    Okay so do i make combine all these vector ratios into a single vector and then prove the 5:6 ratio? When i do this though, the vectors do not result into a 5:6 ratio. How should I attempt this question?
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  2. #2
    MHF Contributor red_dog's Avatar
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    Let \frac{AP}{PE}=k

    \overrightarrow{BP}=\frac{1}{1+k}\overrightarrow{B  A}+\frac{k}{1+k}\overrightarrow{BE}

    \overrightarrow{BF}=\overrightarrow{BA}+\overright  arrow{AF}=\overrightarrow{BA}+\frac{1}{2}\overrigh  tarrow{AD}=

    =\overrightarrow{BA}+\frac{1}{2}\overrightarrow{BC  }=\overrightarrow{BA}+\frac{5}{6}\overrightarrow{B  E}

    But the vectors BP and BF are coliniar, then \overrightarrow{BP}=\alpha\overrightarrow{BF}

    \frac{1}{1+k}\overrightarrow{BA}+\frac{k}{1+k}\ove  rrightarrow{BE}=\alpha\overrightarrow{BA}+\frac{5\  alpha}{6}\overrightarrow{BE}

    \Rightarrow\frac{1}{1+k}=\alpha, \ \frac{k}{1+k}=\frac{5\alpha}{6}\Rightarrow k=\frac{5}{6}
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