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Thread: Vector problem

  1. #1
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    Vector problem

    Let $\displaystyle ABC$ be a triangle with $\displaystyle AB=\vec{a}$ and $\displaystyle AC=\vec{b}$ and $\displaystyle M\in[BC]$, $\displaystyle \frac{BM} {MC}=k$.
    $\displaystyle AM= ?$
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  2. #2
    MHF Contributor red_dog's Avatar
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    We have $\displaystyle \overrightarrow{BM}=k\cdot\overrightarrow{MC}$

    $\displaystyle \overrightarrow{BM}=\overrightarrow{AM}-\overrightarrow{AB}$

    $\displaystyle \overrightarrow{MC}=\overrightarrow{AC}-\overrightarrow{AM}$

    Then, $\displaystyle \overrightarrow{AM}-\overrightarrow{AB}=k\left(\overrightarrow{AC}-\overrightarrow{AM}\right)\Rightarrow$

    $\displaystyle \Rightarrow (1+k)\overrightarrow{AM}=\overrightarrow{AB}+k\ove rrightarrow{AC}\Rightarrow\overrightarrow{AM}=\fra c{1}{1+k}\overrightarrow{AB}+\frac{k}{1+k}\overrig htarrow{AC}\Rightarrow$

    $\displaystyle \overrightarrow{AM}=\frac{1}{1+k}\overrightarrow{a }+\frac{k}{1+k}\overrightarrow{b}$
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  3. #3
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    Quote Originally Posted by sillyme View Post
    Let $\displaystyle ABC$ be a triangle with $\displaystyle AB=\vec{a}$ and $\displaystyle AC=\vec{b}$ and $\displaystyle M\in[BC]$, $\displaystyle \frac{BM} {MC}=k$.
    $\displaystyle AM= ?$
    Your notation is slightly confusing. You say that $\displaystyle AB= \vec{a}$ so "AB" means the vector from A to B but then you use BM and MC as lengths of vector since you are dividing them. I don't know whether the "AM" you want is to be a vector or a length.

    I am going to call the vector from B to C, [tex]\vec{c}[/itex], the vector from B to M, $\displaystyle \vec{m}$, and the vector from A to M, $\displaystyle \vec{x}$ and use the capital letters only for the lengths.

    Since $\displaystyle \frac{BM}{MC}= k$, BM= kMC and BM+ MC= (k+1)MC.
    Then $\displaystyle \vec{m}= (BM/(BM+ MC)\vec{c}= (k/k+1)\vec{c}$. But $\displaystyle \vec{b}= \vec{a}+ \vec{c}$ so $\displaystyle \vec{c}= \vec{b}- \vec{a}$ and then $\displaystyle \vec{m}= (k/(k+1))(\vec{b}- \vec{a})$ so, finally, $\displaystyle \vec{x}= \vec{a}+ \vec{m}= \vec{a}+ (k/(k+1))(\vec{b}- \vec{a}))$$\displaystyle = (1/(k+1)\vec{a}+ (k/(k+1))\vec{b}$

    The distance AM is the length of the vector $\displaystyle \vec{x}$ above.
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  4. #4
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    Thanks

    Thank you red_dog!
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