1. ## Vector problem

Let $\displaystyle ABC$ be a triangle with $\displaystyle AB=\vec{a}$ and $\displaystyle AC=\vec{b}$ and $\displaystyle M\in[BC]$, $\displaystyle \frac{BM} {MC}=k$.
$\displaystyle AM= ?$

2. We have $\displaystyle \overrightarrow{BM}=k\cdot\overrightarrow{MC}$

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

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

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

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

$\displaystyle \overrightarrow{AM}=\frac{1}{1+k}\overrightarrow{a }+\frac{k}{1+k}\overrightarrow{b}$

3. Originally Posted by sillyme
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.

4. ## Thanks

Thank you red_dog!