# Vector problem

• Feb 11th 2009, 06:20 AM
sillyme
Vector problem
Let $ABC$ be a triangle with $AB=\vec{a}$ and $AC=\vec{b}$ and $M\in[BC]$, $\frac{BM} {MC}=k$.
$AM= ?$
• Feb 11th 2009, 07:29 AM
red_dog
We have $\overrightarrow{BM}=k\cdot\overrightarrow{MC}$

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

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

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

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

$\overrightarrow{AM}=\frac{1}{1+k}\overrightarrow{a }+\frac{k}{1+k}\overrightarrow{b}$
• Feb 11th 2009, 07:32 AM
HallsofIvy
Quote:

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

Your notation is slightly confusing. You say that $AB= \vec{a}$ 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, $\vec{m}$, and the vector from A to M, $\vec{x}$ and use the capital letters only for the lengths.

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

The distance AM is the length of the vector $\vec{x}$ above.
• Feb 11th 2009, 07:34 AM
sillyme
Thanks
Thank you red_dog!