# We have DeltaABCand the point M such that vec(BM)=2vec(MC).How to determinate x,y such that vec(AM)=xvec(AB)+yvec(AC)?

May 10, 2017

The answer is $x = \frac{1}{3}$ and $y = \frac{2}{3}$

#### Explanation:

We apply Chasles' relation

$\vec{A B} = \vec{A C} + \vec{C B}$

Therefore,

$\vec{B M} = 2 \vec{M C}$

$\vec{B A} + \vec{A M} = 2 \left(\vec{M A} + \vec{A C}\right)$

$\vec{A M} - 2 \vec{M A} = - \vec{B A} + 2 \vec{A C}$

But,

$\vec{A M} = - \vec{M A}$ and

$\vec{B A} = - \vec{A B}$

So,

$\vec{A M} + 2 \vec{A M} = \vec{A B} + 2 \vec{A C}$

$3 \vec{A M} = \vec{A B} + 2 \vec{A C}$

$\vec{A M} = \frac{1}{3} \vec{A B} + \frac{2}{3} \vec{A C}$

So,

$x = \frac{1}{3}$ and

$y = \frac{2}{3}$

May 10, 2017

$x = \frac{1}{3} , y = \frac{2}{3}$

#### Explanation:

We can define $P \in \left[A B\right]$, and $Q \in \left[A C\right]$ such that

$\left\{\begin{matrix}M = B + \frac{2}{3} \left(C - B\right) \\ P = B + \frac{2}{3} \left(A - B\right) \\ Q = A + \frac{2}{3} \left(C - A\right)\end{matrix}\right.$

and then

$M - A = \left(Q - A\right) + \left(P - A\right)$

or after substituting

$M - A = \frac{2}{3} \left(C - A\right) + \frac{1}{3} \left(B - A\right)$

so

$x = \frac{1}{3} , y = \frac{2}{3}$