Question

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

Answer 1

The answer is `x=1/3` and `y=2/3`

Explanation:

We apply Chasles' relation

`vec(AB)=vec(AC)+vec(CB)`

Therefore,

`vec(BM)=2vec(MC)`

`vec(BA)+vec(AM)=2(vec(MA)+vec(AC))`

`vec(AM)-2vec(MA)=-vec(BA)+2vec(AC)`

But,

`vec(AM)=-vec(MA)` and

`vec(BA)=-vec(AB)`

So,

`vec(AM)+2vec(AM)=vec(AB)+2vec(AC)`

`3vec(AM)=vec(AB)+2vec(AC)`

`vec(AM)=1/3vec(AB)+2/3vec(AC)`

So,

`x=1/3` and

`y=2/3`

Answer 2

`x = 1/3, y = 2/3`

Explanation:

We can define `P in [AB]`, and `Q in [AC]` such that

`{(M = B + 2/3(C-B)),(P=B+2/3(A-B)),(Q=A+2/3(C-A)):}`

and then

`M-A = (Q-A)+(P-A)`

or after substituting

`M-A=2/3(C-A)+1/3(B-A)`

so

`x = 1/3, y = 2/3`