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)#?

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Answer 1 · 2017-05-10T14:14:40

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

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 · 2017-05-10T14:28:49

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

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#