Hey guys I'm sorry this is the third time that I post this exercice , any advice or answers can help me
be ABC a triangle non flat,we define the isobarycentre of the triangle ABC the point G such that :
vector(AG)+vector(BG)+vector(CG)=0
the point G is the centre of gravity of the triangle ABC .
QUESTIONS :
1.the respective points I,J and k are placed in the segments (AB),(BC) and (AC). express the vector AJ according to the vector(AB) and the vector(AC) ?.
2.using the expression: vector(AG)+vector(BG)+vector(CG)=0, express the vector(AG) accrording to the vector(AB) and the vector(AC) ? .
3.show that we can write vector(AG)=Kvector(AJ) "(AG=KAJ) " , for a certain k to be determined ?.
In 1. what do you mean by retrospectively ?
Is J placed so that it lies on the vector AG or is it at some random point on BC, (and then, in which case, how can AJ be a scalar multiple of AG ? )
1.respectively that the point I is placed in the middle of the segment AB and j in the middle of the segment BC and k in the middle of the segment AC.
- for J in the statement he says it's a respective point of segmnet BC so I have to put it.
Then for AJ and AG .I think I should use the relation of chasle .
I want to put an image so you can see the the drawing of the triangle but I can't let me .
So J is the mid-point of BC, why couldn't you say that first time round ?
Since J is the mid point of BC, we can say that CJ = - BJ, (equal in magnitude and opposite in direction).
Triangle law , twice.
AJ = AB + BJ ,
AJ = AC + CJ .
Add them and divide by 2, AJ = (AB + AC)/2 , (the other two vectors have cancelled each other).
Now do a similar thing to get an expression for AG , (using BG + CG = - AG on the way).
Combine the two results, (you should find that k = 2/3).
For AG i did:
GA=-(GA+AB)-(GA+AC)
⇔GA=-GA-AB-AC
⇔-2GA-AB-AC
⇔GA+2GA=-AB-AC
⇔3GA=-AB-AC
⇔GA=-1/3AB-1/3AC
I used GA+GB+GC=0 instead of AG+BG+CG=0
And if we want to deduce that the points A,G and k are aligned how we can prove it ?
It would have been better to have started with AG = AB + BG and AG = AC + CG etc.,
but your final equation is correct and you can make use of it by multiplying throughout by a negative sign.
- GA = - (- 1/3 AB - 1/3 AC) , so AG = (1/3)( AB + AC ).
Now you should see that AJ and AG are both scalar multiples of the same vector AB + AC ,
in which case A G and J must be aligned.
Finally, work out the value of k in order that AG = k.AJ .
So I said :
if AB and AC are collinear ABC are aligned
show that AG and AJ are collinear
goal: we want to get AG =KAJ
mean to express AG and AC. .with the help of AB(AI=AB) and AC(AK=AC)
SO: AJ=IB+BJ
AJ=IA+AB+1/2BC (since BJ=1/2 of BC)
⇔1/3AB+AB+1/2(BA+1/2AC) since BA+AC=BC
⇔1/3AB+AB+1/2AB+1/2AC(because we have multiplied by 1/2 for AB and AC)
⇔5/6AB+1/2AC( I did 1/3+1-1/2)
now for AG I did
AG=IA+AK
⇔(1/3AB)+(1/2AC)
Now I am blocked.
Is it correct for AJ?
