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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 ?.

Guest Jan 4, 2019

#1**+1 **

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__ ? )

Guest Jan 5, 2019

#2**+1 **

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 .

Guest Jan 5, 2019

edited by
Guest
Jan 5, 2019

#3**+1 **

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).

Guest Jan 5, 2019

#4**+1 **

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

Guest Jan 5, 2019

edited by
Guest
Jan 5, 2019

#5**+1 **

And if we want to deduce that the points A,G and k are aligned how we can prove it ?

Guest Jan 5, 2019

#6**0 **

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__ .

Guest Jan 6, 2019

#7**0 **

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?

Guest Jan 6, 2019