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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 mesmiley

 

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 ?.

 Jan 4, 2019
 #1
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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 ? )

 Jan 5, 2019
 #2
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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 .

 Jan 5, 2019
edited by Guest  Jan 5, 2019
 #3
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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).

 Jan 5, 2019
 #4
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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

 Jan 5, 2019
edited by Guest  Jan 5, 2019
 #5
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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
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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
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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?

 Jan 6, 2019
 #8
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For k I said:

AJ=AB+BJ

⇔AJ=AB+1/2BC

AJ=AB+1/2BA+1/2AC

AJ=1/2AB+1/2AC

so 1/2 × 2/3 =1/3 

We have 2/3 × AJ = AG 

which prove that AG=2/3AJ

 Jan 6, 2019
 #9
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Hi ,

Never mind for my answer I found it 

AG=AG

BG=GJ+JB

CG=GJ+JC=0 

SO we have AG+AG+AB+AG+AC

which give:AG=1/3(AB+AC) 

Since J is the middle of BC we can write that AB+AC=AJ

So AG=1/3(AJ)=2/3AJ

SO AG=2/3AJ

 So A,G,J are aligned  

Thank you very much for your help appreciate it ❤😀

 Jan 6, 2019

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