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There is a regular hexagon given to you ABCDEF.It is asked to find the sum of AB + AC+AD+AE+AF vectors.

I am not getting is how will we make the directions of vectors or you can say head and tail of the hexagon vectors.

Done till now :

1 I have AF vectors as the resultant using polygon law.

2Then taken vectors AC,AD,AE and AF(This one is already made as well).

$ What I saw done online and doubts I got.

Q They took OC vector direction to make AC = AO+OC.OC direction is not given then how can they say that.Maybe it is CO as well.enter image description here

If my doubt for OC vector is cleared.I think I can solve it.

Please do share if you have some other method to solve it.

Please help.

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    $\begingroup$ It is a regular hexagon so $OC = AB$ and $OC \parallel AB$. $\triangle ABO$ and $\triangle OCB$ are equilateral triangles. $\endgroup$ – Math Lover Dec 4 '20 at 7:31
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    $\begingroup$ The direction is either same or opposite directions. It cannot be anything else. $\endgroup$ – Math Lover Dec 4 '20 at 7:34
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    $\begingroup$ $\angle ABO = \angle BOC$. So $AB$ and $OC$ are parallel. But even if they were not parallel, the vector addition does not change. $\vec{AC} = \vec{AO} + \vec{OC}$ will still be true. $\endgroup$ – Math Lover Dec 4 '20 at 7:46
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    $\begingroup$ @AnindyaPrithvi ok I will $\endgroup$ – user841124 Dec 4 '20 at 7:47
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    $\begingroup$ You are welcome. $\endgroup$ – Math Lover Dec 4 '20 at 7:49
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Note $AD=2AO$.

$AOCB$ is a parallelogram. So $AC=AO+AB$

$AOEF$ is a parallelogram. $AE=AO+AF$

$ABOF$ is also a parallelogram. So $AB+AF=AO$

Summing, $$AB+AC+AD+AE+AF=4AO+2(AB+AF)=6AO$$

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Note that since we have a regular hexagon,

$$\begin{align} AB + AC + AD + AE + AF &= (AB + AF) + (AC + AE) + AD \\ \\ &= (AB + BO) + [(AB + BC) + (AF + FE)] + AD \\ \\ &= AO + [(AB + BC + AF) + FE] + AD \\ \\ &= AO + [(AB + BC + CD) + AO] + AD \\ \\ &= AO + [AD + AO] + AD \\ \\ &= AO + 2AO + AO + 2AO \\ \\ &= 6AO \end{align}$$

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If $O$ is the centre of the regular hexagon

$$\begin{aligned} AB + \dots +AF&= 5AO + OB +\dots +OF\\ &=5AO-OA+OA+ OB+\dots +OF\\ &=6AO \end{aligned}$$

As the sum $$OA+\dots +OF$$ vanishes.

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  • $\begingroup$ How did you get 5 AO and OB+...OF sir? $\endgroup$ – user841124 Dec 4 '20 at 7:31
  • $\begingroup$ For any points $M,N$, you have $MN =MO+ON$. Apply that 5 times to $AB, AC...$. $\endgroup$ – mathcounterexamples.net Dec 4 '20 at 7:33
  • $\begingroup$ I am just thinking on it. $\endgroup$ – user841124 Dec 4 '20 at 7:35
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    $\begingroup$ And I made a mistake that is now corrected. $\endgroup$ – mathcounterexamples.net Dec 4 '20 at 7:35
  • $\begingroup$ That is exactly my doubt $\endgroup$ – user841124 Dec 4 '20 at 7:36

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