Coordinates of Hexagon Vertices in Base (AC, AD)

[Petrus]

Consider a regular hexagon ABCDEF (in order counterclockwise). Determine the coordinates of AB, AE AND AF (->) in the base (AC, AD) (->)

AB(->)=(_____,_____)
AE(->)=(_____,_____)
AF(->)=(_____,_____)

what I mean with exemple AF(->) positive way from A to F. I have draw a it but I got problem to rewrite
I got AE=EF+FA(->) I am correct?

 

[Petrus]

Re: hexagon, coordinates, base

I could uppload picture from internet but it should be insted of F it should B. In ourder counterclockwise. Well I can rewrite AE=EF+FA (->)

 

[HallsofIvy]

Re: hexagon, coordinates, base

No, you have the order wrong: AE= AF+ FE.

 

[Petrus]

Re: hexagon, coordinates, base

No, you have the order wrong: AE= AF+ FE.

Yeah I forgot to Edit. How do I do for AF and AB? Then I got Also My base.

 

[Petrus]

Re: hexagon, coordinates, base

View attachment 725
So I draw it.
We know from origo to A it is 1 and origo to D it is 1.

 

[Petrus]

I think I have thought wrong...
We got the base AC and AD and I put \(\displaystyle AC=(1,0)\) and \(\displaystyle AD(0,1)\)
I start with AB
We know that \(\displaystyle AC=AB+BC\) ( ->) that means \(\displaystyle AB=AC-BC <=> AB=AC-BC\) and we know that \(\displaystyle AC=(1,0)\) That means \(\displaystyle AB=(1,0)-BC\) But is BC same as from A to origo?

 

[Petrus]

So I did wrong... After a lot reading I think I got correct progress now...
We know it says regular hexagon that means from origo to any point got the length 1.
A circle is 360 degree and we got 8 lines. \(\displaystyle \frac{360}{8}=45\) So we got now (look picture). we know x value is cos and y value is sin so we know
\(\displaystyle A=(1,0)\)
\(\displaystyle B=(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})\)
\(\displaystyle C=(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})\)
\(\displaystyle D=(-1,0)\)
\(\displaystyle E=(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}})\)
\(\displaystyle F=(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}})\)
That means
\(\displaystyle AC=(-\frac{1}{\sqrt{2}}-1,\frac{1}{\sqrt{2}})\)
\(\displaystyle AD=(-2,0)\)
----
\(\displaystyle AB=(\frac{1}{\sqrt{2}}-1,\frac{1}{\sqrt{2}})\)
\(\displaystyle AE=(-\frac{1}{\sqrt{2}}-1,-\frac{1}{\sqrt{2}})\)
\(\displaystyle AF=(\frac{1}{\sqrt{2}}-1,-\frac{1}{\sqrt{2}})\)
So now I got all point but got problem to determine our cordinate with our base. View attachment 727