Given an Equilateral triangle on the $xy$ plane with sides $a$, we pick the point $(0,0)$ to be at the center of the triangle(where the medians meet) and the direction of the $x$ axis to be parallel to one of the sides. Let $(r_1,r_2,r_3)$ be vectors that represent the vectors to vertices or corners of the equilateral triangle. What is the cartesian expression of vectors $ (r_1,r_2,r_3)$? The answer is $r1=(0,\frac{a}{\sqrt3}),r2=(\frac{-a}{2},\frac{-a}{(2\sqrt3)}), r3=(\frac{a}{2},-\frac{a}{(2\sqrt3)}).$
We know that all sides are equal to $a$ , the direction vectors are $\hat i=(1,0)$ and $\hat j=(0,1)$ and one of the sides is parallel to the direction of the $x$ axis so $\hat i = \alpha r_n$
But I don't actually know how to begin with this as we don't have the vectors $r_1,r_2,r_3$ I think it has to be related to the point $O=(0,0)$ and get the vector $\vec{ r_n O}$ I also tried doing $a=a \cos(\frac{\pi}{3}) \cdot \hat i + a\sin(\frac{\pi}{3})\cdot \hat j$ but then I get the same point to all the edges which is not correct.
I would appreciate any help and tips on this problem.
The side parallel to the $x$-axis is perpendicular to some $r_i,$ wlog $r_1.$ So, $r_1=(0,c).$ Also wlog, $c>0$ (rotating the triangle by $\pi$ if necessary), and $r_2,r_3$ are $r_1$ rotated by $\frac{2\pi}3$ and $-\frac{2\pi}3$ respectively (renumbering again the $r_i$'s if necessary). So: $$r_1=(0,c),\quad r_2=\left(-\frac{c\sqrt3}2,-\frac c2\right),\quad r_3=\left(\frac{c\sqrt3}2,-\frac c2\right).$$ There remains to calculate $c:$ $$a^2=\|r_3-r_2\|^2=3c^2$$ hence $c=\frac a{\sqrt3}$ and $$r_1=\left(0,\frac a{\sqrt3}\right),\quad r_2=\left(-\frac a2,-\frac a{2\sqrt3}\right),\quad r_3=\left(\frac a2,-\frac a{2\sqrt3}\right).$$