Calculating Distance on a 3D Complex Eisenstein Lattice

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Hi, all.

I am interested in finding the distance between two vectors in a 3-dimensional complex Eisenstein lattice.

Is it simply an extension of finding the Euclidean distance between two real vectors?

i.e. D=sqrt((x2-x1)^2+(y2-y1)^2+...+(n2-n1)^2)?

Thanks.
 
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Thats the formula for magnitude of a vector. You got to tell me how they are oriented.

If you want to find distance between two paralle vectors, find the cross product of the two vectors which will give you a normal vector, pick points from each vector and make another vector from those points, then find the projection of the the new vector on to the normal vector. Get the magnitude of that projection. You have a distance between two parallel vectors, or any vectors that do not intersect. If they intersect the distance is zero because you have to pick the shortest distance, which is where they intersect, which has zero distance because they intersect.
 
I didn't quite mean the distance between two parallel vectors. I meant to say that the vectors denote a position in the real-complex plane (in three dimensions and given in terms of Eisenstein Integers). I know how to derive the distance between two vectors, but that does not help in this case.
 
The Euclidian distance between x = (x1, ..., xm) and y = (y1, ... , ym) in Rm is defined to be || x - y || = square root ( (x1 - y1)^2 + ... + (xm - ym)^2 )

hope you're still interested in that :)
 
can you please tell more specificly about the points you mentioned.how to take the points and to draw the vector from that point.
please help me i am interested to find distance between two points
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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