Consider the hyperbolic 3-space ( H 3 , d s 2 ) with H 3 := { ( x , y , z ) ∈ R 3 | z &gt; 0 } , d s 2 = d x 2 + d y 2 + d z 2 z 2 Geodesics for this space are circular arcs normal to { z = 0 } and vertical rays normal to { z = 0 } .

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Consider the hyperbolic 3-space   (            H        3    ,  d      s    2    ) with            H        3    :=  {  (  x  ,  y  ,  z  )  ∈            R        3        |    z  &amp;gt;  0  }  ,    d      s    2    =            d              x        2            +      d              y        2            +      d              z        2                    z      2      Geodesics for this space are circular arcs normal to   {  z  =  0  } and vertical rays normal to   {  z  =  0  } .

Cristopher Barrera · Accepted Answer

Step 1A clean way to do this is to work with the hyperboloid model of the hyperbolic space where       H    n   . Then, given two points p, q in the (upper) hyperboloid       H    n    ⊂      R          1      ,      n      ,n, their midpoint is            p      +      q              &amp;lt;      p      +      q      ,      p      +      q      &amp;gt;        .If you know how to go back and forth between the upper half space and the hyperboloid model, you can translate this formula into the upper half space model. (I think it will not be pretty.) See also this question.

anginih86 · Accepted Answer

Step 1Assume that the geodesic passing through the given points       p    1   and       p    2   is the circle with center at   p  =  (  a  ,  b  ,  0  ) and radius r. Then  (  x  −  a      )    2    +  (  y  −  b      )    2    +      z    2    =      r    2  The vectors   u  =  p      p    1   and   v  =  p      p    2   satisfy  (  u  ×  v  )  .  k  =  0Using these two equations, you can evaluate a, b and r. I think(I am not sure) the rest is easy.I remembered that the Euclidean bisector of the segment       p    1        p    2   intersects the plane   z  =  0 at the center of the required geodesic.

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