# projection of a sphere onto a plane

Consider you have a sphere centered at the origin.The sphere has a diameter of $\frac{1}{2} \sqrt{\frac{3}{2}}$. This means that the inscribed cube has an edge of 1.

Take any point from the plane (1,1,0)-perpendicular to Z such that x in [-0.5,0.5], y in [-0.5,0.5] z in [-0.5,0.5]. What is the coordinate of the point on the sphere for which the orthogonal(perpendicular) projection on the plane has the coordinate (x,Y,Z)? It will be (x,Y,what value does Z have?), as the projection is along Z in this case.

How would the equations look if the plane was(1,0,1)-perpendicular to Y axis?

What I am trying is to take 6 projections of the sphere, which will constitute the mapping of the sphere to cube faces.

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Maybe I'm misinterpreting the question here, but wouldn't it just be the points $(x,Y,\pm\sqrt{3/8-x^2-Y^2})$ for the first case and $(x,\pm\sqrt{3/8-x^2-Z^2},Z)$ for the projection along the $Y$-axis? –  Heike Nov 29 '11 at 10:35
the coresponding point on the sphere for the point (0.5,0.5,0) in plane should be (0.5,0.5,+/-0.5). For your equations it is $(0.5,0.5, \frac {1}{2 \sqrt{2}}$, if you take Abs on the value under sqrt –  Ryan Nov 29 '11 at 11:29