Image:Slanted circle.png

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[edit] Summary

In[132]:= <<Graphics`Shapes` In[133]:= Solve[{x^2+y^2+z^2\[Equal]1,x+y+z\[Equal]0},{x,y}] Out[133]= \!\({{x \[Rule] 1\/2\ \((\(-z\) - \@\(2 - 3\ z\^2\))\), y \[Rule] 1\/2\ \((\(-z\) + \@\(2 - 3\ z\^2\))\)}, {x \[Rule] 1\/2\ \((\(-z\) + \@\(2 - 3\ z\^2\))\), y \[Rule] 1\/2\ \((\(-z\) - \@\(2 - 3\ z\^2\))\)}}\) In[134]:= \!\(\(p1[z_] := {1\/2\ \((\(-z\) - \@\(2 - 3\ z\^2\))\), 1\/2\ \((\(-z\) + \@\(2 - 3\ z\^2\))\), z};\)\[IndentingNewLine] \(p2[z_] := {1\/2\ \((\(-z\) + \@\(2 - 3\ z\^2\))\), 1\/2\ \((\(-z\) - \@\(2 - 3\ z\^2\))\), z};\)\[IndentingNewLine] \) In[136]:= \!\(\(Show[{\[IndentingNewLine]WireFrame[ Graphics3D[ Sphere[1, 20, 20]]], \[IndentingNewLine]Graphics3D[{\[IndentingNewLine]\ RGBColor[1, 0, 0], \[IndentingNewLine]Thickness[ 0.009], \[IndentingNewLine]Line[ Table[p1[z], {z, \(-\@\(2\/3\)\), \@\(2\/3\), 0.01}]], \[IndentingNewLine]Line[ Table[p2[z], {z, \(-\@\(2\/3\)\), \@\(2\/3\), 0.01}]], \[IndentingNewLine]Text[ StyleForm["\<x+y+z==0\>", Section, FontColor \[Rule] RGBColor[1, 0, 0]], {1, 0.2, 1}, {\(-1\), 0}]}]\[IndentingNewLine]}, Boxed \[Rule] False, Axes \[Rule] True, ViewPoint -> {8.043, \ \(-2.956\), \ 1.784}, Boxed \[Rule] True, AxesEdge \[Rule] {{1, \(-1\)}, {1, \(-1\)}, {1, 1}}, AxesLabel \[Rule] {StyleForm[x, Section], StyleForm[y, Section], StyleForm[z, Section]}, Ticks \[Rule] {{\(-1\), \(- .5\), 0, .5, 1}, {\(-1\), \(- .5\), 0, .5, 1}, {\(-1\), \(- .5\), 0, .5, 1}}, PlotLabel \[Rule] StyleForm[TraditionalForm[x^2 + y^2 + z^2 \[Equal] 1]\ , Section, FontColor -> RGBColor[0, 0, 0]]]\ ;\)\) </tt>

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I, the copyright holder of this work, hereby grant the permission to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation; with no Invariant Sections, no Front-Cover Texts, and no Back-Cover Texts.

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	Date/Time	Dimensions	User	Comment
current	04:42, 9 May 2007	322×330 (6 KB)	Jakob.scholbach (Talk \| contribs)	(<tt><nowiki> In[132]:= <<Graphics`Shapes` In[133]:= Solve[{x^2+y^2+z^2\[Equal]1,x+y+z\[Equal]0},{x,y}] Out[133]= \!\({{x \[Rule] 1\/2\ \((\(-z\) - \@\(2 - 3\ z\^2\))\), y \[Rule] 1\/2\ \((\(-z\) + \@\(2 - 3\ z\^2\))\)}, {x \[Rule] 1\/2\ \()