Given $x^2+y^2+z^2=1$ and $y-z+2x=-2$, how can I find the intersection curve? Using Geogebra, I find $X=(-0.67,-0.33,0.33)+(-0.26\cos(t)-0.21\sin(t),0.52\cos(t)-0.11\sin(t), -0.53\sin(t))$ How can I do it manually?
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Check this out: https://math.stackexchange.com/questions/943383/determine-circle-of-intersection-of-plane-and-sphere – John Jun 14 '20 at 22:17
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Solve for $y$ in the second equation. Then substitute in the quadratic equation.
You get a quadratic in $x$ and $z.$ This is a circle in plane parallel to the $xz$ plane.
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