I have a plane $$x + y + z = 0$$ and a sphere $$x^2 + y^2 + z^2 = R^2$$. How do I find a parametric equation for the intersection of the plane and sphere? The closest I've gotten was $r = (\sin{t} + \cos{t}, \sin{t} + \cos{t}, \cos{t}) $, but it doesn't seem to be correct.
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Well, indeed it is a circle, but you would have to rotate it around the axes so that it's in the plane ... – Matti P. Dec 04 '19 at 10:57
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How do I rotate it around the axes? – henhao Dec 04 '19 at 11:15
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Well, if you know the $x,y$ and $z$-components of your geometry, you can use rotation matrices (for example) and multiply the point vector by that matrix. – Matti P. Dec 04 '19 at 11:18
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There are at least three items in the handy list of related questions to the right that illustrate how to do this. – amd Dec 04 '19 at 19:56
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From the system given derive that: $(\frac{y}{2}+z) ^2+3\frac{y^2}{4}=\frac{R^2}{4}$, (solve for $x$ the first equation and replace it to the second.) Then set $y=\frac{R}{\sqrt{3}}\cos{t}$ and $z=\frac{R}{2}\sin{t}-\frac{R}{2\sqrt{3}}\cos{t}$. Then from the equation $x+y+z=0$ we have $x=-\frac{R}{2\sqrt{3}}\cos{t}-\frac{R}{2}\sin{t}$ .
dmtri
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