Find the intersection of plane and sphere

Question

If the equation of the sphere is $x^2+y^2+z^2=1$ and the plane is $x+y+z=1$, then how can the equation of a circle be determined from the equations of a sphere and a plane? and what is the parametric form of this circle?

Answer 1

The points $P(1,0,0),\;Q(0,1,0),\;R(0,0,1)$, forming an equilateral triangle, each lie on both the sphere and the plane given.

This is sufficient to determine that the centre of the circle is $C(\frac 13,\frac 13,\frac 13)$ and that the radius of the circle is $$|PC|=\frac {\sqrt{6}}{3}.$$

The normal to the plane of the circle is $$\underline{n}=\left(\begin{matrix}1\\1\\1\end {matrix}\right)$$

Two mutually perpendicular vectors in the plane of the circle are therefore $$\overrightarrow{PC}=\left(\begin{matrix}-\frac 23\\ \frac 13\\ \frac 13\end{matrix}\right)$$ and $$\underline{n}\times\overrightarrow{PC}=\left(\begin{matrix}0\\-1\\1\end{matrix}\right)$$

We therefore have two mutually perpendicular radius vectors in the plane of the circle, namely $$\left(\begin{matrix}-\frac 23\\ \frac 13\\ \frac 13\end{matrix}\right),\:\frac{1}{\sqrt{3}}\left(\begin{matrix}0\\-1\\1\end{matrix}\right)$$

we can now introduce a parameter $\theta$ and the parametric equations of the circle are$$\left(\begin{matrix}x\\y\\z\end{matrix}\right)=\left(\begin{matrix}\frac 13\\ \frac 13\\ \frac 13\end{matrix}\right)+\left(\begin{matrix}-\frac 23\\ \frac 13\\ \frac 13\end{matrix}\right)\cos \theta+\frac{1}{\sqrt{3}}\left(\begin{matrix}0\\-1\\1\end{matrix}\right)\sin\theta$$