Past exam of the mathematical methods I course of the University of Oxford / 2017 / 10 / c

Provided $p,q$ are linearly independent, their linear combinations fill a two-dimensional direction space. Adding $r_0$ translates it to the plane through $r_0$. The unit normal and constant areThen $n\cdot (r_0+t_{1}p+t_{2}q)=c_0$ for every pair of parameters. Conversely a point satisfying $n\cdot r=c_0$ differs from $r_0$ by a vector in $n^{\perp }=\mathrm{span}\{p,q\}$, proving that the two descriptions give exactly the same plane.

For the three points, choose $r_0=(3,1,-2)$, $p=(-4,1,6)$ and $q=(-1,-2,3)$. Their cross product is $(15,6,9)=3(5,2,3)$, which is nonzero. The plane therefore has equationEach supplied point satisfies it. Its unit normal is $(5,2,3)/\sqrt{38}$, up to an overall sign, and $c_0=11/\sqrt{38}$.

Let $s=(x_0,y_0,z_0)$ be the sphere centre. The circle centre is its perpendicular projection onto the plane:For the specific plane this isFor a point on the section, its displacement from $h$ lies in the plane and is perpendicular to $s-h$. The Pythagorean theorem therefore givesA genuine circle requires $d<R$. At $d=R$ the section is a single tangent point, and at $d>R$ the plane misses the sphere. A central cut $d=0$ gives a great circle of radius $R$.