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

For two nonparallel vectors $v_1,v_2\in {\mathbb{R}}^2$, a nontrivial relation $a{v}_1+b{v}_2=0$ would make one a scalar multiple of the other. They are therefore linearly independent. Two independent vectors in a two-dimensional vector space form a basis.

Likewise, three vectors in ${\mathbb{R}}^3$ that are not contained in one plane through the origin cannot obey a nontrivial linear dependence: such a relation would express one as a combination of the other two, or already make that pair dependent. Thus they are independent and form a basis of ${\mathbb{R}}^3$. The plane condition here is linear coplanarity with the origin, as required for vectors, rather than affine coplanarity of three arbitrary points.

Let $\Delta =v_1\cdot (v_2\times v_3)\ne 0$, since the original vectors form a basis. The reciprocal basis isTaking the dot product with each $v_j$ gives $u^i\cdot v_j={\delta }_{ij}$, using orthogonality of a cross product and cyclic symmetry of the scalar triple product. If ${\sum }_i{a}_i{u}^i=0$, taking its dot product with $v_j$ yields $a_j=0$. The three reciprocal vectors are therefore linearly independent and form a basis of ${\mathbb{R}}^3$.

Equivalently, if $V$ has columns $v_j$ and $U$ columns $u^i$, then $U^{T}V=I$, so $U=V^{-T}$. In particular any vector has coordinates $r={\sum }_i(r\cdot u^i)v_i$, even when the original basis is not orthonormal.

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 $R$.