Past exam of the mathematical methods I course of the University of Oxford / 2015 / 10

In one dimension the nonzero vector $b$ already spans the real line. Hence $\mathrm{span}(b,c)=\mathbb{R}$, and translating it by $a$ leavesThis is an affine subspace containing the origin for every allowed choice of the three vectors.

The affine subspace is $S=a+\mathrm{span}(b,c)$. If $b,c$ are linearly independent, their linear span is all of ${\mathbb{R}}^2$, so $S$ is the whole plane. If they are proportional, $S$ is the line through $a$ in direction $b$. This line passes through the origin exactly when $a\in \mathrm{span}(b)$. Because $b$ is nonzero, the set cannot be a single point.

If $b,c$ are linearly independent, $S$ is the plane through $a$ with normal $b\times c$. Its equation isso it contains the origin exactly when $a\cdot (b\times c)=0$. If $b,c$ are proportional, $S$ is the line through $a$ in direction $b$, containing the origin exactly when $a$ is proportional to $b$. The linear span of two nonzero vectors has dimension one or two, so neither a single point nor all of ${\mathbb{R}}^3$ occurs.

For the remaining distance calculation with $b=c$, every point on the line is $a+tb$. The nearest point to $p$ is found by orthogonal projection onto $b$:Indeed, differentiating $|a+tb-p{|}^2$ sets its component along $b$ to zero. Its second derivative is $2|b{|}^2>0$, proving this is the minimum.

A negative value of either radius makes the intersection empty. Assume henceforth $\alpha ,\beta \ge 0$ and put $D=|d|$. If $D=0$, the two spheres are concentric: their intersection is the entire sphere when $\alpha =\beta >0$, the origin when $\alpha =\beta =0$, and empty otherwise.

For $D>0$, subtract their squared equations to obtain the planeThe centre of the circle of intersection of two spheres is $x_0=s\hat{d}$, whereEvery point of that plane can be written $x=x_0+u$ with $u\cdot d=0$. Its remaining equation isThus the intersection is a circle of centre $x_0$ and radius $\rho$ if $|\alpha -\beta |<D<\alpha +\beta$; a single point $x_0$ at either equality; and empty otherwise. These include tangency to a zero-radius sphere. A negative $s$ places the circle centre beyond the first sphere's centre on the side opposite $d$.

In ${\mathbb{R}}^n$ for $n\ge 2$, the same calculation gives an $(n-2)$-dimensional sphere in the perpendicular affine hyperplane when ${\rho }^2>0$, a point when ${\rho }^2=0$, and the empty set otherwise. In two dimensions the positive-radius case consists of two points. In one dimension the perpendicular hyperplane has only one point, so a nonconcentric intersection exists only when ${\rho }^2=0$. Concentric equal-radius sets coincide in every dimension.