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

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.