Show that $\|b \|^2\ (\sin\theta)^2 = \min_x\ \|ax-b\|^2$ where $\theta$ is the angle between $a,\ b$. This is for vectors $a, b \in \mathbb{R}^m$ \ {$0$}.
The second part of the question says: Interpret and illustrate (for $m = 2$) the result from the previous item. Hint: You might find it helpful to do part two before part one.
So I tried to illustrate $||b||^2(\sin\theta)^2$ but got stuck. I understand that the norm of a vector (at least in $2$ dimensions like in this case) is the hypotenuse of the right angle triangle made by the sides of the vector ($b_1$ and $b_2$ in this case). I'm not sure how to factor in the $(\sin \theta)^2$, or how exactly this whole thing would look if I tried to illustrate it. Any help would be appreciated, thanks.
I assume that you are working in a Hilbert space or a finite-dimensional vector space $\mathcal{H}$ over $\mathbb{F}\in\{\mathbb{R},\mathbb{C}\}$ equipped with $\langle\_,\_\rangle$, a positive-definite symmetric bilinear form (for $\mathbb{F}=\mathbb{R}$) or Hermitian sesquilinear form (for $\mathbb{F}=\mathbb{C}$). The norm $\|\_\|$ is induced by $\langle\_,\_\rangle$. In the case $\mathbb{F}=\mathbb{C}$, suppose that $\langle\_,\_\rangle$ is linear in the first argument and antilinear in the second argument.
Let $a$ and $b$ be nonzero elements of $\mathcal{H}$. Your job is to minimize the function $f:\mathbb{F}\to\mathbb{R}_{\geq 0}$ given by $$f(x):=\|ax-b\|^2\text{ for each }x\in\mathbb{F}\,.$$
Note that $$\begin{align}f(x)&=\langle ax-b,ax-b\rangle =\langle ax,ax\rangle-\langle ax,b\rangle-\langle b,ax\rangle+\langle b,b\rangle\\ &=\|a\|^2\,|x|^2-\langle a,b\rangle\,x-\overline{\langle a,b\rangle}\,\bar{x}+\|b\|^2 \\ &=\left|\|a\|x-\frac{\overline{\langle a,b\rangle}}{\|a\|}\right|^2+\left(\frac{\|a\|^2\,\|b\|^2-\big|\langle a,b\rangle\big|^2}{\|a\|^2}\right) \,.\end{align}$$ for all $x\in\mathbb{F}$. Ergo, $$f(x)\geq \frac{\|a\|^2\,\|b\|^2-\big|\langle a,b\rangle\big|^2}{\|a\|^2} \text{ for every }x\in \mathbb{F}\,.$$ The equality holds if and only if $$x=\frac{\overline{\langle a,b\rangle}}{\|a\|\,\|b\|}\,.$$
Finally, if $\mathbb{F}=\mathbb{R}$, then the angle $\theta$ between $a$ and $b$ satisfies $$\cos(\theta)=\frac{\overline{\langle a,b\rangle}}{\|a\|\,\|b\|}=\frac{\langle a,b\rangle}{\|a\|\,\|b\|}\,.$$ That is, $f(x)$ is minimized iff $x=\cos(\theta)$, so the minimum value of $f$ is $$\frac{\|a\|^2\,\|b\|^2-\big|\langle a,b\rangle\big|^2}{\|a\|^2}=\frac{\|a\|^2\,\|b\|^2\big(1-\cos^2(\theta)\big)}{\|a\|^2}=\|b\|^2\,\sin^2(\theta)\,.$$
P.S. The question tells you to deal with the $2$-dimensional case first probably because it suffices to assume that $\mathcal{H}$ is the $\mathbb{F}$-span of $a$ and $b$. In which case, $\dim_\mathbb{F}(\mathcal{H})\in\{1,2\}$, depending on whether $a$ and $b$ are parallel (linearly dependent). In particular, when $\dim_\mathbb{F}(\mathcal{H})\in\{1,2\}$ and $\mathbb{F}=\mathbb{R}$, you can embed $\mathcal{H}$ into a $2$-dimensional Euclidean space, which you should know how to work with.