Geometric/Visual Solution - Shortest Vector for which Dot Product = x + 2y = 5. (Strang P21 1.2.26)

Question

Much as I referenced this same exercise, I'm questing after an exclusively geometric solution.

Question: If $\color{#0070FF}{\vec{v} = (1,2)}$ draw all vectors $\vec{w} = (x,y)$ in the plane $x,y$ with $\color{#0070FF}{\vec{v}}\cdot\vec{w} = \color{#228B22}{x + 2y = 5}$. Which is the shortest $\vec{w}$?

Algebraic solution: It's given that $\mathbf{v \cdot w} = \color{#228B22}{x + 2y = 5}$. Thus, $\mathbf{v}$ forsooth lies on $x + 2y = 5$. We need an upper bound on $|\mathbf{w}|$, thus apply the Cauchy-Schwarz Inequality. $ \mathbf{v \cdot w} \leq |\mathbf{v}| \mid\mathbf{w}| \iff \cfrac{\mathbf{v \cdot w}}{|\mathbf{v}|} \leq |\mathbf{w}| \tag{CS$\neq$}$

Thus, $min \;\mathbf{w} \iff $ (CS$\neq$) becomes an equality $\iff$ $\mathbf{v}, \mathbf{w}$ are collinear $\iff k\mathbf{v} = \mathbf{w}$.

$\large{1.}$ How and why is $k = 1$?

$\large{2.}$ How can I solve this question with but the following picture?
Since $\mathbf{v \cdot w} := 5,$ thus $\mathbf{v \cdot w} > 0 \iff \cos(\text{angle between $\mathbf{v}$ & $\mathbf{w}$})>0$ $ \iff \text{angle between $\mathbf{v}$ & $\mathbf{w}$} \in (0, 90^{\circ}). $ So $\mathbf{w}$ must be right of/above $\color{magenta}{x + 2y = 0}$.

Answer 1

To answer your first question: $$ \mathbf{v \cdot w} = 5 \Longrightarrow \mathbf{v} \cdot (k\mathbf{v}) = 5 \Longrightarrow k \|\mathbf{v}\|^2 = 5 \Longrightarrow k=1 $$

Regarding the second question ...

By definition, we have $\mathbf{v \cdot w} = \|\mathbf{v}\| \cdot \|\mathbf{w}\| \cos\theta$, where $\theta$ is the angle between $\mathbf{v}$ and $\mathbf{w}$.

But we know that $\mathbf{v \cdot w} = 5$, and $\|\mathbf{v}\| = \sqrt 5$, so $$ \|\mathbf{w}\|\, \cos\theta = 5/{\sqrt 5} = \sqrt 5. $$ Now $\|\mathbf{w}\| \cos\theta$ is the length of the projection of $\mathbf{w}$ onto $\mathbf{v}$. So, we are looking for the shortest possible vector $\mathbf{w}$ whose projected length along $\mathbf{v}$ is $\sqrt 5$. It's obvious (I think) that the shortest vector giving the desired projected length must be parallel to $\mathbf{v}$.

In other words, if the length of $\mathbf{w}$ is to be minimized, $\mathbf{w}$ must be a scalar multiple of $\mathbf{v}$, say $\mathbf{w} = k\mathbf{v}$. $\color{green}{\bigstar}$ But $\|\mathbf{v}\| = \sqrt 5$, so, to get a projected length of $\sqrt 5$, the scalar multiplier $k$ must be 1. So $\mathbf{w} = \mathbf{v}$.

Alternative methodology after $\color{green}{\bigstar}$: But $\mathbf{w} = k\mathbf{v} \Longrightarrow \|\mathbf{w}\|\ = k\|\mathbf{v}\|$,
where $\|\mathbf{w}\|\ \geq 0 \Longrightarrow LHS \geq 0 $ as well. So for the shortest $\mathbf{w}$ such that $\mathbf{v \cdot w} = 5$ , we must choose $k = 1$. Finally, because we already determined in the anterior paragraph $\mathbf{w} = k\mathbf{v},$ thus $ \mathbf{w} = \mathbf{v}.$

Answer 2

When one seeketh the vector which nigh the origin, it behests the seeker to look upon circles centered therein. When one considers such perfect shapes one finds manifest the answer which one seeks. Hither, ye must consider circles $x^2+y^2=k$ as these contain the locus of all points distance $\sqrt{k}$ from the origin. Verily, it is clear the circle which takes $\langle 1,2 \rangle$ as its radius is the smallest circle for which the line $x+2y=5$ doth reside.