Hough Transform, Computer Vision
Consider a vector $\mathbf v$ of any dimension. The hyperplanes orthogonal to that vector are given by the equation $r = \mathbf u\cdot \mathbf v$ for various $r$s. That is, the collection of vectors $\mathbf u$ satisfying $r=\mathbf u\cdot\mathbf v$ describe a hyperplane orthogonal to $\mathbf v$. If $\varphi$ is the (acute) angle between $\mathbf u$ and $\mathbf v$ (in the plane that they span, not to be confused with the hyperplane the earlier equation describes), this can be written as $r=|\mathbf u||\mathbf v|\cos\varphi$. From this we can see that $|\mathbf u|$ is minimum when $\varphi = 0$ so that $\cos\varphi=1$, the maximum value of $\cos$. If we choose $\mathbf v$ to be a unit vector, i.e. $|\mathbf v|=1$, then the point on the hyperplane closest to the origin is $\mathbf u=r\mathbf v$ and has distance $r$, i.e. $|\mathbf u|=r$.

In the 2D case, the "hyperplane" is actually a line. A unit vector in the plane has the form $\mathbf v = (\cos\theta,\sin\theta)$ for some $\theta$. The dot product equation describing the line is then $r=\mathbf u\cdot\mathbf v = u_x\cos\theta + u_y\sin\theta$ where $\mathbf u = (u_x,u_y)$. The point on the line closest to the origin is then $r\mathbf v = (r\cos\theta,r\sin\theta)$ which is a vector of length $r$ just as stated in the article. As mentioned in the previous paragraph, if $\mathbf v$ is chosen as a unit vector (and it always can be), then this is true in general: $r$ is the distance of the closest point to the origin of the described hyperplane, and that closest point is $r\mathbf v$.