Why is vertical distance the standard in least squares? When fitting a curve in $\mathbb R^2$ to data points in $\mathbb R^2$ (example), why is each point's vertical distance from the curve squared instead of its shortest (possibly diagonal) distance from the curve?
Ignoring my poorly drawn curves, it seems obvious that  is a worse curve-to-point fit than  even though the red line is shorter in the first image, because you can draw the much shorter blue line instead (which I labeled b in the second image).  Minimizing $b^2$ seems much more important than minimizing $a^2$.
 A: In fact, diagonal distance is used in some cases.
From the operative point of view, the standard "vertical" distance is taken when it is assumed that in the 2D data available, the $x$ has been measured (almost) exactly, while the $y$ is subject to error.
When both $x$ and $y$ are subject to error, under the usual assumptions of independence, near Gaussian distribution, etc. the distance  shall be taken along a line with slope $\sigma_y / \sigma_x$.   
That's called Total least squares.
A: The idea behind many examples of least-squares fit is to use it as a predictive model: given some points $\{(x_1,y_1), \dots, (x_n,y_n)\}$, you find a curve that takes an input $x$ and gives you a predicted value of $f(x)$.
In this case, the vertical distance $|y_1 - f(x_1)|$ is the difference between the actual known value $y_1$ and the predicted value $f(x_1)$. (And $(y_1 - f(x_1))^2$ is the squared error.) This is the actual measurement of error we care about, because it's a prediction of how far off we'll be if we guess $f(x)$ for a different value $x$.
A: Your data points are $(x_i, y_i)$ and your regression line is $y=mx+b.$
The  absolute error in estimation at each point is $|y_i -mx_i-b|$ which is the vertical distance from your data point to your regression line.
