The blow-up $\tilde{X}$ of $\mathbb{A}^3$ at the line $\ell: x_1=x_2=0$ is by definition the closure in $\mathbb{A}^3 \times \mathbb{P}^1$ of the graph of the function $f: \mathbb{A}^3 - \ell \rightarrow \mathbb{P}^1$, that takes a point $(x_1,x_2,x_3)$ to $(x_1:x_2)$. As it turns out, $\tilde{X}$ is the closed set of $\mathbb{A}^3 \times \mathbb{P}^1$ given by the equations $x_1 y_2 = x_2 y_1$. The exceptional set of $\tilde{X}$ is isomorphic to $\mathbb{A}^1 \times \mathbb{P}^1$.
Question: The exceptional set of the blow up of $\mathbb{A}^3$ at a point is isomorphic to $\mathbb{P}^2$, which is interpreted as the directions of all lines in $\mathbb{A}^3$ through that point. What is the analogous interpretation of the blow-up described above? In particular, how can one understand the geometric meaning of its exceptional set?
$\newcommand{\Cpx}{\mathbf{C}}\newcommand{\Proj}{\mathbf{P}}$Over the complex numbers, blowing up a smooth variety $Y \subset X$ may be viewed geometrically as the following procedure:
Let $\pi:\nu \to Y$ denote the normal bundle of $Y$ in $X$, and let $E = \Proj(\nu) \to Y$ be its projectivization (i.e., remove the zero section and quotient by the $\Cpx^{\times}$-action given by scalar multiplication in the fibres). There is a tautological line sub-bundle $\tau \subset \pi^{*}\nu \to E$ whose fibre over a point $[p]$ of $\Proj(\nu)$ is the line represented by $[p]$. (That is, $[p]$ represents a normal line to $Y$ at some point of $Y$, and the fibre of $\tau$ at $[p]$ is that normal line to $Y$.)
It's straightforward to check that:
The complement of the zero section of $\nu$ is biholomorphic to the complement of the zero section of $\tau$.
The total space of $\tau$ is obtained from the total space of $\nu$ by removing the zero section and gluing in the exceptional divisor $E = \Proj(\nu)$.
The preimage in total space of $\tau$ of some point $y$ in $Y$ is obtained by blowing up the normal space $\nu_{y}$ at the origin.
Working locally in the horizontal directions, if $Y = \Cpx^{k} \subset X = \Cpx^{n}$, blowing up $Y$ amounts to blowing up the origin in $\Cpx^{n-k}$ and crossing with $\Cpx^{k}$.