Blowing up a line $L\subset \mathbb A^3$ I am trying to understand blow-ups, so I'd like to get some advises. Let me show you how I started to blow up a line $L\subset \mathbb A^3$. Any hint/comment/correction is highly appreciated.
Suppose $L$ is given parametrically by $x=at,y=bt,z=ct$, so that $x/a=y/b=z/c$. The blow-up $\pi:\textrm{Bl}_L\mathbb A^3\to \mathbb A^3$ is the resolution of the rational map
$$
\phi:\mathbb A^3\dashrightarrow \mathbb P^1
$$
sending $P\mapsto \lambda$, where $H_\lambda\subset \mathbb A^3$ is the unique plane through $L$ and $P$. Hence $$\textrm{Bl}_L\mathbb A^3=\{(P,\lambda)\in \mathbb A^3\times\mathbb P^1\,|\,P\in H_\lambda \}\subset \mathbb A^3\times\mathbb P^1.$$
If $\lambda=(u:v)$, then $H_\lambda$ has equation $u(xb-ya)+v(yc-zb)=0$, thus
$$
\textrm{Bl}_L\mathbb A^3=\{((x,y,z);(u:v))\in \mathbb A^3\times\mathbb P^1\,|\,u(xb-ya)+v(yc-zb)=0 \}. 
$$

Question 1. Is this correct?
Question 2. Can you please give me a hint to find the equations of the
  exceptional divisor $E=\pi^{-1}(L)$?

(I am stuck on question 2, because I know I should add one more equation to $\textrm{Bl}_L\mathbb A^3$ but I cannot figure which: $L$ has two defining equations)
Thanks in advance.
 A: Notice that for any $(x,y,z)\in L,$ we have $xb-ya=yc-zb=0,$ which implies that for every $[u:v]\in\mathbb P^1,$ we have $((x,y,z),[u:v])\in\operatorname{Bl}_L\mathbb A^3.$ These are all the points of the exceptional divisor. To get "an" equation of the exceptional divisor, we need to focus on the two charts of $\operatorname{Bl}_L\mathbb A^3$ separately. In the chart $u=1$ the blowup is cut out by $(xb-ya)+v'(yc-zb)\in\mathbb A^4 = \operatorname{Spec}(k[x,y,z,v']).$ In particular, in the coordinate ring of the blowup, which we write as $k[x',y',z',v']$ we have $x'b-y'a = -v'(y'c-z'b),$ and in this chart, the exceptional divisor is defined by $y'c-z'b=0.$ The same goes, mutatis mutandis, in the other chart.
I find it a little easier to understand as follows. The equations $f_1=xb-ya$ and $f_2=yc-zb$ cut out the line $L$, so we know that the blow up has two charts defined by their coordinate rings $R_1=k[x,y,z][f_2/f_1]\subseteq k(x,y,z)$ and $R_2=k[x,y,z][f_1/f_2]\subseteq k(x,y,z).$ The equations defining $L$ map, under $k[x,y,z]\to R_i,$ to $f_1,(f_2/f_1)f_1$ and $(f_1/f_2)f_2,f_2$ respectively, and the equation of the exceptional divisor in each chart is $f_1=0$ and $f_2=0.$ I encourage you to write this out explicitly in the case $L:y=0,z=0$ is the $x$-axis, so that it looks less like gobbledygook.
