Embeddings of bundles in projective space. Consider the projective variety $X = \mathbb{P}^2$, and the line bundle $\mathcal{O}_X(dH)$ where $H$ is a plane and $d \in \mathbb{N}$.
Let $L$ be the total space of $\mathcal{O}_X(dH)$. I know how to form this in the following way: Let $U_i$ be the standard affine opens of $X$, then
$$
L = \coprod (U_i \times \mathbb{A}^1)/\sim
$$
where the equivalence is given by the glueiing matrices $A_{ij}= (x_j/x_i)^d$.
I am pretty sure the total space is a quasi-projective variety. Can anybody give me an explicit embedding?
For my purposes i am happy with the cased $d \in \{1,2\}$, but a general treatment would be nice.
Thanks.
Edit: Projective changed to quasi projective.
 A: So let me start again. If $\mathcal M$ is a coherent locally free sheaf on $X$, then the total space of the associated vector bundle $M$ is the affine $X$-scheme 
$$M =\mathrm{Spec} (\mathcal{Sym}_{O_X}(\mathcal M^{\vee}))\to X.$$
(the cech means dual.) This is an affine morphism, hence quasi-projective. As $X$ has an ample sheaf, this implies that $M$ admits an open immersion into some $\mathbb P^n_X$ (EGA, II.5.3.3). 
This immersion can be "explicitely" constructed as follows. Let $n\ge 1$ be big enough so that $\mathcal M^{\vee}(n)$ is generated by its global sections. Write 
$$ O_X^m \twoheadrightarrow \mathcal M^{\vee}(n), \quad O_X^3\twoheadrightarrow O_X(n)$$
(If $T_0, T_1, T_2$ is a basis of $H^0(X, O_X(1))$, then $O_X(n)$ is generated by $T_0^n, T_1^n, T_2^n$). 
Then 
$$\mathrm{Spec} (\mathcal{Sym}_{O_X}(\mathcal M^{\vee})) \hookrightarrow \mathbb P(\mathcal M^{\vee}\oplus O_X)\simeq {\mathbb P}(\mathcal M^{\vee}(n)\oplus O_X(n)) \hookrightarrow {\mathbb P}(O_X^{m+3})={\mathbb P}^{m+2}\times X.$$ 
When $\mathcal M=O_X(d)$, we can take $n=d$ and $m=1$. Hence an immersion
$$L \hookrightarrow {\mathbb P}^{3}\times X.$$
Now to write the map in your chart: over $U_i$, the map is 
$$(\lambda_i, [x_0, x_1, x_2])\mapsto ([\lambda_i x_i^d, x_0^d, x_1^d, x_2^d], [x_0, x_1, x_2]).$$
Le glueing map on $U_i\cap U_j$ is $\lambda_i x_i^d=\lambda_j x_j^d$. 
===================================== 
Below is the embedding for the total space of $O_X(-d)$. 
Let $e_1,\dots, e_n$ be a basis of $H^0(X, O_X(d))$. This means that the map
$$O_X^n\to O_X(d), \quad (f_1,\dots, f_n)\mapsto \sum_i e_if_i$$ 
is surjective. So $L$ is a closed subvariety of $\mathbb A^n\times X$ and is therefore quasi-projective. 
This holds for any quasi-projective variety $X$ and any line bundle $L$ on $X$ generated by its global sections. 
EDIT Please forget the followings lines. 
It can't be. Because otherwise you would have a non-constant morphism from $\mathbb P^3$ (your total spae has dimension $3$) to $\mathbb P^2$. But such a morphism doesn't exist (this is an exercice in Hartshorne: there is no non-constant morphism from $\mathbb P^n$ to $\mathbb P^m$ if $n>m$). 
A: If $E$ is a vector bundle of rank $r$ on a projective variety $X$, then the total space of $E$ is a quasi-projective variety : this is a vast generalization of what you are asking.  
And yet the proof is easy: the vector bundle $E$ has a projective completion $\bar E=\mathbb P(E\oplus \theta)$ where $\theta=X\times \mathbb A_k^1$ denotes the trivial vector bundle of rank one on $X$.
This bundle is a locally trivial bundle $p:\bar E\to X$ with fiber $\mathbb P^r$ and is a projective variety.
And finally the original vector bundle $E$ admits of an open embedding $$E\stackrel {\cong}{\to}\mathbb P(E\oplus 1) \stackrel {\text {open}}\subset \mathbb P(E\oplus \theta)=\bar E$$ into the projective variety $\bar E$, which shows that $E$ is indeed quasi-projective.  
