TITLE: Embeddings of bundles in projective space.
QUESTION [5 upvotes]: 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.

REPLY [4 votes]: 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.