TITLE: Lines in projective space
QUESTION [47 upvotes]: I have the following definitions:
Given a vector space $V$ over a field $k$, we can define the projective space $\mathbb P V = (V \backslash \{0\}) / \sim $ where $\sim$ identifies all points that lie on the same line through the origin. 
A projective subspace $\mathbb P W$ of $\mathbb P V$ is of the form $\pi(W \backslash \{0\})$, where $\pi$ is the residue class map and $W$ is a vector subspace of $V$. Define $\mathrm{dim} (\mathbb P V) = \mathrm{dim}( V )- 1$. A line in $\mathbb P V$ is a $1$-dimensional projective subspace.

Now I'm finding it difficult to visualise what a line in projective space actually is. I can understand why any two lines in a projective plane intersect. Suppose I'm in $\mathbb P^3$ and want to write 'an equation' for the line that goes through the points $ p = (1:0:0:0)$ and $q = (a:b:c:d)$. How could I do that? Does my question even make sense? I'm concerned because $\mathbb P V$ isn't actually a vector space, so can I think of points inside it as vectors?
Thanks

REPLY [6 votes]: Every linear subspace in $V$ is intersection of hyperplanes. Projecting down to $\Bbb P(V)$ you obtain that every linear subspace in $\Bbb P(V)$ is intersection of hyperplanes.
If $\dim V=4$, i.e. $\dim\Bbb P(V)=3$, a line $\ell\subseteq\Bbb P(V)$ corresponds to a $2$-dimensional subspace $U\subset V$ which can always be obtained as intesection of two $3$-dimensional subspaces, $U=W_1\cap W_2$, obviously not uniquely determined. Hence $\ell=\pi(W_1)\cap\pi(W_2)$.
If $\ell$ is given as the line through two points $A$ and $B$ you can always find some $W_1$ and $W_2$ as above knowing that an hyperplane in $V$ has equation
$$
a_1X_1+a_2X_2+a_3X_3+a_4X_4=0\qquad\qquad(\ast)
$$
and imposing that the coordinates of $A$ and $B$ satisfy this equation. You thus obtain a homogeneous linear system of two equations in 4 variables (the coefficients of $(\star)$) wich you can certainly solve.