# Lines projective space

I have a question concerning the answer of Georges Elencwajg in Lines in projective space

There he states that the line $\overline {AB}=\mathbb P(\Lambda)\subset \mathbb P^n$ has its points of the form $[ua_0+vb_0:\ldots :ua_n+vb_n] \quad (u,v \in k,$ not both zero ).

My question is: Why? I can't see the connection betwwen this form and the vectors which spane a plane, I would really appreciate if someone could explain this a little bit better to me.

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Let's imagine this for $\mathbb P^2$. Your setup is something like this: the affine point $(x, y)$ corresponds to the projective point with homogenous coordinates $(x, y, 1)$. This assumes the standard embedding with last coordinate $z=1$; other choices are of course possible. For homogenous coordinates, any multiple of the coordinate vector denotes the same point, so $(\lambda x, \lambda y, \lambda)$ is the same, as long as $\lambda\neq0$. To reverse the process, you dehomogenize a vector $(x, y, z)$ by first dividing by the last component to obtain $(x/z, y/z, 1)$ and the omitting the last component to end up with $(x/z, y/z)$.
Now let us imagine all of this in $\mathbb R^3$. A single homogenous coordinate vector is a point in space, but as you are not really interested in that particular vector, but instead in all non-zero multiples of that vector, you beter imagine this as a line through the origin of $\mathbb R^3$. So a point in $\mathbb P^2$ corresponds to a 1-dimensional linear subspace (i.e. a line through the origin) of $\mathbb R^3$. To turn this back into an affine point, you intersect that line with the plane at $z=1$. That point of intersection is exactly the multiple of your coordinate vector which has $1$ as it's last coordinate, i.e. it is $(x/z, y/z, 1)$.
You can perform the same considerations for objects of one dimension higher: a line in $\mathbb P^2$ is a 2-dimensional subspace (i.e. a plane through the origin) of $\mathbb R^3$. The affine interpretation of that plane can be obtained by intersecting said plane with the plane $z=1$, which yields de desired line.
So now in the $\mathbb R^3$ image, you are given two lines through the origin, and are interested in finding the plane through the origin which contains them both. This is simply the span of the directions of the lines. Intersecting that plane with $z=1$ will therefore give the line connecting the two projective points. This carries over to higher dimensions, although things become harder to visualize.
This page has a Java applet which will visualize the line connecting two points, both in the $z=1$ plane and in the enclosing $\mathbb R^3$ space. It also draws the normal of the plane, which can be computed using the cross product. The text is all in German, but the applet itself should be useful enough.
Thanks for your answer, it helped me a lot. Referring to the java application: Since $[g]=[A\times B]$ it follows that $\overline {AB}=\mathbb P(\Lambda)\subset \mathbb P^n$ has its points of the form $[ua_0+vb_0:\ldots :ua_n+vb_n] \quad (u,v \in k,$ not both zero ). – Alexander Jan 12 '13 at 14:28