# Vector equation of a line (or something)

So there's something in my linear algebra textbook that I just can't understand.

Here goes: (keep in mind that the textbook is not in English so my translation may not be the best):

The line l can be determined by a point P that's on that line and a vector n that's orthogonal to the line. Let's pick the point O as the center of a coordinate system. Then, the point X is on the line l if and only if PX is orthogonal to n, meaning:

$PX \cdot n = 0$

Using the position vectors of the points P and X we get:

$r \cdot n = p$

where p is a constant and $r = OX$

And here's the illustration that follows (redrawn by me, but it's exactly like in the textbook):

Things I don't get here:

What the hell is p (the small one)? How is this even an equation of a line?

There is also a part about parametric representation of the line which, oddly, comes before this part in the textbook, and that one I understood. But this I just can't get through my head, it seems as if there is a lack of information or something. Anyhow I've googled this a lot and couldn't find anything like it. In the textbook this is called 'vector equation of a line' whereas when I google that I get the parametric equation of a line which is a different thing I suppose. I would gladly skip this but the problem is that it is used as a step in a proof for the distance between a point and a line. So if anyone could explain me what's going on here, that would be nice.

• Consider Vector addition and consider the triangle $OPX$ : we have that vector $PX$ is the difference of vectors $OP$ and $OX$. – Mauro ALLEGRANZA May 25 '18 at 13:50
• Thus $PX \cdot n= (OP-X) \cdot n=0$. An thus : $OP \cdot n= X \cdot n$. – Mauro ALLEGRANZA May 25 '18 at 13:52

This is indeed the equation of a line. $$r\cdot n=p$$ (note the little dot $\cdot$ is a dot product, a product of vectors that gives a scalar)
in coordinates, let $r=(x,y),n=(n_x,n_y)$, this equation is $$x n_x+y n_y=p$$
If the normal vector to the line, $n$, has length 1, $|n|=\sqrt{n_x^2+n_y^2}=1$, then $p$ represents the distance from the line to the origin. We can see this by considering the point $Q$ on the line which is closest to the origin (it would be close to $P$ in your figure). For this point, $OQ$ is parallel to $n$, so $OP \cdot n=|OP||n|=|OP|=p$.
• There is some point, say $Q$, on the line such that $OQ$ is parallel to $n$, and this is the closest point to the origin. It seemed like $P$ was intended to be this point. I'll edit. – Wouter May 25 '18 at 14:29