0
$\begingroup$

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):

enter image description here

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.

Thanks in advance.

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

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 $OP \cdot n=|OP||n|=|OP|=p$.

$\endgroup$
  • $\begingroup$ Okay so I haven't redrawn the figure exactly as it is, but I've taken a quick look at the original one now. It seems that OP is NOT parallel to n. It just turned out to look so in my figure by accident, sorry about that. $\endgroup$ – Koy May 25 '18 at 13:53
  • $\begingroup$ My guess is that P is supposed to be an arbitrary point on the line. $\endgroup$ – Koy May 25 '18 at 13:53
  • $\begingroup$ 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. $\endgroup$ – Wouter May 25 '18 at 14:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.