Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If I have the following vector { x, y, z, w }, what role does w have in calculations such as dot product, cross product, vector addition / subtraction, etc. I'm just a bit confused as to whether it should be included in calculations or not.

share|improve this question
2  
What do you mean? Are you talking about 3 dimensional projective space? –  Berci Nov 30 '12 at 11:14
    
Why does it make a difference if it's projective, or orthagonal, etc? I'm looking at this from a computer science point of view. –  Mark Ingram Nov 30 '12 at 12:01
add comment

2 Answers 2

The set of $n$-tuples of a given set $S$ is denoted by $S^n$.

The homogeneous coordinates of the 3d projective space are $4$-tuples of real numbers, (not all of them are zero), i.e. elements of $\Bbb R^4\setminus\{(0,0,0,0)\}$.

If $P=(x,y,z,w)$ is a $4$-tuple of real numbers, then it represents the following point of the projective space:

  • If $w\ne 0$, then the ordinary point with coordinates $\big(\displaystyle\frac xw,\frac yw,\frac zw \big)$.
  • If $w=0$, then an ideal point (pt in the infinity) "lying" in (the end of both directions of) all the lines in the ordinary 3d Euclidean space which have direction vector $(x,y,z)$.

We have addition and multiplication by scalar in $\Bbb R^4$, equally including all $4$ coordinates, and is very useful, but it doesn't give back the addition of ordinary vectors of the Euclidean 3d space! (Basically because, $\lambda x$ represents the same point as $x\in\Bbb R^4$.)

share|improve this answer
add comment

Real projective $3$-space $P^3$ is the space of all real quadruples ${\bf p}=(x,y,z,w)\ne(0,0,0,0)$ modulo the equivalence relation $${\bf p}\sim{\bf p}'\quad\Leftrightarrow\quad {\bf p'}=\lambda{\bf p}\quad {\rm for\ some} \ \lambda\ne0\ .$$ In this space there is no such thing as addition, scalar product, or vector product, but the notions of line and plane, the intersecting of two planes, or a line and a plane, make sense.

If you are working in ordinary $(x,y,z)$-space $R^3$ of three dimensions, for certain problems it makes sense to immerse $R^3$ into $P^3$ by means of the map $$\psi:\quad(x,y,z)\mapsto (x,y,z,1)/_\sim\ .$$ This means that your working space $R^3$ is interpreted as hyperplane $w=1$ in $R^4$, and the point $(x,y,z)$ is mapped onto the equivalence class of $(x,y,z,1)$, i.e., the ray in $R^4$ containing all points $\lambda(x,y,z,1)$, $\ \lambda\ne0$.

Now $\psi(R^3)$ is not all of $P^3$: Points of the form $(x,y,z,0)/_\sim\in P^3$ are not attained by $\psi$. These points of $P^3$ correspond to "points at infinity" of ordinary $R^3$. Historically the whole setup has the exact reason to enable the correct handling of these "points at infinity".

When you are given a point $\hat{\bf p}\in P^3$ by its homogeneous coordinates $(x,y,z,w)$, i.e., if $\hat{\bf p}=(x',y',z',w)/_\sim$, and if $w\ne0$ then you can get back the corresponding point $(x,y,z)=\psi^{-1}(\hat{\bf p})$ in your working space $R^3$ by means of $$x={x'\over w},\quad y={y'\over w},\quad z={z'\over w}\ .$$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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