<p>So, a <a href="https://en.wikipedia.org/wiki/Projective_plane" rel="nofollow noreferrer">projective plane</a> is a plane constructed such that parallel lines intersect at infinity. The plane defined by the equation (x, y, 1) is not projective. Just draw two parallel lines on that plane, and you will find that they never intersect, even under homogeneous transformations.</p>
<p>Now, homogeneous transformations can <em>create</em> projective planes. But such a plane would not have a fixed homogeneous coordinate of 1, because the homogeneous transform  with a coordinate of 1 doesn't change the other coordinates. A projective plane in homogeneous space needs to be able to represent points at infinity, and that only happens with a homogeneous coordinate of zero.</p>
<p>So if a plane in homogeneous space does not pass through the 0 homogeneous coordinate, it <em>cannot</em> represent a projective plane.</p>
<p>Since the rest of your question is based on the assumption that (x, y, 1) is projective, it's kind of irrelevant.</p>

CC BY-SA 4.0
2021-11-21T20:33:57.730Z
so what about (x, y, z, 1) in projective space?
S. M.
CC BY-SA 4.0
2021-11-21T20:47:29.450Z
your answer isn't according to my question. I am very much unsatisfied with your answer.
S. M.
CC BY-SA 4.0
2021-11-22T00:00:30.813Z
@B-spline: "*so what about (x, y, z, 1) in projective space?*" If (x, y, 1) isn't a projective plane, then why would you think that (x, y, z, 1) is a "projective space"? "*I am very much unsatisfied with your answer.*" My answer is correct. It's not a projective plane, and the rest of your question is predicated on the assumption that it is. So the rest of your question is effectively moot.
Nicol Bolas
CC BY-SA 4.0
2021-11-21T20:26:18.300Z
2021-11-22T00:00:54.517Z
2021-11-22T00:00:54.517Z
Nicol Bolas
<p>We know that any 2D point <span class="math-container">$(x, y) $</span>which represents as 3D homogeneous coordinates is of the form <span class="math-container">$(x, y, 1)$</span> which is the points of projective plane <span class="math-container">$P^2.$</span></p>
<p>If I use the same concepts for 3D points  <span class="math-container">$(x, y, z)$</span> which represents as 4D homogeneous coordinates is of the form <span class="math-container">$(x, y, z, 1).$</span></p>
<p>My question is <span class="math-container">$(x, y, z, 1)$</span> could be the points of 3D projective space?</p>
<p>My second question is always homogenous coordinates are the points of projective space ?</p>
<p>N. B. -- <span class="math-container">$P^2$</span> is projective plane which contains the points of <span class="math-container">$\mathbb{R^2}$</span>(<span class="math-container">$\mathbb{R^2}$</span> points in <span class="math-container">$P^2$</span> can be represents as <span class="math-container">$(x, y, 1)$</span> ) and ideal points, is of the form <span class="math-container">$(x, y, 0)$</span>  which is called points at infinity.</p>

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2021-11-22T00:26:39.590Z
Is $P^2$ supposed to be the real projective plane $RP^2$?
lightxbulb
CC BY-SA 4.0
2021-11-22T03:14:49.283Z
@lightxbulb explicitly $P^2=\mathbb{R^2}$+ ideal points.
S. M.
CC BY-SA 4.0
2021-11-22T06:32:05.093Z
Is the $+$ supposed to be union? Define "ideal points". Better yet - post the reference that you are using which employs this $P^2$ notation.
lightxbulb
CC BY-SA 4.0
2021-11-22T07:01:26.093Z
@lightxbulb ideal points is of form (x, y, 0)  which is point at infinity.
S. M.
CC BY-SA 4.0
2021-11-22T07:53:00.240Z
@B-spline: "*$\mathbb{R^2}$ points in $P^2$ can be represents as $(x, y, 1)$*" This is the part that I don't see is true. The way a homogeneous space works, those are just the points of $\mathbb{R^2}$ as transformed into a homogeneous space. The *plane* defined by having a Z component of 1 is *not* a projective plane. The lines in (x, y, 1) are the same lines as in (x, y). Parallel lines don't intersect.
Nicol Bolas
CC BY-SA 4.0
2021-11-22T08:20:41.943Z
@Nicol [See this](https://medium.com/@daniel.j.lenton/part-i-projective-geometry-in-2d-b1ca26d5fa2a)
S. M.
CC BY-SA 4.0
2021-11-22T10:00:23.653Z
The medium article explains your confusion. They **define** the term projective plane to be the plane $(x,y,1)$. It is true by their definition. This is usually not how the term is used as it refers to the real projective plane which is not just $(x,y,1)$. I could not find the definition of the medium article's author anywhere else (including the book that they cite). But by their definition $(x,y,z,1)$ would correspond to their definition of projective plane in 3D. I advise against using this terminology as I believe it is nonstandard and confusing (conflicts with the original term).
lightxbulb
CC BY-SA 4.0
2021-11-23T05:29:35.857Z
It is a thing, though. It's an **_Affine_** plane. That's cool, too, right?
luser droog
<3d><transformations><projections><2d>
3D homogenous coordinates versus 4D homogenous coordinates
S. M.
# 3D homogenous coordinates versus 4D homogenous coordinates

(Asked by: username_0 on 2021-11-21)

We know that any 2D point $(x, y) $which represents as 3D homogeneous coordinates is of the form $(x, y, 1)$ which is the points of projective plane $P^2.$

If I use the same concepts for 3D points $(x, y, z)$ which represents as 4D homogeneous coordinates is of the form $(x, y, z, 1).$

My question is $(x, y, z, 1)$ could be the points of 3D projective space?

My second question is always homogenous coordinates are the points of projective space ?

N. B. -- $P^2$ is projective plane which contains the points of $\mathbb{R^2}$($\mathbb{R^2}$ points in $P^2$ can be represents as $(x, y, 1)$ ) and ideal points, is of the form $(x, y, 0)$ which is called points at infinity.

**Tags:** 3d, transformations, projections, 2d

> **Comments:**
> **username_1**: Is $P^2$ supposed to be the real projective plane $RP^2$?  
> **username_0**: @username_1 explicitly $P^2=\mathbb{R^2}$+ ideal points.  
> **username_1**: Is the $+$ supposed to be union? Define "ideal points". Better yet - post the reference that you are using which employs this $P^2$ notation.  
> **username_0**: @username_1 ideal points is of form (x, y, 0)  which is point at infinity.  
> **username_2**: @B-spline: "*$\mathbb{R^2}$ points in $P^2$ can be represents as $(x, y, 1)$*" This is the part that I don't see is true. The way a homogeneous space works, those are just the points of $\mathbb{R^2}$ as transformed into a homogeneous space. The *plane* defined by having a Z component of 1 is *not* a projective plane. The lines in (x, y, 1) are the same lines as in (x, y). Parallel lines don't intersect.  
> **username_0**: @username_2 [See this](https://medium.com/@daniel.j.lenton/part-i-projective-geometry-in-2d-b1ca26d5fa2a)  
> **username_1**: The medium article explains your confusion. They **define** the term projective plane to be the plane $(x,y,1)$. It is true by their definition. This is usually not how the term is used as it refers to the real projective plane which is not just $(x,y,1)$. I could not find the definition of the medium article's author anywhere else (including the book that they cite). But by their definition $(x,y,z,1)$ would correspond to their definition of projective plane in 3D. I advise against using this terminology as I believe it is nonstandard and confusing (conflicts with the original term).  
> **username_3**: It is a thing, though. It's an **_Affine_** plane. That's cool, too, right?  

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## 1 answers

### Answer by username_2 on 2021-11-21. Score: 1 | Accepted answer

So, a [projective plane](https://en.wikipedia.org/wiki/Projective_plane) is a plane constructed such that parallel lines intersect at infinity. The plane defined by the equation (x, y, 1) is not projective. Just draw two parallel lines on that plane, and you will find that they never intersect, even under homogeneous transformations.

Now, homogeneous transformations can *create* projective planes. But such a plane would not have a fixed homogeneous coordinate of 1, because the homogeneous transform with a coordinate of 1 doesn't change the other coordinates. A projective plane in homogeneous space needs to be able to represent points at infinity, and that only happens with a homogeneous coordinate of zero.

So if a plane in homogeneous space does not pass through the 0 homogeneous coordinate, it *cannot* represent a projective plane.

Since the rest of your question is based on the assumption that (x, y, 1) is projective, it's kind of irrelevant.

> **Comments:**
> **username_0**: so what about (x, y, z, 1) in projective space?  
> **username_0**: your answer isn't according to my question. I am very much unsatisfied with your answer.  
> **username_2**: @B-spline: "*so what about (x, y, z, 1) in projective space?*" If (x, y, 1) isn't a projective plane, then why would you think that (x, y, z, 1) is a "projective space"? "*I am very much unsatisfied with your answer.*" My answer is correct. It's not a projective plane, and the rest of your question is predicated on the assumption that it is. So the rest of your question is effectively moot.  

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