<p>I suppose in some respects it's a matter of perspective (no pun intended).  The ordered triple <span class="math-container">$(x,y,w)$</span> is a point in a 3-dimensional <a href="https://en.wikipedia.org/wiki/Projective_space" rel="nofollow noreferrer">projective space</a> that is mapped (projected) to a 2-dimensional point in the Euclidean plane: <span class="math-container">$(x/w, y/w).$</span>  Given that, <span class="math-container">$(x,y,1)$</span> would be 2-dimensional plane in that 3-dimensional space (or a specific point if <span class="math-container">$x$</span> and <span class="math-container">$y$</span> are fixed) that maps to <span class="math-container">$(x,y)$</span> in the standard Euclidean 2-dimensional plane.</p>

CC BY-SA 4.0
2021-11-19T22:13:33.553Z
(x,y,1) is a projective plane points, so how it is points in 3d?
S. M.
CC BY-SA 4.0
2021-11-19T22:21:21.757Z
(x,y,1) has two degree of freedom,, so how it is 3d?
S. M.
CC BY-SA 4.0
2021-11-19T23:05:11.450Z
@B-spline: "*(x,y,1) has two degree of freedom*" It's not a function; it's a point in space. Points don't have "degrees of freedom"; they have coordinates. A point with 3 coordinates is said to be three dimensional.
Nicol Bolas
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2021-11-20T08:07:16.003Z
@NicolBolas There's nothing wrong with calling $x,y$ degrees of freedom as long as they remain variables and are not constants.
lightxbulb
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2021-11-20T13:57:03.673Z
@B-spline `(x,y,1) is a projective plane points, so how it is points in 3d?` To be more precise the triplet is an element of $\mathbb{R}^3$. If you're just given a triplet you cannot tell whether it's 2D homogeneous coordinates. If you know that the triplet $(x,y,z)$ is in homogeneous coordinates however, then you know that it corresponds to the point $(x/z, y/z)$ in 2D. Note that this also doesn't contradict the fact that $(x,y,z)$ is an element in $\mathbb{R}^3$ regardless of whether it's in 2D homogeneous coordinates or whether it's a typical 3D point/vector.
lightxbulb
CC BY-SA 4.0
2021-11-19T22:05:27.147Z
2021-11-19T23:05:18.070Z
2021-11-19T23:05:18.070Z
GrapefruitIsAwesome
<p>If you have <span class="math-container">$(x,y,z) \in \mathbb{R}^3$</span> and you relate it to <span class="math-container">$(x/z, y/z) \in \mathbb{R}^2$</span> then you have interpreted <span class="math-container">$(x,y,z)$</span> as one possible representation of the 2D vector <span class="math-container">$(x/z, y/z)$</span> in homogeneous coordinates. The special case of <span class="math-container">$z=0$</span> corresponds to &quot;points at infinity&quot; or directions. If you have  <span class="math-container">$(x,y,z,w) \in \mathbb{R}^4$</span> and you relate it to <span class="math-container">$(x/w,y/w,z/w) \in \mathbb{R}^3$</span> then you have interpreted <span class="math-container">$(x,y,z,w)$</span> as one possible representation of the 3D vector <span class="math-container">$(x/w, y/w, z/w)$</span> in homogeneous coordinates. From this it becomes clear that to represent an <span class="math-container">$N$</span>-dimensional point in homogeneous coordinates you need <span class="math-container">$N+1$</span> components.</p>
<p>To clarify what I mean by one possible representation, consider the following example:</p>
<p><span class="math-container">$$(3,7) \equiv (3,7,1) \equiv (6,14,2),$$</span></p>
<p>where the second and third vectors are different representations of the same 2D vector in homogeneous coordinates. In general, if you have a 2D vector <span class="math-container">$(X,Y)$</span>, then for any <span class="math-container">$\lambda\ne 0$</span>, the vector <span class="math-container">$(\lambda X, \lambda Y, \lambda)$</span> is a representation of <span class="math-container">$(X,Y)$</span> in homogeneous coordinates. So you really get an equivalence class of points in <span class="math-container">$N+1$</span> dimensions corresponding to a specific point in <span class="math-container">$N$</span> dimensions. Notably this equivalence class forms a line through <span class="math-container">$(0,0,0)$</span> and <span class="math-container">$(X,Y,1)$</span> in the <span class="math-container">$N+1$</span> dimensional space.</p>

CC BY-SA 4.0
2021-11-20T18:10:25.540Z
So (x, y) is 2D points and (x, y, 1) is It's 3D homogeneous coordinates?
S. M.
CC BY-SA 4.0
2021-11-20T18:11:41.247Z
@B-spline What is $A,B,C,D$ here, and what is the relation to the current question.  If $(x,y)$ is a 2D point then $(x,y,1)$ is one possible version of it in homogeneous coordinates. But also $(2x,2y,2)$ is a version of it in homogeneous coordinates. For any $\lambda\ne 0$, $(\lambda x, \lambda y, \lambda)$ is a representation of $(x,y)$ in homogeneous coordinates.
lightxbulb
CC BY-SA 4.0
2021-11-20T18:13:32.640Z
So what we said the homogenous coordinates (x, y, 1)  ? 2d or 3d?
S. M.
CC BY-SA 4.0
2021-11-20T18:15:55.453Z
@B-spline $(\lambda x, \lambda y,\lambda),\, \lambda \ne 0$ is in $\mathbb{R}^3$, while $(x, y)$ is in $\mathbb{R}^2$. However $(\lambda x, \lambda y, \lambda)$ are the homogeneous coordinates of $(x,y)$.
lightxbulb
CC BY-SA 4.0
2021-11-20T18:19:35.170Z
@bulb you don't say directly which invites confusion... You just tell  (x, y, 1)  is 2d homogenous coordinates or 3d homogenous coordinates? In one line nothing else!!!
S. M.
CC BY-SA 4.0
2021-11-20T18:21:52.813Z
@B-spline $(x,y,1)$ is a 3D point and at the same time the representation of the 2D point $(x,y)$ in homogeneous coordinates. The two statements are not contradictory. It is however not the representation of a 3D point in homogeneous coordinates, since for that you would need a 4D point.
lightxbulb
CC BY-SA 4.0
2021-11-21T02:51:06.317Z
@B-spline For a concise answer, consider "*homogenous*" to mean `+ 1`. **(_x_,_y_,1)** represents a 2D(+1) point. The fact that it's *homogenous* adds one. (Does that help, or am I making things worse??)
luser droog
CC BY-SA 4.0
2021-11-20T17:36:05.483Z
2021-11-20T17:36:05.483Z
lightxbulb
<p>We know that <span class="math-container">$(x, y, 1)$</span> are the homogenous coordinates of a 2D point <span class="math-container">$(x, y)$</span>. <span class="math-container">$(x, y, 1)$</span> has 2 degrees of freedom. That's why we should call it 2D homogenous coordinates. But many websites say it's 3D homogenous coordinates.</p>
<p>My question is: What is right? <span class="math-container">$(x, y, 1)$</span> are 2D homogenous coordinates or 3D homogenous coordinates?</p>

CC BY-SA 4.0
2021-11-20T08:04:26.523Z
It's 2D homogeneous coordinates. 3D homogeneous coordinates would be $(x,y,z,w)$.
lightxbulb
CC BY-SA 4.0
2021-11-20T12:22:16.973Z
Can you give a better context?  Interpreting the meaning of a "vector" is allllll about how the values are being interpreted, and interpretation is alllll about context: Consider this vec4(1,3,2,1); Its a vector, no its a point, no its a plane, no its a quaternion, no its a bivector.... It's all those things, but which one makes sense? Well that is all about context. So asking if it some random vec3 is 2D or 3D without giving a context is pretty much meaningless.
pmw1234
<3d><geometry><projections>
$(x, y, 1)$ is 2D homogenous coordinates or 3D homogenous coordinates?
S. M.
# $(x, y, 1)$ is 2D homogenous coordinates or 3D homogenous coordinates?

(Asked by: username_0 on 2021-11-19)

We know that $(x, y, 1)$ are the homogenous coordinates of a 2D point $(x, y)$. $(x, y, 1)$ has 2 degrees of freedom. That's why we should call it 2D homogenous coordinates. But many websites say it's 3D homogenous coordinates.

My question is: What is right? $(x, y, 1)$ are 2D homogenous coordinates or 3D homogenous coordinates?

**Tags:** 3d, geometry, projections

> **Comments:**
> **username_1**: It's 2D homogeneous coordinates. 3D homogeneous coordinates would be $(x,y,z,w)$.  
> **username_2**: Can you give a better context?  Interpreting the meaning of a "vector" is allllll about how the values are being interpreted, and interpretation is alllll about context: Consider this vec4(1,3,2,1); Its a vector, no its a point, no its a plane, no its a quaternion, no its a bivector.... It's all those things, but which one makes sense? Well that is all about context. So asking if it some random vec3 is 2D or 3D without giving a context is pretty much meaningless.  

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## 2 answers

### Answer by username_1 on 2021-11-20. Score: 4 | Accepted answer

If you have $(x,y,z) \in \mathbb{R}^3$ and you relate it to $(x/z, y/z) \in \mathbb{R}^2$ then you have interpreted $(x,y,z)$ as one possible representation of the 2D vector $(x/z, y/z)$ in homogeneous coordinates. The special case of $z=0$ corresponds to "points at infinity" or directions. If you have $(x,y,z,w) \in \mathbb{R}^4$ and you relate it to $(x/w,y/w,z/w) \in \mathbb{R}^3$ then you have interpreted $(x,y,z,w)$ as one possible representation of the 3D vector $(x/w, y/w, z/w)$ in homogeneous coordinates. From this it becomes clear that to represent an $N$-dimensional point in homogeneous coordinates you need $N+1$ components.

To clarify what I mean by one possible representation, consider the following example:

$$(3,7) \equiv (3,7,1) \equiv (6,14,2),$$

where the second and third vectors are different representations of the same 2D vector in homogeneous coordinates. In general, if you have a 2D vector $(X,Y)$, then for any $\lambda\ne 0$, the vector $(\lambda X, \lambda Y, \lambda)$ is a representation of $(X,Y)$ in homogeneous coordinates. So you really get an equivalence class of points in $N+1$ dimensions corresponding to a specific point in $N$ dimensions. Notably this equivalence class forms a line through $(0,0,0)$ and $(X,Y,1)$ in the $N+1$ dimensional space.

> **Comments:**
> **username_0**: So (x, y) is 2D points and (x, y, 1) is It's 3D homogeneous coordinates?  
> **username_1**: @B-spline What is $A,B,C,D$ here, and what is the relation to the current question.  If $(x,y)$ is a 2D point then $(x,y,1)$ is one possible version of it in homogeneous coordinates. But also $(2x,2y,2)$ is a version of it in homogeneous coordinates. For any $\lambda\ne 0$, $(\lambda x, \lambda y, \lambda)$ is a representation of $(x,y)$ in homogeneous coordinates.  
> **username_0**: So what we said the homogenous coordinates (x, y, 1)  ? 2d or 3d?  
> **username_1**: @B-spline $(\lambda x, \lambda y,\lambda),\, \lambda \ne 0$ is in $\mathbb{R}^3$, while $(x, y)$ is in $\mathbb{R}^2$. However $(\lambda x, \lambda y, \lambda)$ are the homogeneous coordinates of $(x,y)$.  
> **username_0**: @bulb you don't say directly which invites confusion... You just tell  (x, y, 1)  is 2d homogenous coordinates or 3d homogenous coordinates? In one line nothing else!!!  
> **username_1**: @B-spline $(x,y,1)$ is a 3D point and at the same time the representation of the 2D point $(x,y)$ in homogeneous coordinates. The two statements are not contradictory. It is however not the representation of a 3D point in homogeneous coordinates, since for that you would need a 4D point.  
> **username_5**: @B-spline For a concise answer, consider "*homogenous*" to mean `+ 1`. **(_x_,_y_,1)** represents a 2D(+1) point. The fact that it's *homogenous* adds one. (Does that help, or am I making things worse??)  

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### Answer by username_3 on 2021-11-19. Score: 1

I suppose in some respects it's a matter of perspective (no pun intended). The ordered triple $(x,y,w)$ is a point in a 3-dimensional [projective space](https://en.wikipedia.org/wiki/Projective_space) that is mapped (projected) to a 2-dimensional point in the Euclidean plane: $(x/w, y/w).$ Given that, $(x,y,1)$ would be 2-dimensional plane in that 3-dimensional space (or a specific point if $x$ and $y$ are fixed) that maps to $(x,y)$ in the standard Euclidean 2-dimensional plane.

> **Comments:**
> **username_0**: (x,y,1) is a projective plane points, so how it is points in 3d?  
> **username_0**: (x,y,1) has two degree of freedom,, so how it is 3d?  
> **username_4**: @B-spline: "*(x,y,1) has two degree of freedom*" It's not a function; it's a point in space. Points don't have "degrees of freedom"; they have coordinates. A point with 3 coordinates is said to be three dimensional.  
> **username_1**: @username_4 There's nothing wrong with calling $x,y$ degrees of freedom as long as they remain variables and are not constants.  
> **username_1**: @B-spline `(x,y,1) is a projective plane points, so how it is points in 3d?` To be more precise the triplet is an element of $\mathbb{R}^3$. If you're just given a triplet you cannot tell whether it's 2D homogeneous coordinates. If you know that the triplet $(x,y,z)$ is in homogeneous coordinates however, then you know that it corresponds to the point $(x/z, y/z)$ in 2D. Note that this also doesn't contradict the fact that $(x,y,z)$ is an element in $\mathbb{R}^3$ regardless of whether it's in 2D homogeneous coordinates or whether it's a typical 3D point/vector.  

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