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Can 3D co-ordinates be transferred into 2D co-ordinates?

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1

$begingroup$

Is it possible to transform co-ordinates $(a,b,c)$ into $(x,y) $ such that $(x,y)$ is unique for each $(a,b,c)$ ? $a, b, c, x, y$ are in $BbbR$ .

geometry linear-transformations laplace-transform transformation plane-geometry

share|cite|improve this question

edited 2 days ago

dmtri

1,6082521

asked 2 days ago

BJKShahBJKShah

84

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Just to clarify, do you want the map from $Bbb R^3$ to $Bbb R^2$ to be invertible, or just exist? As BJKShah, the former is possible but not if it's linear, whereas the latter is possible even then (such a map is an example of a projection).
  $endgroup$
  – J.G.
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I want the conversion to be possible in both directions.( How to convert former as you said , any links?) Thanks
  $endgroup$
  – BJKShah
  2 days ago
  

  
  
  
  

add a comment | 

1

$begingroup$

Is it possible to transform co-ordinates $(a,b,c)$ into $(x,y) $ such that $(x,y)$ is unique for each $(a,b,c)$ ? $a, b, c, x, y$ are in $BbbR$ .

geometry linear-transformations laplace-transform transformation plane-geometry

share|cite|improve this question

edited 2 days ago

dmtri

1,6082521

asked 2 days ago

BJKShahBJKShah

84

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Just to clarify, do you want the map from $Bbb R^3$ to $Bbb R^2$ to be invertible, or just exist? As BJKShah, the former is possible but not if it's linear, whereas the latter is possible even then (such a map is an example of a projection).
  $endgroup$
  – J.G.
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I want the conversion to be possible in both directions.( How to convert former as you said , any links?) Thanks
  $endgroup$
  – BJKShah
  2 days ago
  

  
  
  
  

add a comment | 

$begingroup$

Is it possible to transform co-ordinates $(a,b,c)$ into $(x,y) $ such that $(x,y)$ is unique for each $(a,b,c)$ ? $a, b, c, x, y$ are in $BbbR$ .

geometry linear-transformations laplace-transform transformation plane-geometry

share|cite|improve this question

edited 2 days ago

dmtri

1,6082521

asked 2 days ago

BJKShahBJKShah

84

$endgroup$

share|cite|improve this question

edited 2 days ago

dmtri

1,6082521

asked 2 days ago

BJKShahBJKShah

84

share|cite|improve this question

edited 2 days ago

dmtri

1,6082521

asked 2 days ago

BJKShahBJKShah

84

edited 2 days ago

dmtri

1,6082521

edited 2 days ago

dmtri

1,6082521

asked 2 days ago

BJKShahBJKShah

84

asked 2 days ago

BJKShahBJKShah

84

2 Answers
2

active

oldest

votes

1

$begingroup$

It is possible, but it's not pretty, and as VENKITESH says, it's certainly not linear. Here is one way: define $$f(a,b,c)=(x,y)$$ where $x=a$, and $y$ is the alternating decimal expansion of $b$ and $c$. What that means is this:

Take the non-terminating decimal expansions of $b$ and $c$; if necessary, insert leading zeroes so that the integral parts have the same length. For instance, if $b=1234.345$ and $c=76.987698798ldots$, we convert $b$ to its non-terminating form $b=colorred1234.344999999ldots$, and we write $c$ as $colorblue0076.987698798ldots$ to make the lengths before the decimal point equal.

Now we can write $y$ by taking digits from $b$ and $c$ in turn:

$$y=colorred1colorblue0colorred2colorblue0colorred3colorblue7colorred4colorblue6.colorred3colorblue9colorred4colorblue8colorred4colorblue7colorred9colorblue6colorred9colorblue9colorred9colorblue8colorred9colorblue7colorred9colorblue9colorred9colorblue8ldots$$

This is just one way. By the way, the name alternating decimal expansion is something I just made up, so don't take it too seriously.

Edited to add: I see you changed your criteria in a comment. This mapping is an injection, but not a surjection, so it doesn't have an inverse. You can make it surjective, but it's messy, and I'm not going to do it here.

Edited to add: As the OP points out in a comment, I have not define the mapping for $b,cle 0$. My bad! But you can fix this by using the decimal expansions of $e^b$ and $e^c$, which are always strictly greater than zero.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

TonyKTonyK

43k356135

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  (1,2,1)=(1,21) but (1,-2,-1) and (1,-2,1)=(?) Thanks for the example.
  $endgroup$
  – BJKShah
  2 days ago
  

  
  
  
  

add a comment | 

3

$begingroup$

It's is possible to have a bijection between $mathbbR^3$ and $mathbbR^2$, since it can be shown that $mathbbR$ and $mathbbR^2$, and hence every $mathbbR^n$ has the same cardinality. But a linear or affine isomorphism is not possible if the dimensions are different.

share|cite|improve this answer

edited 2 days ago

tatan

5,79762760

answered 2 days ago

VENKITESHVENKITESH

36817

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can you please explain the last sentence a bit more?
  $endgroup$
  – BJKShah
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Right. A map f:R^nto R^m will be linear if f(u+v)=f(u)+f(v) and f(au)=af(u) for all scalars a and vectors u,v. We can show that if such an f is a bijection, then m=n. This follows from what is called the Rank Nullity Theorem. An affine map is of the form f+c, for a linear map f and a constant vector c. Again if an affine map is bijective, then we can show that m=n.
  $endgroup$
  – VENKITESH
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Any good books for studying this?
  $endgroup$
  – BJKShah
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Calculus and Analytic Geometry by Thomas and Finney, for coordinate geometry, and Linear Algebra by Gilbert Strang for linear algebra.
  $endgroup$
  – VENKITESH
  2 days ago
  

  
  
  
  

add a comment | 

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1

$begingroup$

It is possible, but it's not pretty, and as VENKITESH says, it's certainly not linear. Here is one way: define $$f(a,b,c)=(x,y)$$ where $x=a$, and $y$ is the alternating decimal expansion of $b$ and $c$. What that means is this:

Take the non-terminating decimal expansions of $b$ and $c$; if necessary, insert leading zeroes so that the integral parts have the same length. For instance, if $b=1234.345$ and $c=76.987698798ldots$, we convert $b$ to its non-terminating form $b=colorred1234.344999999ldots$, and we write $c$ as $colorblue0076.987698798ldots$ to make the lengths before the decimal point equal.

Now we can write $y$ by taking digits from $b$ and $c$ in turn:

$$y=colorred1colorblue0colorred2colorblue0colorred3colorblue7colorred4colorblue6.colorred3colorblue9colorred4colorblue8colorred4colorblue7colorred9colorblue6colorred9colorblue9colorred9colorblue8colorred9colorblue7colorred9colorblue9colorred9colorblue8ldots$$

This is just one way. By the way, the name alternating decimal expansion is something I just made up, so don't take it too seriously.

Edited to add: I see you changed your criteria in a comment. This mapping is an injection, but not a surjection, so it doesn't have an inverse. You can make it surjective, but it's messy, and I'm not going to do it here.

Edited to add: As the OP points out in a comment, I have not define the mapping for $b,cle 0$. My bad! But you can fix this by using the decimal expansions of $e^b$ and $e^c$, which are always strictly greater than zero.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

TonyKTonyK

43k356135

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  (1,2,1)=(1,21) but (1,-2,-1) and (1,-2,1)=(?) Thanks for the example.
  $endgroup$
  – BJKShah
  2 days ago
  

  
  
  
  

add a comment | 

1

$begingroup$

It is possible, but it's not pretty, and as VENKITESH says, it's certainly not linear. Here is one way: define $$f(a,b,c)=(x,y)$$ where $x=a$, and $y$ is the alternating decimal expansion of $b$ and $c$. What that means is this:

Take the non-terminating decimal expansions of $b$ and $c$; if necessary, insert leading zeroes so that the integral parts have the same length. For instance, if $b=1234.345$ and $c=76.987698798ldots$, we convert $b$ to its non-terminating form $b=colorred1234.344999999ldots$, and we write $c$ as $colorblue0076.987698798ldots$ to make the lengths before the decimal point equal.

Now we can write $y$ by taking digits from $b$ and $c$ in turn:

$$y=colorred1colorblue0colorred2colorblue0colorred3colorblue7colorred4colorblue6.colorred3colorblue9colorred4colorblue8colorred4colorblue7colorred9colorblue6colorred9colorblue9colorred9colorblue8colorred9colorblue7colorred9colorblue9colorred9colorblue8ldots$$

This is just one way. By the way, the name alternating decimal expansion is something I just made up, so don't take it too seriously.

Edited to add: I see you changed your criteria in a comment. This mapping is an injection, but not a surjection, so it doesn't have an inverse. You can make it surjective, but it's messy, and I'm not going to do it here.

Edited to add: As the OP points out in a comment, I have not define the mapping for $b,cle 0$. My bad! But you can fix this by using the decimal expansions of $e^b$ and $e^c$, which are always strictly greater than zero.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

TonyKTonyK

43k356135

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  (1,2,1)=(1,21) but (1,-2,-1) and (1,-2,1)=(?) Thanks for the example.
  $endgroup$
  – BJKShah
  2 days ago
  

  
  
  
  

add a comment | 

$begingroup$

It is possible, but it's not pretty, and as VENKITESH says, it's certainly not linear. Here is one way: define $$f(a,b,c)=(x,y)$$ where $x=a$, and $y$ is the alternating decimal expansion of $b$ and $c$. What that means is this:

Take the non-terminating decimal expansions of $b$ and $c$; if necessary, insert leading zeroes so that the integral parts have the same length. For instance, if $b=1234.345$ and $c=76.987698798ldots$, we convert $b$ to its non-terminating form $b=colorred1234.344999999ldots$, and we write $c$ as $colorblue0076.987698798ldots$ to make the lengths before the decimal point equal.

Now we can write $y$ by taking digits from $b$ and $c$ in turn:

$$y=colorred1colorblue0colorred2colorblue0colorred3colorblue7colorred4colorblue6.colorred3colorblue9colorred4colorblue8colorred4colorblue7colorred9colorblue6colorred9colorblue9colorred9colorblue8colorred9colorblue7colorred9colorblue9colorred9colorblue8ldots$$

This is just one way. By the way, the name alternating decimal expansion is something I just made up, so don't take it too seriously.

Edited to add: I see you changed your criteria in a comment. This mapping is an injection, but not a surjection, so it doesn't have an inverse. You can make it surjective, but it's messy, and I'm not going to do it here.

Edited to add: As the OP points out in a comment, I have not define the mapping for $b,cle 0$. My bad! But you can fix this by using the decimal expansions of $e^b$ and $e^c$, which are always strictly greater than zero.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

TonyKTonyK

43k356135

$endgroup$

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

TonyKTonyK

43k356135

answered 2 days ago

TonyKTonyK

43k356135

answered 2 days ago

TonyKTonyK

43k356135

3

$begingroup$

It's is possible to have a bijection between $mathbbR^3$ and $mathbbR^2$, since it can be shown that $mathbbR$ and $mathbbR^2$, and hence every $mathbbR^n$ has the same cardinality. But a linear or affine isomorphism is not possible if the dimensions are different.

share|cite|improve this answer

edited 2 days ago

tatan

5,79762760

answered 2 days ago

VENKITESHVENKITESH

36817

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can you please explain the last sentence a bit more?
  $endgroup$
  – BJKShah
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Right. A map f:R^nto R^m will be linear if f(u+v)=f(u)+f(v) and f(au)=af(u) for all scalars a and vectors u,v. We can show that if such an f is a bijection, then m=n. This follows from what is called the Rank Nullity Theorem. An affine map is of the form f+c, for a linear map f and a constant vector c. Again if an affine map is bijective, then we can show that m=n.
  $endgroup$
  – VENKITESH
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Any good books for studying this?
  $endgroup$
  – BJKShah
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Calculus and Analytic Geometry by Thomas and Finney, for coordinate geometry, and Linear Algebra by Gilbert Strang for linear algebra.
  $endgroup$
  – VENKITESH
  2 days ago
  

  
  
  
  

add a comment | 

3

$begingroup$

It's is possible to have a bijection between $mathbbR^3$ and $mathbbR^2$, since it can be shown that $mathbbR$ and $mathbbR^2$, and hence every $mathbbR^n$ has the same cardinality. But a linear or affine isomorphism is not possible if the dimensions are different.

share|cite|improve this answer

edited 2 days ago

tatan

5,79762760

answered 2 days ago

VENKITESHVENKITESH

36817

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can you please explain the last sentence a bit more?
  $endgroup$
  – BJKShah
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Right. A map f:R^nto R^m will be linear if f(u+v)=f(u)+f(v) and f(au)=af(u) for all scalars a and vectors u,v. We can show that if such an f is a bijection, then m=n. This follows from what is called the Rank Nullity Theorem. An affine map is of the form f+c, for a linear map f and a constant vector c. Again if an affine map is bijective, then we can show that m=n.
  $endgroup$
  – VENKITESH
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Any good books for studying this?
  $endgroup$
  – BJKShah
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Calculus and Analytic Geometry by Thomas and Finney, for coordinate geometry, and Linear Algebra by Gilbert Strang for linear algebra.
  $endgroup$
  – VENKITESH
  2 days ago
  

  
  
  
  

add a comment | 

$begingroup$

It's is possible to have a bijection between $mathbbR^3$ and $mathbbR^2$, since it can be shown that $mathbbR$ and $mathbbR^2$, and hence every $mathbbR^n$ has the same cardinality. But a linear or affine isomorphism is not possible if the dimensions are different.

share|cite|improve this answer

edited 2 days ago

tatan

5,79762760

answered 2 days ago

VENKITESHVENKITESH

36817

$endgroup$

share|cite|improve this answer

edited 2 days ago

tatan

5,79762760

answered 2 days ago

VENKITESHVENKITESH

36817

edited 2 days ago

tatan

5,79762760

edited 2 days ago

tatan

5,79762760

answered 2 days ago

VENKITESHVENKITESH

36817

answered 2 days ago

VENKITESHVENKITESH

36817