How to check if lines in homogenous coordinates are distinct?
1
$begingroup$
In $mathbb{P}^3$ the 4 lines in homogeneous coordinates are given:
$$g = [1:-1:1], h = [1:1:0], k = [3:-3:-1], l = [1:0:0]$$
Show that g, h, k, l are all pairwise distinct and that each three of them build the sides of a triangle.
I don't quite understand how we can check if two lines are distinct in homogeneous coordinates.
geometry
asked Dec 3 '18 at 14:20
ladyeli555ladyeli555
92
$endgroup$
|
show 6 more comments
1
$begingroup$
In $mathbb{P}^3$ the 4 lines in homogeneous coordinates are given:
$$g = [1:-1:1], h = [1:1:0], k = [3:-3:-1], l = [1:0:0]$$
Show that g, h, k, l are all pairwise distinct and that each three of them build the sides of a triangle.
I don't quite understand how we can check if two lines are distinct in homogeneous coordinates.
geometry
asked Dec 3 '18 at 14:20
ladyeli555ladyeli555
92
$endgroup$
$begingroup$
could you please clarify what you mean by lines? in my opinion those are just 4 points in the projective space (if you wish, lines through the origin). However, if you interpret them as actual lines in $mathbb{k}^n$ we get that all go through the origin, and hence the second part of the question makes no sense. Are you sure you did not miss some variable in the definition? especially since lines in homogenous coordinates sounds like lines in $mathbb{P}^n$
$endgroup$
– Enkidu
Dec 3 '18 at 14:31
$begingroup$
Yes, I am sorry for not specifying. I edited the question. The lines are in $mathbb{P}^3$.
$endgroup$
– ladyeli555
Dec 3 '18 at 14:39
$begingroup$
they are however still points in $mathbb{P}^3$ not lines in $mathbb{P}^3$.
$endgroup$
– Enkidu
Dec 3 '18 at 14:40
$begingroup$
You can have points and lines in $mathbb{P}^3$. If you calculate the connection line between two points you get a line in this form. And in this case, we have lines.
$endgroup$
– ladyeli555
Dec 3 '18 at 14:50
$begingroup$
so you mean that you want to consider the 6 lines defined by those points? then please make that clear! also, is it now 4 or 6, since 4 points define 6 lines. I am sorry for pestering, but this question is far from clearly stated.
$endgroup$
– Enkidu
Dec 3 '18 at 14:53
|
show 6 more comments
1
1
1
$begingroup$
In $mathbb{P}^3$ the 4 lines in homogeneous coordinates are given:
$$g = [1:-1:1], h = [1:1:0], k = [3:-3:-1], l = [1:0:0]$$
Show that g, h, k, l are all pairwise distinct and that each three of them build the sides of a triangle.
I don't quite understand how we can check if two lines are distinct in homogeneous coordinates.
geometry
asked Dec 3 '18 at 14:20
ladyeli555ladyeli555
92
$endgroup$
In $mathbb{P}^3$ the 4 lines in homogeneous coordinates are given:
$$g = [1:-1:1], h = [1:1:0], k = [3:-3:-1], l = [1:0:0]$$
Show that g, h, k, l are all pairwise distinct and that each three of them build the sides of a triangle.
I don't quite understand how we can check if two lines are distinct in homogeneous coordinates.
geometry
geometry
asked Dec 3 '18 at 14:20
ladyeli555ladyeli555
92
asked Dec 3 '18 at 14:20
ladyeli555ladyeli555
92
edited Dec 3 '18 at 14:38
ladyeli555
asked Dec 3 '18 at 14:20
ladyeli555ladyeli555
92
asked Dec 3 '18 at 14:20
ladyeli555ladyeli555
92
asked Dec 3 '18 at 14:20
ladyeli555ladyeli555
92
92
$begingroup$
could you please clarify what you mean by lines? in my opinion those are just 4 points in the projective space (if you wish, lines through the origin). However, if you interpret them as actual lines in $mathbb{k}^n$ we get that all go through the origin, and hence the second part of the question makes no sense. Are you sure you did not miss some variable in the definition? especially since lines in homogenous coordinates sounds like lines in $mathbb{P}^n$
$endgroup$
– Enkidu
Dec 3 '18 at 14:31
$begingroup$
Yes, I am sorry for not specifying. I edited the question. The lines are in $mathbb{P}^3$.
$endgroup$
– ladyeli555
Dec 3 '18 at 14:39
$begingroup$
they are however still points in $mathbb{P}^3$ not lines in $mathbb{P}^3$.
$endgroup$
– Enkidu
Dec 3 '18 at 14:40
$begingroup$
You can have points and lines in $mathbb{P}^3$. If you calculate the connection line between two points you get a line in this form. And in this case, we have lines.
$endgroup$
– ladyeli555
Dec 3 '18 at 14:50
$begingroup$
so you mean that you want to consider the 6 lines defined by those points? then please make that clear! also, is it now 4 or 6, since 4 points define 6 lines. I am sorry for pestering, but this question is far from clearly stated.
$endgroup$
– Enkidu
Dec 3 '18 at 14:53
|
show 6 more comments
$begingroup$
could you please clarify what you mean by lines? in my opinion those are just 4 points in the projective space (if you wish, lines through the origin). However, if you interpret them as actual lines in $mathbb{k}^n$ we get that all go through the origin, and hence the second part of the question makes no sense. Are you sure you did not miss some variable in the definition? especially since lines in homogenous coordinates sounds like lines in $mathbb{P}^n$
$endgroup$
– Enkidu
Dec 3 '18 at 14:31
$begingroup$
Yes, I am sorry for not specifying. I edited the question. The lines are in $mathbb{P}^3$.
$endgroup$
– ladyeli555
Dec 3 '18 at 14:39
$begingroup$
they are however still points in $mathbb{P}^3$ not lines in $mathbb{P}^3$.
$endgroup$
– Enkidu
Dec 3 '18 at 14:40
$begingroup$
You can have points and lines in $mathbb{P}^3$. If you calculate the connection line between two points you get a line in this form. And in this case, we have lines.
$endgroup$
– ladyeli555
Dec 3 '18 at 14:50
$begingroup$
so you mean that you want to consider the 6 lines defined by those points? then please make that clear! also, is it now 4 or 6, since 4 points define 6 lines. I am sorry for pestering, but this question is far from clearly stated.
$endgroup$
– Enkidu
Dec 3 '18 at 14:53
$begingroup$
could you please clarify what you mean by lines? in my opinion those are just 4 points in the projective space (if you wish, lines through the origin). However, if you interpret them as actual lines in $mathbb{k}^n$ we get that all go through the origin, and hence the second part of the question makes no sense. Are you sure you did not miss some variable in the definition? especially since lines in homogenous coordinates sounds like lines in $mathbb{P}^n$
$endgroup$
– Enkidu
Dec 3 '18 at 14:31
$begingroup$
could you please clarify what you mean by lines? in my opinion those are just 4 points in the projective space (if you wish, lines through the origin). However, if you interpret them as actual lines in $mathbb{k}^n$ we get that all go through the origin, and hence the second part of the question makes no sense. Are you sure you did not miss some variable in the definition? especially since lines in homogenous coordinates sounds like lines in $mathbb{P}^n$
$endgroup$
– Enkidu
Dec 3 '18 at 14:31
$begingroup$
Yes, I am sorry for not specifying. I edited the question. The lines are in $mathbb{P}^3$.
$endgroup$
– ladyeli555
Dec 3 '18 at 14:39
$begingroup$
Yes, I am sorry for not specifying. I edited the question. The lines are in $mathbb{P}^3$.
$endgroup$
– ladyeli555
Dec 3 '18 at 14:39
$begingroup$
they are however still points in $mathbb{P}^3$ not lines in $mathbb{P}^3$.
$endgroup$
– Enkidu
Dec 3 '18 at 14:40
$begingroup$
they are however still points in $mathbb{P}^3$ not lines in $mathbb{P}^3$.
$endgroup$
– Enkidu
Dec 3 '18 at 14:40
$begingroup$
You can have points and lines in $mathbb{P}^3$. If you calculate the connection line between two points you get a line in this form. And in this case, we have lines.
$endgroup$
– ladyeli555
Dec 3 '18 at 14:50
$begingroup$
You can have points and lines in $mathbb{P}^3$. If you calculate the connection line between two points you get a line in this form. And in this case, we have lines.
$endgroup$
– ladyeli555
Dec 3 '18 at 14:50
$begingroup$
so you mean that you want to consider the 6 lines defined by those points? then please make that clear! also, is it now 4 or 6, since 4 points define 6 lines. I am sorry for pestering, but this question is far from clearly stated.
$endgroup$
– Enkidu
Dec 3 '18 at 14:53
$begingroup$
so you mean that you want to consider the 6 lines defined by those points? then please make that clear! also, is it now 4 or 6, since 4 points define 6 lines. I am sorry for pestering, but this question is far from clearly stated.
$endgroup$
– Enkidu
Dec 3 '18 at 14:53
|
show 6 more comments
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$begingroup$
could you please clarify what you mean by lines? in my opinion those are just 4 points in the projective space (if you wish, lines through the origin). However, if you interpret them as actual lines in $mathbb{k}^n$ we get that all go through the origin, and hence the second part of the question makes no sense. Are you sure you did not miss some variable in the definition? especially since lines in homogenous coordinates sounds like lines in $mathbb{P}^n$
$endgroup$
– Enkidu
Dec 3 '18 at 14:31
$begingroup$
Yes, I am sorry for not specifying. I edited the question. The lines are in $mathbb{P}^3$.
$endgroup$
– ladyeli555
Dec 3 '18 at 14:39
$begingroup$
they are however still points in $mathbb{P}^3$ not lines in $mathbb{P}^3$.
$endgroup$
– Enkidu
Dec 3 '18 at 14:40
$begingroup$
You can have points and lines in $mathbb{P}^3$. If you calculate the connection line between two points you get a line in this form. And in this case, we have lines.
$endgroup$
– ladyeli555
Dec 3 '18 at 14:50
$begingroup$
so you mean that you want to consider the 6 lines defined by those points? then please make that clear! also, is it now 4 or 6, since 4 points define 6 lines. I am sorry for pestering, but this question is far from clearly stated.
$endgroup$
– Enkidu
Dec 3 '18 at 14:53