Is it possible to define a line with a single polynomial equation?

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Is it possible to come up with a polynomial $F(x,y,z)$ such that the solutions to the equation $F(x,y,z)=0$ are all points that lie on a given line?
geometry polynomials
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Is it possible to come up with a polynomial $F(x,y,z)$ such that the solutions to the equation $F(x,y,z)=0$ are all points that lie on a given line?
geometry polynomials
4
Yes. For example $(a_1x+b_1y+c_1z-d_1)^2+(a_2x+b_2y+c_2z-d_2)^2=0$ gives the intersection of two planes.
– Thomas Andrews
Jan 31 at 19:27
hint: $a_i^T x = b_i$ for all $i$ iff $sum_i (a_i^T x - b_i)^2 = 0$.
– user251257
Jan 31 at 19:28
@ThomasAndrews How could I go about proving that? I had never seen that before.
– sepehr78
Jan 31 at 19:52
2
Hint: If $U,V$ are real numbers, then $U^2+V^2=0$ if and only if $U=0$ and $V=0$.
– Thomas Andrews
Jan 31 at 20:02
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up vote
0
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up vote
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down vote
favorite
Is it possible to come up with a polynomial $F(x,y,z)$ such that the solutions to the equation $F(x,y,z)=0$ are all points that lie on a given line?
geometry polynomials
Is it possible to come up with a polynomial $F(x,y,z)$ such that the solutions to the equation $F(x,y,z)=0$ are all points that lie on a given line?
geometry polynomials
geometry polynomials
asked Jan 31 at 19:25
sepehr78
555
555
4
Yes. For example $(a_1x+b_1y+c_1z-d_1)^2+(a_2x+b_2y+c_2z-d_2)^2=0$ gives the intersection of two planes.
– Thomas Andrews
Jan 31 at 19:27
hint: $a_i^T x = b_i$ for all $i$ iff $sum_i (a_i^T x - b_i)^2 = 0$.
– user251257
Jan 31 at 19:28
@ThomasAndrews How could I go about proving that? I had never seen that before.
– sepehr78
Jan 31 at 19:52
2
Hint: If $U,V$ are real numbers, then $U^2+V^2=0$ if and only if $U=0$ and $V=0$.
– Thomas Andrews
Jan 31 at 20:02
add a comment |Â
4
Yes. For example $(a_1x+b_1y+c_1z-d_1)^2+(a_2x+b_2y+c_2z-d_2)^2=0$ gives the intersection of two planes.
– Thomas Andrews
Jan 31 at 19:27
hint: $a_i^T x = b_i$ for all $i$ iff $sum_i (a_i^T x - b_i)^2 = 0$.
– user251257
Jan 31 at 19:28
@ThomasAndrews How could I go about proving that? I had never seen that before.
– sepehr78
Jan 31 at 19:52
2
Hint: If $U,V$ are real numbers, then $U^2+V^2=0$ if and only if $U=0$ and $V=0$.
– Thomas Andrews
Jan 31 at 20:02
4
4
Yes. For example $(a_1x+b_1y+c_1z-d_1)^2+(a_2x+b_2y+c_2z-d_2)^2=0$ gives the intersection of two planes.
– Thomas Andrews
Jan 31 at 19:27
Yes. For example $(a_1x+b_1y+c_1z-d_1)^2+(a_2x+b_2y+c_2z-d_2)^2=0$ gives the intersection of two planes.
– Thomas Andrews
Jan 31 at 19:27
hint: $a_i^T x = b_i$ for all $i$ iff $sum_i (a_i^T x - b_i)^2 = 0$.
– user251257
Jan 31 at 19:28
hint: $a_i^T x = b_i$ for all $i$ iff $sum_i (a_i^T x - b_i)^2 = 0$.
– user251257
Jan 31 at 19:28
@ThomasAndrews How could I go about proving that? I had never seen that before.
– sepehr78
Jan 31 at 19:52
@ThomasAndrews How could I go about proving that? I had never seen that before.
– sepehr78
Jan 31 at 19:52
2
2
Hint: If $U,V$ are real numbers, then $U^2+V^2=0$ if and only if $U=0$ and $V=0$.
– Thomas Andrews
Jan 31 at 20:02
Hint: If $U,V$ are real numbers, then $U^2+V^2=0$ if and only if $U=0$ and $V=0$.
– Thomas Andrews
Jan 31 at 20:02
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2 Answers
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Yes.
For example $(a_1x+b_1y+c_1z−d_1)^2+(a_2x+b_2y+c_2z−d_2)^2=0$ gives the intersection of two planes.
Proof sketch: If $u,v$ are real numbers, then $u^2+v^2=0$ if and only if $u=0$ and $v=0$
(See Thomas Andrews' comment, which gives the correct answer)
add a comment |Â
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1
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If the solution set is a line, then the distance from any member of the solution set to the line is zero.
In other words, find the equation of a cylinder of radius $r$ centered on the line, then set $r=0$.
So, for instance, the solution set $y^2 + z^2 = 0$ will be the line parameterized by the lines $x = t, y= z = 0.$
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Yes.
For example $(a_1x+b_1y+c_1z−d_1)^2+(a_2x+b_2y+c_2z−d_2)^2=0$ gives the intersection of two planes.
Proof sketch: If $u,v$ are real numbers, then $u^2+v^2=0$ if and only if $u=0$ and $v=0$
(See Thomas Andrews' comment, which gives the correct answer)
add a comment |Â
up vote
1
down vote
accepted
Yes.
For example $(a_1x+b_1y+c_1z−d_1)^2+(a_2x+b_2y+c_2z−d_2)^2=0$ gives the intersection of two planes.
Proof sketch: If $u,v$ are real numbers, then $u^2+v^2=0$ if and only if $u=0$ and $v=0$
(See Thomas Andrews' comment, which gives the correct answer)
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Yes.
For example $(a_1x+b_1y+c_1z−d_1)^2+(a_2x+b_2y+c_2z−d_2)^2=0$ gives the intersection of two planes.
Proof sketch: If $u,v$ are real numbers, then $u^2+v^2=0$ if and only if $u=0$ and $v=0$
(See Thomas Andrews' comment, which gives the correct answer)
Yes.
For example $(a_1x+b_1y+c_1z−d_1)^2+(a_2x+b_2y+c_2z−d_2)^2=0$ gives the intersection of two planes.
Proof sketch: If $u,v$ are real numbers, then $u^2+v^2=0$ if and only if $u=0$ and $v=0$
(See Thomas Andrews' comment, which gives the correct answer)
answered Sep 10 at 10:57
André Aichert
513
513
add a comment |Â
add a comment |Â
up vote
1
down vote
If the solution set is a line, then the distance from any member of the solution set to the line is zero.
In other words, find the equation of a cylinder of radius $r$ centered on the line, then set $r=0$.
So, for instance, the solution set $y^2 + z^2 = 0$ will be the line parameterized by the lines $x = t, y= z = 0.$
add a comment |Â
up vote
1
down vote
If the solution set is a line, then the distance from any member of the solution set to the line is zero.
In other words, find the equation of a cylinder of radius $r$ centered on the line, then set $r=0$.
So, for instance, the solution set $y^2 + z^2 = 0$ will be the line parameterized by the lines $x = t, y= z = 0.$
add a comment |Â
up vote
1
down vote
up vote
1
down vote
If the solution set is a line, then the distance from any member of the solution set to the line is zero.
In other words, find the equation of a cylinder of radius $r$ centered on the line, then set $r=0$.
So, for instance, the solution set $y^2 + z^2 = 0$ will be the line parameterized by the lines $x = t, y= z = 0.$
If the solution set is a line, then the distance from any member of the solution set to the line is zero.
In other words, find the equation of a cylinder of radius $r$ centered on the line, then set $r=0$.
So, for instance, the solution set $y^2 + z^2 = 0$ will be the line parameterized by the lines $x = t, y= z = 0.$
edited Jan 31 at 20:08


amWhy
190k27221433
190k27221433
answered Jan 31 at 19:31
Acccumulation
5,8642616
5,8642616
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4
Yes. For example $(a_1x+b_1y+c_1z-d_1)^2+(a_2x+b_2y+c_2z-d_2)^2=0$ gives the intersection of two planes.
– Thomas Andrews
Jan 31 at 19:27
hint: $a_i^T x = b_i$ for all $i$ iff $sum_i (a_i^T x - b_i)^2 = 0$.
– user251257
Jan 31 at 19:28
@ThomasAndrews How could I go about proving that? I had never seen that before.
– sepehr78
Jan 31 at 19:52
2
Hint: If $U,V$ are real numbers, then $U^2+V^2=0$ if and only if $U=0$ and $V=0$.
– Thomas Andrews
Jan 31 at 20:02