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?






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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















up vote
0
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







share|cite|improve this question

















  • 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













up vote
0
down vote

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up vote
0
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







share|cite|improve this question













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






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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













  • 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











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)






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    up vote
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    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 $x = t, y= z = 0.$






    share|cite|improve this answer






















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      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)






      share|cite|improve this answer
























        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)






        share|cite|improve this answer






















          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)






          share|cite|improve this 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)







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Sep 10 at 10:57









          André Aichert

          513




          513




















              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 $x = t, y= z = 0.$






              share|cite|improve this answer


























                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 $x = t, y= z = 0.$






                share|cite|improve this answer
























                  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 $x = t, y= z = 0.$






                  share|cite|improve this answer














                  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 $x = t, y= z = 0.$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 31 at 20:08









                  amWhy

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                  answered Jan 31 at 19:31









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