Proving if two planes are coincident












0












$begingroup$


How would I prove if the two planes below are coincident or not?



$x+2y-5z=1$



$2x-3y+z=3$



Do I need to prove that one equation can equal the other?










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$endgroup$












  • $begingroup$
    They are not as both have different set of solutions
    $endgroup$
    – Bhaskar Vashishth
    Aug 9 '15 at 11:58










  • $begingroup$
    Two planes are coincident iff they have infinite solutions.
    $endgroup$
    – Harshal Gajjar
    Aug 9 '15 at 12:26










  • $begingroup$
    @HarshalGajjar Two planes can meet in a line, which has an infinite number of points. In most of the spaces we are interested in it is impossible for two planes to meet in a non-zero finite number of points.
    $endgroup$
    – Mark Bennet
    Aug 9 '15 at 12:51










  • $begingroup$
    @MarkBennet Oh sorry, to correct: two planes are coincident iff the have infinite solutions with varying x,y,z. Pls correct if wrong. :)
    $endgroup$
    – Harshal Gajjar
    Aug 9 '15 at 12:53










  • $begingroup$
    Also, @MarkBennet For only two planes: will only the ratio(s) of constants and coefficients decide whether the planes are coincident?
    $endgroup$
    – Harshal Gajjar
    Aug 9 '15 at 12:56
















0












$begingroup$


How would I prove if the two planes below are coincident or not?



$x+2y-5z=1$



$2x-3y+z=3$



Do I need to prove that one equation can equal the other?










share|cite|improve this question









$endgroup$












  • $begingroup$
    They are not as both have different set of solutions
    $endgroup$
    – Bhaskar Vashishth
    Aug 9 '15 at 11:58










  • $begingroup$
    Two planes are coincident iff they have infinite solutions.
    $endgroup$
    – Harshal Gajjar
    Aug 9 '15 at 12:26










  • $begingroup$
    @HarshalGajjar Two planes can meet in a line, which has an infinite number of points. In most of the spaces we are interested in it is impossible for two planes to meet in a non-zero finite number of points.
    $endgroup$
    – Mark Bennet
    Aug 9 '15 at 12:51










  • $begingroup$
    @MarkBennet Oh sorry, to correct: two planes are coincident iff the have infinite solutions with varying x,y,z. Pls correct if wrong. :)
    $endgroup$
    – Harshal Gajjar
    Aug 9 '15 at 12:53










  • $begingroup$
    Also, @MarkBennet For only two planes: will only the ratio(s) of constants and coefficients decide whether the planes are coincident?
    $endgroup$
    – Harshal Gajjar
    Aug 9 '15 at 12:56














0












0








0





$begingroup$


How would I prove if the two planes below are coincident or not?



$x+2y-5z=1$



$2x-3y+z=3$



Do I need to prove that one equation can equal the other?










share|cite|improve this question









$endgroup$




How would I prove if the two planes below are coincident or not?



$x+2y-5z=1$



$2x-3y+z=3$



Do I need to prove that one equation can equal the other?







vector-spaces






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Aug 9 '15 at 11:55









stuartstuart

1916




1916












  • $begingroup$
    They are not as both have different set of solutions
    $endgroup$
    – Bhaskar Vashishth
    Aug 9 '15 at 11:58










  • $begingroup$
    Two planes are coincident iff they have infinite solutions.
    $endgroup$
    – Harshal Gajjar
    Aug 9 '15 at 12:26










  • $begingroup$
    @HarshalGajjar Two planes can meet in a line, which has an infinite number of points. In most of the spaces we are interested in it is impossible for two planes to meet in a non-zero finite number of points.
    $endgroup$
    – Mark Bennet
    Aug 9 '15 at 12:51










  • $begingroup$
    @MarkBennet Oh sorry, to correct: two planes are coincident iff the have infinite solutions with varying x,y,z. Pls correct if wrong. :)
    $endgroup$
    – Harshal Gajjar
    Aug 9 '15 at 12:53










  • $begingroup$
    Also, @MarkBennet For only two planes: will only the ratio(s) of constants and coefficients decide whether the planes are coincident?
    $endgroup$
    – Harshal Gajjar
    Aug 9 '15 at 12:56


















  • $begingroup$
    They are not as both have different set of solutions
    $endgroup$
    – Bhaskar Vashishth
    Aug 9 '15 at 11:58










  • $begingroup$
    Two planes are coincident iff they have infinite solutions.
    $endgroup$
    – Harshal Gajjar
    Aug 9 '15 at 12:26










  • $begingroup$
    @HarshalGajjar Two planes can meet in a line, which has an infinite number of points. In most of the spaces we are interested in it is impossible for two planes to meet in a non-zero finite number of points.
    $endgroup$
    – Mark Bennet
    Aug 9 '15 at 12:51










  • $begingroup$
    @MarkBennet Oh sorry, to correct: two planes are coincident iff the have infinite solutions with varying x,y,z. Pls correct if wrong. :)
    $endgroup$
    – Harshal Gajjar
    Aug 9 '15 at 12:53










  • $begingroup$
    Also, @MarkBennet For only two planes: will only the ratio(s) of constants and coefficients decide whether the planes are coincident?
    $endgroup$
    – Harshal Gajjar
    Aug 9 '15 at 12:56
















$begingroup$
They are not as both have different set of solutions
$endgroup$
– Bhaskar Vashishth
Aug 9 '15 at 11:58




$begingroup$
They are not as both have different set of solutions
$endgroup$
– Bhaskar Vashishth
Aug 9 '15 at 11:58












$begingroup$
Two planes are coincident iff they have infinite solutions.
$endgroup$
– Harshal Gajjar
Aug 9 '15 at 12:26




$begingroup$
Two planes are coincident iff they have infinite solutions.
$endgroup$
– Harshal Gajjar
Aug 9 '15 at 12:26












$begingroup$
@HarshalGajjar Two planes can meet in a line, which has an infinite number of points. In most of the spaces we are interested in it is impossible for two planes to meet in a non-zero finite number of points.
$endgroup$
– Mark Bennet
Aug 9 '15 at 12:51




$begingroup$
@HarshalGajjar Two planes can meet in a line, which has an infinite number of points. In most of the spaces we are interested in it is impossible for two planes to meet in a non-zero finite number of points.
$endgroup$
– Mark Bennet
Aug 9 '15 at 12:51












$begingroup$
@MarkBennet Oh sorry, to correct: two planes are coincident iff the have infinite solutions with varying x,y,z. Pls correct if wrong. :)
$endgroup$
– Harshal Gajjar
Aug 9 '15 at 12:53




$begingroup$
@MarkBennet Oh sorry, to correct: two planes are coincident iff the have infinite solutions with varying x,y,z. Pls correct if wrong. :)
$endgroup$
– Harshal Gajjar
Aug 9 '15 at 12:53












$begingroup$
Also, @MarkBennet For only two planes: will only the ratio(s) of constants and coefficients decide whether the planes are coincident?
$endgroup$
– Harshal Gajjar
Aug 9 '15 at 12:56




$begingroup$
Also, @MarkBennet For only two planes: will only the ratio(s) of constants and coefficients decide whether the planes are coincident?
$endgroup$
– Harshal Gajjar
Aug 9 '15 at 12:56










4 Answers
4






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oldest

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0












$begingroup$

Notice that $(0,0,3)$ belongs to the second plane, but not the first, as $0+2times 0-5 times 3=-15$






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$endgroup$





















    0












    $begingroup$

    By definition of coincident planes: The ratio of coefficient of respective variable$(x,y,z)$ must be all equal.Also this ratio should be equal to ratio of constant terms. In short, one of the equation of plane should be equal to some constant times the other equation .
    In your question the 2 planes are not coincident.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      Let $P_{1}: x+2y-5z=1$ and



      $P_{2}: 2x-3y+z=3$



      You can rewrite the plane equations in their vectorial form, i.e.



      $P_{1}: (1,2,-5) cdot (x-1,y,z)=0$;



      $P_{2}: (2,-3,1) cdot (x,y,z-3)=0$



      Note that $(1,2,-5)$ and $(2,-3,1)$ are linearly independent, so the planes will not be parallel, then there's an intersection between them, given by the line wich its direction vector is $(1,2,-5) times (2,-3,1)$






      share|cite|improve this answer









      $endgroup$





















        0












        $begingroup$

        The normals are $(1,2,-5)$ and $(2,-3,1)$. These aren't parallel, so neither are the planes and they meet in a line.



        If by coincident you meant the same plane, then you just need to check if one equation is a multiple of the other. The answer is no. (Of course if they were, then the normals would be multiples of each other as well).






        share|cite|improve this answer











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          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          0












          $begingroup$

          Notice that $(0,0,3)$ belongs to the second plane, but not the first, as $0+2times 0-5 times 3=-15$






          share|cite|improve this answer









          $endgroup$


















            0












            $begingroup$

            Notice that $(0,0,3)$ belongs to the second plane, but not the first, as $0+2times 0-5 times 3=-15$






            share|cite|improve this answer









            $endgroup$
















              0












              0








              0





              $begingroup$

              Notice that $(0,0,3)$ belongs to the second plane, but not the first, as $0+2times 0-5 times 3=-15$






              share|cite|improve this answer









              $endgroup$



              Notice that $(0,0,3)$ belongs to the second plane, but not the first, as $0+2times 0-5 times 3=-15$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Aug 9 '15 at 11:58









              mich95mich95

              6,88011126




              6,88011126























                  0












                  $begingroup$

                  By definition of coincident planes: The ratio of coefficient of respective variable$(x,y,z)$ must be all equal.Also this ratio should be equal to ratio of constant terms. In short, one of the equation of plane should be equal to some constant times the other equation .
                  In your question the 2 planes are not coincident.






                  share|cite|improve this answer









                  $endgroup$


















                    0












                    $begingroup$

                    By definition of coincident planes: The ratio of coefficient of respective variable$(x,y,z)$ must be all equal.Also this ratio should be equal to ratio of constant terms. In short, one of the equation of plane should be equal to some constant times the other equation .
                    In your question the 2 planes are not coincident.






                    share|cite|improve this answer









                    $endgroup$
















                      0












                      0








                      0





                      $begingroup$

                      By definition of coincident planes: The ratio of coefficient of respective variable$(x,y,z)$ must be all equal.Also this ratio should be equal to ratio of constant terms. In short, one of the equation of plane should be equal to some constant times the other equation .
                      In your question the 2 planes are not coincident.






                      share|cite|improve this answer









                      $endgroup$



                      By definition of coincident planes: The ratio of coefficient of respective variable$(x,y,z)$ must be all equal.Also this ratio should be equal to ratio of constant terms. In short, one of the equation of plane should be equal to some constant times the other equation .
                      In your question the 2 planes are not coincident.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Dec 1 '18 at 5:42









                      Atharva KathaleAtharva Kathale

                      889




                      889























                          0












                          $begingroup$

                          Let $P_{1}: x+2y-5z=1$ and



                          $P_{2}: 2x-3y+z=3$



                          You can rewrite the plane equations in their vectorial form, i.e.



                          $P_{1}: (1,2,-5) cdot (x-1,y,z)=0$;



                          $P_{2}: (2,-3,1) cdot (x,y,z-3)=0$



                          Note that $(1,2,-5)$ and $(2,-3,1)$ are linearly independent, so the planes will not be parallel, then there's an intersection between them, given by the line wich its direction vector is $(1,2,-5) times (2,-3,1)$






                          share|cite|improve this answer









                          $endgroup$


















                            0












                            $begingroup$

                            Let $P_{1}: x+2y-5z=1$ and



                            $P_{2}: 2x-3y+z=3$



                            You can rewrite the plane equations in their vectorial form, i.e.



                            $P_{1}: (1,2,-5) cdot (x-1,y,z)=0$;



                            $P_{2}: (2,-3,1) cdot (x,y,z-3)=0$



                            Note that $(1,2,-5)$ and $(2,-3,1)$ are linearly independent, so the planes will not be parallel, then there's an intersection between them, given by the line wich its direction vector is $(1,2,-5) times (2,-3,1)$






                            share|cite|improve this answer









                            $endgroup$
















                              0












                              0








                              0





                              $begingroup$

                              Let $P_{1}: x+2y-5z=1$ and



                              $P_{2}: 2x-3y+z=3$



                              You can rewrite the plane equations in their vectorial form, i.e.



                              $P_{1}: (1,2,-5) cdot (x-1,y,z)=0$;



                              $P_{2}: (2,-3,1) cdot (x,y,z-3)=0$



                              Note that $(1,2,-5)$ and $(2,-3,1)$ are linearly independent, so the planes will not be parallel, then there's an intersection between them, given by the line wich its direction vector is $(1,2,-5) times (2,-3,1)$






                              share|cite|improve this answer









                              $endgroup$



                              Let $P_{1}: x+2y-5z=1$ and



                              $P_{2}: 2x-3y+z=3$



                              You can rewrite the plane equations in their vectorial form, i.e.



                              $P_{1}: (1,2,-5) cdot (x-1,y,z)=0$;



                              $P_{2}: (2,-3,1) cdot (x,y,z-3)=0$



                              Note that $(1,2,-5)$ and $(2,-3,1)$ are linearly independent, so the planes will not be parallel, then there's an intersection between them, given by the line wich its direction vector is $(1,2,-5) times (2,-3,1)$







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Dec 1 '18 at 6:26









                              rowcolrowcol

                              734




                              734























                                  0












                                  $begingroup$

                                  The normals are $(1,2,-5)$ and $(2,-3,1)$. These aren't parallel, so neither are the planes and they meet in a line.



                                  If by coincident you meant the same plane, then you just need to check if one equation is a multiple of the other. The answer is no. (Of course if they were, then the normals would be multiples of each other as well).






                                  share|cite|improve this answer











                                  $endgroup$


















                                    0












                                    $begingroup$

                                    The normals are $(1,2,-5)$ and $(2,-3,1)$. These aren't parallel, so neither are the planes and they meet in a line.



                                    If by coincident you meant the same plane, then you just need to check if one equation is a multiple of the other. The answer is no. (Of course if they were, then the normals would be multiples of each other as well).






                                    share|cite|improve this answer











                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      The normals are $(1,2,-5)$ and $(2,-3,1)$. These aren't parallel, so neither are the planes and they meet in a line.



                                      If by coincident you meant the same plane, then you just need to check if one equation is a multiple of the other. The answer is no. (Of course if they were, then the normals would be multiples of each other as well).






                                      share|cite|improve this answer











                                      $endgroup$



                                      The normals are $(1,2,-5)$ and $(2,-3,1)$. These aren't parallel, so neither are the planes and they meet in a line.



                                      If by coincident you meant the same plane, then you just need to check if one equation is a multiple of the other. The answer is no. (Of course if they were, then the normals would be multiples of each other as well).







                                      share|cite|improve this answer














                                      share|cite|improve this answer



                                      share|cite|improve this answer








                                      edited Jan 4 at 5:39

























                                      answered Jan 4 at 3:50









                                      Chris CusterChris Custer

                                      11.3k3824




                                      11.3k3824






























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