Proving if two planes are coincident
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
asked Aug 9 '15 at 11:55
stuartstuart
1916
add a comment |
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
asked Aug 9 '15 at 11:55
stuartstuart
1916
They are not as both have different set of solutions
– Bhaskar Vashishth
Aug 9 '15 at 11:58
Two planes are coincident iff they have infinite solutions.
– Harshal Gajjar
Aug 9 '15 at 12:26
@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.
– Mark Bennet
Aug 9 '15 at 12:51
@MarkBennet Oh sorry, to correct: two planes are coincident iff the have infinite solutions with varying x,y,z. Pls correct if wrong. :)
– Harshal Gajjar
Aug 9 '15 at 12:53
Also, @MarkBennet For only two planes: will only the ratio(s) of constants and coefficients decide whether the planes are coincident?
– Harshal Gajjar
Aug 9 '15 at 12:56
add a comment |
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
asked Aug 9 '15 at 11:55
stuartstuart
1916
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
vector-spaces
asked Aug 9 '15 at 11:55
stuartstuart
1916
asked Aug 9 '15 at 11:55
stuartstuart
1916
asked Aug 9 '15 at 11:55
stuartstuart
1916
asked Aug 9 '15 at 11:55
stuartstuart
1916
asked Aug 9 '15 at 11:55
stuartstuart
1916
1916
They are not as both have different set of solutions
– Bhaskar Vashishth
Aug 9 '15 at 11:58
Two planes are coincident iff they have infinite solutions.
– Harshal Gajjar
Aug 9 '15 at 12:26
@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.
– Mark Bennet
Aug 9 '15 at 12:51
@MarkBennet Oh sorry, to correct: two planes are coincident iff the have infinite solutions with varying x,y,z. Pls correct if wrong. :)
– Harshal Gajjar
Aug 9 '15 at 12:53
Also, @MarkBennet For only two planes: will only the ratio(s) of constants and coefficients decide whether the planes are coincident?
– Harshal Gajjar
Aug 9 '15 at 12:56
add a comment |
They are not as both have different set of solutions
– Bhaskar Vashishth
Aug 9 '15 at 11:58
Two planes are coincident iff they have infinite solutions.
– Harshal Gajjar
Aug 9 '15 at 12:26
@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.
– Mark Bennet
Aug 9 '15 at 12:51
@MarkBennet Oh sorry, to correct: two planes are coincident iff the have infinite solutions with varying x,y,z. Pls correct if wrong. :)
– Harshal Gajjar
Aug 9 '15 at 12:53
Also, @MarkBennet For only two planes: will only the ratio(s) of constants and coefficients decide whether the planes are coincident?
– Harshal Gajjar
Aug 9 '15 at 12:56
They are not as both have different set of solutions
– Bhaskar Vashishth
Aug 9 '15 at 11:58
They are not as both have different set of solutions
– Bhaskar Vashishth
Aug 9 '15 at 11:58
Two planes are coincident iff they have infinite solutions.
– Harshal Gajjar
Aug 9 '15 at 12:26
Two planes are coincident iff they have infinite solutions.
– Harshal Gajjar
Aug 9 '15 at 12:26
@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.
– Mark Bennet
Aug 9 '15 at 12:51
@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.
– Mark Bennet
Aug 9 '15 at 12:51
@MarkBennet Oh sorry, to correct: two planes are coincident iff the have infinite solutions with varying x,y,z. Pls correct if wrong. :)
– Harshal Gajjar
Aug 9 '15 at 12:53
@MarkBennet Oh sorry, to correct: two planes are coincident iff the have infinite solutions with varying x,y,z. Pls correct if wrong. :)
– Harshal Gajjar
Aug 9 '15 at 12:53
Also, @MarkBennet For only two planes: will only the ratio(s) of constants and coefficients decide whether the planes are coincident?
– Harshal Gajjar
Aug 9 '15 at 12:56
Also, @MarkBennet For only two planes: will only the ratio(s) of constants and coefficients decide whether the planes are coincident?
– Harshal Gajjar
Aug 9 '15 at 12:56
add a comment |
4 Answers
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Notice that $(0,0,3)$ belongs to the second plane, but not the first, as $0+2times 0-5 times 3=-15$
answered Aug 9 '15 at 11:58
mich95mich95
6,88011126
add a comment |
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.
answered Dec 1 '18 at 5:42
Atharva KathaleAtharva Kathale
889
add a comment |
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 $(1,2,-5) times (2,-3,1)$
answered Dec 1 '18 at 6:26
rowcolrowcol
734
add a comment |
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).
answered Jan 4 at 3:50
Chris CusterChris Custer
11.1k3824
add a comment |
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4 Answers
4
active
oldest
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4 Answers
4
active
oldest
votes
active
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active
oldest
votes
Notice that $(0,0,3)$ belongs to the second plane, but not the first, as $0+2times 0-5 times 3=-15$
answered Aug 9 '15 at 11:58
mich95mich95
6,88011126
add a comment |
Notice that $(0,0,3)$ belongs to the second plane, but not the first, as $0+2times 0-5 times 3=-15$
answered Aug 9 '15 at 11:58
mich95mich95
6,88011126
add a comment |
Notice that $(0,0,3)$ belongs to the second plane, but not the first, as $0+2times 0-5 times 3=-15$
answered Aug 9 '15 at 11:58
mich95mich95
6,88011126
Notice that $(0,0,3)$ belongs to the second plane, but not the first, as $0+2times 0-5 times 3=-15$
answered Aug 9 '15 at 11:58
mich95mich95
6,88011126
answered Aug 9 '15 at 11:58
mich95mich95
6,88011126
answered Aug 9 '15 at 11:58
mich95mich95
6,88011126
answered Aug 9 '15 at 11:58
mich95mich95
6,88011126
6,88011126
add a comment |
add a comment |
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.
answered Dec 1 '18 at 5:42
Atharva KathaleAtharva Kathale
889
add a comment |
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.
answered Dec 1 '18 at 5:42
Atharva KathaleAtharva Kathale
889
add a comment |
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.
answered Dec 1 '18 at 5:42
Atharva KathaleAtharva Kathale
889
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.
answered Dec 1 '18 at 5:42
Atharva KathaleAtharva Kathale
889
answered Dec 1 '18 at 5:42
Atharva KathaleAtharva Kathale
889
answered Dec 1 '18 at 5:42
Atharva KathaleAtharva Kathale
889
answered Dec 1 '18 at 5:42
Atharva KathaleAtharva Kathale
889
889
add a comment |
add a comment |
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 $(1,2,-5) times (2,-3,1)$
answered Dec 1 '18 at 6:26
rowcolrowcol
734
add a comment |
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 $(1,2,-5) times (2,-3,1)$
answered Dec 1 '18 at 6:26
rowcolrowcol
734
add a comment |
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 $(1,2,-5) times (2,-3,1)$
answered Dec 1 '18 at 6:26
rowcolrowcol
734
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 $(1,2,-5) times (2,-3,1)$
answered Dec 1 '18 at 6:26
rowcolrowcol
734
answered Dec 1 '18 at 6:26
rowcolrowcol
734
answered Dec 1 '18 at 6:26
rowcolrowcol
734
answered Dec 1 '18 at 6:26
rowcolrowcol
734
734
add a comment |
add a comment |
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).
answered Jan 4 at 3:50
Chris CusterChris Custer
11.1k3824
add a comment |
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).
answered Jan 4 at 3:50
Chris CusterChris Custer
11.1k3824
add a comment |
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).
answered Jan 4 at 3:50
Chris CusterChris Custer
11.1k3824
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).
answered Jan 4 at 3:50
Chris CusterChris Custer
11.1k3824
edited Jan 4 at 5:39
answered Jan 4 at 3:50
Chris CusterChris Custer
11.1k3824
answered Jan 4 at 3:50
Chris CusterChris Custer
11.1k3824
answered Jan 4 at 3:50
Chris CusterChris Custer
11.1k3824
11.1k3824
add a comment |
add a comment |
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They are not as both have different set of solutions
– Bhaskar Vashishth
Aug 9 '15 at 11:58
Two planes are coincident iff they have infinite solutions.
– Harshal Gajjar
Aug 9 '15 at 12:26
@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.
– Mark Bennet
Aug 9 '15 at 12:51
@MarkBennet Oh sorry, to correct: two planes are coincident iff the have infinite solutions with varying x,y,z. Pls correct if wrong. :)
– Harshal Gajjar
Aug 9 '15 at 12:53
Also, @MarkBennet For only two planes: will only the ratio(s) of constants and coefficients decide whether the planes are coincident?
– Harshal Gajjar
Aug 9 '15 at 12:56