How can $2x -5y +z = 3$ be equation of a line? The 2019 Stack Overflow Developer Survey Results Are InEuclid / Hilbert: “Two lines parallel to a third line are parallel to each other.”model for intersection of two circles in the complex projective planeIf a plane contains one line and intersects another one elsewhere, then the two lines are not coplanarequation of plane containing the line of intersection between two planes and a pointExercise: Stereometry > Lines and planes > Parallel lines and planesEquation of a $ 3$rd plane - two points and parallel to the line of intersection of two planesEquation of a plane passing through intersection of two planes and parallel to a given line.Intuition on the skew Line distancesHow can affine plane extended of projective plane?Somehow, I'm getting the equation of a plane when setting equal the equations of two other, non-parallel planes
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How can $2x -5y +z = 3$ be equation of a line?
The 2019 Stack Overflow Developer Survey Results Are InEuclid / Hilbert: “Two lines parallel to a third line are parallel to each other.”model for intersection of two circles in the complex projective planeIf a plane contains one line and intersects another one elsewhere, then the two lines are not coplanarequation of plane containing the line of intersection between two planes and a pointExercise: Stereometry > Lines and planes > Parallel lines and planesEquation of a $ 3$rd plane - two points and parallel to the line of intersection of two planesEquation of a plane passing through intersection of two planes and parallel to a given line.Intuition on the skew Line distancesHow can affine plane extended of projective plane?Somehow, I'm getting the equation of a plane when setting equal the equations of two other, non-parallel planes
$begingroup$
The equation of the plane containing the lines $2x - 5y + z = 3$, $x + y + 4z = 5$ and parallel to the plane $x +3y + 6z = 7$? I actually have the solution to this question, but I don't understand it. I am a bit new to 3d geometry. Isn't equation of a plane always of form $ax+by+cz=d$? Aren't the two lines mentioned in the question actually planes? The solution I have uses the two equations for lines as equations for planes in family of planes (i.e., $P_1 +kP_2=0$).
Basically my doubt is: why are $boldsymbol2x - 5y + z = 3$ and $boldsymbolx + y + 4z = 5$ called lines when they are obviously equations of planes? Is there something I'm missing here?
geometry plane-geometry
edited Mar 24 at 6:47
Rócherz
3,0263823
asked Mar 24 at 6:17
HemaHema
6621213
$endgroup$
add a comment |
$begingroup$
The equation of the plane containing the lines $2x - 5y + z = 3$, $x + y + 4z = 5$ and parallel to the plane $x +3y + 6z = 7$? I actually have the solution to this question, but I don't understand it. I am a bit new to 3d geometry. Isn't equation of a plane always of form $ax+by+cz=d$? Aren't the two lines mentioned in the question actually planes? The solution I have uses the two equations for lines as equations for planes in family of planes (i.e., $P_1 +kP_2=0$).
Basically my doubt is: why are $boldsymbol2x - 5y + z = 3$ and $boldsymbolx + y + 4z = 5$ called lines when they are obviously equations of planes? Is there something I'm missing here?
geometry plane-geometry
edited Mar 24 at 6:47
Rócherz
3,0263823
asked Mar 24 at 6:17
HemaHema
6621213
$endgroup$
2
$begingroup$
Should the word "lines" in your first sentence simply be "line"?
$endgroup$
– Lord Shark the Unknown
Mar 24 at 6:21
add a comment |
$begingroup$
The equation of the plane containing the lines $2x - 5y + z = 3$, $x + y + 4z = 5$ and parallel to the plane $x +3y + 6z = 7$? I actually have the solution to this question, but I don't understand it. I am a bit new to 3d geometry. Isn't equation of a plane always of form $ax+by+cz=d$? Aren't the two lines mentioned in the question actually planes? The solution I have uses the two equations for lines as equations for planes in family of planes (i.e., $P_1 +kP_2=0$).
Basically my doubt is: why are $boldsymbol2x - 5y + z = 3$ and $boldsymbolx + y + 4z = 5$ called lines when they are obviously equations of planes? Is there something I'm missing here?
geometry plane-geometry
edited Mar 24 at 6:47
Rócherz
3,0263823
asked Mar 24 at 6:17
HemaHema
6621213
$endgroup$
The equation of the plane containing the lines $2x - 5y + z = 3$, $x + y + 4z = 5$ and parallel to the plane $x +3y + 6z = 7$? I actually have the solution to this question, but I don't understand it. I am a bit new to 3d geometry. Isn't equation of a plane always of form $ax+by+cz=d$? Aren't the two lines mentioned in the question actually planes? The solution I have uses the two equations for lines as equations for planes in family of planes (i.e., $P_1 +kP_2=0$).
Basically my doubt is: why are $boldsymbol2x - 5y + z = 3$ and $boldsymbolx + y + 4z = 5$ called lines when they are obviously equations of planes? Is there something I'm missing here?
geometry plane-geometry
geometry plane-geometry
edited Mar 24 at 6:47
Rócherz
3,0263823
asked Mar 24 at 6:17
HemaHema
6621213
edited Mar 24 at 6:47
Rócherz
3,0263823
asked Mar 24 at 6:17
HemaHema
6621213
edited Mar 24 at 6:47
Rócherz
3,0263823
edited Mar 24 at 6:47
Rócherz
3,0263823
edited Mar 24 at 6:47
Rócherz
3,0263823
3,0263823
asked Mar 24 at 6:17
HemaHema
6621213
asked Mar 24 at 6:17
HemaHema
6621213
asked Mar 24 at 6:17
HemaHema
6621213
6621213
2
$begingroup$
Should the word "lines" in your first sentence simply be "line"?
$endgroup$
– Lord Shark the Unknown
Mar 24 at 6:21
add a comment |
2
$begingroup$
Should the word "lines" in your first sentence simply be "line"?
$endgroup$
– Lord Shark the Unknown
Mar 24 at 6:21
2
2
$begingroup$
Should the word "lines" in your first sentence simply be "line"?
$endgroup$
– Lord Shark the Unknown
Mar 24 at 6:21
$begingroup$
Should the word "lines" in your first sentence simply be "line"?
$endgroup$
– Lord Shark the Unknown
Mar 24 at 6:21
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Actually they are referring to the line formed by intersection of the two planes$ 2x−5y+z=3$and $x+y+4z=5$. There is mistake in understanding the question
answered Mar 24 at 6:50
TojrahTojrah
4036
$endgroup$
add a comment |
$begingroup$
They should not be called lines, but planes. You are absolutely right. I am pretty sure that they made a mistake.
answered Mar 24 at 6:20
BadAtGeometryBadAtGeometry
306215
$endgroup$
add a comment |
$begingroup$
The only way I can see of making sense of this question is to read it as:
"[Find the] equation of the plane containing the line of intersection of the planes $ 2x - 5y + z = 3 $ and $ x + y + 4z = 5 $, and parallel to the plane $ x +3y + 6z = 7 $."
Since the line of intersection of the first two planes happens to be parallel to the third, this problem has a unique solution. It doesn't help to merely replace the word "lines" with "planes" in the original statement, because the expression "the equation of the plane containing the planes $ 2x - 5y + z = 3 $ and $ x + y + 4z = 5 $, ..." doesn't make any sense. No such plane exists.
answered Mar 24 at 6:50
lonza leggieralonza leggiera
1,42928
$endgroup$
add a comment |
$begingroup$
Maybe the start of the question said the line (singular, not plural) determined by $2x - 5y + z = 3, x + y + 4z = 5$, which is correct because it denotes the intersection (you want both equations to hold simultaneously) of two non-parallel planes in $mathbbR^3$, so these determine a unique line.
You are right that one equation $ax+by+cz+d$ determines a plane in $mathbbR^d$ but two of these equations that hold simultaneously typically determine a line (or the set is empty, as in two parallel planes, or a plane when the two equations denote the same plane). This is only true in $mathbbR^3$..
answered Mar 24 at 6:48
Henno BrandsmaHenno Brandsma
116k349127
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Actually they are referring to the line formed by intersection of the two planes$ 2x−5y+z=3$and $x+y+4z=5$. There is mistake in understanding the question
answered Mar 24 at 6:50
TojrahTojrah
4036
$endgroup$
add a comment |
$begingroup$
Actually they are referring to the line formed by intersection of the two planes$ 2x−5y+z=3$and $x+y+4z=5$. There is mistake in understanding the question
answered Mar 24 at 6:50
TojrahTojrah
4036
$endgroup$
add a comment |
$begingroup$
Actually they are referring to the line formed by intersection of the two planes$ 2x−5y+z=3$and $x+y+4z=5$. There is mistake in understanding the question
answered Mar 24 at 6:50
TojrahTojrah
4036
$endgroup$
Actually they are referring to the line formed by intersection of the two planes$ 2x−5y+z=3$and $x+y+4z=5$. There is mistake in understanding the question
answered Mar 24 at 6:50
TojrahTojrah
4036
answered Mar 24 at 6:50
TojrahTojrah
4036
answered Mar 24 at 6:50
TojrahTojrah
4036
answered Mar 24 at 6:50
TojrahTojrah
4036
4036
add a comment |
add a comment |
$begingroup$
They should not be called lines, but planes. You are absolutely right. I am pretty sure that they made a mistake.
answered Mar 24 at 6:20
BadAtGeometryBadAtGeometry
306215
$endgroup$
add a comment |
$begingroup$
They should not be called lines, but planes. You are absolutely right. I am pretty sure that they made a mistake.
answered Mar 24 at 6:20
BadAtGeometryBadAtGeometry
306215
$endgroup$
add a comment |
$begingroup$
They should not be called lines, but planes. You are absolutely right. I am pretty sure that they made a mistake.
answered Mar 24 at 6:20
BadAtGeometryBadAtGeometry
306215
$endgroup$
They should not be called lines, but planes. You are absolutely right. I am pretty sure that they made a mistake.
answered Mar 24 at 6:20
BadAtGeometryBadAtGeometry
306215
answered Mar 24 at 6:20
BadAtGeometryBadAtGeometry
306215
answered Mar 24 at 6:20
BadAtGeometryBadAtGeometry
306215
answered Mar 24 at 6:20
BadAtGeometryBadAtGeometry
306215
306215
add a comment |
add a comment |
$begingroup$
The only way I can see of making sense of this question is to read it as:
"[Find the] equation of the plane containing the line of intersection of the planes $ 2x - 5y + z = 3 $ and $ x + y + 4z = 5 $, and parallel to the plane $ x +3y + 6z = 7 $."
Since the line of intersection of the first two planes happens to be parallel to the third, this problem has a unique solution. It doesn't help to merely replace the word "lines" with "planes" in the original statement, because the expression "the equation of the plane containing the planes $ 2x - 5y + z = 3 $ and $ x + y + 4z = 5 $, ..." doesn't make any sense. No such plane exists.
answered Mar 24 at 6:50
lonza leggieralonza leggiera
1,42928
$endgroup$
add a comment |
$begingroup$
The only way I can see of making sense of this question is to read it as:
"[Find the] equation of the plane containing the line of intersection of the planes $ 2x - 5y + z = 3 $ and $ x + y + 4z = 5 $, and parallel to the plane $ x +3y + 6z = 7 $."
Since the line of intersection of the first two planes happens to be parallel to the third, this problem has a unique solution. It doesn't help to merely replace the word "lines" with "planes" in the original statement, because the expression "the equation of the plane containing the planes $ 2x - 5y + z = 3 $ and $ x + y + 4z = 5 $, ..." doesn't make any sense. No such plane exists.
answered Mar 24 at 6:50
lonza leggieralonza leggiera
1,42928
$endgroup$
add a comment |
$begingroup$
The only way I can see of making sense of this question is to read it as:
"[Find the] equation of the plane containing the line of intersection of the planes $ 2x - 5y + z = 3 $ and $ x + y + 4z = 5 $, and parallel to the plane $ x +3y + 6z = 7 $."
Since the line of intersection of the first two planes happens to be parallel to the third, this problem has a unique solution. It doesn't help to merely replace the word "lines" with "planes" in the original statement, because the expression "the equation of the plane containing the planes $ 2x - 5y + z = 3 $ and $ x + y + 4z = 5 $, ..." doesn't make any sense. No such plane exists.
answered Mar 24 at 6:50
lonza leggieralonza leggiera
1,42928
$endgroup$
The only way I can see of making sense of this question is to read it as:
"[Find the] equation of the plane containing the line of intersection of the planes $ 2x - 5y + z = 3 $ and $ x + y + 4z = 5 $, and parallel to the plane $ x +3y + 6z = 7 $."
Since the line of intersection of the first two planes happens to be parallel to the third, this problem has a unique solution. It doesn't help to merely replace the word "lines" with "planes" in the original statement, because the expression "the equation of the plane containing the planes $ 2x - 5y + z = 3 $ and $ x + y + 4z = 5 $, ..." doesn't make any sense. No such plane exists.
answered Mar 24 at 6:50
lonza leggieralonza leggiera
1,42928
answered Mar 24 at 6:50
lonza leggieralonza leggiera
1,42928
answered Mar 24 at 6:50
lonza leggieralonza leggiera
1,42928
answered Mar 24 at 6:50
lonza leggieralonza leggiera
1,42928
1,42928
add a comment |
add a comment |
$begingroup$
Maybe the start of the question said the line (singular, not plural) determined by $2x - 5y + z = 3, x + y + 4z = 5$, which is correct because it denotes the intersection (you want both equations to hold simultaneously) of two non-parallel planes in $mathbbR^3$, so these determine a unique line.
You are right that one equation $ax+by+cz+d$ determines a plane in $mathbbR^d$ but two of these equations that hold simultaneously typically determine a line (or the set is empty, as in two parallel planes, or a plane when the two equations denote the same plane). This is only true in $mathbbR^3$..
answered Mar 24 at 6:48
Henno BrandsmaHenno Brandsma
116k349127
$endgroup$
add a comment |
$begingroup$
Maybe the start of the question said the line (singular, not plural) determined by $2x - 5y + z = 3, x + y + 4z = 5$, which is correct because it denotes the intersection (you want both equations to hold simultaneously) of two non-parallel planes in $mathbbR^3$, so these determine a unique line.
You are right that one equation $ax+by+cz+d$ determines a plane in $mathbbR^d$ but two of these equations that hold simultaneously typically determine a line (or the set is empty, as in two parallel planes, or a plane when the two equations denote the same plane). This is only true in $mathbbR^3$..
answered Mar 24 at 6:48
Henno BrandsmaHenno Brandsma
116k349127
$endgroup$
add a comment |
$begingroup$
Maybe the start of the question said the line (singular, not plural) determined by $2x - 5y + z = 3, x + y + 4z = 5$, which is correct because it denotes the intersection (you want both equations to hold simultaneously) of two non-parallel planes in $mathbbR^3$, so these determine a unique line.
You are right that one equation $ax+by+cz+d$ determines a plane in $mathbbR^d$ but two of these equations that hold simultaneously typically determine a line (or the set is empty, as in two parallel planes, or a plane when the two equations denote the same plane). This is only true in $mathbbR^3$..
answered Mar 24 at 6:48
Henno BrandsmaHenno Brandsma
116k349127
$endgroup$
Maybe the start of the question said the line (singular, not plural) determined by $2x - 5y + z = 3, x + y + 4z = 5$, which is correct because it denotes the intersection (you want both equations to hold simultaneously) of two non-parallel planes in $mathbbR^3$, so these determine a unique line.
You are right that one equation $ax+by+cz+d$ determines a plane in $mathbbR^d$ but two of these equations that hold simultaneously typically determine a line (or the set is empty, as in two parallel planes, or a plane when the two equations denote the same plane). This is only true in $mathbbR^3$..
answered Mar 24 at 6:48
Henno BrandsmaHenno Brandsma
116k349127
answered Mar 24 at 6:48
Henno BrandsmaHenno Brandsma
116k349127
answered Mar 24 at 6:48
Henno BrandsmaHenno Brandsma
116k349127
answered Mar 24 at 6:48
Henno BrandsmaHenno Brandsma
116k349127
116k349127
add a comment |
add a comment |
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$begingroup$
Should the word "lines" in your first sentence simply be "line"?
$endgroup$
– Lord Shark the Unknown
Mar 24 at 6:21