Why is $x+2y+3z=0$ a plane?
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I took this lecture from Sir Denis at OCW about multivariable calculus where he explains vectors and he gives this question at the end of the lecture?
$x+2y+3z = 0$ is a plane.
But I don't really understand why is it a plane, granted the vectors $(x, y, z)$ and $(1, 2, 3)$ are perpendicular as their dot product is $0$ which implies the angle between them is $90^circ$.
But if the angle is $90^circ$, it should be a perpendicular, not a plane.
What am I missing?
linear-algebra geometry vectors euclidean-geometry
asked Apr 20 at 9:37
Daksh Miglani
1227
 |Â
show 5 more comments
up vote
-1
down vote
favorite
I took this lecture from Sir Denis at OCW about multivariable calculus where he explains vectors and he gives this question at the end of the lecture?
$x+2y+3z = 0$ is a plane.
But I don't really understand why is it a plane, granted the vectors $(x, y, z)$ and $(1, 2, 3)$ are perpendicular as their dot product is $0$ which implies the angle between them is $90^circ$.
But if the angle is $90^circ$, it should be a perpendicular, not a plane.
What am I missing?
linear-algebra geometry vectors euclidean-geometry
asked Apr 20 at 9:37
Daksh Miglani
1227
1
See Equation of the plane : "the equation $ax+by+cz+d=0$ represents a plane having the vector $n = (a, b, c)$ as a normal. This familiar equation for a plane is called the general form of the equation of the plane."
– Mauro ALLEGRANZA
Apr 20 at 9:39
1
Probably you miss the fact that the vector $(1,2,3)$ is not in plane but is instead orthogonal to it.
– user
Apr 20 at 9:41
1
And a plane is uniquely determined by its normal.
– Mauro ALLEGRANZA
Apr 20 at 9:45
1
@MauroALLEGRANZA okay, from what i can grasp. I think you're saying just like the general equation of a circle in 2d, ax+by+cz+d=0 in 3d is the general equation of a plane and whenever an equation is represented as such, it means that the graph of it will be a plane right?
– Daksh Miglani
Apr 20 at 9:45
1
Yes; the set of 3D points $(x,y,z)$ that satisfy the equation belong to the plane perpendicular to vector $(a,b,c)$.
– Mauro ALLEGRANZA
Apr 20 at 9:49
 |Â
show 5 more comments
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
I took this lecture from Sir Denis at OCW about multivariable calculus where he explains vectors and he gives this question at the end of the lecture?
$x+2y+3z = 0$ is a plane.
But I don't really understand why is it a plane, granted the vectors $(x, y, z)$ and $(1, 2, 3)$ are perpendicular as their dot product is $0$ which implies the angle between them is $90^circ$.
But if the angle is $90^circ$, it should be a perpendicular, not a plane.
What am I missing?
linear-algebra geometry vectors euclidean-geometry
asked Apr 20 at 9:37
Daksh Miglani
1227
I took this lecture from Sir Denis at OCW about multivariable calculus where he explains vectors and he gives this question at the end of the lecture?
$x+2y+3z = 0$ is a plane.
But I don't really understand why is it a plane, granted the vectors $(x, y, z)$ and $(1, 2, 3)$ are perpendicular as their dot product is $0$ which implies the angle between them is $90^circ$.
But if the angle is $90^circ$, it should be a perpendicular, not a plane.
What am I missing?
linear-algebra geometry vectors euclidean-geometry
asked Apr 20 at 9:37
Daksh Miglani
1227
edited Aug 13 at 10:56
asked Apr 20 at 9:37
Daksh Miglani
1227
asked Apr 20 at 9:37
Daksh Miglani
1227
asked Apr 20 at 9:37
Daksh Miglani
1227
1227
1
See Equation of the plane : "the equation $ax+by+cz+d=0$ represents a plane having the vector $n = (a, b, c)$ as a normal. This familiar equation for a plane is called the general form of the equation of the plane."
– Mauro ALLEGRANZA
Apr 20 at 9:39
1
Probably you miss the fact that the vector $(1,2,3)$ is not in plane but is instead orthogonal to it.
– user
Apr 20 at 9:41
1
And a plane is uniquely determined by its normal.
– Mauro ALLEGRANZA
Apr 20 at 9:45
1
@MauroALLEGRANZA okay, from what i can grasp. I think you're saying just like the general equation of a circle in 2d, ax+by+cz+d=0 in 3d is the general equation of a plane and whenever an equation is represented as such, it means that the graph of it will be a plane right?
– Daksh Miglani
Apr 20 at 9:45
1
Yes; the set of 3D points $(x,y,z)$ that satisfy the equation belong to the plane perpendicular to vector $(a,b,c)$.
– Mauro ALLEGRANZA
Apr 20 at 9:49
 |Â
show 5 more comments
1
See Equation of the plane : "the equation $ax+by+cz+d=0$ represents a plane having the vector $n = (a, b, c)$ as a normal. This familiar equation for a plane is called the general form of the equation of the plane."
– Mauro ALLEGRANZA
Apr 20 at 9:39
1
Probably you miss the fact that the vector $(1,2,3)$ is not in plane but is instead orthogonal to it.
– user
Apr 20 at 9:41
1
And a plane is uniquely determined by its normal.
– Mauro ALLEGRANZA
Apr 20 at 9:45
1
@MauroALLEGRANZA okay, from what i can grasp. I think you're saying just like the general equation of a circle in 2d, ax+by+cz+d=0 in 3d is the general equation of a plane and whenever an equation is represented as such, it means that the graph of it will be a plane right?
– Daksh Miglani
Apr 20 at 9:45
1
Yes; the set of 3D points $(x,y,z)$ that satisfy the equation belong to the plane perpendicular to vector $(a,b,c)$.
– Mauro ALLEGRANZA
Apr 20 at 9:49
1
1
See Equation of the plane : "the equation $ax+by+cz+d=0$ represents a plane having the vector $n = (a, b, c)$ as a normal. This familiar equation for a plane is called the general form of the equation of the plane."
– Mauro ALLEGRANZA
Apr 20 at 9:39
See Equation of the plane : "the equation $ax+by+cz+d=0$ represents a plane having the vector $n = (a, b, c)$ as a normal. This familiar equation for a plane is called the general form of the equation of the plane."
– Mauro ALLEGRANZA
Apr 20 at 9:39
1
1
Probably you miss the fact that the vector $(1,2,3)$ is not in plane but is instead orthogonal to it.
– user
Apr 20 at 9:41
Probably you miss the fact that the vector $(1,2,3)$ is not in plane but is instead orthogonal to it.
– user
Apr 20 at 9:41
1
1
And a plane is uniquely determined by its normal.
– Mauro ALLEGRANZA
Apr 20 at 9:45
And a plane is uniquely determined by its normal.
– Mauro ALLEGRANZA
Apr 20 at 9:45
1
1
@MauroALLEGRANZA okay, from what i can grasp. I think you're saying just like the general equation of a circle in 2d, ax+by+cz+d=0 in 3d is the general equation of a plane and whenever an equation is represented as such, it means that the graph of it will be a plane right?
– Daksh Miglani
Apr 20 at 9:45
@MauroALLEGRANZA okay, from what i can grasp. I think you're saying just like the general equation of a circle in 2d, ax+by+cz+d=0 in 3d is the general equation of a plane and whenever an equation is represented as such, it means that the graph of it will be a plane right?
– Daksh Miglani
Apr 20 at 9:45
1
1
Yes; the set of 3D points $(x,y,z)$ that satisfy the equation belong to the plane perpendicular to vector $(a,b,c)$.
– Mauro ALLEGRANZA
Apr 20 at 9:49
Yes; the set of 3D points $(x,y,z)$ that satisfy the equation belong to the plane perpendicular to vector $(a,b,c)$.
– Mauro ALLEGRANZA
Apr 20 at 9:49
 |Â
show 5 more comments
3 Answers
3
active
oldest
votes
up vote
3
down vote
accepted
It is the set of all vectors (in all directions) which are orthogonal to $(1,2,3)$. Therefore, it's a plane.
For instance, the set of all vectors orthogonal to $(0,0,1)$ is the plane $z=0$.
answered Apr 20 at 9:38
José Carlos Santos
116k1699178
add a comment |Â
up vote
2
down vote
Since $x=-2y-3z$ each point vector $(x,y,z) = y(-2,1,0)+z(-3,0,1)$ in that set is linear combination of vectors $a= (-2,1,0)$ and $b=(-3,0,1)$. So every point in that set is in a plane spanned with these two vectors.
answered Apr 20 at 9:43
greedoid
26.8k93575
add a comment |Â
up vote
1
down vote
You say "it should be a perpendicular". But a perpendicular to what ?
The answer is a perpendicular to the vector $(1,2,3)$, which is also a perpendicular to the line of direction $(1,2,3)$, and this is a plane.
To prove that the locus is not a single line, one can exhibit three points that are not aligned, such as $(0,0,0), (1,1,-1)$ and $(2,-1,0)$. Clearly, the vectors that they form two by two are not proportional to each other.
answered Apr 20 at 9:47
Yves Daoust
112k665205
yea perpendicular to the vector, correct.
– Daksh Miglani
Apr 20 at 9:48
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
It is the set of all vectors (in all directions) which are orthogonal to $(1,2,3)$. Therefore, it's a plane.
For instance, the set of all vectors orthogonal to $(0,0,1)$ is the plane $z=0$.
answered Apr 20 at 9:38
José Carlos Santos
116k1699178
add a comment |Â
up vote
3
down vote
accepted
It is the set of all vectors (in all directions) which are orthogonal to $(1,2,3)$. Therefore, it's a plane.
For instance, the set of all vectors orthogonal to $(0,0,1)$ is the plane $z=0$.
answered Apr 20 at 9:38
José Carlos Santos
116k1699178
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
It is the set of all vectors (in all directions) which are orthogonal to $(1,2,3)$. Therefore, it's a plane.
For instance, the set of all vectors orthogonal to $(0,0,1)$ is the plane $z=0$.
answered Apr 20 at 9:38
José Carlos Santos
116k1699178
It is the set of all vectors (in all directions) which are orthogonal to $(1,2,3)$. Therefore, it's a plane.
For instance, the set of all vectors orthogonal to $(0,0,1)$ is the plane $z=0$.
answered Apr 20 at 9:38
José Carlos Santos
116k1699178
answered Apr 20 at 9:38
José Carlos Santos
116k1699178
answered Apr 20 at 9:38
José Carlos Santos
116k1699178
answered Apr 20 at 9:38
José Carlos Santos
116k1699178
116k1699178
add a comment |Â
add a comment |Â
up vote
2
down vote
Since $x=-2y-3z$ each point vector $(x,y,z) = y(-2,1,0)+z(-3,0,1)$ in that set is linear combination of vectors $a= (-2,1,0)$ and $b=(-3,0,1)$. So every point in that set is in a plane spanned with these two vectors.
answered Apr 20 at 9:43
greedoid
26.8k93575
add a comment |Â
up vote
2
down vote
Since $x=-2y-3z$ each point vector $(x,y,z) = y(-2,1,0)+z(-3,0,1)$ in that set is linear combination of vectors $a= (-2,1,0)$ and $b=(-3,0,1)$. So every point in that set is in a plane spanned with these two vectors.
answered Apr 20 at 9:43
greedoid
26.8k93575
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Since $x=-2y-3z$ each point vector $(x,y,z) = y(-2,1,0)+z(-3,0,1)$ in that set is linear combination of vectors $a= (-2,1,0)$ and $b=(-3,0,1)$. So every point in that set is in a plane spanned with these two vectors.
answered Apr 20 at 9:43
greedoid
26.8k93575
Since $x=-2y-3z$ each point vector $(x,y,z) = y(-2,1,0)+z(-3,0,1)$ in that set is linear combination of vectors $a= (-2,1,0)$ and $b=(-3,0,1)$. So every point in that set is in a plane spanned with these two vectors.
answered Apr 20 at 9:43
greedoid
26.8k93575
edited Apr 20 at 11:35
answered Apr 20 at 9:43
greedoid
26.8k93575
answered Apr 20 at 9:43
greedoid
26.8k93575
answered Apr 20 at 9:43
greedoid
26.8k93575
26.8k93575
add a comment |Â
add a comment |Â
up vote
1
down vote
You say "it should be a perpendicular". But a perpendicular to what ?
The answer is a perpendicular to the vector $(1,2,3)$, which is also a perpendicular to the line of direction $(1,2,3)$, and this is a plane.
To prove that the locus is not a single line, one can exhibit three points that are not aligned, such as $(0,0,0), (1,1,-1)$ and $(2,-1,0)$. Clearly, the vectors that they form two by two are not proportional to each other.
answered Apr 20 at 9:47
Yves Daoust
112k665205
yea perpendicular to the vector, correct.
– Daksh Miglani
Apr 20 at 9:48
add a comment |Â
up vote
1
down vote
You say "it should be a perpendicular". But a perpendicular to what ?
The answer is a perpendicular to the vector $(1,2,3)$, which is also a perpendicular to the line of direction $(1,2,3)$, and this is a plane.
To prove that the locus is not a single line, one can exhibit three points that are not aligned, such as $(0,0,0), (1,1,-1)$ and $(2,-1,0)$. Clearly, the vectors that they form two by two are not proportional to each other.
answered Apr 20 at 9:47
Yves Daoust
112k665205
yea perpendicular to the vector, correct.
– Daksh Miglani
Apr 20 at 9:48
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You say "it should be a perpendicular". But a perpendicular to what ?
The answer is a perpendicular to the vector $(1,2,3)$, which is also a perpendicular to the line of direction $(1,2,3)$, and this is a plane.
To prove that the locus is not a single line, one can exhibit three points that are not aligned, such as $(0,0,0), (1,1,-1)$ and $(2,-1,0)$. Clearly, the vectors that they form two by two are not proportional to each other.
answered Apr 20 at 9:47
Yves Daoust
112k665205
You say "it should be a perpendicular". But a perpendicular to what ?
The answer is a perpendicular to the vector $(1,2,3)$, which is also a perpendicular to the line of direction $(1,2,3)$, and this is a plane.
To prove that the locus is not a single line, one can exhibit three points that are not aligned, such as $(0,0,0), (1,1,-1)$ and $(2,-1,0)$. Clearly, the vectors that they form two by two are not proportional to each other.
answered Apr 20 at 9:47
Yves Daoust
112k665205
edited Apr 20 at 9:53
answered Apr 20 at 9:47
Yves Daoust
112k665205
answered Apr 20 at 9:47
Yves Daoust
112k665205
answered Apr 20 at 9:47
Yves Daoust
112k665205
112k665205
yea perpendicular to the vector, correct.
– Daksh Miglani
Apr 20 at 9:48
add a comment |Â
yea perpendicular to the vector, correct.
– Daksh Miglani
Apr 20 at 9:48
yea perpendicular to the vector, correct.
– Daksh Miglani
Apr 20 at 9:48
yea perpendicular to the vector, correct.
– Daksh Miglani
Apr 20 at 9:48
add a comment |Â
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1
See Equation of the plane : "the equation $ax+by+cz+d=0$ represents a plane having the vector $n = (a, b, c)$ as a normal. This familiar equation for a plane is called the general form of the equation of the plane."
– Mauro ALLEGRANZA
Apr 20 at 9:39
1
Probably you miss the fact that the vector $(1,2,3)$ is not in plane but is instead orthogonal to it.
– user
Apr 20 at 9:41
1
And a plane is uniquely determined by its normal.
– Mauro ALLEGRANZA
Apr 20 at 9:45
1
@MauroALLEGRANZA okay, from what i can grasp. I think you're saying just like the general equation of a circle in 2d, ax+by+cz+d=0 in 3d is the general equation of a plane and whenever an equation is represented as such, it means that the graph of it will be a plane right?
– Daksh Miglani
Apr 20 at 9:45
1
Yes; the set of 3D points $(x,y,z)$ that satisfy the equation belong to the plane perpendicular to vector $(a,b,c)$.
– Mauro ALLEGRANZA
Apr 20 at 9:49