Orientation of a normal vector of a plane
$begingroup$
I found this question:
vector normal to a plane
It seems to be related to a problem I'm struggling with, but I need to know what is the rule for the normal vector's orientation. Assuming we want a normal unit vector to a plane, it could still go two ways. How is this solution oriented and why?
vectors analytic-geometry 3d
asked Jan 15 at 17:19
KotlopouKotlopou
261
$endgroup$
|
show 1 more comment
$begingroup$
I found this question:
vector normal to a plane
It seems to be related to a problem I'm struggling with, but I need to know what is the rule for the normal vector's orientation. Assuming we want a normal unit vector to a plane, it could still go two ways. How is this solution oriented and why?
vectors analytic-geometry 3d
asked Jan 15 at 17:19
KotlopouKotlopou
261
$endgroup$
$begingroup$
You want the $textbf{outward}$ facing normal vector.
$endgroup$
– Adam Higgins
Jan 15 at 17:23
$begingroup$
What does outward mean? The vector starts at the plane, so both ways are out.
$endgroup$
– Kotlopou
Jan 15 at 17:26
$begingroup$
Use the right hand rule. Imagine putting the index finger and middle finger at right angles to each other within the plane, and your thumb at right angles to both. The direction of your thumb is the outward direction.
$endgroup$
– Adam Higgins
Jan 15 at 17:35
$begingroup$
You can still do that two ways, swapping the index and middle fingers, so that the thumb may point in either direction.
$endgroup$
– Kotlopou
Jan 15 at 17:40
$begingroup$
Clearly, by multiplying the equation by -1, we get the opposite normal vector, but the same plane. So orientation follows from the equation, not the plane.
$endgroup$
– Kotlopou
Jan 15 at 18:25
|
show 1 more comment
$begingroup$
I found this question:
vector normal to a plane
It seems to be related to a problem I'm struggling with, but I need to know what is the rule for the normal vector's orientation. Assuming we want a normal unit vector to a plane, it could still go two ways. How is this solution oriented and why?
vectors analytic-geometry 3d
asked Jan 15 at 17:19
KotlopouKotlopou
261
$endgroup$
I found this question:
vector normal to a plane
It seems to be related to a problem I'm struggling with, but I need to know what is the rule for the normal vector's orientation. Assuming we want a normal unit vector to a plane, it could still go two ways. How is this solution oriented and why?
vectors analytic-geometry 3d
vectors analytic-geometry 3d
asked Jan 15 at 17:19
KotlopouKotlopou
261
asked Jan 15 at 17:19
KotlopouKotlopou
261
asked Jan 15 at 17:19
KotlopouKotlopou
261
asked Jan 15 at 17:19
KotlopouKotlopou
261
asked Jan 15 at 17:19
KotlopouKotlopou
261
261
$begingroup$
You want the $textbf{outward}$ facing normal vector.
$endgroup$
– Adam Higgins
Jan 15 at 17:23
$begingroup$
What does outward mean? The vector starts at the plane, so both ways are out.
$endgroup$
– Kotlopou
Jan 15 at 17:26
$begingroup$
Use the right hand rule. Imagine putting the index finger and middle finger at right angles to each other within the plane, and your thumb at right angles to both. The direction of your thumb is the outward direction.
$endgroup$
– Adam Higgins
Jan 15 at 17:35
$begingroup$
You can still do that two ways, swapping the index and middle fingers, so that the thumb may point in either direction.
$endgroup$
– Kotlopou
Jan 15 at 17:40
$begingroup$
Clearly, by multiplying the equation by -1, we get the opposite normal vector, but the same plane. So orientation follows from the equation, not the plane.
$endgroup$
– Kotlopou
Jan 15 at 18:25
|
show 1 more comment
$begingroup$
You want the $textbf{outward}$ facing normal vector.
$endgroup$
– Adam Higgins
Jan 15 at 17:23
$begingroup$
What does outward mean? The vector starts at the plane, so both ways are out.
$endgroup$
– Kotlopou
Jan 15 at 17:26
$begingroup$
Use the right hand rule. Imagine putting the index finger and middle finger at right angles to each other within the plane, and your thumb at right angles to both. The direction of your thumb is the outward direction.
$endgroup$
– Adam Higgins
Jan 15 at 17:35
$begingroup$
You can still do that two ways, swapping the index and middle fingers, so that the thumb may point in either direction.
$endgroup$
– Kotlopou
Jan 15 at 17:40
$begingroup$
Clearly, by multiplying the equation by -1, we get the opposite normal vector, but the same plane. So orientation follows from the equation, not the plane.
$endgroup$
– Kotlopou
Jan 15 at 18:25
$begingroup$
You want the $textbf{outward}$ facing normal vector.
$endgroup$
– Adam Higgins
Jan 15 at 17:23
$begingroup$
You want the $textbf{outward}$ facing normal vector.
$endgroup$
– Adam Higgins
Jan 15 at 17:23
$begingroup$
What does outward mean? The vector starts at the plane, so both ways are out.
$endgroup$
– Kotlopou
Jan 15 at 17:26
$begingroup$
What does outward mean? The vector starts at the plane, so both ways are out.
$endgroup$
– Kotlopou
Jan 15 at 17:26
$begingroup$
Use the right hand rule. Imagine putting the index finger and middle finger at right angles to each other within the plane, and your thumb at right angles to both. The direction of your thumb is the outward direction.
$endgroup$
– Adam Higgins
Jan 15 at 17:35
$begingroup$
Use the right hand rule. Imagine putting the index finger and middle finger at right angles to each other within the plane, and your thumb at right angles to both. The direction of your thumb is the outward direction.
$endgroup$
– Adam Higgins
Jan 15 at 17:35
$begingroup$
You can still do that two ways, swapping the index and middle fingers, so that the thumb may point in either direction.
$endgroup$
– Kotlopou
Jan 15 at 17:40
$begingroup$
You can still do that two ways, swapping the index and middle fingers, so that the thumb may point in either direction.
$endgroup$
– Kotlopou
Jan 15 at 17:40
$begingroup$
Clearly, by multiplying the equation by -1, we get the opposite normal vector, but the same plane. So orientation follows from the equation, not the plane.
$endgroup$
– Kotlopou
Jan 15 at 18:25
$begingroup$
Clearly, by multiplying the equation by -1, we get the opposite normal vector, but the same plane. So orientation follows from the equation, not the plane.
$endgroup$
– Kotlopou
Jan 15 at 18:25
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
A plane divides 3D space into two halfspaces (halves of the full space; one half is on one side of the plane, the other is on the other).
Because the surface normal is perpendicular to the plane, it always points to one halfspace. Essentially, the normal distinguishes the two halfspaces, by always pointing to exactly one of them.
Let's say we have a plane described by its unit normal vector $hat{n}$, and the signed distance $d$ from origin to the plane. Point $vec{p}$ is on the plane if and only if
$$bbox{ vec{p} cdot hat{n} - d = 0 }$$
This is just the special case for the signed distance between any point $vec{p}$ and the plane described by $hat{n}$ and $d$:
$$bbox{ L = vec{p} cdot hat{n} - d }$$
or, if you prefer the traditional Cartesian component form, $L = x n_x + y n_y + z n_z - d$, where $hat{n} = ( n_x , n_y , n_z )$ and $vec{p} = ( x , y , z )$.
Compare those signed distances $L$ to the real line. Effectively, the direction of the plane normal determines where the positive axis of the signed distances to that plane is.
(And $d$ is just the offset from coordinate system origin to the zero point on that line, measured along that line. In fact, if $hat{n}$ is not an unit vector, $d$ measures the distance in units of $leftlVerthat{n}rightrVert_2$, i.e. in units of the Euclidean length of the normal vector. Which is why I think the analog I used above is apt.)
If we only knew the locus of points on the plane (i.e., where the plane is), without the normal, we could only have positive distances to the plane, since we'd have no way of telling one side from another.
answered Jan 16 at 6:50
Nominal AnimalNominal Animal
7,1232617
$endgroup$
$begingroup$
Can you somehow describe the particular halfspace the vector points into by looking at the equation?
$endgroup$
– Kotlopou
Jan 16 at 8:29
$begingroup$
@Kotlopou: Well, the halfspace the normal vector $hat{n}$ points into is the locus of points $vec{p}$ for which $vec{p}cdothat{n} gt d$. You can identify $hat{n}$ and $d$ by looking at the equation, trivially. The important part is that the division into halfspaces is not affected at all if you negate both $hat{n}$ and $d$: it only changes the halfspace the normal vector points to. Thus, it is an arbitrary choice, that does not affect the halfspaces themselves in any way. It just gives you a way to distinguish between the two, if you want to.
$endgroup$
– Nominal Animal
Jan 16 at 9:14
add a comment |
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$begingroup$
A plane divides 3D space into two halfspaces (halves of the full space; one half is on one side of the plane, the other is on the other).
Because the surface normal is perpendicular to the plane, it always points to one halfspace. Essentially, the normal distinguishes the two halfspaces, by always pointing to exactly one of them.
Let's say we have a plane described by its unit normal vector $hat{n}$, and the signed distance $d$ from origin to the plane. Point $vec{p}$ is on the plane if and only if
$$bbox{ vec{p} cdot hat{n} - d = 0 }$$
This is just the special case for the signed distance between any point $vec{p}$ and the plane described by $hat{n}$ and $d$:
$$bbox{ L = vec{p} cdot hat{n} - d }$$
or, if you prefer the traditional Cartesian component form, $L = x n_x + y n_y + z n_z - d$, where $hat{n} = ( n_x , n_y , n_z )$ and $vec{p} = ( x , y , z )$.
Compare those signed distances $L$ to the real line. Effectively, the direction of the plane normal determines where the positive axis of the signed distances to that plane is.
(And $d$ is just the offset from coordinate system origin to the zero point on that line, measured along that line. In fact, if $hat{n}$ is not an unit vector, $d$ measures the distance in units of $leftlVerthat{n}rightrVert_2$, i.e. in units of the Euclidean length of the normal vector. Which is why I think the analog I used above is apt.)
If we only knew the locus of points on the plane (i.e., where the plane is), without the normal, we could only have positive distances to the plane, since we'd have no way of telling one side from another.
answered Jan 16 at 6:50
Nominal AnimalNominal Animal
7,1232617
$endgroup$
$begingroup$
Can you somehow describe the particular halfspace the vector points into by looking at the equation?
$endgroup$
– Kotlopou
Jan 16 at 8:29
$begingroup$
@Kotlopou: Well, the halfspace the normal vector $hat{n}$ points into is the locus of points $vec{p}$ for which $vec{p}cdothat{n} gt d$. You can identify $hat{n}$ and $d$ by looking at the equation, trivially. The important part is that the division into halfspaces is not affected at all if you negate both $hat{n}$ and $d$: it only changes the halfspace the normal vector points to. Thus, it is an arbitrary choice, that does not affect the halfspaces themselves in any way. It just gives you a way to distinguish between the two, if you want to.
$endgroup$
– Nominal Animal
Jan 16 at 9:14
add a comment |
$begingroup$
A plane divides 3D space into two halfspaces (halves of the full space; one half is on one side of the plane, the other is on the other).
Because the surface normal is perpendicular to the plane, it always points to one halfspace. Essentially, the normal distinguishes the two halfspaces, by always pointing to exactly one of them.
Let's say we have a plane described by its unit normal vector $hat{n}$, and the signed distance $d$ from origin to the plane. Point $vec{p}$ is on the plane if and only if
$$bbox{ vec{p} cdot hat{n} - d = 0 }$$
This is just the special case for the signed distance between any point $vec{p}$ and the plane described by $hat{n}$ and $d$:
$$bbox{ L = vec{p} cdot hat{n} - d }$$
or, if you prefer the traditional Cartesian component form, $L = x n_x + y n_y + z n_z - d$, where $hat{n} = ( n_x , n_y , n_z )$ and $vec{p} = ( x , y , z )$.
Compare those signed distances $L$ to the real line. Effectively, the direction of the plane normal determines where the positive axis of the signed distances to that plane is.
(And $d$ is just the offset from coordinate system origin to the zero point on that line, measured along that line. In fact, if $hat{n}$ is not an unit vector, $d$ measures the distance in units of $leftlVerthat{n}rightrVert_2$, i.e. in units of the Euclidean length of the normal vector. Which is why I think the analog I used above is apt.)
If we only knew the locus of points on the plane (i.e., where the plane is), without the normal, we could only have positive distances to the plane, since we'd have no way of telling one side from another.
answered Jan 16 at 6:50
Nominal AnimalNominal Animal
7,1232617
$endgroup$
$begingroup$
Can you somehow describe the particular halfspace the vector points into by looking at the equation?
$endgroup$
– Kotlopou
Jan 16 at 8:29
$begingroup$
@Kotlopou: Well, the halfspace the normal vector $hat{n}$ points into is the locus of points $vec{p}$ for which $vec{p}cdothat{n} gt d$. You can identify $hat{n}$ and $d$ by looking at the equation, trivially. The important part is that the division into halfspaces is not affected at all if you negate both $hat{n}$ and $d$: it only changes the halfspace the normal vector points to. Thus, it is an arbitrary choice, that does not affect the halfspaces themselves in any way. It just gives you a way to distinguish between the two, if you want to.
$endgroup$
– Nominal Animal
Jan 16 at 9:14
add a comment |
$begingroup$
A plane divides 3D space into two halfspaces (halves of the full space; one half is on one side of the plane, the other is on the other).
Because the surface normal is perpendicular to the plane, it always points to one halfspace. Essentially, the normal distinguishes the two halfspaces, by always pointing to exactly one of them.
Let's say we have a plane described by its unit normal vector $hat{n}$, and the signed distance $d$ from origin to the plane. Point $vec{p}$ is on the plane if and only if
$$bbox{ vec{p} cdot hat{n} - d = 0 }$$
This is just the special case for the signed distance between any point $vec{p}$ and the plane described by $hat{n}$ and $d$:
$$bbox{ L = vec{p} cdot hat{n} - d }$$
or, if you prefer the traditional Cartesian component form, $L = x n_x + y n_y + z n_z - d$, where $hat{n} = ( n_x , n_y , n_z )$ and $vec{p} = ( x , y , z )$.
Compare those signed distances $L$ to the real line. Effectively, the direction of the plane normal determines where the positive axis of the signed distances to that plane is.
(And $d$ is just the offset from coordinate system origin to the zero point on that line, measured along that line. In fact, if $hat{n}$ is not an unit vector, $d$ measures the distance in units of $leftlVerthat{n}rightrVert_2$, i.e. in units of the Euclidean length of the normal vector. Which is why I think the analog I used above is apt.)
If we only knew the locus of points on the plane (i.e., where the plane is), without the normal, we could only have positive distances to the plane, since we'd have no way of telling one side from another.
answered Jan 16 at 6:50
Nominal AnimalNominal Animal
7,1232617
$endgroup$
A plane divides 3D space into two halfspaces (halves of the full space; one half is on one side of the plane, the other is on the other).
Because the surface normal is perpendicular to the plane, it always points to one halfspace. Essentially, the normal distinguishes the two halfspaces, by always pointing to exactly one of them.
Let's say we have a plane described by its unit normal vector $hat{n}$, and the signed distance $d$ from origin to the plane. Point $vec{p}$ is on the plane if and only if
$$bbox{ vec{p} cdot hat{n} - d = 0 }$$
This is just the special case for the signed distance between any point $vec{p}$ and the plane described by $hat{n}$ and $d$:
$$bbox{ L = vec{p} cdot hat{n} - d }$$
or, if you prefer the traditional Cartesian component form, $L = x n_x + y n_y + z n_z - d$, where $hat{n} = ( n_x , n_y , n_z )$ and $vec{p} = ( x , y , z )$.
Compare those signed distances $L$ to the real line. Effectively, the direction of the plane normal determines where the positive axis of the signed distances to that plane is.
(And $d$ is just the offset from coordinate system origin to the zero point on that line, measured along that line. In fact, if $hat{n}$ is not an unit vector, $d$ measures the distance in units of $leftlVerthat{n}rightrVert_2$, i.e. in units of the Euclidean length of the normal vector. Which is why I think the analog I used above is apt.)
If we only knew the locus of points on the plane (i.e., where the plane is), without the normal, we could only have positive distances to the plane, since we'd have no way of telling one side from another.
answered Jan 16 at 6:50
Nominal AnimalNominal Animal
7,1232617
answered Jan 16 at 6:50
Nominal AnimalNominal Animal
7,1232617
answered Jan 16 at 6:50
Nominal AnimalNominal Animal
7,1232617
answered Jan 16 at 6:50
Nominal AnimalNominal Animal
7,1232617
7,1232617
$begingroup$
Can you somehow describe the particular halfspace the vector points into by looking at the equation?
$endgroup$
– Kotlopou
Jan 16 at 8:29
$begingroup$
@Kotlopou: Well, the halfspace the normal vector $hat{n}$ points into is the locus of points $vec{p}$ for which $vec{p}cdothat{n} gt d$. You can identify $hat{n}$ and $d$ by looking at the equation, trivially. The important part is that the division into halfspaces is not affected at all if you negate both $hat{n}$ and $d$: it only changes the halfspace the normal vector points to. Thus, it is an arbitrary choice, that does not affect the halfspaces themselves in any way. It just gives you a way to distinguish between the two, if you want to.
$endgroup$
– Nominal Animal
Jan 16 at 9:14
add a comment |
$begingroup$
Can you somehow describe the particular halfspace the vector points into by looking at the equation?
$endgroup$
– Kotlopou
Jan 16 at 8:29
$begingroup$
@Kotlopou: Well, the halfspace the normal vector $hat{n}$ points into is the locus of points $vec{p}$ for which $vec{p}cdothat{n} gt d$. You can identify $hat{n}$ and $d$ by looking at the equation, trivially. The important part is that the division into halfspaces is not affected at all if you negate both $hat{n}$ and $d$: it only changes the halfspace the normal vector points to. Thus, it is an arbitrary choice, that does not affect the halfspaces themselves in any way. It just gives you a way to distinguish between the two, if you want to.
$endgroup$
– Nominal Animal
Jan 16 at 9:14
$begingroup$
Can you somehow describe the particular halfspace the vector points into by looking at the equation?
$endgroup$
– Kotlopou
Jan 16 at 8:29
$begingroup$
Can you somehow describe the particular halfspace the vector points into by looking at the equation?
$endgroup$
– Kotlopou
Jan 16 at 8:29
$begingroup$
@Kotlopou: Well, the halfspace the normal vector $hat{n}$ points into is the locus of points $vec{p}$ for which $vec{p}cdothat{n} gt d$. You can identify $hat{n}$ and $d$ by looking at the equation, trivially. The important part is that the division into halfspaces is not affected at all if you negate both $hat{n}$ and $d$: it only changes the halfspace the normal vector points to. Thus, it is an arbitrary choice, that does not affect the halfspaces themselves in any way. It just gives you a way to distinguish between the two, if you want to.
$endgroup$
– Nominal Animal
Jan 16 at 9:14
$begingroup$
@Kotlopou: Well, the halfspace the normal vector $hat{n}$ points into is the locus of points $vec{p}$ for which $vec{p}cdothat{n} gt d$. You can identify $hat{n}$ and $d$ by looking at the equation, trivially. The important part is that the division into halfspaces is not affected at all if you negate both $hat{n}$ and $d$: it only changes the halfspace the normal vector points to. Thus, it is an arbitrary choice, that does not affect the halfspaces themselves in any way. It just gives you a way to distinguish between the two, if you want to.
$endgroup$
– Nominal Animal
Jan 16 at 9:14
add a comment |
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$begingroup$
You want the $textbf{outward}$ facing normal vector.
$endgroup$
– Adam Higgins
Jan 15 at 17:23
$begingroup$
What does outward mean? The vector starts at the plane, so both ways are out.
$endgroup$
– Kotlopou
Jan 15 at 17:26
$begingroup$
Use the right hand rule. Imagine putting the index finger and middle finger at right angles to each other within the plane, and your thumb at right angles to both. The direction of your thumb is the outward direction.
$endgroup$
– Adam Higgins
Jan 15 at 17:35
$begingroup$
You can still do that two ways, swapping the index and middle fingers, so that the thumb may point in either direction.
$endgroup$
– Kotlopou
Jan 15 at 17:40
$begingroup$
Clearly, by multiplying the equation by -1, we get the opposite normal vector, but the same plane. So orientation follows from the equation, not the plane.
$endgroup$
– Kotlopou
Jan 15 at 18:25