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What is the 'meaning' behind $r·n = a·n$
What does the dot product of two vectors represent?What are all the ways to get an equation of a plane?How to know when a line is parallel to the xz-planeStereographic projections - equation of a plane questionVisually understanding the formula for the distance from a point to plane.Flux intergrals/Finding normals and their dot productsDistance between point and plane - why use the dot product?How to find the equation of a plane in the form $ax + by + cz = d$ which contains two lines, $l$ and $m$?What is the meaning of the inner product of a vector and a gradient?How can a normal vector and a vector on the plane give an equation of the plane?Taking the Dot product of two normalized vectors (unit vectors) vs the dot product of two non-normalized vectors
$begingroup$
I was confused about what the dot product represents and I think I have grasped that now from this post What does the dot product of two vectors represent? .
However, I still cannot understand what $r·n = a·n$ represents in the same way. I am not sure if this is the universal notation for it, but I was told that '$r$' is any point on a plane $(x,y,z)$ and '$n$' is the normal to the plane with '$a$' being any known point on the plane. I can't understand why the dot product of the normal and ($O$ - any point on the plane) would be the same as the dot product of the normal and ($O$ - one particular point on the plane) as ($O$ - a point on the plane) would be a different vector every time where n stays constant surely resulting in a different dot product.
Probably haven't articulated my misunderstanding too well but any help appreciated.
vectors
edited Mar 17 at 13:02
dmtri
1,7582521
asked Mar 17 at 12:49
TommyTommy
83
$endgroup$
add a comment |
$begingroup$
I was confused about what the dot product represents and I think I have grasped that now from this post What does the dot product of two vectors represent? .
However, I still cannot understand what $r·n = a·n$ represents in the same way. I am not sure if this is the universal notation for it, but I was told that '$r$' is any point on a plane $(x,y,z)$ and '$n$' is the normal to the plane with '$a$' being any known point on the plane. I can't understand why the dot product of the normal and ($O$ - any point on the plane) would be the same as the dot product of the normal and ($O$ - one particular point on the plane) as ($O$ - a point on the plane) would be a different vector every time where n stays constant surely resulting in a different dot product.
Probably haven't articulated my misunderstanding too well but any help appreciated.
vectors
edited Mar 17 at 13:02
dmtri
1,7582521
asked Mar 17 at 12:49
TommyTommy
83
$endgroup$
1
$begingroup$
The equation $vec r cdot vec n = vec a cdot vec n$ defines the plane containing the point with position vector $vec a$ and having normal $vec a$, when $vec r$ is taken (as per usual notation) to stand for the position vector of an arbitrary point $(x, y, z)$. For any point $(x, y, z)$ in the plane, so that $r$ is its position vector, then $vec r - vec a$ is a vector lying in the plane, and therefore perpendicular to $vec n$ (a normal to the plane). Thus, $vec n cdot (vec r - vec a) = 0$.
$endgroup$
– M. Vinay
Mar 17 at 12:55
add a comment |
$begingroup$
I was confused about what the dot product represents and I think I have grasped that now from this post What does the dot product of two vectors represent? .
However, I still cannot understand what $r·n = a·n$ represents in the same way. I am not sure if this is the universal notation for it, but I was told that '$r$' is any point on a plane $(x,y,z)$ and '$n$' is the normal to the plane with '$a$' being any known point on the plane. I can't understand why the dot product of the normal and ($O$ - any point on the plane) would be the same as the dot product of the normal and ($O$ - one particular point on the plane) as ($O$ - a point on the plane) would be a different vector every time where n stays constant surely resulting in a different dot product.
Probably haven't articulated my misunderstanding too well but any help appreciated.
vectors
edited Mar 17 at 13:02
dmtri
1,7582521
asked Mar 17 at 12:49
TommyTommy
83
$endgroup$
I was confused about what the dot product represents and I think I have grasped that now from this post What does the dot product of two vectors represent? .
However, I still cannot understand what $r·n = a·n$ represents in the same way. I am not sure if this is the universal notation for it, but I was told that '$r$' is any point on a plane $(x,y,z)$ and '$n$' is the normal to the plane with '$a$' being any known point on the plane. I can't understand why the dot product of the normal and ($O$ - any point on the plane) would be the same as the dot product of the normal and ($O$ - one particular point on the plane) as ($O$ - a point on the plane) would be a different vector every time where n stays constant surely resulting in a different dot product.
Probably haven't articulated my misunderstanding too well but any help appreciated.
vectors
vectors
edited Mar 17 at 13:02
dmtri
1,7582521
asked Mar 17 at 12:49
TommyTommy
83
edited Mar 17 at 13:02
dmtri
1,7582521
asked Mar 17 at 12:49
TommyTommy
83
edited Mar 17 at 13:02
dmtri
1,7582521
edited Mar 17 at 13:02
dmtri
1,7582521
edited Mar 17 at 13:02
dmtri
1,7582521
1,7582521
asked Mar 17 at 12:49
TommyTommy
83
asked Mar 17 at 12:49
TommyTommy
83
asked Mar 17 at 12:49
TommyTommy
83
83
1
$begingroup$
The equation $vec r cdot vec n = vec a cdot vec n$ defines the plane containing the point with position vector $vec a$ and having normal $vec a$, when $vec r$ is taken (as per usual notation) to stand for the position vector of an arbitrary point $(x, y, z)$. For any point $(x, y, z)$ in the plane, so that $r$ is its position vector, then $vec r - vec a$ is a vector lying in the plane, and therefore perpendicular to $vec n$ (a normal to the plane). Thus, $vec n cdot (vec r - vec a) = 0$.
$endgroup$
– M. Vinay
Mar 17 at 12:55
add a comment |
1
$begingroup$
The equation $vec r cdot vec n = vec a cdot vec n$ defines the plane containing the point with position vector $vec a$ and having normal $vec a$, when $vec r$ is taken (as per usual notation) to stand for the position vector of an arbitrary point $(x, y, z)$. For any point $(x, y, z)$ in the plane, so that $r$ is its position vector, then $vec r - vec a$ is a vector lying in the plane, and therefore perpendicular to $vec n$ (a normal to the plane). Thus, $vec n cdot (vec r - vec a) = 0$.
$endgroup$
– M. Vinay
Mar 17 at 12:55
1
1
$begingroup$
The equation $vec r cdot vec n = vec a cdot vec n$ defines the plane containing the point with position vector $vec a$ and having normal $vec a$, when $vec r$ is taken (as per usual notation) to stand for the position vector of an arbitrary point $(x, y, z)$. For any point $(x, y, z)$ in the plane, so that $r$ is its position vector, then $vec r - vec a$ is a vector lying in the plane, and therefore perpendicular to $vec n$ (a normal to the plane). Thus, $vec n cdot (vec r - vec a) = 0$.
$endgroup$
– M. Vinay
Mar 17 at 12:55
$begingroup$
The equation $vec r cdot vec n = vec a cdot vec n$ defines the plane containing the point with position vector $vec a$ and having normal $vec a$, when $vec r$ is taken (as per usual notation) to stand for the position vector of an arbitrary point $(x, y, z)$. For any point $(x, y, z)$ in the plane, so that $r$ is its position vector, then $vec r - vec a$ is a vector lying in the plane, and therefore perpendicular to $vec n$ (a normal to the plane). Thus, $vec n cdot (vec r - vec a) = 0$.
$endgroup$
– M. Vinay
Mar 17 at 12:55
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Another way to visualize the equation is that if you construct an alternative orthonormal basis in $mathbb R^3,$
with basis vectors $mathbf e_1,$ $mathbf e_2,$ and $mathbf e_3,$
you can get the coordinates of any vector $mathbf r$ over the alternative basis by taking the dot product of $mathbf r$ with each basis vector.
For example, the first coordinate is
$mathbf r cdot mathbf e_1.$
Given a unit vector $mathbf n$ normal to some plane, you can make it part of an orthonormal basis by setting $mathbf e_1=mathbf n.$
The other two basis vectors then are parallel to the plane.
If you take an arbitrary vector $mathbf a$ and then take all vectors $mathbf r$ such that $mathbf r cdot mathbf n= mathbf a cdot mathbf n,$
those vectors describe a plane through $mathbf a$ parallel to your initial plane in the same way that all vectors with first coordinate $k$ describe a plane parallel to the plane $x=1.$
If $mathbf a$ is the position vector of a point in your initial plane then the “parallel” plane is simply the plane you started with.
To prove that this all works, you would probably need to do some algebraic manipulations. But once you have proved that it works, you can visualize $mathbf rcdot mathbf n$ as one of the coordinates of $mathbf r$ in an orthonormal basis with $mathbf n$ as one of the basis vectors without thinking again about the algebra you used in the proof.
answered Mar 17 at 13:54
David KDavid K
55.4k344120
$endgroup$
$begingroup$
Not too sure how - considering I've had to google 90% of what you wrote - but that does actually make a good amount of sense. Think I was considering the equation in completely the wrong way so thank you for clearing that up.
$endgroup$
– Tommy
Mar 17 at 14:39
add a comment |
$begingroup$
If you rearrange it, it is $$(mathbfr - mathbfa)cdotmathbfn = 0.$$
Since $mathbfr - mathbfa$ is the vector joining $mathbfa$ to $mathbfr$, this equation says that $mathbfn$ is orthogonal (perpendicular) to the vector joining $mathbfa$ to $mathbfr$ for any $mathbfr$ on the plane. If you draw a diagram, you should be able to intuitively see that this is the case.
answered Mar 17 at 12:54
Minus One-TwelfthMinus One-Twelfth
2,823413
$endgroup$
$begingroup$
Yes we were taught that to begin with and I understand that. My issue was I couldn't understand what the final form represented or is it a case of the first form is very easily explainable and then you just have to accept that the rearranged form is true as a result of it following from the original form?
$endgroup$
– Tommy
Mar 17 at 13:00
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Another way to visualize the equation is that if you construct an alternative orthonormal basis in $mathbb R^3,$
with basis vectors $mathbf e_1,$ $mathbf e_2,$ and $mathbf e_3,$
you can get the coordinates of any vector $mathbf r$ over the alternative basis by taking the dot product of $mathbf r$ with each basis vector.
For example, the first coordinate is
$mathbf r cdot mathbf e_1.$
Given a unit vector $mathbf n$ normal to some plane, you can make it part of an orthonormal basis by setting $mathbf e_1=mathbf n.$
The other two basis vectors then are parallel to the plane.
If you take an arbitrary vector $mathbf a$ and then take all vectors $mathbf r$ such that $mathbf r cdot mathbf n= mathbf a cdot mathbf n,$
those vectors describe a plane through $mathbf a$ parallel to your initial plane in the same way that all vectors with first coordinate $k$ describe a plane parallel to the plane $x=1.$
If $mathbf a$ is the position vector of a point in your initial plane then the “parallel” plane is simply the plane you started with.
To prove that this all works, you would probably need to do some algebraic manipulations. But once you have proved that it works, you can visualize $mathbf rcdot mathbf n$ as one of the coordinates of $mathbf r$ in an orthonormal basis with $mathbf n$ as one of the basis vectors without thinking again about the algebra you used in the proof.
answered Mar 17 at 13:54
David KDavid K
55.4k344120
$endgroup$
$begingroup$
Not too sure how - considering I've had to google 90% of what you wrote - but that does actually make a good amount of sense. Think I was considering the equation in completely the wrong way so thank you for clearing that up.
$endgroup$
– Tommy
Mar 17 at 14:39
add a comment |
$begingroup$
Another way to visualize the equation is that if you construct an alternative orthonormal basis in $mathbb R^3,$
with basis vectors $mathbf e_1,$ $mathbf e_2,$ and $mathbf e_3,$
you can get the coordinates of any vector $mathbf r$ over the alternative basis by taking the dot product of $mathbf r$ with each basis vector.
For example, the first coordinate is
$mathbf r cdot mathbf e_1.$
Given a unit vector $mathbf n$ normal to some plane, you can make it part of an orthonormal basis by setting $mathbf e_1=mathbf n.$
The other two basis vectors then are parallel to the plane.
If you take an arbitrary vector $mathbf a$ and then take all vectors $mathbf r$ such that $mathbf r cdot mathbf n= mathbf a cdot mathbf n,$
those vectors describe a plane through $mathbf a$ parallel to your initial plane in the same way that all vectors with first coordinate $k$ describe a plane parallel to the plane $x=1.$
If $mathbf a$ is the position vector of a point in your initial plane then the “parallel” plane is simply the plane you started with.
To prove that this all works, you would probably need to do some algebraic manipulations. But once you have proved that it works, you can visualize $mathbf rcdot mathbf n$ as one of the coordinates of $mathbf r$ in an orthonormal basis with $mathbf n$ as one of the basis vectors without thinking again about the algebra you used in the proof.
answered Mar 17 at 13:54
David KDavid K
55.4k344120
$endgroup$
$begingroup$
Not too sure how - considering I've had to google 90% of what you wrote - but that does actually make a good amount of sense. Think I was considering the equation in completely the wrong way so thank you for clearing that up.
$endgroup$
– Tommy
Mar 17 at 14:39
add a comment |
$begingroup$
Another way to visualize the equation is that if you construct an alternative orthonormal basis in $mathbb R^3,$
with basis vectors $mathbf e_1,$ $mathbf e_2,$ and $mathbf e_3,$
you can get the coordinates of any vector $mathbf r$ over the alternative basis by taking the dot product of $mathbf r$ with each basis vector.
For example, the first coordinate is
$mathbf r cdot mathbf e_1.$
Given a unit vector $mathbf n$ normal to some plane, you can make it part of an orthonormal basis by setting $mathbf e_1=mathbf n.$
The other two basis vectors then are parallel to the plane.
If you take an arbitrary vector $mathbf a$ and then take all vectors $mathbf r$ such that $mathbf r cdot mathbf n= mathbf a cdot mathbf n,$
those vectors describe a plane through $mathbf a$ parallel to your initial plane in the same way that all vectors with first coordinate $k$ describe a plane parallel to the plane $x=1.$
If $mathbf a$ is the position vector of a point in your initial plane then the “parallel” plane is simply the plane you started with.
To prove that this all works, you would probably need to do some algebraic manipulations. But once you have proved that it works, you can visualize $mathbf rcdot mathbf n$ as one of the coordinates of $mathbf r$ in an orthonormal basis with $mathbf n$ as one of the basis vectors without thinking again about the algebra you used in the proof.
answered Mar 17 at 13:54
David KDavid K
55.4k344120
$endgroup$
Another way to visualize the equation is that if you construct an alternative orthonormal basis in $mathbb R^3,$
with basis vectors $mathbf e_1,$ $mathbf e_2,$ and $mathbf e_3,$
you can get the coordinates of any vector $mathbf r$ over the alternative basis by taking the dot product of $mathbf r$ with each basis vector.
For example, the first coordinate is
$mathbf r cdot mathbf e_1.$
Given a unit vector $mathbf n$ normal to some plane, you can make it part of an orthonormal basis by setting $mathbf e_1=mathbf n.$
The other two basis vectors then are parallel to the plane.
If you take an arbitrary vector $mathbf a$ and then take all vectors $mathbf r$ such that $mathbf r cdot mathbf n= mathbf a cdot mathbf n,$
those vectors describe a plane through $mathbf a$ parallel to your initial plane in the same way that all vectors with first coordinate $k$ describe a plane parallel to the plane $x=1.$
If $mathbf a$ is the position vector of a point in your initial plane then the “parallel” plane is simply the plane you started with.
To prove that this all works, you would probably need to do some algebraic manipulations. But once you have proved that it works, you can visualize $mathbf rcdot mathbf n$ as one of the coordinates of $mathbf r$ in an orthonormal basis with $mathbf n$ as one of the basis vectors without thinking again about the algebra you used in the proof.
answered Mar 17 at 13:54
David KDavid K
55.4k344120
answered Mar 17 at 13:54
David KDavid K
55.4k344120
answered Mar 17 at 13:54
David KDavid K
55.4k344120
answered Mar 17 at 13:54
David KDavid K
55.4k344120
55.4k344120
$begingroup$
Not too sure how - considering I've had to google 90% of what you wrote - but that does actually make a good amount of sense. Think I was considering the equation in completely the wrong way so thank you for clearing that up.
$endgroup$
– Tommy
Mar 17 at 14:39
add a comment |
$begingroup$
Not too sure how - considering I've had to google 90% of what you wrote - but that does actually make a good amount of sense. Think I was considering the equation in completely the wrong way so thank you for clearing that up.
$endgroup$
– Tommy
Mar 17 at 14:39
$begingroup$
Not too sure how - considering I've had to google 90% of what you wrote - but that does actually make a good amount of sense. Think I was considering the equation in completely the wrong way so thank you for clearing that up.
$endgroup$
– Tommy
Mar 17 at 14:39
$begingroup$
Not too sure how - considering I've had to google 90% of what you wrote - but that does actually make a good amount of sense. Think I was considering the equation in completely the wrong way so thank you for clearing that up.
$endgroup$
– Tommy
Mar 17 at 14:39
add a comment |
$begingroup$
If you rearrange it, it is $$(mathbfr - mathbfa)cdotmathbfn = 0.$$
Since $mathbfr - mathbfa$ is the vector joining $mathbfa$ to $mathbfr$, this equation says that $mathbfn$ is orthogonal (perpendicular) to the vector joining $mathbfa$ to $mathbfr$ for any $mathbfr$ on the plane. If you draw a diagram, you should be able to intuitively see that this is the case.
answered Mar 17 at 12:54
Minus One-TwelfthMinus One-Twelfth
2,823413
$endgroup$
$begingroup$
Yes we were taught that to begin with and I understand that. My issue was I couldn't understand what the final form represented or is it a case of the first form is very easily explainable and then you just have to accept that the rearranged form is true as a result of it following from the original form?
$endgroup$
– Tommy
Mar 17 at 13:00
add a comment |
$begingroup$
If you rearrange it, it is $$(mathbfr - mathbfa)cdotmathbfn = 0.$$
Since $mathbfr - mathbfa$ is the vector joining $mathbfa$ to $mathbfr$, this equation says that $mathbfn$ is orthogonal (perpendicular) to the vector joining $mathbfa$ to $mathbfr$ for any $mathbfr$ on the plane. If you draw a diagram, you should be able to intuitively see that this is the case.
answered Mar 17 at 12:54
Minus One-TwelfthMinus One-Twelfth
2,823413
$endgroup$
$begingroup$
Yes we were taught that to begin with and I understand that. My issue was I couldn't understand what the final form represented or is it a case of the first form is very easily explainable and then you just have to accept that the rearranged form is true as a result of it following from the original form?
$endgroup$
– Tommy
Mar 17 at 13:00
add a comment |
$begingroup$
If you rearrange it, it is $$(mathbfr - mathbfa)cdotmathbfn = 0.$$
Since $mathbfr - mathbfa$ is the vector joining $mathbfa$ to $mathbfr$, this equation says that $mathbfn$ is orthogonal (perpendicular) to the vector joining $mathbfa$ to $mathbfr$ for any $mathbfr$ on the plane. If you draw a diagram, you should be able to intuitively see that this is the case.
answered Mar 17 at 12:54
Minus One-TwelfthMinus One-Twelfth
2,823413
$endgroup$
If you rearrange it, it is $$(mathbfr - mathbfa)cdotmathbfn = 0.$$
Since $mathbfr - mathbfa$ is the vector joining $mathbfa$ to $mathbfr$, this equation says that $mathbfn$ is orthogonal (perpendicular) to the vector joining $mathbfa$ to $mathbfr$ for any $mathbfr$ on the plane. If you draw a diagram, you should be able to intuitively see that this is the case.
answered Mar 17 at 12:54
Minus One-TwelfthMinus One-Twelfth
2,823413
answered Mar 17 at 12:54
Minus One-TwelfthMinus One-Twelfth
2,823413
answered Mar 17 at 12:54
Minus One-TwelfthMinus One-Twelfth
2,823413
answered Mar 17 at 12:54
Minus One-TwelfthMinus One-Twelfth
2,823413
2,823413
$begingroup$
Yes we were taught that to begin with and I understand that. My issue was I couldn't understand what the final form represented or is it a case of the first form is very easily explainable and then you just have to accept that the rearranged form is true as a result of it following from the original form?
$endgroup$
– Tommy
Mar 17 at 13:00
add a comment |
$begingroup$
Yes we were taught that to begin with and I understand that. My issue was I couldn't understand what the final form represented or is it a case of the first form is very easily explainable and then you just have to accept that the rearranged form is true as a result of it following from the original form?
$endgroup$
– Tommy
Mar 17 at 13:00
$begingroup$
Yes we were taught that to begin with and I understand that. My issue was I couldn't understand what the final form represented or is it a case of the first form is very easily explainable and then you just have to accept that the rearranged form is true as a result of it following from the original form?
$endgroup$
– Tommy
Mar 17 at 13:00
$begingroup$
Yes we were taught that to begin with and I understand that. My issue was I couldn't understand what the final form represented or is it a case of the first form is very easily explainable and then you just have to accept that the rearranged form is true as a result of it following from the original form?
$endgroup$
– Tommy
Mar 17 at 13:00
add a comment |
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Required, but never shown
Required, but never shown
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The equation $vec r cdot vec n = vec a cdot vec n$ defines the plane containing the point with position vector $vec a$ and having normal $vec a$, when $vec r$ is taken (as per usual notation) to stand for the position vector of an arbitrary point $(x, y, z)$. For any point $(x, y, z)$ in the plane, so that $r$ is its position vector, then $vec r - vec a$ is a vector lying in the plane, and therefore perpendicular to $vec n$ (a normal to the plane). Thus, $vec n cdot (vec r - vec a) = 0$.
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– M. Vinay
Mar 17 at 12:55