General method to find the perpendicular distance between a plane and a point.
$begingroup$
Here's the question I'm puzzling over:
$textbf{Find the perpendicular distance of the point } (p, q, r) textbf{ from the plane } \ax + by + cz = d.$
I tried bringing in the idea of a dot product and attempted to get going with solving the problem, but I'm heading nowhere. That is:
$text{The direction vector of the normal of the plane } = (atextbf{i}+btextbf{j}+ctextbf{k}) text{, where } \textbf{i}, textbf{ j},textbf{ k } text{ are unit vectors.}$
This dotted with the direction vector of the point (position vector, precisely) should equal 0. Am I right? And will this method work even?
vector-spaces vectors 3d
asked Dec 14 '18 at 4:56
RamanaRamana
17210
$endgroup$
add a comment |
$begingroup$
Here's the question I'm puzzling over:
$textbf{Find the perpendicular distance of the point } (p, q, r) textbf{ from the plane } \ax + by + cz = d.$
I tried bringing in the idea of a dot product and attempted to get going with solving the problem, but I'm heading nowhere. That is:
$text{The direction vector of the normal of the plane } = (atextbf{i}+btextbf{j}+ctextbf{k}) text{, where } \textbf{i}, textbf{ j},textbf{ k } text{ are unit vectors.}$
This dotted with the direction vector of the point (position vector, precisely) should equal 0. Am I right? And will this method work even?
vector-spaces vectors 3d
asked Dec 14 '18 at 4:56
RamanaRamana
17210
$endgroup$
add a comment |
$begingroup$
Here's the question I'm puzzling over:
$textbf{Find the perpendicular distance of the point } (p, q, r) textbf{ from the plane } \ax + by + cz = d.$
I tried bringing in the idea of a dot product and attempted to get going with solving the problem, but I'm heading nowhere. That is:
$text{The direction vector of the normal of the plane } = (atextbf{i}+btextbf{j}+ctextbf{k}) text{, where } \textbf{i}, textbf{ j},textbf{ k } text{ are unit vectors.}$
This dotted with the direction vector of the point (position vector, precisely) should equal 0. Am I right? And will this method work even?
vector-spaces vectors 3d
asked Dec 14 '18 at 4:56
RamanaRamana
17210
$endgroup$
Here's the question I'm puzzling over:
$textbf{Find the perpendicular distance of the point } (p, q, r) textbf{ from the plane } \ax + by + cz = d.$
I tried bringing in the idea of a dot product and attempted to get going with solving the problem, but I'm heading nowhere. That is:
$text{The direction vector of the normal of the plane } = (atextbf{i}+btextbf{j}+ctextbf{k}) text{, where } \textbf{i}, textbf{ j},textbf{ k } text{ are unit vectors.}$
This dotted with the direction vector of the point (position vector, precisely) should equal 0. Am I right? And will this method work even?
vector-spaces vectors 3d
vector-spaces vectors 3d
asked Dec 14 '18 at 4:56
RamanaRamana
17210
asked Dec 14 '18 at 4:56
RamanaRamana
17210
asked Dec 14 '18 at 4:56
RamanaRamana
17210
asked Dec 14 '18 at 4:56
RamanaRamana
17210
asked Dec 14 '18 at 4:56
RamanaRamana
17210
17210
add a comment |
add a comment |
2 Answers
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$begingroup$
Let
$$bbox {
vec{n} = ( a , b , c ) = left [ begin{matrix} a \ b \ c end{matrix} right ] }, quad bbox {
vec{v} = ( p , q , r ) = left [ begin{matrix} p \ q \ r end{matrix} right ] }, quad bbox {
vec{w} = ( x , y , z ) = left [ begin{matrix} x \ y \ z end{matrix} right ] }$$
where $vec{n}$ is the plane normal vector, $vec{v}$ is the point those distance $L$ to the plane we want to find out, and $vec{w}$ is just an example point. The equation of the plane is then
$$bbox{ vec{n} cdot vec{w} = d }$$
The signed distance $l$ between the plane and point $vec{v}$, in units of plane normal length, is
$$bbox{ l = vec{n} cdot vec{v} - d }$$
because both $d$ (the signed distance between origin and the plane) and $l$ (the signed distance between the point and the plane) are measured in the same direction, $vec{n}$.
To find the actual distance $L$ between the point $vec{v}$ and the plane, we must divide the absolute value of $l$ by the length of the normal vector $vec{n}$:
$$bbox[#fffff7, 1em] { L = frac{leftlvert vec{n} cdot vec{v} - d rightrvert }{leftlVert vec{n} rightrVert} = frac{ leftlvert a p + b q + c r - d rightrvert}{sqrt{a^2 + b^2 + c^2}} }$$
For a more detailed derivation, look at the Wolfram Mathworld Point-Plane Distance article.
answered Dec 15 '18 at 13:47
Nominal AnimalNominal Animal
7,1332617
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$begingroup$
If $(x,y,z)$ is the point in the plane which has shortest distance to $(p,q,r)$ then $(x-p,y-q,z-r)$ must be along the normal, i.e. $(x-p,y-q,z-r)=alpha (a,b,c)$ for some $alpha$. Thus $ x=p+alpha a$ etc. Use the equation of the plane to find $alpha$ and then compute $sqrt {(x-p)^{2}+(y-q)^{2}+(z-r)^{2}}$.
answered Dec 14 '18 at 5:55
Kavi Rama MurthyKavi Rama Murthy
73.3k53170
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let
$$bbox {
vec{n} = ( a , b , c ) = left [ begin{matrix} a \ b \ c end{matrix} right ] }, quad bbox {
vec{v} = ( p , q , r ) = left [ begin{matrix} p \ q \ r end{matrix} right ] }, quad bbox {
vec{w} = ( x , y , z ) = left [ begin{matrix} x \ y \ z end{matrix} right ] }$$
where $vec{n}$ is the plane normal vector, $vec{v}$ is the point those distance $L$ to the plane we want to find out, and $vec{w}$ is just an example point. The equation of the plane is then
$$bbox{ vec{n} cdot vec{w} = d }$$
The signed distance $l$ between the plane and point $vec{v}$, in units of plane normal length, is
$$bbox{ l = vec{n} cdot vec{v} - d }$$
because both $d$ (the signed distance between origin and the plane) and $l$ (the signed distance between the point and the plane) are measured in the same direction, $vec{n}$.
To find the actual distance $L$ between the point $vec{v}$ and the plane, we must divide the absolute value of $l$ by the length of the normal vector $vec{n}$:
$$bbox[#fffff7, 1em] { L = frac{leftlvert vec{n} cdot vec{v} - d rightrvert }{leftlVert vec{n} rightrVert} = frac{ leftlvert a p + b q + c r - d rightrvert}{sqrt{a^2 + b^2 + c^2}} }$$
For a more detailed derivation, look at the Wolfram Mathworld Point-Plane Distance article.
answered Dec 15 '18 at 13:47
Nominal AnimalNominal Animal
7,1332617
$endgroup$
add a comment |
$begingroup$
Let
$$bbox {
vec{n} = ( a , b , c ) = left [ begin{matrix} a \ b \ c end{matrix} right ] }, quad bbox {
vec{v} = ( p , q , r ) = left [ begin{matrix} p \ q \ r end{matrix} right ] }, quad bbox {
vec{w} = ( x , y , z ) = left [ begin{matrix} x \ y \ z end{matrix} right ] }$$
where $vec{n}$ is the plane normal vector, $vec{v}$ is the point those distance $L$ to the plane we want to find out, and $vec{w}$ is just an example point. The equation of the plane is then
$$bbox{ vec{n} cdot vec{w} = d }$$
The signed distance $l$ between the plane and point $vec{v}$, in units of plane normal length, is
$$bbox{ l = vec{n} cdot vec{v} - d }$$
because both $d$ (the signed distance between origin and the plane) and $l$ (the signed distance between the point and the plane) are measured in the same direction, $vec{n}$.
To find the actual distance $L$ between the point $vec{v}$ and the plane, we must divide the absolute value of $l$ by the length of the normal vector $vec{n}$:
$$bbox[#fffff7, 1em] { L = frac{leftlvert vec{n} cdot vec{v} - d rightrvert }{leftlVert vec{n} rightrVert} = frac{ leftlvert a p + b q + c r - d rightrvert}{sqrt{a^2 + b^2 + c^2}} }$$
For a more detailed derivation, look at the Wolfram Mathworld Point-Plane Distance article.
answered Dec 15 '18 at 13:47
Nominal AnimalNominal Animal
7,1332617
$endgroup$
add a comment |
$begingroup$
Let
$$bbox {
vec{n} = ( a , b , c ) = left [ begin{matrix} a \ b \ c end{matrix} right ] }, quad bbox {
vec{v} = ( p , q , r ) = left [ begin{matrix} p \ q \ r end{matrix} right ] }, quad bbox {
vec{w} = ( x , y , z ) = left [ begin{matrix} x \ y \ z end{matrix} right ] }$$
where $vec{n}$ is the plane normal vector, $vec{v}$ is the point those distance $L$ to the plane we want to find out, and $vec{w}$ is just an example point. The equation of the plane is then
$$bbox{ vec{n} cdot vec{w} = d }$$
The signed distance $l$ between the plane and point $vec{v}$, in units of plane normal length, is
$$bbox{ l = vec{n} cdot vec{v} - d }$$
because both $d$ (the signed distance between origin and the plane) and $l$ (the signed distance between the point and the plane) are measured in the same direction, $vec{n}$.
To find the actual distance $L$ between the point $vec{v}$ and the plane, we must divide the absolute value of $l$ by the length of the normal vector $vec{n}$:
$$bbox[#fffff7, 1em] { L = frac{leftlvert vec{n} cdot vec{v} - d rightrvert }{leftlVert vec{n} rightrVert} = frac{ leftlvert a p + b q + c r - d rightrvert}{sqrt{a^2 + b^2 + c^2}} }$$
For a more detailed derivation, look at the Wolfram Mathworld Point-Plane Distance article.
answered Dec 15 '18 at 13:47
Nominal AnimalNominal Animal
7,1332617
$endgroup$
Let
$$bbox {
vec{n} = ( a , b , c ) = left [ begin{matrix} a \ b \ c end{matrix} right ] }, quad bbox {
vec{v} = ( p , q , r ) = left [ begin{matrix} p \ q \ r end{matrix} right ] }, quad bbox {
vec{w} = ( x , y , z ) = left [ begin{matrix} x \ y \ z end{matrix} right ] }$$
where $vec{n}$ is the plane normal vector, $vec{v}$ is the point those distance $L$ to the plane we want to find out, and $vec{w}$ is just an example point. The equation of the plane is then
$$bbox{ vec{n} cdot vec{w} = d }$$
The signed distance $l$ between the plane and point $vec{v}$, in units of plane normal length, is
$$bbox{ l = vec{n} cdot vec{v} - d }$$
because both $d$ (the signed distance between origin and the plane) and $l$ (the signed distance between the point and the plane) are measured in the same direction, $vec{n}$.
To find the actual distance $L$ between the point $vec{v}$ and the plane, we must divide the absolute value of $l$ by the length of the normal vector $vec{n}$:
$$bbox[#fffff7, 1em] { L = frac{leftlvert vec{n} cdot vec{v} - d rightrvert }{leftlVert vec{n} rightrVert} = frac{ leftlvert a p + b q + c r - d rightrvert}{sqrt{a^2 + b^2 + c^2}} }$$
For a more detailed derivation, look at the Wolfram Mathworld Point-Plane Distance article.
answered Dec 15 '18 at 13:47
Nominal AnimalNominal Animal
7,1332617
answered Dec 15 '18 at 13:47
Nominal AnimalNominal Animal
7,1332617
answered Dec 15 '18 at 13:47
Nominal AnimalNominal Animal
7,1332617
answered Dec 15 '18 at 13:47
Nominal AnimalNominal Animal
7,1332617
7,1332617
add a comment |
add a comment |
$begingroup$
If $(x,y,z)$ is the point in the plane which has shortest distance to $(p,q,r)$ then $(x-p,y-q,z-r)$ must be along the normal, i.e. $(x-p,y-q,z-r)=alpha (a,b,c)$ for some $alpha$. Thus $ x=p+alpha a$ etc. Use the equation of the plane to find $alpha$ and then compute $sqrt {(x-p)^{2}+(y-q)^{2}+(z-r)^{2}}$.
answered Dec 14 '18 at 5:55
Kavi Rama MurthyKavi Rama Murthy
73.3k53170
$endgroup$
add a comment |
$begingroup$
If $(x,y,z)$ is the point in the plane which has shortest distance to $(p,q,r)$ then $(x-p,y-q,z-r)$ must be along the normal, i.e. $(x-p,y-q,z-r)=alpha (a,b,c)$ for some $alpha$. Thus $ x=p+alpha a$ etc. Use the equation of the plane to find $alpha$ and then compute $sqrt {(x-p)^{2}+(y-q)^{2}+(z-r)^{2}}$.
answered Dec 14 '18 at 5:55
Kavi Rama MurthyKavi Rama Murthy
73.3k53170
$endgroup$
add a comment |
$begingroup$
If $(x,y,z)$ is the point in the plane which has shortest distance to $(p,q,r)$ then $(x-p,y-q,z-r)$ must be along the normal, i.e. $(x-p,y-q,z-r)=alpha (a,b,c)$ for some $alpha$. Thus $ x=p+alpha a$ etc. Use the equation of the plane to find $alpha$ and then compute $sqrt {(x-p)^{2}+(y-q)^{2}+(z-r)^{2}}$.
answered Dec 14 '18 at 5:55
Kavi Rama MurthyKavi Rama Murthy
73.3k53170
$endgroup$
If $(x,y,z)$ is the point in the plane which has shortest distance to $(p,q,r)$ then $(x-p,y-q,z-r)$ must be along the normal, i.e. $(x-p,y-q,z-r)=alpha (a,b,c)$ for some $alpha$. Thus $ x=p+alpha a$ etc. Use the equation of the plane to find $alpha$ and then compute $sqrt {(x-p)^{2}+(y-q)^{2}+(z-r)^{2}}$.
answered Dec 14 '18 at 5:55
Kavi Rama MurthyKavi Rama Murthy
73.3k53170
answered Dec 14 '18 at 5:55
Kavi Rama MurthyKavi Rama Murthy
73.3k53170
answered Dec 14 '18 at 5:55
Kavi Rama MurthyKavi Rama Murthy
73.3k53170
answered Dec 14 '18 at 5:55
Kavi Rama MurthyKavi Rama Murthy
73.3k53170
73.3k53170
add a comment |
add a comment |
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