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The many ways in which to express a plane
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There are many ways to express a plane of $R^3$. I am focusing on two of them.
The first is the cartesian equation $Ax + By + Cz + D = 0$.
The second is to give two direction vectors $u$ and $v$ and a point $P$ of the plane.
My question is: how can I obtain two ortogonal direction vectors $u$ and $v$ and a point $P$ from the cartesian equation $Ax + By + Cz + D = 0$? How can I obtain the cartesian equation from the direction vectors and a point of the plane?
linear-algebra geometry algebraic-geometry
edited Nov 18 at 19:16
KReiser
9,02711233
asked Nov 18 at 19:08
Asghabard
215
add a comment |
up vote
2
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favorite
There are many ways to express a plane of $R^3$. I am focusing on two of them.
The first is the cartesian equation $Ax + By + Cz + D = 0$.
The second is to give two direction vectors $u$ and $v$ and a point $P$ of the plane.
My question is: how can I obtain two ortogonal direction vectors $u$ and $v$ and a point $P$ from the cartesian equation $Ax + By + Cz + D = 0$? How can I obtain the cartesian equation from the direction vectors and a point of the plane?
linear-algebra geometry algebraic-geometry
edited Nov 18 at 19:16
KReiser
9,02711233
asked Nov 18 at 19:08
Asghabard
215
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
There are many ways to express a plane of $R^3$. I am focusing on two of them.
The first is the cartesian equation $Ax + By + Cz + D = 0$.
The second is to give two direction vectors $u$ and $v$ and a point $P$ of the plane.
My question is: how can I obtain two ortogonal direction vectors $u$ and $v$ and a point $P$ from the cartesian equation $Ax + By + Cz + D = 0$? How can I obtain the cartesian equation from the direction vectors and a point of the plane?
linear-algebra geometry algebraic-geometry
edited Nov 18 at 19:16
KReiser
9,02711233
asked Nov 18 at 19:08
Asghabard
215
There are many ways to express a plane of $R^3$. I am focusing on two of them.
The first is the cartesian equation $Ax + By + Cz + D = 0$.
The second is to give two direction vectors $u$ and $v$ and a point $P$ of the plane.
My question is: how can I obtain two ortogonal direction vectors $u$ and $v$ and a point $P$ from the cartesian equation $Ax + By + Cz + D = 0$? How can I obtain the cartesian equation from the direction vectors and a point of the plane?
linear-algebra geometry algebraic-geometry
linear-algebra geometry algebraic-geometry
edited Nov 18 at 19:16
KReiser
9,02711233
asked Nov 18 at 19:08
Asghabard
215
edited Nov 18 at 19:16
KReiser
9,02711233
asked Nov 18 at 19:08
Asghabard
215
edited Nov 18 at 19:16
KReiser
9,02711233
edited Nov 18 at 19:16
KReiser
9,02711233
edited Nov 18 at 19:16
KReiser
9,02711233
9,02711233
asked Nov 18 at 19:08
Asghabard
215
asked Nov 18 at 19:08
Asghabard
215
asked Nov 18 at 19:08
Asghabard
215
215
add a comment |
add a comment |
5 Answers
5
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up vote
2
down vote
$$(a,b,c)=vec n$$ is a vector normal to the plane.
From the knowledge of the Cartesian equation, choose the vector among $(0,c,-b), (-c,0,a)$ and $(b,-a,0)$ with the largest norm (do this to avoid degeneracies). This gives you a first vector perpendicular to $vec n$, let $vec u$. Then set $vec v=vec ntimesvec u$, and you have your second vector.
For the point, you can project the origin orthogonally onto the plane, i.e. find $lambda$ such that $lambda(a,b,c)$ fulfills the plane equation. This yields
$$lambda(a^2+b^2+c^2)+d=0.$$
The converse is easier.
Compute
$$(a,b,c)=vec utimesvec v$$ and expand
$$a(x-x_P)+b(y-y_P)+c(z-z_P)=0.$$
answered 2 days ago
Yves Daoust
121k668216
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From $$Ax+By+Cz+D=0$$
you get first the normal vector to the plane $n=(A,B,C)$.
then you can take
$$u=(0,C,-B)$$
and
$v$ as the vectorial product of $n$ by $u$.
To get the cartesian equation from two vectors $u,v$ and a point $P$,
$$det(PM,u,v)=0$$
with $M=(x,y,z)$.
answered Nov 18 at 19:14
hamam_Abdallah
36.5k21533
Really thanks for the help, but I need two specifications. Firstly, I understand where $n$ come from and the use of the vectorial product to gain $v$, but why $u$ is egual to $(0,C,-B)$? Second: in the formula $det(PM, u, v)=0$ the product $PM$ is a vectorial product?
– Asghabard
2 days ago
For planes of the form $Ax + D = 0$ this procedure gives $u = v = (0, 0, 0)$.
– Travis
2 days ago
@Travis In this case $n=(A 0,0),u=(0,1,0),v=(0,0,1)$.
– hamam_Abdallah
2 days ago
add a comment |
up vote
1
down vote
As other answers point out, you know the normal vector $n=[A,B,C]$. Suppose you have one vector $uperp n$, $|u|neq 0$, then clearly you can find $vperp u$, $vperp n$ using $v=utimes n$.
So the problem reduces to finding a single nonzero vector perpendicular to $n$.
In 3D, there is no single (non-branching) formula that will do this: see this question.
answered 2 days ago
Wouter
5,84521435
add a comment |
up vote
0
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We can write $Ax+By+Cz+D=0$ as:
$$(A,B,C)cdot (x,y,z) = -D$$
That is, all points $(x,y,z)$ that have the same dot product with $(A,B,C)$, which is $-D$. The shortest vector for which this is the case, is the one that is a parallel to $(A,B,C)$. It means that this vector must be perpendicular to the plane to achieve that. Therefore the vector $mathbf n$:
$$mathbf n = frac{(A,B,C)}{sqrt{A^2+B^2+C^2}}$$
is the normal vector of the plane with length $1$.
Consequently, we can write the equation as:
$$mathbf n cdot (x,y,z) = -frac{D}{sqrt{A^2+B^2+C^2}} = d$$
To find a vector $mathbf P$ in the plane, any vector with $mathbf n cdot mathbf P = d$ will do. That's because the plane consists of all points with the same dot product with $mathbf n$, which is $d$. We can pick the one that represents the shortest distance:
$$mathbf P = dmathbf n = -frac{D}{A^2+B^2+C^2}(A,B,C)$$
And since $mathbf n$ is a unit vector, the distance of the origin to the plane must be $|d|=frac{|D|}{sqrt{A^2+B^2+C^2}}$.
To find 2 vectors in the plane, we need 2 independent vectors that are perpendicular to the normal vector $(A,B,C)$. It suffices if the dot product is $0$. Without loss of generality, let's assume that $Ane 0$. Then we can pick:
$$mathbf u =(B,-A,0)quadtext{and}quad mathbf v = (C,0,-A)$$
answered 2 days ago
I like Serena
3,1981718
1
Thanks @Travis. Fixed it by assuming without loss of generality that $Ane 0$.
– I like Serena
2 days ago
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Going one direction:
Given $Ax + By + Cz + D = 0$
$(B,-A, 0)$ is a vector in the plane
$(A,B,frac {A^2 + B^2}{C})$ is a vector in the plane orthogonal to the first.
$(0,0,-frac {D}{C})$ is a point in the plane
Of course, this is just one way to find orthogonal vectors and a point in the plane.
Going the other direction....
given, vectors in the plane $u,v$ and point in the plane $P$
$utimes v = N$ is normal vector in the plane.
Suppose the components are $N = (N_x,N_y,N_z)$
then $N_x x + N_y y+ N_z z - Ncdot P = 0$ will be an equation for the plane.
answered 2 days ago
Doug M
42.6k31752
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
$$(a,b,c)=vec n$$ is a vector normal to the plane.
From the knowledge of the Cartesian equation, choose the vector among $(0,c,-b), (-c,0,a)$ and $(b,-a,0)$ with the largest norm (do this to avoid degeneracies). This gives you a first vector perpendicular to $vec n$, let $vec u$. Then set $vec v=vec ntimesvec u$, and you have your second vector.
For the point, you can project the origin orthogonally onto the plane, i.e. find $lambda$ such that $lambda(a,b,c)$ fulfills the plane equation. This yields
$$lambda(a^2+b^2+c^2)+d=0.$$
The converse is easier.
Compute
$$(a,b,c)=vec utimesvec v$$ and expand
$$a(x-x_P)+b(y-y_P)+c(z-z_P)=0.$$
answered 2 days ago
Yves Daoust
121k668216
add a comment |
up vote
2
down vote
$$(a,b,c)=vec n$$ is a vector normal to the plane.
From the knowledge of the Cartesian equation, choose the vector among $(0,c,-b), (-c,0,a)$ and $(b,-a,0)$ with the largest norm (do this to avoid degeneracies). This gives you a first vector perpendicular to $vec n$, let $vec u$. Then set $vec v=vec ntimesvec u$, and you have your second vector.
For the point, you can project the origin orthogonally onto the plane, i.e. find $lambda$ such that $lambda(a,b,c)$ fulfills the plane equation. This yields
$$lambda(a^2+b^2+c^2)+d=0.$$
The converse is easier.
Compute
$$(a,b,c)=vec utimesvec v$$ and expand
$$a(x-x_P)+b(y-y_P)+c(z-z_P)=0.$$
answered 2 days ago
Yves Daoust
121k668216
add a comment |
up vote
2
down vote
up vote
2
down vote
$$(a,b,c)=vec n$$ is a vector normal to the plane.
From the knowledge of the Cartesian equation, choose the vector among $(0,c,-b), (-c,0,a)$ and $(b,-a,0)$ with the largest norm (do this to avoid degeneracies). This gives you a first vector perpendicular to $vec n$, let $vec u$. Then set $vec v=vec ntimesvec u$, and you have your second vector.
For the point, you can project the origin orthogonally onto the plane, i.e. find $lambda$ such that $lambda(a,b,c)$ fulfills the plane equation. This yields
$$lambda(a^2+b^2+c^2)+d=0.$$
The converse is easier.
Compute
$$(a,b,c)=vec utimesvec v$$ and expand
$$a(x-x_P)+b(y-y_P)+c(z-z_P)=0.$$
answered 2 days ago
Yves Daoust
121k668216
$$(a,b,c)=vec n$$ is a vector normal to the plane.
From the knowledge of the Cartesian equation, choose the vector among $(0,c,-b), (-c,0,a)$ and $(b,-a,0)$ with the largest norm (do this to avoid degeneracies). This gives you a first vector perpendicular to $vec n$, let $vec u$. Then set $vec v=vec ntimesvec u$, and you have your second vector.
For the point, you can project the origin orthogonally onto the plane, i.e. find $lambda$ such that $lambda(a,b,c)$ fulfills the plane equation. This yields
$$lambda(a^2+b^2+c^2)+d=0.$$
The converse is easier.
Compute
$$(a,b,c)=vec utimesvec v$$ and expand
$$a(x-x_P)+b(y-y_P)+c(z-z_P)=0.$$
answered 2 days ago
Yves Daoust
121k668216
edited 2 days ago
answered 2 days ago
Yves Daoust
121k668216
answered 2 days ago
Yves Daoust
121k668216
answered 2 days ago
Yves Daoust
121k668216
121k668216
add a comment |
add a comment |
up vote
1
down vote
From $$Ax+By+Cz+D=0$$
you get first the normal vector to the plane $n=(A,B,C)$.
then you can take
$$u=(0,C,-B)$$
and
$v$ as the vectorial product of $n$ by $u$.
To get the cartesian equation from two vectors $u,v$ and a point $P$,
$$det(PM,u,v)=0$$
with $M=(x,y,z)$.
answered Nov 18 at 19:14
hamam_Abdallah
36.5k21533
Really thanks for the help, but I need two specifications. Firstly, I understand where $n$ come from and the use of the vectorial product to gain $v$, but why $u$ is egual to $(0,C,-B)$? Second: in the formula $det(PM, u, v)=0$ the product $PM$ is a vectorial product?
– Asghabard
2 days ago
For planes of the form $Ax + D = 0$ this procedure gives $u = v = (0, 0, 0)$.
– Travis
2 days ago
@Travis In this case $n=(A 0,0),u=(0,1,0),v=(0,0,1)$.
– hamam_Abdallah
2 days ago
add a comment |
up vote
1
down vote
From $$Ax+By+Cz+D=0$$
you get first the normal vector to the plane $n=(A,B,C)$.
then you can take
$$u=(0,C,-B)$$
and
$v$ as the vectorial product of $n$ by $u$.
To get the cartesian equation from two vectors $u,v$ and a point $P$,
$$det(PM,u,v)=0$$
with $M=(x,y,z)$.
answered Nov 18 at 19:14
hamam_Abdallah
36.5k21533
Really thanks for the help, but I need two specifications. Firstly, I understand where $n$ come from and the use of the vectorial product to gain $v$, but why $u$ is egual to $(0,C,-B)$? Second: in the formula $det(PM, u, v)=0$ the product $PM$ is a vectorial product?
– Asghabard
2 days ago
For planes of the form $Ax + D = 0$ this procedure gives $u = v = (0, 0, 0)$.
– Travis
2 days ago
@Travis In this case $n=(A 0,0),u=(0,1,0),v=(0,0,1)$.
– hamam_Abdallah
2 days ago
add a comment |
up vote
1
down vote
up vote
1
down vote
From $$Ax+By+Cz+D=0$$
you get first the normal vector to the plane $n=(A,B,C)$.
then you can take
$$u=(0,C,-B)$$
and
$v$ as the vectorial product of $n$ by $u$.
To get the cartesian equation from two vectors $u,v$ and a point $P$,
$$det(PM,u,v)=0$$
with $M=(x,y,z)$.
answered Nov 18 at 19:14
hamam_Abdallah
36.5k21533
From $$Ax+By+Cz+D=0$$
you get first the normal vector to the plane $n=(A,B,C)$.
then you can take
$$u=(0,C,-B)$$
and
$v$ as the vectorial product of $n$ by $u$.
To get the cartesian equation from two vectors $u,v$ and a point $P$,
$$det(PM,u,v)=0$$
with $M=(x,y,z)$.
answered Nov 18 at 19:14
hamam_Abdallah
36.5k21533
answered Nov 18 at 19:14
hamam_Abdallah
36.5k21533
answered Nov 18 at 19:14
hamam_Abdallah
36.5k21533
answered Nov 18 at 19:14
hamam_Abdallah
36.5k21533
36.5k21533
Really thanks for the help, but I need two specifications. Firstly, I understand where $n$ come from and the use of the vectorial product to gain $v$, but why $u$ is egual to $(0,C,-B)$? Second: in the formula $det(PM, u, v)=0$ the product $PM$ is a vectorial product?
– Asghabard
2 days ago
For planes of the form $Ax + D = 0$ this procedure gives $u = v = (0, 0, 0)$.
– Travis
2 days ago
@Travis In this case $n=(A 0,0),u=(0,1,0),v=(0,0,1)$.
– hamam_Abdallah
2 days ago
add a comment |
Really thanks for the help, but I need two specifications. Firstly, I understand where $n$ come from and the use of the vectorial product to gain $v$, but why $u$ is egual to $(0,C,-B)$? Second: in the formula $det(PM, u, v)=0$ the product $PM$ is a vectorial product?
– Asghabard
2 days ago
For planes of the form $Ax + D = 0$ this procedure gives $u = v = (0, 0, 0)$.
– Travis
2 days ago
@Travis In this case $n=(A 0,0),u=(0,1,0),v=(0,0,1)$.
– hamam_Abdallah
2 days ago
Really thanks for the help, but I need two specifications. Firstly, I understand where $n$ come from and the use of the vectorial product to gain $v$, but why $u$ is egual to $(0,C,-B)$? Second: in the formula $det(PM, u, v)=0$ the product $PM$ is a vectorial product?
– Asghabard
2 days ago
Really thanks for the help, but I need two specifications. Firstly, I understand where $n$ come from and the use of the vectorial product to gain $v$, but why $u$ is egual to $(0,C,-B)$? Second: in the formula $det(PM, u, v)=0$ the product $PM$ is a vectorial product?
– Asghabard
2 days ago
For planes of the form $Ax + D = 0$ this procedure gives $u = v = (0, 0, 0)$.
– Travis
2 days ago
For planes of the form $Ax + D = 0$ this procedure gives $u = v = (0, 0, 0)$.
– Travis
2 days ago
@Travis In this case $n=(A 0,0),u=(0,1,0),v=(0,0,1)$.
– hamam_Abdallah
2 days ago
@Travis In this case $n=(A 0,0),u=(0,1,0),v=(0,0,1)$.
– hamam_Abdallah
2 days ago
add a comment |
up vote
1
down vote
As other answers point out, you know the normal vector $n=[A,B,C]$. Suppose you have one vector $uperp n$, $|u|neq 0$, then clearly you can find $vperp u$, $vperp n$ using $v=utimes n$.
So the problem reduces to finding a single nonzero vector perpendicular to $n$.
In 3D, there is no single (non-branching) formula that will do this: see this question.
answered 2 days ago
Wouter
5,84521435
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up vote
1
down vote
As other answers point out, you know the normal vector $n=[A,B,C]$. Suppose you have one vector $uperp n$, $|u|neq 0$, then clearly you can find $vperp u$, $vperp n$ using $v=utimes n$.
So the problem reduces to finding a single nonzero vector perpendicular to $n$.
In 3D, there is no single (non-branching) formula that will do this: see this question.
answered 2 days ago
Wouter
5,84521435
add a comment |
up vote
1
down vote
up vote
1
down vote
As other answers point out, you know the normal vector $n=[A,B,C]$. Suppose you have one vector $uperp n$, $|u|neq 0$, then clearly you can find $vperp u$, $vperp n$ using $v=utimes n$.
So the problem reduces to finding a single nonzero vector perpendicular to $n$.
In 3D, there is no single (non-branching) formula that will do this: see this question.
answered 2 days ago
Wouter
5,84521435
As other answers point out, you know the normal vector $n=[A,B,C]$. Suppose you have one vector $uperp n$, $|u|neq 0$, then clearly you can find $vperp u$, $vperp n$ using $v=utimes n$.
So the problem reduces to finding a single nonzero vector perpendicular to $n$.
In 3D, there is no single (non-branching) formula that will do this: see this question.
answered 2 days ago
Wouter
5,84521435
answered 2 days ago
Wouter
5,84521435
answered 2 days ago
Wouter
5,84521435
answered 2 days ago
Wouter
5,84521435
5,84521435
add a comment |
add a comment |
up vote
0
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We can write $Ax+By+Cz+D=0$ as:
$$(A,B,C)cdot (x,y,z) = -D$$
That is, all points $(x,y,z)$ that have the same dot product with $(A,B,C)$, which is $-D$. The shortest vector for which this is the case, is the one that is a parallel to $(A,B,C)$. It means that this vector must be perpendicular to the plane to achieve that. Therefore the vector $mathbf n$:
$$mathbf n = frac{(A,B,C)}{sqrt{A^2+B^2+C^2}}$$
is the normal vector of the plane with length $1$.
Consequently, we can write the equation as:
$$mathbf n cdot (x,y,z) = -frac{D}{sqrt{A^2+B^2+C^2}} = d$$
To find a vector $mathbf P$ in the plane, any vector with $mathbf n cdot mathbf P = d$ will do. That's because the plane consists of all points with the same dot product with $mathbf n$, which is $d$. We can pick the one that represents the shortest distance:
$$mathbf P = dmathbf n = -frac{D}{A^2+B^2+C^2}(A,B,C)$$
And since $mathbf n$ is a unit vector, the distance of the origin to the plane must be $|d|=frac{|D|}{sqrt{A^2+B^2+C^2}}$.
To find 2 vectors in the plane, we need 2 independent vectors that are perpendicular to the normal vector $(A,B,C)$. It suffices if the dot product is $0$. Without loss of generality, let's assume that $Ane 0$. Then we can pick:
$$mathbf u =(B,-A,0)quadtext{and}quad mathbf v = (C,0,-A)$$
answered 2 days ago
I like Serena
3,1981718
1
Thanks @Travis. Fixed it by assuming without loss of generality that $Ane 0$.
– I like Serena
2 days ago
add a comment |
up vote
0
down vote
We can write $Ax+By+Cz+D=0$ as:
$$(A,B,C)cdot (x,y,z) = -D$$
That is, all points $(x,y,z)$ that have the same dot product with $(A,B,C)$, which is $-D$. The shortest vector for which this is the case, is the one that is a parallel to $(A,B,C)$. It means that this vector must be perpendicular to the plane to achieve that. Therefore the vector $mathbf n$:
$$mathbf n = frac{(A,B,C)}{sqrt{A^2+B^2+C^2}}$$
is the normal vector of the plane with length $1$.
Consequently, we can write the equation as:
$$mathbf n cdot (x,y,z) = -frac{D}{sqrt{A^2+B^2+C^2}} = d$$
To find a vector $mathbf P$ in the plane, any vector with $mathbf n cdot mathbf P = d$ will do. That's because the plane consists of all points with the same dot product with $mathbf n$, which is $d$. We can pick the one that represents the shortest distance:
$$mathbf P = dmathbf n = -frac{D}{A^2+B^2+C^2}(A,B,C)$$
And since $mathbf n$ is a unit vector, the distance of the origin to the plane must be $|d|=frac{|D|}{sqrt{A^2+B^2+C^2}}$.
To find 2 vectors in the plane, we need 2 independent vectors that are perpendicular to the normal vector $(A,B,C)$. It suffices if the dot product is $0$. Without loss of generality, let's assume that $Ane 0$. Then we can pick:
$$mathbf u =(B,-A,0)quadtext{and}quad mathbf v = (C,0,-A)$$
answered 2 days ago
I like Serena
3,1981718
1
Thanks @Travis. Fixed it by assuming without loss of generality that $Ane 0$.
– I like Serena
2 days ago
add a comment |
up vote
0
down vote
up vote
0
down vote
We can write $Ax+By+Cz+D=0$ as:
$$(A,B,C)cdot (x,y,z) = -D$$
That is, all points $(x,y,z)$ that have the same dot product with $(A,B,C)$, which is $-D$. The shortest vector for which this is the case, is the one that is a parallel to $(A,B,C)$. It means that this vector must be perpendicular to the plane to achieve that. Therefore the vector $mathbf n$:
$$mathbf n = frac{(A,B,C)}{sqrt{A^2+B^2+C^2}}$$
is the normal vector of the plane with length $1$.
Consequently, we can write the equation as:
$$mathbf n cdot (x,y,z) = -frac{D}{sqrt{A^2+B^2+C^2}} = d$$
To find a vector $mathbf P$ in the plane, any vector with $mathbf n cdot mathbf P = d$ will do. That's because the plane consists of all points with the same dot product with $mathbf n$, which is $d$. We can pick the one that represents the shortest distance:
$$mathbf P = dmathbf n = -frac{D}{A^2+B^2+C^2}(A,B,C)$$
And since $mathbf n$ is a unit vector, the distance of the origin to the plane must be $|d|=frac{|D|}{sqrt{A^2+B^2+C^2}}$.
To find 2 vectors in the plane, we need 2 independent vectors that are perpendicular to the normal vector $(A,B,C)$. It suffices if the dot product is $0$. Without loss of generality, let's assume that $Ane 0$. Then we can pick:
$$mathbf u =(B,-A,0)quadtext{and}quad mathbf v = (C,0,-A)$$
answered 2 days ago
I like Serena
3,1981718
We can write $Ax+By+Cz+D=0$ as:
$$(A,B,C)cdot (x,y,z) = -D$$
That is, all points $(x,y,z)$ that have the same dot product with $(A,B,C)$, which is $-D$. The shortest vector for which this is the case, is the one that is a parallel to $(A,B,C)$. It means that this vector must be perpendicular to the plane to achieve that. Therefore the vector $mathbf n$:
$$mathbf n = frac{(A,B,C)}{sqrt{A^2+B^2+C^2}}$$
is the normal vector of the plane with length $1$.
Consequently, we can write the equation as:
$$mathbf n cdot (x,y,z) = -frac{D}{sqrt{A^2+B^2+C^2}} = d$$
To find a vector $mathbf P$ in the plane, any vector with $mathbf n cdot mathbf P = d$ will do. That's because the plane consists of all points with the same dot product with $mathbf n$, which is $d$. We can pick the one that represents the shortest distance:
$$mathbf P = dmathbf n = -frac{D}{A^2+B^2+C^2}(A,B,C)$$
And since $mathbf n$ is a unit vector, the distance of the origin to the plane must be $|d|=frac{|D|}{sqrt{A^2+B^2+C^2}}$.
To find 2 vectors in the plane, we need 2 independent vectors that are perpendicular to the normal vector $(A,B,C)$. It suffices if the dot product is $0$. Without loss of generality, let's assume that $Ane 0$. Then we can pick:
$$mathbf u =(B,-A,0)quadtext{and}quad mathbf v = (C,0,-A)$$
answered 2 days ago
I like Serena
3,1981718
edited 2 days ago
answered 2 days ago
I like Serena
3,1981718
answered 2 days ago
I like Serena
3,1981718
answered 2 days ago
I like Serena
3,1981718
3,1981718
1
Thanks @Travis. Fixed it by assuming without loss of generality that $Ane 0$.
– I like Serena
2 days ago
add a comment |
1
Thanks @Travis. Fixed it by assuming without loss of generality that $Ane 0$.
– I like Serena
2 days ago
1
1
Thanks @Travis. Fixed it by assuming without loss of generality that $Ane 0$.
– I like Serena
2 days ago
Thanks @Travis. Fixed it by assuming without loss of generality that $Ane 0$.
– I like Serena
2 days ago
add a comment |
up vote
0
down vote
Going one direction:
Given $Ax + By + Cz + D = 0$
$(B,-A, 0)$ is a vector in the plane
$(A,B,frac {A^2 + B^2}{C})$ is a vector in the plane orthogonal to the first.
$(0,0,-frac {D}{C})$ is a point in the plane
Of course, this is just one way to find orthogonal vectors and a point in the plane.
Going the other direction....
given, vectors in the plane $u,v$ and point in the plane $P$
$utimes v = N$ is normal vector in the plane.
Suppose the components are $N = (N_x,N_y,N_z)$
then $N_x x + N_y y+ N_z z - Ncdot P = 0$ will be an equation for the plane.
answered 2 days ago
Doug M
42.6k31752
add a comment |
up vote
0
down vote
Going one direction:
Given $Ax + By + Cz + D = 0$
$(B,-A, 0)$ is a vector in the plane
$(A,B,frac {A^2 + B^2}{C})$ is a vector in the plane orthogonal to the first.
$(0,0,-frac {D}{C})$ is a point in the plane
Of course, this is just one way to find orthogonal vectors and a point in the plane.
Going the other direction....
given, vectors in the plane $u,v$ and point in the plane $P$
$utimes v = N$ is normal vector in the plane.
Suppose the components are $N = (N_x,N_y,N_z)$
then $N_x x + N_y y+ N_z z - Ncdot P = 0$ will be an equation for the plane.
answered 2 days ago
Doug M
42.6k31752
add a comment |
up vote
0
down vote
up vote
0
down vote
Going one direction:
Given $Ax + By + Cz + D = 0$
$(B,-A, 0)$ is a vector in the plane
$(A,B,frac {A^2 + B^2}{C})$ is a vector in the plane orthogonal to the first.
$(0,0,-frac {D}{C})$ is a point in the plane
Of course, this is just one way to find orthogonal vectors and a point in the plane.
Going the other direction....
given, vectors in the plane $u,v$ and point in the plane $P$
$utimes v = N$ is normal vector in the plane.
Suppose the components are $N = (N_x,N_y,N_z)$
then $N_x x + N_y y+ N_z z - Ncdot P = 0$ will be an equation for the plane.
answered 2 days ago
Doug M
42.6k31752
Going one direction:
Given $Ax + By + Cz + D = 0$
$(B,-A, 0)$ is a vector in the plane
$(A,B,frac {A^2 + B^2}{C})$ is a vector in the plane orthogonal to the first.
$(0,0,-frac {D}{C})$ is a point in the plane
Of course, this is just one way to find orthogonal vectors and a point in the plane.
Going the other direction....
given, vectors in the plane $u,v$ and point in the plane $P$
$utimes v = N$ is normal vector in the plane.
Suppose the components are $N = (N_x,N_y,N_z)$
then $N_x x + N_y y+ N_z z - Ncdot P = 0$ will be an equation for the plane.
answered 2 days ago
Doug M
42.6k31752
answered 2 days ago
Doug M
42.6k31752
answered 2 days ago
Doug M
42.6k31752
answered 2 days ago
Doug M
42.6k31752
42.6k31752
add a comment |
add a comment |
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