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Show that any plane whose normal lies on cone $(b+c)x^2+(c+a)y^2+(a+b)z^2=0$ cuts the surface $ax^2+by^2+cz^2=1$ is rectangular hyperbola
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Show that any plane whose normal lies on cone $(b+c)x^2+(c+a)y^2+(a+b)z^2=0$ cuts the surface $ax^2+by^2+cz^2=1$ is rectangular hyperbola
My attempt:
let $frac xl = frac ym = frac zn$ be normal of plane $lx+my+nz = 0$. Now the normal is generator of given cone. So we get condition $(b+c)l^2+(c+a)m^2+(a+b)n^2=0$ --->(1)
THe given plane cuts the conicoid $ax^2+by^2+cz^2=1$.
The intersecting conic :
Eliminating z between plane equation and conicoid equation
$z = frac-lx-myn $ Subsittuting this in conicoid equation we get,
$ax^2+by^2+cfrac(-lx-my)^2n^2 =1 $ --->(2)
This represents rectangular hyperbola if
coefficient of $x^2$ + coefficient of $y^2$ = 0
$(2)$ implies
$a+b+cfracl^2n^2+cfracm^2n^2 =0 \ fbox $n^2(a+b) + c(l^2+m^2) = 0$$ for conic to be rectangular hyperbola
If i eliminate x, i get $fbox $l^2(b+c)+a(m^2+n^2)=0$$ for conic to be rectangular hyperbola
If i eliminate y, i get $fbox $m^2(c+a)+b(n^2+l^2)=0$$ for conic to be rectangular hyperbola
But what i have is equation (1) with me for conic to be rectangular hyperbola
How to proceed further?
analytic-geometry conic-sections 3d solid-geometry convex-cone
edited Jul 21 at 13:58
Key Flex
4,406525
asked Jul 15 at 12:17
Magneto
795213
add a comment |Â
up vote
0
down vote
favorite
Show that any plane whose normal lies on cone $(b+c)x^2+(c+a)y^2+(a+b)z^2=0$ cuts the surface $ax^2+by^2+cz^2=1$ is rectangular hyperbola
My attempt:
let $frac xl = frac ym = frac zn$ be normal of plane $lx+my+nz = 0$. Now the normal is generator of given cone. So we get condition $(b+c)l^2+(c+a)m^2+(a+b)n^2=0$ --->(1)
THe given plane cuts the conicoid $ax^2+by^2+cz^2=1$.
The intersecting conic :
Eliminating z between plane equation and conicoid equation
$z = frac-lx-myn $ Subsittuting this in conicoid equation we get,
$ax^2+by^2+cfrac(-lx-my)^2n^2 =1 $ --->(2)
This represents rectangular hyperbola if
coefficient of $x^2$ + coefficient of $y^2$ = 0
$(2)$ implies
$a+b+cfracl^2n^2+cfracm^2n^2 =0 \ fbox $n^2(a+b) + c(l^2+m^2) = 0$$ for conic to be rectangular hyperbola
If i eliminate x, i get $fbox $l^2(b+c)+a(m^2+n^2)=0$$ for conic to be rectangular hyperbola
If i eliminate y, i get $fbox $m^2(c+a)+b(n^2+l^2)=0$$ for conic to be rectangular hyperbola
But what i have is equation (1) with me for conic to be rectangular hyperbola
How to proceed further?
analytic-geometry conic-sections 3d solid-geometry convex-cone
edited Jul 21 at 13:58
Key Flex
4,406525
asked Jul 15 at 12:17
Magneto
795213
Eliminating $z$ from the equations gives you the equation of a projection of the intersection curve onto the $x$-$y$ plane, not the curve itself.
– amd
Jul 17 at 21:07
@amd yes that is what i got - projections on xy, yz, zx ... how to go ahead? pls guide
– Magneto
Jul 18 at 8:50
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Show that any plane whose normal lies on cone $(b+c)x^2+(c+a)y^2+(a+b)z^2=0$ cuts the surface $ax^2+by^2+cz^2=1$ is rectangular hyperbola
My attempt:
let $frac xl = frac ym = frac zn$ be normal of plane $lx+my+nz = 0$. Now the normal is generator of given cone. So we get condition $(b+c)l^2+(c+a)m^2+(a+b)n^2=0$ --->(1)
THe given plane cuts the conicoid $ax^2+by^2+cz^2=1$.
The intersecting conic :
Eliminating z between plane equation and conicoid equation
$z = frac-lx-myn $ Subsittuting this in conicoid equation we get,
$ax^2+by^2+cfrac(-lx-my)^2n^2 =1 $ --->(2)
This represents rectangular hyperbola if
coefficient of $x^2$ + coefficient of $y^2$ = 0
$(2)$ implies
$a+b+cfracl^2n^2+cfracm^2n^2 =0 \ fbox $n^2(a+b) + c(l^2+m^2) = 0$$ for conic to be rectangular hyperbola
If i eliminate x, i get $fbox $l^2(b+c)+a(m^2+n^2)=0$$ for conic to be rectangular hyperbola
If i eliminate y, i get $fbox $m^2(c+a)+b(n^2+l^2)=0$$ for conic to be rectangular hyperbola
But what i have is equation (1) with me for conic to be rectangular hyperbola
How to proceed further?
analytic-geometry conic-sections 3d solid-geometry convex-cone
edited Jul 21 at 13:58
Key Flex
4,406525
asked Jul 15 at 12:17
Magneto
795213
Show that any plane whose normal lies on cone $(b+c)x^2+(c+a)y^2+(a+b)z^2=0$ cuts the surface $ax^2+by^2+cz^2=1$ is rectangular hyperbola
My attempt:
let $frac xl = frac ym = frac zn$ be normal of plane $lx+my+nz = 0$. Now the normal is generator of given cone. So we get condition $(b+c)l^2+(c+a)m^2+(a+b)n^2=0$ --->(1)
THe given plane cuts the conicoid $ax^2+by^2+cz^2=1$.
The intersecting conic :
Eliminating z between plane equation and conicoid equation
$z = frac-lx-myn $ Subsittuting this in conicoid equation we get,
$ax^2+by^2+cfrac(-lx-my)^2n^2 =1 $ --->(2)
This represents rectangular hyperbola if
coefficient of $x^2$ + coefficient of $y^2$ = 0
$(2)$ implies
$a+b+cfracl^2n^2+cfracm^2n^2 =0 \ fbox $n^2(a+b) + c(l^2+m^2) = 0$$ for conic to be rectangular hyperbola
If i eliminate x, i get $fbox $l^2(b+c)+a(m^2+n^2)=0$$ for conic to be rectangular hyperbola
If i eliminate y, i get $fbox $m^2(c+a)+b(n^2+l^2)=0$$ for conic to be rectangular hyperbola
But what i have is equation (1) with me for conic to be rectangular hyperbola
How to proceed further?
analytic-geometry conic-sections 3d solid-geometry convex-cone
edited Jul 21 at 13:58
Key Flex
4,406525
asked Jul 15 at 12:17
Magneto
795213
edited Jul 21 at 13:58
Key Flex
4,406525
edited Jul 21 at 13:58
Key Flex
4,406525
edited Jul 21 at 13:58
Key Flex
4,406525
4,406525
asked Jul 15 at 12:17
Magneto
795213
asked Jul 15 at 12:17
Magneto
795213
asked Jul 15 at 12:17
Magneto
795213
795213
Eliminating $z$ from the equations gives you the equation of a projection of the intersection curve onto the $x$-$y$ plane, not the curve itself.
– amd
Jul 17 at 21:07
@amd yes that is what i got - projections on xy, yz, zx ... how to go ahead? pls guide
– Magneto
Jul 18 at 8:50
add a comment |Â
Eliminating $z$ from the equations gives you the equation of a projection of the intersection curve onto the $x$-$y$ plane, not the curve itself.
– amd
Jul 17 at 21:07
@amd yes that is what i got - projections on xy, yz, zx ... how to go ahead? pls guide
– Magneto
Jul 18 at 8:50
Eliminating $z$ from the equations gives you the equation of a projection of the intersection curve onto the $x$-$y$ plane, not the curve itself.
– amd
Jul 17 at 21:07
Eliminating $z$ from the equations gives you the equation of a projection of the intersection curve onto the $x$-$y$ plane, not the curve itself.
– amd
Jul 17 at 21:07
@amd yes that is what i got - projections on xy, yz, zx ... how to go ahead? pls guide
– Magneto
Jul 18 at 8:50
@amd yes that is what i got - projections on xy, yz, zx ... how to go ahead? pls guide
– Magneto
Jul 18 at 8:50
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
First we will make some required transformations to assure that the formula $(b+c)x^2+(c+a)y^2+(a+b)z^2=0;$ represents a real cone.
A real cone with circular base over the $x times y$ plane and vertice at the origin should be represented as
$$
Kto m x^2+n y^2+p z^2 = 0
$$
with $m = n$ and $m > 0, n > 0, p < 0$
so we will choose $a,b,c$ such that
$$
b+c=m\
a+c=n\
a+b=p
$$
in this representation a line $L$
$$
Lto P = rhovec v to
left{
beginarrayrcl
x = rho alpha\
y = rho beta\
z = rho gamma
endarray
right.;;;; P = (x,y,z)
$$
is such that $L in K$ when
$$
malpha^2+nbeta^2+pgamma^2=0
$$
Now with the help of $vec v$ and $vec w = (malpha,nbeta,pgamma)$ Here $left < vec v, vec wright> = 0$, we can build a plane $Pi$
$$
Pito P = lambda vec w + mu left(vec v times vec wright)
$$
as required. Given now the surface
$$
Sto ax^2+by^2+cy^2-1=0equiv frac12 x^2 (-m+n+p)+frac12 y^2 (m-n+p)+frac12 z^2 (m+n-p)-1 = 0
$$
the intersection $Scap Pi$ is obtained
$$
left(ScircPiright)(lambda,mu) = frac12 left(-4 (m-n-p) (alpha lambda m+beta gamma mu (n-p))^2+4 (m+n-p) (alpha beta mu (m-n)+gamma lambda p)^2+4 (m-n+p)
(alpha gamma mu (p-m)+beta lambda n)^2-2right)= 0
$$
but we can introduce a slight normalization to simplify the representation by considering
$$
malpha^2+nbeta^2+pgamma^2=0\
alpha^2+beta^2+gamma^2 = 1\
beta = beta_0 = 0\
m = n\
$$
and thus we have the intersection curve on $Pi$ as
$$
2np^2lambda^2-2np^2mu^2-1=0
$$
which is a rectangular hyperbola.
answered Jul 21 at 12:54
Cesareo
5,7922412
i did not understand following -- (1) Any 2 points on line will be $(m alpha , nbeta, p gamma), ( alpha , beta, gamma)$ these form 2 points on normal to plane. But these apparantly seem perpendicular. How is that possible? (2)I did not understand how did u write Surface equaiton and plane equation. Plane equatin = $vec n/(vec r- vec r_0)=0$ . Here $vec v cross vec w $ gives line that is in given plane. Then i did not understand meaning of $w+v cross w$
– Magneto
Jul 21 at 13:40
@Magneto $L in K Rightarrow rho^2(malpha^2+nbeta^2+pgamma^2)=0$ Here $vec v = (alpha,beta,gamma)$ has the line direction and $vec w =(malpha,nbeta,pgamma)$ is orthogonal to $vec v$ then the plane $Pito P = lambda vec v + mu vec v timesvec w$ by construction, is the required plane.
– Cesareo
Jul 21 at 14:12
@Magneto I didn't understand the formula you wrote $vec n /(vec r-vec r_0)$ ??
– Cesareo
Jul 21 at 14:15
i understood the proof now. Thank you. Too good. But can u tell me where i went wrong? i could not find out. Pls see it once. It helps in rectifying my mistake
– Magneto
Jul 21 at 14:22
@Magneto I think that the real problem is to establish the conditions in $a,b,c$ for $(b+c)x^2+(c+a)y^2+(a+b)z^2=0$ to be representing a real cone. Try your way including before those conditions.
– Cesareo
Jul 21 at 15:07
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
First we will make some required transformations to assure that the formula $(b+c)x^2+(c+a)y^2+(a+b)z^2=0;$ represents a real cone.
A real cone with circular base over the $x times y$ plane and vertice at the origin should be represented as
$$
Kto m x^2+n y^2+p z^2 = 0
$$
with $m = n$ and $m > 0, n > 0, p < 0$
so we will choose $a,b,c$ such that
$$
b+c=m\
a+c=n\
a+b=p
$$
in this representation a line $L$
$$
Lto P = rhovec v to
left{
beginarrayrcl
x = rho alpha\
y = rho beta\
z = rho gamma
endarray
right.;;;; P = (x,y,z)
$$
is such that $L in K$ when
$$
malpha^2+nbeta^2+pgamma^2=0
$$
Now with the help of $vec v$ and $vec w = (malpha,nbeta,pgamma)$ Here $left < vec v, vec wright> = 0$, we can build a plane $Pi$
$$
Pito P = lambda vec w + mu left(vec v times vec wright)
$$
as required. Given now the surface
$$
Sto ax^2+by^2+cy^2-1=0equiv frac12 x^2 (-m+n+p)+frac12 y^2 (m-n+p)+frac12 z^2 (m+n-p)-1 = 0
$$
the intersection $Scap Pi$ is obtained
$$
left(ScircPiright)(lambda,mu) = frac12 left(-4 (m-n-p) (alpha lambda m+beta gamma mu (n-p))^2+4 (m+n-p) (alpha beta mu (m-n)+gamma lambda p)^2+4 (m-n+p)
(alpha gamma mu (p-m)+beta lambda n)^2-2right)= 0
$$
but we can introduce a slight normalization to simplify the representation by considering
$$
malpha^2+nbeta^2+pgamma^2=0\
alpha^2+beta^2+gamma^2 = 1\
beta = beta_0 = 0\
m = n\
$$
and thus we have the intersection curve on $Pi$ as
$$
2np^2lambda^2-2np^2mu^2-1=0
$$
which is a rectangular hyperbola.
answered Jul 21 at 12:54
Cesareo
5,7922412
i did not understand following -- (1) Any 2 points on line will be $(m alpha , nbeta, p gamma), ( alpha , beta, gamma)$ these form 2 points on normal to plane. But these apparantly seem perpendicular. How is that possible? (2)I did not understand how did u write Surface equaiton and plane equation. Plane equatin = $vec n/(vec r- vec r_0)=0$ . Here $vec v cross vec w $ gives line that is in given plane. Then i did not understand meaning of $w+v cross w$
– Magneto
Jul 21 at 13:40
@Magneto $L in K Rightarrow rho^2(malpha^2+nbeta^2+pgamma^2)=0$ Here $vec v = (alpha,beta,gamma)$ has the line direction and $vec w =(malpha,nbeta,pgamma)$ is orthogonal to $vec v$ then the plane $Pito P = lambda vec v + mu vec v timesvec w$ by construction, is the required plane.
– Cesareo
Jul 21 at 14:12
@Magneto I didn't understand the formula you wrote $vec n /(vec r-vec r_0)$ ??
– Cesareo
Jul 21 at 14:15
i understood the proof now. Thank you. Too good. But can u tell me where i went wrong? i could not find out. Pls see it once. It helps in rectifying my mistake
– Magneto
Jul 21 at 14:22
@Magneto I think that the real problem is to establish the conditions in $a,b,c$ for $(b+c)x^2+(c+a)y^2+(a+b)z^2=0$ to be representing a real cone. Try your way including before those conditions.
– Cesareo
Jul 21 at 15:07
 |Â
show 1 more comment
up vote
1
down vote
accepted
First we will make some required transformations to assure that the formula $(b+c)x^2+(c+a)y^2+(a+b)z^2=0;$ represents a real cone.
A real cone with circular base over the $x times y$ plane and vertice at the origin should be represented as
$$
Kto m x^2+n y^2+p z^2 = 0
$$
with $m = n$ and $m > 0, n > 0, p < 0$
so we will choose $a,b,c$ such that
$$
b+c=m\
a+c=n\
a+b=p
$$
in this representation a line $L$
$$
Lto P = rhovec v to
left{
beginarrayrcl
x = rho alpha\
y = rho beta\
z = rho gamma
endarray
right.;;;; P = (x,y,z)
$$
is such that $L in K$ when
$$
malpha^2+nbeta^2+pgamma^2=0
$$
Now with the help of $vec v$ and $vec w = (malpha,nbeta,pgamma)$ Here $left < vec v, vec wright> = 0$, we can build a plane $Pi$
$$
Pito P = lambda vec w + mu left(vec v times vec wright)
$$
as required. Given now the surface
$$
Sto ax^2+by^2+cy^2-1=0equiv frac12 x^2 (-m+n+p)+frac12 y^2 (m-n+p)+frac12 z^2 (m+n-p)-1 = 0
$$
the intersection $Scap Pi$ is obtained
$$
left(ScircPiright)(lambda,mu) = frac12 left(-4 (m-n-p) (alpha lambda m+beta gamma mu (n-p))^2+4 (m+n-p) (alpha beta mu (m-n)+gamma lambda p)^2+4 (m-n+p)
(alpha gamma mu (p-m)+beta lambda n)^2-2right)= 0
$$
but we can introduce a slight normalization to simplify the representation by considering
$$
malpha^2+nbeta^2+pgamma^2=0\
alpha^2+beta^2+gamma^2 = 1\
beta = beta_0 = 0\
m = n\
$$
and thus we have the intersection curve on $Pi$ as
$$
2np^2lambda^2-2np^2mu^2-1=0
$$
which is a rectangular hyperbola.
answered Jul 21 at 12:54
Cesareo
5,7922412
i did not understand following -- (1) Any 2 points on line will be $(m alpha , nbeta, p gamma), ( alpha , beta, gamma)$ these form 2 points on normal to plane. But these apparantly seem perpendicular. How is that possible? (2)I did not understand how did u write Surface equaiton and plane equation. Plane equatin = $vec n/(vec r- vec r_0)=0$ . Here $vec v cross vec w $ gives line that is in given plane. Then i did not understand meaning of $w+v cross w$
– Magneto
Jul 21 at 13:40
@Magneto $L in K Rightarrow rho^2(malpha^2+nbeta^2+pgamma^2)=0$ Here $vec v = (alpha,beta,gamma)$ has the line direction and $vec w =(malpha,nbeta,pgamma)$ is orthogonal to $vec v$ then the plane $Pito P = lambda vec v + mu vec v timesvec w$ by construction, is the required plane.
– Cesareo
Jul 21 at 14:12
@Magneto I didn't understand the formula you wrote $vec n /(vec r-vec r_0)$ ??
– Cesareo
Jul 21 at 14:15
i understood the proof now. Thank you. Too good. But can u tell me where i went wrong? i could not find out. Pls see it once. It helps in rectifying my mistake
– Magneto
Jul 21 at 14:22
@Magneto I think that the real problem is to establish the conditions in $a,b,c$ for $(b+c)x^2+(c+a)y^2+(a+b)z^2=0$ to be representing a real cone. Try your way including before those conditions.
– Cesareo
Jul 21 at 15:07
 |Â
show 1 more comment
up vote
1
down vote
accepted
up vote
1
down vote
accepted
First we will make some required transformations to assure that the formula $(b+c)x^2+(c+a)y^2+(a+b)z^2=0;$ represents a real cone.
A real cone with circular base over the $x times y$ plane and vertice at the origin should be represented as
$$
Kto m x^2+n y^2+p z^2 = 0
$$
with $m = n$ and $m > 0, n > 0, p < 0$
so we will choose $a,b,c$ such that
$$
b+c=m\
a+c=n\
a+b=p
$$
in this representation a line $L$
$$
Lto P = rhovec v to
left{
beginarrayrcl
x = rho alpha\
y = rho beta\
z = rho gamma
endarray
right.;;;; P = (x,y,z)
$$
is such that $L in K$ when
$$
malpha^2+nbeta^2+pgamma^2=0
$$
Now with the help of $vec v$ and $vec w = (malpha,nbeta,pgamma)$ Here $left < vec v, vec wright> = 0$, we can build a plane $Pi$
$$
Pito P = lambda vec w + mu left(vec v times vec wright)
$$
as required. Given now the surface
$$
Sto ax^2+by^2+cy^2-1=0equiv frac12 x^2 (-m+n+p)+frac12 y^2 (m-n+p)+frac12 z^2 (m+n-p)-1 = 0
$$
the intersection $Scap Pi$ is obtained
$$
left(ScircPiright)(lambda,mu) = frac12 left(-4 (m-n-p) (alpha lambda m+beta gamma mu (n-p))^2+4 (m+n-p) (alpha beta mu (m-n)+gamma lambda p)^2+4 (m-n+p)
(alpha gamma mu (p-m)+beta lambda n)^2-2right)= 0
$$
but we can introduce a slight normalization to simplify the representation by considering
$$
malpha^2+nbeta^2+pgamma^2=0\
alpha^2+beta^2+gamma^2 = 1\
beta = beta_0 = 0\
m = n\
$$
and thus we have the intersection curve on $Pi$ as
$$
2np^2lambda^2-2np^2mu^2-1=0
$$
which is a rectangular hyperbola.
answered Jul 21 at 12:54
Cesareo
5,7922412
First we will make some required transformations to assure that the formula $(b+c)x^2+(c+a)y^2+(a+b)z^2=0;$ represents a real cone.
A real cone with circular base over the $x times y$ plane and vertice at the origin should be represented as
$$
Kto m x^2+n y^2+p z^2 = 0
$$
with $m = n$ and $m > 0, n > 0, p < 0$
so we will choose $a,b,c$ such that
$$
b+c=m\
a+c=n\
a+b=p
$$
in this representation a line $L$
$$
Lto P = rhovec v to
left{
beginarrayrcl
x = rho alpha\
y = rho beta\
z = rho gamma
endarray
right.;;;; P = (x,y,z)
$$
is such that $L in K$ when
$$
malpha^2+nbeta^2+pgamma^2=0
$$
Now with the help of $vec v$ and $vec w = (malpha,nbeta,pgamma)$ Here $left < vec v, vec wright> = 0$, we can build a plane $Pi$
$$
Pito P = lambda vec w + mu left(vec v times vec wright)
$$
as required. Given now the surface
$$
Sto ax^2+by^2+cy^2-1=0equiv frac12 x^2 (-m+n+p)+frac12 y^2 (m-n+p)+frac12 z^2 (m+n-p)-1 = 0
$$
the intersection $Scap Pi$ is obtained
$$
left(ScircPiright)(lambda,mu) = frac12 left(-4 (m-n-p) (alpha lambda m+beta gamma mu (n-p))^2+4 (m+n-p) (alpha beta mu (m-n)+gamma lambda p)^2+4 (m-n+p)
(alpha gamma mu (p-m)+beta lambda n)^2-2right)= 0
$$
but we can introduce a slight normalization to simplify the representation by considering
$$
malpha^2+nbeta^2+pgamma^2=0\
alpha^2+beta^2+gamma^2 = 1\
beta = beta_0 = 0\
m = n\
$$
and thus we have the intersection curve on $Pi$ as
$$
2np^2lambda^2-2np^2mu^2-1=0
$$
which is a rectangular hyperbola.
answered Jul 21 at 12:54
Cesareo
5,7922412
edited Jul 21 at 14:16
answered Jul 21 at 12:54
Cesareo
5,7922412
answered Jul 21 at 12:54
Cesareo
5,7922412
answered Jul 21 at 12:54
Cesareo
5,7922412
5,7922412
i did not understand following -- (1) Any 2 points on line will be $(m alpha , nbeta, p gamma), ( alpha , beta, gamma)$ these form 2 points on normal to plane. But these apparantly seem perpendicular. How is that possible? (2)I did not understand how did u write Surface equaiton and plane equation. Plane equatin = $vec n/(vec r- vec r_0)=0$ . Here $vec v cross vec w $ gives line that is in given plane. Then i did not understand meaning of $w+v cross w$
– Magneto
Jul 21 at 13:40
@Magneto $L in K Rightarrow rho^2(malpha^2+nbeta^2+pgamma^2)=0$ Here $vec v = (alpha,beta,gamma)$ has the line direction and $vec w =(malpha,nbeta,pgamma)$ is orthogonal to $vec v$ then the plane $Pito P = lambda vec v + mu vec v timesvec w$ by construction, is the required plane.
– Cesareo
Jul 21 at 14:12
@Magneto I didn't understand the formula you wrote $vec n /(vec r-vec r_0)$ ??
– Cesareo
Jul 21 at 14:15
i understood the proof now. Thank you. Too good. But can u tell me where i went wrong? i could not find out. Pls see it once. It helps in rectifying my mistake
– Magneto
Jul 21 at 14:22
@Magneto I think that the real problem is to establish the conditions in $a,b,c$ for $(b+c)x^2+(c+a)y^2+(a+b)z^2=0$ to be representing a real cone. Try your way including before those conditions.
– Cesareo
Jul 21 at 15:07
 |Â
show 1 more comment
i did not understand following -- (1) Any 2 points on line will be $(m alpha , nbeta, p gamma), ( alpha , beta, gamma)$ these form 2 points on normal to plane. But these apparantly seem perpendicular. How is that possible? (2)I did not understand how did u write Surface equaiton and plane equation. Plane equatin = $vec n/(vec r- vec r_0)=0$ . Here $vec v cross vec w $ gives line that is in given plane. Then i did not understand meaning of $w+v cross w$
– Magneto
Jul 21 at 13:40
@Magneto $L in K Rightarrow rho^2(malpha^2+nbeta^2+pgamma^2)=0$ Here $vec v = (alpha,beta,gamma)$ has the line direction and $vec w =(malpha,nbeta,pgamma)$ is orthogonal to $vec v$ then the plane $Pito P = lambda vec v + mu vec v timesvec w$ by construction, is the required plane.
– Cesareo
Jul 21 at 14:12
@Magneto I didn't understand the formula you wrote $vec n /(vec r-vec r_0)$ ??
– Cesareo
Jul 21 at 14:15
i understood the proof now. Thank you. Too good. But can u tell me where i went wrong? i could not find out. Pls see it once. It helps in rectifying my mistake
– Magneto
Jul 21 at 14:22
@Magneto I think that the real problem is to establish the conditions in $a,b,c$ for $(b+c)x^2+(c+a)y^2+(a+b)z^2=0$ to be representing a real cone. Try your way including before those conditions.
– Cesareo
Jul 21 at 15:07
i did not understand following -- (1) Any 2 points on line will be $(m alpha , nbeta, p gamma), ( alpha , beta, gamma)$ these form 2 points on normal to plane. But these apparantly seem perpendicular. How is that possible? (2)I did not understand how did u write Surface equaiton and plane equation. Plane equatin = $vec n/(vec r- vec r_0)=0$ . Here $vec v cross vec w $ gives line that is in given plane. Then i did not understand meaning of $w+v cross w$
– Magneto
Jul 21 at 13:40
i did not understand following -- (1) Any 2 points on line will be $(m alpha , nbeta, p gamma), ( alpha , beta, gamma)$ these form 2 points on normal to plane. But these apparantly seem perpendicular. How is that possible? (2)I did not understand how did u write Surface equaiton and plane equation. Plane equatin = $vec n/(vec r- vec r_0)=0$ . Here $vec v cross vec w $ gives line that is in given plane. Then i did not understand meaning of $w+v cross w$
– Magneto
Jul 21 at 13:40
@Magneto $L in K Rightarrow rho^2(malpha^2+nbeta^2+pgamma^2)=0$ Here $vec v = (alpha,beta,gamma)$ has the line direction and $vec w =(malpha,nbeta,pgamma)$ is orthogonal to $vec v$ then the plane $Pito P = lambda vec v + mu vec v timesvec w$ by construction, is the required plane.
– Cesareo
Jul 21 at 14:12
@Magneto $L in K Rightarrow rho^2(malpha^2+nbeta^2+pgamma^2)=0$ Here $vec v = (alpha,beta,gamma)$ has the line direction and $vec w =(malpha,nbeta,pgamma)$ is orthogonal to $vec v$ then the plane $Pito P = lambda vec v + mu vec v timesvec w$ by construction, is the required plane.
– Cesareo
Jul 21 at 14:12
@Magneto I didn't understand the formula you wrote $vec n /(vec r-vec r_0)$ ??
– Cesareo
Jul 21 at 14:15
@Magneto I didn't understand the formula you wrote $vec n /(vec r-vec r_0)$ ??
– Cesareo
Jul 21 at 14:15
i understood the proof now. Thank you. Too good. But can u tell me where i went wrong? i could not find out. Pls see it once. It helps in rectifying my mistake
– Magneto
Jul 21 at 14:22
i understood the proof now. Thank you. Too good. But can u tell me where i went wrong? i could not find out. Pls see it once. It helps in rectifying my mistake
– Magneto
Jul 21 at 14:22
@Magneto I think that the real problem is to establish the conditions in $a,b,c$ for $(b+c)x^2+(c+a)y^2+(a+b)z^2=0$ to be representing a real cone. Try your way including before those conditions.
– Cesareo
Jul 21 at 15:07
@Magneto I think that the real problem is to establish the conditions in $a,b,c$ for $(b+c)x^2+(c+a)y^2+(a+b)z^2=0$ to be representing a real cone. Try your way including before those conditions.
– Cesareo
Jul 21 at 15:07
 |Â
show 1 more comment
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Eliminating $z$ from the equations gives you the equation of a projection of the intersection curve onto the $x$-$y$ plane, not the curve itself.
– amd
Jul 17 at 21:07
@amd yes that is what i got - projections on xy, yz, zx ... how to go ahead? pls guide
– Magneto
Jul 18 at 8:50