Mixing
What is the intersection of a plane and a sphere? Is it necessarily a circle? Can it be an ellipse?
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 3.27,Exer 3.28
(Exer 3.27) Consider the plane $H$ determined by the equation $x + y -z = 0$. What is a unit normal vector to $H$? Compute the image of $E:=Hcap mathbb S^2$ under the stereographic projection $Phi$.
Asked here and I answered agreeing with OP that the image is the circle $T: = $, but now I ask:
Q1. What's the relevance of asking about the unit normal vector?
I computed the unit normal vectors to be $[1,1,-1]fracpm 1sqrt3$. I observe their terminal points to be on the unit sphere.
Q2. Is $(2)$ below wrong?
$$forall textplanes J, exists textcircle C in mathbb S^2 : C = J cap mathbb S^2 tag2$$
To clarify, compare that (2) is different from (1) below. We know that $$forall textcircles C in mathbb S^2, exists textplane J : C = J cap mathbb S^2. tag1$$
Q3. Do I go wrong in attempting to disprove (2) as follows?
I think $(2)$ contradicts Exer 3.27: I observe that $E$, whose image is the circle $T$, is both
the intersection (WA) of a plane and $mathbb S^2$ and yet
an ellipse (WA) --> Or is it? Sphere-plane intersection seems to be circle (Wiki).
If $(2)$ is true, then for $J=H, exists C in mathbb S^2: C = H cap mathbb S^2 =: E ↯$.
Q3. How is Exer 3.27 consistent with the succeeding Exer 3.28?
-
(Exer 3.28) Prove that every circle in $hatmathbb C$ is the image of some circle in $mathbb S^2$ under stereographic projection $Phi$.
-
I rephrase:
$$forall R in hatmathbb C, exists C in mathbb S^2 : Phi(C)=R$$
-
Now consider $T$, the circle in Exer 3.27 s.t. $Phi(E)=T$.
It seems that by Exer 3.28, for $R= T, exists C in mathbb S^2 : Phi(C)=T$.
$$therefore, Phi(E) = Phi(C) textwhile E ne C text, but Phi textis bijective ↯ ?$$
complex-analysis geometry multivariable-calculus vector-analysis
asked Jul 30 at 22:06
BCLC
6,99821973
add a comment |Â
up vote
0
down vote
favorite
A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 3.27,Exer 3.28
(Exer 3.27) Consider the plane $H$ determined by the equation $x + y -z = 0$. What is a unit normal vector to $H$? Compute the image of $E:=Hcap mathbb S^2$ under the stereographic projection $Phi$.
Asked here and I answered agreeing with OP that the image is the circle $T: = $, but now I ask:
Q1. What's the relevance of asking about the unit normal vector?
I computed the unit normal vectors to be $[1,1,-1]fracpm 1sqrt3$. I observe their terminal points to be on the unit sphere.
Q2. Is $(2)$ below wrong?
$$forall textplanes J, exists textcircle C in mathbb S^2 : C = J cap mathbb S^2 tag2$$
To clarify, compare that (2) is different from (1) below. We know that $$forall textcircles C in mathbb S^2, exists textplane J : C = J cap mathbb S^2. tag1$$
Q3. Do I go wrong in attempting to disprove (2) as follows?
I think $(2)$ contradicts Exer 3.27: I observe that $E$, whose image is the circle $T$, is both
the intersection (WA) of a plane and $mathbb S^2$ and yet
an ellipse (WA) --> Or is it? Sphere-plane intersection seems to be circle (Wiki).
If $(2)$ is true, then for $J=H, exists C in mathbb S^2: C = H cap mathbb S^2 =: E ↯$.
Q3. How is Exer 3.27 consistent with the succeeding Exer 3.28?
-
(Exer 3.28) Prove that every circle in $hatmathbb C$ is the image of some circle in $mathbb S^2$ under stereographic projection $Phi$.
-
I rephrase:
$$forall R in hatmathbb C, exists C in mathbb S^2 : Phi(C)=R$$
-
Now consider $T$, the circle in Exer 3.27 s.t. $Phi(E)=T$.
It seems that by Exer 3.28, for $R= T, exists C in mathbb S^2 : Phi(C)=T$.
$$therefore, Phi(E) = Phi(C) textwhile E ne C text, but Phi textis bijective ↯ ?$$
complex-analysis geometry multivariable-calculus vector-analysis
asked Jul 30 at 22:06
BCLC
6,99821973
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 3.27,Exer 3.28
(Exer 3.27) Consider the plane $H$ determined by the equation $x + y -z = 0$. What is a unit normal vector to $H$? Compute the image of $E:=Hcap mathbb S^2$ under the stereographic projection $Phi$.
Asked here and I answered agreeing with OP that the image is the circle $T: = $, but now I ask:
Q1. What's the relevance of asking about the unit normal vector?
I computed the unit normal vectors to be $[1,1,-1]fracpm 1sqrt3$. I observe their terminal points to be on the unit sphere.
Q2. Is $(2)$ below wrong?
$$forall textplanes J, exists textcircle C in mathbb S^2 : C = J cap mathbb S^2 tag2$$
To clarify, compare that (2) is different from (1) below. We know that $$forall textcircles C in mathbb S^2, exists textplane J : C = J cap mathbb S^2. tag1$$
Q3. Do I go wrong in attempting to disprove (2) as follows?
I think $(2)$ contradicts Exer 3.27: I observe that $E$, whose image is the circle $T$, is both
the intersection (WA) of a plane and $mathbb S^2$ and yet
an ellipse (WA) --> Or is it? Sphere-plane intersection seems to be circle (Wiki).
If $(2)$ is true, then for $J=H, exists C in mathbb S^2: C = H cap mathbb S^2 =: E ↯$.
Q3. How is Exer 3.27 consistent with the succeeding Exer 3.28?
-
(Exer 3.28) Prove that every circle in $hatmathbb C$ is the image of some circle in $mathbb S^2$ under stereographic projection $Phi$.
-
I rephrase:
$$forall R in hatmathbb C, exists C in mathbb S^2 : Phi(C)=R$$
-
Now consider $T$, the circle in Exer 3.27 s.t. $Phi(E)=T$.
It seems that by Exer 3.28, for $R= T, exists C in mathbb S^2 : Phi(C)=T$.
$$therefore, Phi(E) = Phi(C) textwhile E ne C text, but Phi textis bijective ↯ ?$$
complex-analysis geometry multivariable-calculus vector-analysis
asked Jul 30 at 22:06
BCLC
6,99821973
A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 3.27,Exer 3.28
(Exer 3.27) Consider the plane $H$ determined by the equation $x + y -z = 0$. What is a unit normal vector to $H$? Compute the image of $E:=Hcap mathbb S^2$ under the stereographic projection $Phi$.
Asked here and I answered agreeing with OP that the image is the circle $T: = $, but now I ask:
Q1. What's the relevance of asking about the unit normal vector?
I computed the unit normal vectors to be $[1,1,-1]fracpm 1sqrt3$. I observe their terminal points to be on the unit sphere.
Q2. Is $(2)$ below wrong?
$$forall textplanes J, exists textcircle C in mathbb S^2 : C = J cap mathbb S^2 tag2$$
To clarify, compare that (2) is different from (1) below. We know that $$forall textcircles C in mathbb S^2, exists textplane J : C = J cap mathbb S^2. tag1$$
Q3. Do I go wrong in attempting to disprove (2) as follows?
I think $(2)$ contradicts Exer 3.27: I observe that $E$, whose image is the circle $T$, is both
the intersection (WA) of a plane and $mathbb S^2$ and yet
an ellipse (WA) --> Or is it? Sphere-plane intersection seems to be circle (Wiki).
If $(2)$ is true, then for $J=H, exists C in mathbb S^2: C = H cap mathbb S^2 =: E ↯$.
Q3. How is Exer 3.27 consistent with the succeeding Exer 3.28?
-
(Exer 3.28) Prove that every circle in $hatmathbb C$ is the image of some circle in $mathbb S^2$ under stereographic projection $Phi$.
-
I rephrase:
$$forall R in hatmathbb C, exists C in mathbb S^2 : Phi(C)=R$$
-
Now consider $T$, the circle in Exer 3.27 s.t. $Phi(E)=T$.
It seems that by Exer 3.28, for $R= T, exists C in mathbb S^2 : Phi(C)=T$.
$$therefore, Phi(E) = Phi(C) textwhile E ne C text, but Phi textis bijective ↯ ?$$
complex-analysis geometry multivariable-calculus vector-analysis
asked Jul 30 at 22:06
BCLC
6,99821973
edited 6 hours ago
asked Jul 30 at 22:06
BCLC
6,99821973
asked Jul 30 at 22:06
BCLC
6,99821973
asked Jul 30 at 22:06
BCLC
6,99821973
6,99821973
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
0
down vote
As already verified, all intersections of a circle and plane are circles either geodesic/great or small. I avoid symbols in the answer here.
If 2-parameter surface normal coincides with 1-parameter circle normal then the intersection is a great circle. The intersection is a geodesic.
From differential geometry standpoint $psi$ is between arc and meridian. Cylindrical coordinates $(r,z)$ Clairaut constant derivative $dfracd(r sin psi)dz=0$ forms with respect to an arbitrary North-South polar axis.
If the 2-parameter surface normal does not coincide with 1-parameter circle normal, but makes an angle $gamma$ of relative latitude then the intersection is a small circle. Intersection is like a parallel circle but not an equator.
Clairaut constant derivative $dfracd(r sin psi)dz$ is a constant $ =tan gamma$ forms with respect to an arbitrary North-South polar axis.
In stereographic projection all non-equatorial sections of a plane passing through North pole of the sphere at angle $gamma$ to North-South polar plane are small circles.
edited 6 hours ago
BCLC
6,99821973
answered 7 hours ago
Narasimham
20.2k51957
Narasimham, is $E$ indeed an ellipse?
– BCLC
6 hours ago
Narasimham, I posted an answer. How is it please?
– BCLC
5 hours ago
add a comment |Â
up vote
0
down vote
(Q1) I suspect some relationship is being illustrated. The 1st 2 coordinates in $[1,1,-1]$ are the centre of the circle $Phi(E)$ while the norm of $[1,1,-1]$ s the radius of $Phi(E)$.
(Q2) Yes if the planes intersect $mathbb S^2$ or $emptyset$ is considered a circle in $mathbb S^2$.
(Q3), (Q4) E is not an ellipse. It is a circle in 3D (see here too) parametrised as (WA)
$$x =sqrtfrac 2 3 cos[t]$$
$$y = -sqrtfrac 2 4 sin[t] - sqrtfrac 1 6 cos[t]$$
$$z = -sqrtfrac 2 4 sin[t] + sqrtfrac 1 6 cos[t]$$
Observe that while $x^2+y^2+xy=frac12$ is an ellipse, $x^2+y^2+xy=frac12 wedge z= x+y$ is a circle.
answered 5 hours ago
BCLC
6,99821973
1
I guess your major confusion was WA saying ellipse on the intersection. When you set $z=x+y$ into $x^2+y^2+z^2=1$ you get indeed an ellipse, but this is the projection of the intersection onto $xy$-plane, i.e. all the coordinates $(x,y)$ that satisfy both equations.
– A.Γ.
2 hours ago
@A.Γ. I was thinking projection but unable to express precisely. Thanks! Any idea for (Q1) please?
– BCLC
2 hours ago
1
To see that all intersections are circles: 1. Intersection of a plane $z=textconst$ and the sphere is a circle (or empty set). 2. Any plane can be rotated to become $z=textconst$, and sphere is rotation invariant.
– A.Γ.
2 hours ago
@A.Γ. cool. Thanks. How about Q1 please: what's the purpose of unit normal vector?
– BCLC
2 hours ago
1
Is the image of $Phi$ the plane $z=0$ or $z=-1$ (there are different conventions)?
– A.Γ.
1 hour ago
 |Â
show 1 more comment
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
As already verified, all intersections of a circle and plane are circles either geodesic/great or small. I avoid symbols in the answer here.
If 2-parameter surface normal coincides with 1-parameter circle normal then the intersection is a great circle. The intersection is a geodesic.
From differential geometry standpoint $psi$ is between arc and meridian. Cylindrical coordinates $(r,z)$ Clairaut constant derivative $dfracd(r sin psi)dz=0$ forms with respect to an arbitrary North-South polar axis.
If the 2-parameter surface normal does not coincide with 1-parameter circle normal, but makes an angle $gamma$ of relative latitude then the intersection is a small circle. Intersection is like a parallel circle but not an equator.
Clairaut constant derivative $dfracd(r sin psi)dz$ is a constant $ =tan gamma$ forms with respect to an arbitrary North-South polar axis.
In stereographic projection all non-equatorial sections of a plane passing through North pole of the sphere at angle $gamma$ to North-South polar plane are small circles.
edited 6 hours ago
BCLC
6,99821973
answered 7 hours ago
Narasimham
20.2k51957
Narasimham, is $E$ indeed an ellipse?
– BCLC
6 hours ago
Narasimham, I posted an answer. How is it please?
– BCLC
5 hours ago
add a comment |Â
up vote
0
down vote
As already verified, all intersections of a circle and plane are circles either geodesic/great or small. I avoid symbols in the answer here.
If 2-parameter surface normal coincides with 1-parameter circle normal then the intersection is a great circle. The intersection is a geodesic.
From differential geometry standpoint $psi$ is between arc and meridian. Cylindrical coordinates $(r,z)$ Clairaut constant derivative $dfracd(r sin psi)dz=0$ forms with respect to an arbitrary North-South polar axis.
If the 2-parameter surface normal does not coincide with 1-parameter circle normal, but makes an angle $gamma$ of relative latitude then the intersection is a small circle. Intersection is like a parallel circle but not an equator.
Clairaut constant derivative $dfracd(r sin psi)dz$ is a constant $ =tan gamma$ forms with respect to an arbitrary North-South polar axis.
In stereographic projection all non-equatorial sections of a plane passing through North pole of the sphere at angle $gamma$ to North-South polar plane are small circles.
edited 6 hours ago
BCLC
6,99821973
answered 7 hours ago
Narasimham
20.2k51957
Narasimham, is $E$ indeed an ellipse?
– BCLC
6 hours ago
Narasimham, I posted an answer. How is it please?
– BCLC
5 hours ago
add a comment |Â
up vote
0
down vote
up vote
0
down vote
As already verified, all intersections of a circle and plane are circles either geodesic/great or small. I avoid symbols in the answer here.
If 2-parameter surface normal coincides with 1-parameter circle normal then the intersection is a great circle. The intersection is a geodesic.
From differential geometry standpoint $psi$ is between arc and meridian. Cylindrical coordinates $(r,z)$ Clairaut constant derivative $dfracd(r sin psi)dz=0$ forms with respect to an arbitrary North-South polar axis.
If the 2-parameter surface normal does not coincide with 1-parameter circle normal, but makes an angle $gamma$ of relative latitude then the intersection is a small circle. Intersection is like a parallel circle but not an equator.
Clairaut constant derivative $dfracd(r sin psi)dz$ is a constant $ =tan gamma$ forms with respect to an arbitrary North-South polar axis.
In stereographic projection all non-equatorial sections of a plane passing through North pole of the sphere at angle $gamma$ to North-South polar plane are small circles.
edited 6 hours ago
BCLC
6,99821973
answered 7 hours ago
Narasimham
20.2k51957
As already verified, all intersections of a circle and plane are circles either geodesic/great or small. I avoid symbols in the answer here.
If 2-parameter surface normal coincides with 1-parameter circle normal then the intersection is a great circle. The intersection is a geodesic.
From differential geometry standpoint $psi$ is between arc and meridian. Cylindrical coordinates $(r,z)$ Clairaut constant derivative $dfracd(r sin psi)dz=0$ forms with respect to an arbitrary North-South polar axis.
If the 2-parameter surface normal does not coincide with 1-parameter circle normal, but makes an angle $gamma$ of relative latitude then the intersection is a small circle. Intersection is like a parallel circle but not an equator.
Clairaut constant derivative $dfracd(r sin psi)dz$ is a constant $ =tan gamma$ forms with respect to an arbitrary North-South polar axis.
In stereographic projection all non-equatorial sections of a plane passing through North pole of the sphere at angle $gamma$ to North-South polar plane are small circles.
edited 6 hours ago
BCLC
6,99821973
answered 7 hours ago
Narasimham
20.2k51957
edited 6 hours ago
BCLC
6,99821973
edited 6 hours ago
BCLC
6,99821973
edited 6 hours ago
BCLC
6,99821973
6,99821973
answered 7 hours ago
Narasimham
20.2k51957
answered 7 hours ago
Narasimham
20.2k51957
answered 7 hours ago
Narasimham
20.2k51957
20.2k51957
Narasimham, is $E$ indeed an ellipse?
– BCLC
6 hours ago
Narasimham, I posted an answer. How is it please?
– BCLC
5 hours ago
add a comment |Â
Narasimham, is $E$ indeed an ellipse?
– BCLC
6 hours ago
Narasimham, I posted an answer. How is it please?
– BCLC
5 hours ago
Narasimham, is $E$ indeed an ellipse?
– BCLC
6 hours ago
Narasimham, is $E$ indeed an ellipse?
– BCLC
6 hours ago
Narasimham, I posted an answer. How is it please?
– BCLC
5 hours ago
Narasimham, I posted an answer. How is it please?
– BCLC
5 hours ago
add a comment |Â
up vote
0
down vote
(Q1) I suspect some relationship is being illustrated. The 1st 2 coordinates in $[1,1,-1]$ are the centre of the circle $Phi(E)$ while the norm of $[1,1,-1]$ s the radius of $Phi(E)$.
(Q2) Yes if the planes intersect $mathbb S^2$ or $emptyset$ is considered a circle in $mathbb S^2$.
(Q3), (Q4) E is not an ellipse. It is a circle in 3D (see here too) parametrised as (WA)
$$x =sqrtfrac 2 3 cos[t]$$
$$y = -sqrtfrac 2 4 sin[t] - sqrtfrac 1 6 cos[t]$$
$$z = -sqrtfrac 2 4 sin[t] + sqrtfrac 1 6 cos[t]$$
Observe that while $x^2+y^2+xy=frac12$ is an ellipse, $x^2+y^2+xy=frac12 wedge z= x+y$ is a circle.
answered 5 hours ago
BCLC
6,99821973
1
I guess your major confusion was WA saying ellipse on the intersection. When you set $z=x+y$ into $x^2+y^2+z^2=1$ you get indeed an ellipse, but this is the projection of the intersection onto $xy$-plane, i.e. all the coordinates $(x,y)$ that satisfy both equations.
– A.Γ.
2 hours ago
@A.Γ. I was thinking projection but unable to express precisely. Thanks! Any idea for (Q1) please?
– BCLC
2 hours ago
1
To see that all intersections are circles: 1. Intersection of a plane $z=textconst$ and the sphere is a circle (or empty set). 2. Any plane can be rotated to become $z=textconst$, and sphere is rotation invariant.
– A.Γ.
2 hours ago
@A.Γ. cool. Thanks. How about Q1 please: what's the purpose of unit normal vector?
– BCLC
2 hours ago
1
Is the image of $Phi$ the plane $z=0$ or $z=-1$ (there are different conventions)?
– A.Γ.
1 hour ago
 |Â
show 1 more comment
up vote
0
down vote
(Q1) I suspect some relationship is being illustrated. The 1st 2 coordinates in $[1,1,-1]$ are the centre of the circle $Phi(E)$ while the norm of $[1,1,-1]$ s the radius of $Phi(E)$.
(Q2) Yes if the planes intersect $mathbb S^2$ or $emptyset$ is considered a circle in $mathbb S^2$.
(Q3), (Q4) E is not an ellipse. It is a circle in 3D (see here too) parametrised as (WA)
$$x =sqrtfrac 2 3 cos[t]$$
$$y = -sqrtfrac 2 4 sin[t] - sqrtfrac 1 6 cos[t]$$
$$z = -sqrtfrac 2 4 sin[t] + sqrtfrac 1 6 cos[t]$$
Observe that while $x^2+y^2+xy=frac12$ is an ellipse, $x^2+y^2+xy=frac12 wedge z= x+y$ is a circle.
answered 5 hours ago
BCLC
6,99821973
1
I guess your major confusion was WA saying ellipse on the intersection. When you set $z=x+y$ into $x^2+y^2+z^2=1$ you get indeed an ellipse, but this is the projection of the intersection onto $xy$-plane, i.e. all the coordinates $(x,y)$ that satisfy both equations.
– A.Γ.
2 hours ago
@A.Γ. I was thinking projection but unable to express precisely. Thanks! Any idea for (Q1) please?
– BCLC
2 hours ago
1
To see that all intersections are circles: 1. Intersection of a plane $z=textconst$ and the sphere is a circle (or empty set). 2. Any plane can be rotated to become $z=textconst$, and sphere is rotation invariant.
– A.Γ.
2 hours ago
@A.Γ. cool. Thanks. How about Q1 please: what's the purpose of unit normal vector?
– BCLC
2 hours ago
1
Is the image of $Phi$ the plane $z=0$ or $z=-1$ (there are different conventions)?
– A.Γ.
1 hour ago
 |Â
show 1 more comment
up vote
0
down vote
up vote
0
down vote
(Q1) I suspect some relationship is being illustrated. The 1st 2 coordinates in $[1,1,-1]$ are the centre of the circle $Phi(E)$ while the norm of $[1,1,-1]$ s the radius of $Phi(E)$.
(Q2) Yes if the planes intersect $mathbb S^2$ or $emptyset$ is considered a circle in $mathbb S^2$.
(Q3), (Q4) E is not an ellipse. It is a circle in 3D (see here too) parametrised as (WA)
$$x =sqrtfrac 2 3 cos[t]$$
$$y = -sqrtfrac 2 4 sin[t] - sqrtfrac 1 6 cos[t]$$
$$z = -sqrtfrac 2 4 sin[t] + sqrtfrac 1 6 cos[t]$$
Observe that while $x^2+y^2+xy=frac12$ is an ellipse, $x^2+y^2+xy=frac12 wedge z= x+y$ is a circle.
answered 5 hours ago
BCLC
6,99821973
(Q1) I suspect some relationship is being illustrated. The 1st 2 coordinates in $[1,1,-1]$ are the centre of the circle $Phi(E)$ while the norm of $[1,1,-1]$ s the radius of $Phi(E)$.
(Q2) Yes if the planes intersect $mathbb S^2$ or $emptyset$ is considered a circle in $mathbb S^2$.
(Q3), (Q4) E is not an ellipse. It is a circle in 3D (see here too) parametrised as (WA)
$$x =sqrtfrac 2 3 cos[t]$$
$$y = -sqrtfrac 2 4 sin[t] - sqrtfrac 1 6 cos[t]$$
$$z = -sqrtfrac 2 4 sin[t] + sqrtfrac 1 6 cos[t]$$
Observe that while $x^2+y^2+xy=frac12$ is an ellipse, $x^2+y^2+xy=frac12 wedge z= x+y$ is a circle.
answered 5 hours ago
BCLC
6,99821973
edited 2 hours ago
answered 5 hours ago
BCLC
6,99821973
answered 5 hours ago
BCLC
6,99821973
answered 5 hours ago
BCLC
6,99821973
6,99821973
1
I guess your major confusion was WA saying ellipse on the intersection. When you set $z=x+y$ into $x^2+y^2+z^2=1$ you get indeed an ellipse, but this is the projection of the intersection onto $xy$-plane, i.e. all the coordinates $(x,y)$ that satisfy both equations.
– A.Γ.
2 hours ago
@A.Γ. I was thinking projection but unable to express precisely. Thanks! Any idea for (Q1) please?
– BCLC
2 hours ago
1
To see that all intersections are circles: 1. Intersection of a plane $z=textconst$ and the sphere is a circle (or empty set). 2. Any plane can be rotated to become $z=textconst$, and sphere is rotation invariant.
– A.Γ.
2 hours ago
@A.Γ. cool. Thanks. How about Q1 please: what's the purpose of unit normal vector?
– BCLC
2 hours ago
1
Is the image of $Phi$ the plane $z=0$ or $z=-1$ (there are different conventions)?
– A.Γ.
1 hour ago
 |Â
show 1 more comment
1
I guess your major confusion was WA saying ellipse on the intersection. When you set $z=x+y$ into $x^2+y^2+z^2=1$ you get indeed an ellipse, but this is the projection of the intersection onto $xy$-plane, i.e. all the coordinates $(x,y)$ that satisfy both equations.
– A.Γ.
2 hours ago
@A.Γ. I was thinking projection but unable to express precisely. Thanks! Any idea for (Q1) please?
– BCLC
2 hours ago
1
To see that all intersections are circles: 1. Intersection of a plane $z=textconst$ and the sphere is a circle (or empty set). 2. Any plane can be rotated to become $z=textconst$, and sphere is rotation invariant.
– A.Γ.
2 hours ago
@A.Γ. cool. Thanks. How about Q1 please: what's the purpose of unit normal vector?
– BCLC
2 hours ago
1
Is the image of $Phi$ the plane $z=0$ or $z=-1$ (there are different conventions)?
– A.Γ.
1 hour ago
1
1
I guess your major confusion was WA saying ellipse on the intersection. When you set $z=x+y$ into $x^2+y^2+z^2=1$ you get indeed an ellipse, but this is the projection of the intersection onto $xy$-plane, i.e. all the coordinates $(x,y)$ that satisfy both equations.
– A.Γ.
2 hours ago
I guess your major confusion was WA saying ellipse on the intersection. When you set $z=x+y$ into $x^2+y^2+z^2=1$ you get indeed an ellipse, but this is the projection of the intersection onto $xy$-plane, i.e. all the coordinates $(x,y)$ that satisfy both equations.
– A.Γ.
2 hours ago
@A.Γ. I was thinking projection but unable to express precisely. Thanks! Any idea for (Q1) please?
– BCLC
2 hours ago
@A.Γ. I was thinking projection but unable to express precisely. Thanks! Any idea for (Q1) please?
– BCLC
2 hours ago
1
1
To see that all intersections are circles: 1. Intersection of a plane $z=textconst$ and the sphere is a circle (or empty set). 2. Any plane can be rotated to become $z=textconst$, and sphere is rotation invariant.
– A.Γ.
2 hours ago
To see that all intersections are circles: 1. Intersection of a plane $z=textconst$ and the sphere is a circle (or empty set). 2. Any plane can be rotated to become $z=textconst$, and sphere is rotation invariant.
– A.Γ.
2 hours ago
@A.Γ. cool. Thanks. How about Q1 please: what's the purpose of unit normal vector?
– BCLC
2 hours ago
@A.Γ. cool. Thanks. How about Q1 please: what's the purpose of unit normal vector?
– BCLC
2 hours ago
1
1
Is the image of $Phi$ the plane $z=0$ or $z=-1$ (there are different conventions)?
– A.Γ.
1 hour ago
Is the image of $Phi$ the plane $z=0$ or $z=-1$ (there are different conventions)?
– A.Γ.
1 hour ago
 |Â
show 1 more comment
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2867457%2fwhat-is-the-intersection-of-a-plane-and-a-sphere-is-it-necessarily-a-circle-ca%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password