Polar plane of a pole
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Polar plane of a pole A of a sphere is locus of all points R where line through A meets points P and Q on sphere such that 2/AR = 1/AP + 1/AQ. How does polar plane of a pole of a sphere look like? Please provide a 3D view
analytic-geometry
asked Aug 20 at 6:47
abcdmath
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Polar plane of a pole A of a sphere is locus of all points R where line through A meets points P and Q on sphere such that 2/AR = 1/AP + 1/AQ. How does polar plane of a pole of a sphere look like? Please provide a 3D view
analytic-geometry
asked Aug 20 at 6:47
abcdmath
11
If $A$ is a point of a sphere and a line through $A$ meets the sphere at points $P$ and $Q$, then $P=Alor Q=A$ (with a conjunction $P=Q=A$ in a case of the line tangent to the sphere). This implies either $AP=0$ or $AQ=0$ which makes the given condition meaningless – you can't divide $1$ by zero...
– CiaPan
Aug 20 at 7:48
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up vote
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Polar plane of a pole A of a sphere is locus of all points R where line through A meets points P and Q on sphere such that 2/AR = 1/AP + 1/AQ. How does polar plane of a pole of a sphere look like? Please provide a 3D view
analytic-geometry
asked Aug 20 at 6:47
abcdmath
11
Polar plane of a pole A of a sphere is locus of all points R where line through A meets points P and Q on sphere such that 2/AR = 1/AP + 1/AQ. How does polar plane of a pole of a sphere look like? Please provide a 3D view
analytic-geometry
asked Aug 20 at 6:47
abcdmath
11
asked Aug 20 at 6:47
abcdmath
11
asked Aug 20 at 6:47
abcdmath
11
asked Aug 20 at 6:47
abcdmath
11
11
If $A$ is a point of a sphere and a line through $A$ meets the sphere at points $P$ and $Q$, then $P=Alor Q=A$ (with a conjunction $P=Q=A$ in a case of the line tangent to the sphere). This implies either $AP=0$ or $AQ=0$ which makes the given condition meaningless – you can't divide $1$ by zero...
– CiaPan
Aug 20 at 7:48
add a comment |Â
If $A$ is a point of a sphere and a line through $A$ meets the sphere at points $P$ and $Q$, then $P=Alor Q=A$ (with a conjunction $P=Q=A$ in a case of the line tangent to the sphere). This implies either $AP=0$ or $AQ=0$ which makes the given condition meaningless – you can't divide $1$ by zero...
– CiaPan
Aug 20 at 7:48
If $A$ is a point of a sphere and a line through $A$ meets the sphere at points $P$ and $Q$, then $P=Alor Q=A$ (with a conjunction $P=Q=A$ in a case of the line tangent to the sphere). This implies either $AP=0$ or $AQ=0$ which makes the given condition meaningless – you can't divide $1$ by zero...
– CiaPan
Aug 20 at 7:48
If $A$ is a point of a sphere and a line through $A$ meets the sphere at points $P$ and $Q$, then $P=Alor Q=A$ (with a conjunction $P=Q=A$ in a case of the line tangent to the sphere). This implies either $AP=0$ or $AQ=0$ which makes the given condition meaningless – you can't divide $1$ by zero...
– CiaPan
Aug 20 at 7:48
add a comment |Â
1 Answer
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HINT. Consider the particular case when $P$ is the point on the sphere nearest to $A$. By symmetry, the polar plane must be perpendicular to line $AP$.
For a 3D view, I'd suggest you to use GeoGebra.
answered Aug 20 at 17:46
Aretino
21.9k21442
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
HINT. Consider the particular case when $P$ is the point on the sphere nearest to $A$. By symmetry, the polar plane must be perpendicular to line $AP$.
For a 3D view, I'd suggest you to use GeoGebra.
answered Aug 20 at 17:46
Aretino
21.9k21442
add a comment |Â
up vote
0
down vote
HINT. Consider the particular case when $P$ is the point on the sphere nearest to $A$. By symmetry, the polar plane must be perpendicular to line $AP$.
For a 3D view, I'd suggest you to use GeoGebra.
answered Aug 20 at 17:46
Aretino
21.9k21442
add a comment |Â
up vote
0
down vote
up vote
0
down vote
HINT. Consider the particular case when $P$ is the point on the sphere nearest to $A$. By symmetry, the polar plane must be perpendicular to line $AP$.
For a 3D view, I'd suggest you to use GeoGebra.
answered Aug 20 at 17:46
Aretino
21.9k21442
HINT. Consider the particular case when $P$ is the point on the sphere nearest to $A$. By symmetry, the polar plane must be perpendicular to line $AP$.
For a 3D view, I'd suggest you to use GeoGebra.
answered Aug 20 at 17:46
Aretino
21.9k21442
answered Aug 20 at 17:46
Aretino
21.9k21442
answered Aug 20 at 17:46
Aretino
21.9k21442
answered Aug 20 at 17:46
Aretino
21.9k21442
21.9k21442
add a comment |Â
add a comment |Â
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If $A$ is a point of a sphere and a line through $A$ meets the sphere at points $P$ and $Q$, then $P=Alor Q=A$ (with a conjunction $P=Q=A$ in a case of the line tangent to the sphere). This implies either $AP=0$ or $AQ=0$ which makes the given condition meaningless – you can't divide $1$ by zero...
– CiaPan
Aug 20 at 7:48