Smallest close sphere containing $V$
1
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Suppose I have a finite set of vectors $V = {v_1, v_2, ..., v_n} subset mathbb{R}^3$. What is the best way to find the smallest close sphere containing $V$?
vectors spheres
asked Dec 11 '18 at 2:57
davegautdavegaut
63
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show 2 more comments
1
$begingroup$
Suppose I have a finite set of vectors $V = {v_1, v_2, ..., v_n} subset mathbb{R}^3$. What is the best way to find the smallest close sphere containing $V$?
vectors spheres
asked Dec 11 '18 at 2:57
davegautdavegaut
63
$endgroup$
$begingroup$
Try it with two vectors, then three. Generalization should be clear.
$endgroup$
– zoidberg
Dec 11 '18 at 3:00
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Did you understand the hint? Why not using the largest vector as determining the diameter of the sphere?
$endgroup$
– Moti
Dec 11 '18 at 6:50
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@Moti, do you mean the largest vector, or the largest distance among pairs of vectors? I don't think any will work. E.g. think of the four vertices of a regular tetrahedron placed far from the origin of coordinates. The largest vector will be too large, and the largest distance among pairs of vertices would be the side of the tetrahedron, which is too small to be the diameter of the enclosing sphere.
$endgroup$
– mlerma54
Dec 11 '18 at 15:31
$begingroup$
To the best of my knowledge a vector has direction and size - no location.
$endgroup$
– Moti
Dec 12 '18 at 16:05
$begingroup$
@Moti, as elements of $mathbb{R}^3$ they are just triples of real numbers, and could be used to represent points in 3-dim space.
$endgroup$
– mlerma54
Dec 12 '18 at 22:38
|
show 2 more comments
1
1
1
$begingroup$
Suppose I have a finite set of vectors $V = {v_1, v_2, ..., v_n} subset mathbb{R}^3$. What is the best way to find the smallest close sphere containing $V$?
vectors spheres
asked Dec 11 '18 at 2:57
davegautdavegaut
63
$endgroup$
Suppose I have a finite set of vectors $V = {v_1, v_2, ..., v_n} subset mathbb{R}^3$. What is the best way to find the smallest close sphere containing $V$?
vectors spheres
vectors spheres
asked Dec 11 '18 at 2:57
davegautdavegaut
63
asked Dec 11 '18 at 2:57
davegautdavegaut
63
asked Dec 11 '18 at 2:57
davegautdavegaut
63
asked Dec 11 '18 at 2:57
davegautdavegaut
63
asked Dec 11 '18 at 2:57
davegautdavegaut
63
63
$begingroup$
Try it with two vectors, then three. Generalization should be clear.
$endgroup$
– zoidberg
Dec 11 '18 at 3:00
$begingroup$
Did you understand the hint? Why not using the largest vector as determining the diameter of the sphere?
$endgroup$
– Moti
Dec 11 '18 at 6:50
$begingroup$
@Moti, do you mean the largest vector, or the largest distance among pairs of vectors? I don't think any will work. E.g. think of the four vertices of a regular tetrahedron placed far from the origin of coordinates. The largest vector will be too large, and the largest distance among pairs of vertices would be the side of the tetrahedron, which is too small to be the diameter of the enclosing sphere.
$endgroup$
– mlerma54
Dec 11 '18 at 15:31
$begingroup$
To the best of my knowledge a vector has direction and size - no location.
$endgroup$
– Moti
Dec 12 '18 at 16:05
$begingroup$
@Moti, as elements of $mathbb{R}^3$ they are just triples of real numbers, and could be used to represent points in 3-dim space.
$endgroup$
– mlerma54
Dec 12 '18 at 22:38
|
show 2 more comments
$begingroup$
Try it with two vectors, then three. Generalization should be clear.
$endgroup$
– zoidberg
Dec 11 '18 at 3:00
$begingroup$
Did you understand the hint? Why not using the largest vector as determining the diameter of the sphere?
$endgroup$
– Moti
Dec 11 '18 at 6:50
$begingroup$
@Moti, do you mean the largest vector, or the largest distance among pairs of vectors? I don't think any will work. E.g. think of the four vertices of a regular tetrahedron placed far from the origin of coordinates. The largest vector will be too large, and the largest distance among pairs of vertices would be the side of the tetrahedron, which is too small to be the diameter of the enclosing sphere.
$endgroup$
– mlerma54
Dec 11 '18 at 15:31
$begingroup$
To the best of my knowledge a vector has direction and size - no location.
$endgroup$
– Moti
Dec 12 '18 at 16:05
$begingroup$
@Moti, as elements of $mathbb{R}^3$ they are just triples of real numbers, and could be used to represent points in 3-dim space.
$endgroup$
– mlerma54
Dec 12 '18 at 22:38
$begingroup$
Try it with two vectors, then three. Generalization should be clear.
$endgroup$
– zoidberg
Dec 11 '18 at 3:00
$begingroup$
Try it with two vectors, then three. Generalization should be clear.
$endgroup$
– zoidberg
Dec 11 '18 at 3:00
$begingroup$
Did you understand the hint? Why not using the largest vector as determining the diameter of the sphere?
$endgroup$
– Moti
Dec 11 '18 at 6:50
$begingroup$
Did you understand the hint? Why not using the largest vector as determining the diameter of the sphere?
$endgroup$
– Moti
Dec 11 '18 at 6:50
$begingroup$
@Moti, do you mean the largest vector, or the largest distance among pairs of vectors? I don't think any will work. E.g. think of the four vertices of a regular tetrahedron placed far from the origin of coordinates. The largest vector will be too large, and the largest distance among pairs of vertices would be the side of the tetrahedron, which is too small to be the diameter of the enclosing sphere.
$endgroup$
– mlerma54
Dec 11 '18 at 15:31
$begingroup$
@Moti, do you mean the largest vector, or the largest distance among pairs of vectors? I don't think any will work. E.g. think of the four vertices of a regular tetrahedron placed far from the origin of coordinates. The largest vector will be too large, and the largest distance among pairs of vertices would be the side of the tetrahedron, which is too small to be the diameter of the enclosing sphere.
$endgroup$
– mlerma54
Dec 11 '18 at 15:31
$begingroup$
To the best of my knowledge a vector has direction and size - no location.
$endgroup$
– Moti
Dec 12 '18 at 16:05
$begingroup$
To the best of my knowledge a vector has direction and size - no location.
$endgroup$
– Moti
Dec 12 '18 at 16:05
$begingroup$
@Moti, as elements of $mathbb{R}^3$ they are just triples of real numbers, and could be used to represent points in 3-dim space.
$endgroup$
– mlerma54
Dec 12 '18 at 22:38
$begingroup$
@Moti, as elements of $mathbb{R}^3$ they are just triples of real numbers, and could be used to represent points in 3-dim space.
$endgroup$
– mlerma54
Dec 12 '18 at 22:38
|
show 2 more comments
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$begingroup$
Try it with two vectors, then three. Generalization should be clear.
$endgroup$
– zoidberg
Dec 11 '18 at 3:00
$begingroup$
Did you understand the hint? Why not using the largest vector as determining the diameter of the sphere?
$endgroup$
– Moti
Dec 11 '18 at 6:50
$begingroup$
@Moti, do you mean the largest vector, or the largest distance among pairs of vectors? I don't think any will work. E.g. think of the four vertices of a regular tetrahedron placed far from the origin of coordinates. The largest vector will be too large, and the largest distance among pairs of vertices would be the side of the tetrahedron, which is too small to be the diameter of the enclosing sphere.
$endgroup$
– mlerma54
Dec 11 '18 at 15:31
$begingroup$
To the best of my knowledge a vector has direction and size - no location.
$endgroup$
– Moti
Dec 12 '18 at 16:05
$begingroup$
@Moti, as elements of $mathbb{R}^3$ they are just triples of real numbers, and could be used to represent points in 3-dim space.
$endgroup$
– mlerma54
Dec 12 '18 at 22:38