understanding of vector and point The Next CEO of Stack OverflowQuestion in Do Carmo 1-2Simple question regarding orthogonalityQuestion on Do CarmoClosest point of parameterized curve has orthogonal position vector to tangentDo we substitute the point in question in direction vector for parametric equation of 3d vectors?A question about parametrized differentiable curves.Problem understanding the indicatrix tangent definitionDifferential map of velocity vectorAbout the definition of curvatureStrong/weak tangents and limit positions, with rigor
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understanding of vector and point
The Next CEO of Stack OverflowQuestion in Do Carmo 1-2Simple question regarding orthogonalityQuestion on Do CarmoClosest point of parameterized curve has orthogonal position vector to tangentDo we substitute the point in question in direction vector for parametric equation of 3d vectors?A question about parametrized differentiable curves.Problem understanding the indicatrix tangent definitionDifferential map of velocity vectorAbout the definition of curvatureStrong/weak tangents and limit positions, with rigor
$begingroup$
I just have some confusions about basic understandings in vectors and points. For example, if $$alpha(t)=(x(t),y(t),z(t))$$ is a parametrized curve from $I$ to $mathbbR^3$, then for a specific point $t_0 in I$ , $alpha(t_0)$ represents a vector or a point in $mathbbR^3$ ? Why? I think $alpha(t_0)$ is a point on $mathbbR^3$, but I saw $alpha(t_0)$ described as a position vector in Do Carmo's book.
Thank you!
calculus differential-geometry
asked Mar 18 at 19:32
jf1997jf1997
32
$endgroup$
add a comment |
$begingroup$
I just have some confusions about basic understandings in vectors and points. For example, if $$alpha(t)=(x(t),y(t),z(t))$$ is a parametrized curve from $I$ to $mathbbR^3$, then for a specific point $t_0 in I$ , $alpha(t_0)$ represents a vector or a point in $mathbbR^3$ ? Why? I think $alpha(t_0)$ is a point on $mathbbR^3$, but I saw $alpha(t_0)$ described as a position vector in Do Carmo's book.
Thank you!
calculus differential-geometry
asked Mar 18 at 19:32
jf1997jf1997
32
$endgroup$
add a comment |
$begingroup$
I just have some confusions about basic understandings in vectors and points. For example, if $$alpha(t)=(x(t),y(t),z(t))$$ is a parametrized curve from $I$ to $mathbbR^3$, then for a specific point $t_0 in I$ , $alpha(t_0)$ represents a vector or a point in $mathbbR^3$ ? Why? I think $alpha(t_0)$ is a point on $mathbbR^3$, but I saw $alpha(t_0)$ described as a position vector in Do Carmo's book.
Thank you!
calculus differential-geometry
asked Mar 18 at 19:32
jf1997jf1997
32
$endgroup$
I just have some confusions about basic understandings in vectors and points. For example, if $$alpha(t)=(x(t),y(t),z(t))$$ is a parametrized curve from $I$ to $mathbbR^3$, then for a specific point $t_0 in I$ , $alpha(t_0)$ represents a vector or a point in $mathbbR^3$ ? Why? I think $alpha(t_0)$ is a point on $mathbbR^3$, but I saw $alpha(t_0)$ described as a position vector in Do Carmo's book.
Thank you!
calculus differential-geometry
calculus differential-geometry
asked Mar 18 at 19:32
jf1997jf1997
32
asked Mar 18 at 19:32
jf1997jf1997
32
asked Mar 18 at 19:32
jf1997jf1997
32
asked Mar 18 at 19:32
jf1997jf1997
32
asked Mar 18 at 19:32
jf1997jf1997
32
32
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Assuming you are given a coordinate system in $R^3$, there exist a "one to one correspondence" between the set of all points in $R^3$ and the set of three dimensional vectors. So we can interpret the $alpha(t)= (x(t), y(t), z(t))$, for each t, to be either a point or the vector that, if we were to have its base at (0, 0, 0) would have its tip at the point (x(t), y(t), z(t)).
I am not saying that points and vector are the same thing! For example, given the vector (x, y, z), which would represent the vector with end at (0, 0, 0) and tip at the point (x, y, z), we can [b]move[/b] that vector so that its end is at (a, b, c) and its tip is at (a+ x, b+ y, c+ z) but it is the same vector. You can move vectors but you can't move points.
(Some texts use for vectors specifically to distinguish vectors from points (x, y, z).)
answered Mar 18 at 19:51
user247327user247327
11.6k1516
$endgroup$
add a comment |
$begingroup$
The idea is that some things are essentially the same in mathematics -- isomorphic is the term of art.
This is also the case with vectors and points in Euclidean space. In particular, if we always take the position vector (i.e., those originating at the origin of coordinates) as the representative of each equivalence class of vectors, then there is an isomorphism between the points and vectors of that space, so that we may regard them as essentially the same thing.
How we think of them in a particular case would depend on which is more convenient for the problem at hand.
By the way, note that there is a difference between a curve and its trace.
answered Mar 18 at 20:11
AllawonderAllawonder
1
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Assuming you are given a coordinate system in $R^3$, there exist a "one to one correspondence" between the set of all points in $R^3$ and the set of three dimensional vectors. So we can interpret the $alpha(t)= (x(t), y(t), z(t))$, for each t, to be either a point or the vector that, if we were to have its base at (0, 0, 0) would have its tip at the point (x(t), y(t), z(t)).
I am not saying that points and vector are the same thing! For example, given the vector (x, y, z), which would represent the vector with end at (0, 0, 0) and tip at the point (x, y, z), we can [b]move[/b] that vector so that its end is at (a, b, c) and its tip is at (a+ x, b+ y, c+ z) but it is the same vector. You can move vectors but you can't move points.
(Some texts use for vectors specifically to distinguish vectors from points (x, y, z).)
answered Mar 18 at 19:51
user247327user247327
11.6k1516
$endgroup$
add a comment |
$begingroup$
Assuming you are given a coordinate system in $R^3$, there exist a "one to one correspondence" between the set of all points in $R^3$ and the set of three dimensional vectors. So we can interpret the $alpha(t)= (x(t), y(t), z(t))$, for each t, to be either a point or the vector that, if we were to have its base at (0, 0, 0) would have its tip at the point (x(t), y(t), z(t)).
I am not saying that points and vector are the same thing! For example, given the vector (x, y, z), which would represent the vector with end at (0, 0, 0) and tip at the point (x, y, z), we can [b]move[/b] that vector so that its end is at (a, b, c) and its tip is at (a+ x, b+ y, c+ z) but it is the same vector. You can move vectors but you can't move points.
(Some texts use for vectors specifically to distinguish vectors from points (x, y, z).)
answered Mar 18 at 19:51
user247327user247327
11.6k1516
$endgroup$
add a comment |
$begingroup$
Assuming you are given a coordinate system in $R^3$, there exist a "one to one correspondence" between the set of all points in $R^3$ and the set of three dimensional vectors. So we can interpret the $alpha(t)= (x(t), y(t), z(t))$, for each t, to be either a point or the vector that, if we were to have its base at (0, 0, 0) would have its tip at the point (x(t), y(t), z(t)).
I am not saying that points and vector are the same thing! For example, given the vector (x, y, z), which would represent the vector with end at (0, 0, 0) and tip at the point (x, y, z), we can [b]move[/b] that vector so that its end is at (a, b, c) and its tip is at (a+ x, b+ y, c+ z) but it is the same vector. You can move vectors but you can't move points.
(Some texts use for vectors specifically to distinguish vectors from points (x, y, z).)
answered Mar 18 at 19:51
user247327user247327
11.6k1516
$endgroup$
Assuming you are given a coordinate system in $R^3$, there exist a "one to one correspondence" between the set of all points in $R^3$ and the set of three dimensional vectors. So we can interpret the $alpha(t)= (x(t), y(t), z(t))$, for each t, to be either a point or the vector that, if we were to have its base at (0, 0, 0) would have its tip at the point (x(t), y(t), z(t)).
I am not saying that points and vector are the same thing! For example, given the vector (x, y, z), which would represent the vector with end at (0, 0, 0) and tip at the point (x, y, z), we can [b]move[/b] that vector so that its end is at (a, b, c) and its tip is at (a+ x, b+ y, c+ z) but it is the same vector. You can move vectors but you can't move points.
(Some texts use for vectors specifically to distinguish vectors from points (x, y, z).)
answered Mar 18 at 19:51
user247327user247327
11.6k1516
answered Mar 18 at 19:51
user247327user247327
11.6k1516
answered Mar 18 at 19:51
user247327user247327
11.6k1516
answered Mar 18 at 19:51
user247327user247327
11.6k1516
11.6k1516
add a comment |
add a comment |
$begingroup$
The idea is that some things are essentially the same in mathematics -- isomorphic is the term of art.
This is also the case with vectors and points in Euclidean space. In particular, if we always take the position vector (i.e., those originating at the origin of coordinates) as the representative of each equivalence class of vectors, then there is an isomorphism between the points and vectors of that space, so that we may regard them as essentially the same thing.
How we think of them in a particular case would depend on which is more convenient for the problem at hand.
By the way, note that there is a difference between a curve and its trace.
answered Mar 18 at 20:11
AllawonderAllawonder
1
$endgroup$
add a comment |
$begingroup$
The idea is that some things are essentially the same in mathematics -- isomorphic is the term of art.
This is also the case with vectors and points in Euclidean space. In particular, if we always take the position vector (i.e., those originating at the origin of coordinates) as the representative of each equivalence class of vectors, then there is an isomorphism between the points and vectors of that space, so that we may regard them as essentially the same thing.
How we think of them in a particular case would depend on which is more convenient for the problem at hand.
By the way, note that there is a difference between a curve and its trace.
answered Mar 18 at 20:11
AllawonderAllawonder
1
$endgroup$
add a comment |
$begingroup$
The idea is that some things are essentially the same in mathematics -- isomorphic is the term of art.
This is also the case with vectors and points in Euclidean space. In particular, if we always take the position vector (i.e., those originating at the origin of coordinates) as the representative of each equivalence class of vectors, then there is an isomorphism between the points and vectors of that space, so that we may regard them as essentially the same thing.
How we think of them in a particular case would depend on which is more convenient for the problem at hand.
By the way, note that there is a difference between a curve and its trace.
answered Mar 18 at 20:11
AllawonderAllawonder
1
$endgroup$
The idea is that some things are essentially the same in mathematics -- isomorphic is the term of art.
This is also the case with vectors and points in Euclidean space. In particular, if we always take the position vector (i.e., those originating at the origin of coordinates) as the representative of each equivalence class of vectors, then there is an isomorphism between the points and vectors of that space, so that we may regard them as essentially the same thing.
How we think of them in a particular case would depend on which is more convenient for the problem at hand.
By the way, note that there is a difference between a curve and its trace.
answered Mar 18 at 20:11
AllawonderAllawonder
1
answered Mar 18 at 20:11
AllawonderAllawonder
1
answered Mar 18 at 20:11
AllawonderAllawonder
1
answered Mar 18 at 20:11
AllawonderAllawonder
1
1
add a comment |
add a comment |
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