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Are there multiple types of “components” in the study of vectors?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Are parallel vectors always scalar multiple of each others?Vectors components that are not contra or covariant?What does the vector whose components are products of components of two vectors represent?Are there different types of vectors?vectors multiple and parallel rule.Determine the vectors of componentsAre there are different types of algebra?Multiple vector rejection from multiple vectors - possible with matrix notation?Solving vector equations. (Vectors are not with components)How to determine what type of function when there are multiple types
$begingroup$
In class we were taught how to write any vector in "component form" as $(a, b)$, where $a$ is "change in x" and $b$ is "change in y."
However, yesterday we were also taught how to "decompose" a vector into its "components," involving methods like projection:
What, if anything, is the relationship between these two apparently different definitions of a "vector component"? Or is there in fact no difference? I'm confused why the terminology is used in both contexts, and in turn I think that means I don't entirely understand the underlying concepts themselves.
linear-algebra algebra-precalculus vector-spaces vectors
asked Apr 2 at 15:30
Will Will
565
$endgroup$
add a comment |
$begingroup$
In class we were taught how to write any vector in "component form" as $(a, b)$, where $a$ is "change in x" and $b$ is "change in y."
However, yesterday we were also taught how to "decompose" a vector into its "components," involving methods like projection:
What, if anything, is the relationship between these two apparently different definitions of a "vector component"? Or is there in fact no difference? I'm confused why the terminology is used in both contexts, and in turn I think that means I don't entirely understand the underlying concepts themselves.
linear-algebra algebra-precalculus vector-spaces vectors
asked Apr 2 at 15:30
Will Will
565
$endgroup$
2
$begingroup$
You are right that there is variation in how the notion of components is applied to vector spaces. An introductory class in linear algebra will present the example of Cartesian coordinates and "change of basis" matrices, which effectively give the tools for translating one system of components to another.
$endgroup$
– hardmath
Apr 2 at 15:39
add a comment |
$begingroup$
In class we were taught how to write any vector in "component form" as $(a, b)$, where $a$ is "change in x" and $b$ is "change in y."
However, yesterday we were also taught how to "decompose" a vector into its "components," involving methods like projection:
What, if anything, is the relationship between these two apparently different definitions of a "vector component"? Or is there in fact no difference? I'm confused why the terminology is used in both contexts, and in turn I think that means I don't entirely understand the underlying concepts themselves.
linear-algebra algebra-precalculus vector-spaces vectors
asked Apr 2 at 15:30
Will Will
565
$endgroup$
In class we were taught how to write any vector in "component form" as $(a, b)$, where $a$ is "change in x" and $b$ is "change in y."
However, yesterday we were also taught how to "decompose" a vector into its "components," involving methods like projection:
What, if anything, is the relationship between these two apparently different definitions of a "vector component"? Or is there in fact no difference? I'm confused why the terminology is used in both contexts, and in turn I think that means I don't entirely understand the underlying concepts themselves.
linear-algebra algebra-precalculus vector-spaces vectors
linear-algebra algebra-precalculus vector-spaces vectors
asked Apr 2 at 15:30
Will Will
565
asked Apr 2 at 15:30
Will Will
565
asked Apr 2 at 15:30
Will Will
565
asked Apr 2 at 15:30
Will Will
565
asked Apr 2 at 15:30
Will Will
565
565
2
$begingroup$
You are right that there is variation in how the notion of components is applied to vector spaces. An introductory class in linear algebra will present the example of Cartesian coordinates and "change of basis" matrices, which effectively give the tools for translating one system of components to another.
$endgroup$
– hardmath
Apr 2 at 15:39
add a comment |
2
$begingroup$
You are right that there is variation in how the notion of components is applied to vector spaces. An introductory class in linear algebra will present the example of Cartesian coordinates and "change of basis" matrices, which effectively give the tools for translating one system of components to another.
$endgroup$
– hardmath
Apr 2 at 15:39
2
2
$begingroup$
You are right that there is variation in how the notion of components is applied to vector spaces. An introductory class in linear algebra will present the example of Cartesian coordinates and "change of basis" matrices, which effectively give the tools for translating one system of components to another.
$endgroup$
– hardmath
Apr 2 at 15:39
$begingroup$
You are right that there is variation in how the notion of components is applied to vector spaces. An introductory class in linear algebra will present the example of Cartesian coordinates and "change of basis" matrices, which effectively give the tools for translating one system of components to another.
$endgroup$
– hardmath
Apr 2 at 15:39
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It is actually the same thing.
In the first usage of the word, you find that the components you get are exactly equal to the $x$ and $y$ co-ordinates of the vector $u$. You have found the components of $u$ in the $x$ direction and in the $y$ direction, but to save breath, we don't say this in full.
In the second usage, it is perhaps not intuitively clear how the $x$ and $y$ components of $u$ combine to give the $v$ component of $u$. The dot product formula gives this. In fact, if you changed your co-ordinate system from units of $x$ and $y$ into units of $v$ and $v_perp$, where $v_perp$ is perpendicular to $v$, then you'd get $$u=av+bv_perptag1$$ for some numbers $a,b$. Then $a$ is the component of $u$ in the direction of $v$. In this new co-ordinate system, (called a basis), one can write $u=(a,b)$ as a short-hand for the expression $(1)$. Then, this looks exactly like the first usage of the word components.
answered Apr 2 at 15:46
John DoeJohn Doe
12.2k11340
$endgroup$
add a comment |
$begingroup$
In general, the "component" of a vector in a certain direction is to what extent that vector is pointing in that direction.
So the usual notation $vecv=(x,y)$ says the vector is pointing "$x$ units" in the direction of the $x$-axis, and "$y$ units" in the direction of the $y$-axis. Notice that $x$ and $y$ are also the projection of $vecv$ onto the usual axes. So $vecv=(proj_(1,0)(vecv) , , ,proj_(0,1)(vecv))$
Similarly in your examples, the component of $vecv$ in the direction of some vector $vecu$ is the projection $proj_vecu(vecv)$
answered Apr 2 at 15:45
NazimJNazimJ
890110
$endgroup$
add a comment |
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$begingroup$
It is actually the same thing.
In the first usage of the word, you find that the components you get are exactly equal to the $x$ and $y$ co-ordinates of the vector $u$. You have found the components of $u$ in the $x$ direction and in the $y$ direction, but to save breath, we don't say this in full.
In the second usage, it is perhaps not intuitively clear how the $x$ and $y$ components of $u$ combine to give the $v$ component of $u$. The dot product formula gives this. In fact, if you changed your co-ordinate system from units of $x$ and $y$ into units of $v$ and $v_perp$, where $v_perp$ is perpendicular to $v$, then you'd get $$u=av+bv_perptag1$$ for some numbers $a,b$. Then $a$ is the component of $u$ in the direction of $v$. In this new co-ordinate system, (called a basis), one can write $u=(a,b)$ as a short-hand for the expression $(1)$. Then, this looks exactly like the first usage of the word components.
answered Apr 2 at 15:46
John DoeJohn Doe
12.2k11340
$endgroup$
add a comment |
$begingroup$
It is actually the same thing.
In the first usage of the word, you find that the components you get are exactly equal to the $x$ and $y$ co-ordinates of the vector $u$. You have found the components of $u$ in the $x$ direction and in the $y$ direction, but to save breath, we don't say this in full.
In the second usage, it is perhaps not intuitively clear how the $x$ and $y$ components of $u$ combine to give the $v$ component of $u$. The dot product formula gives this. In fact, if you changed your co-ordinate system from units of $x$ and $y$ into units of $v$ and $v_perp$, where $v_perp$ is perpendicular to $v$, then you'd get $$u=av+bv_perptag1$$ for some numbers $a,b$. Then $a$ is the component of $u$ in the direction of $v$. In this new co-ordinate system, (called a basis), one can write $u=(a,b)$ as a short-hand for the expression $(1)$. Then, this looks exactly like the first usage of the word components.
answered Apr 2 at 15:46
John DoeJohn Doe
12.2k11340
$endgroup$
add a comment |
$begingroup$
It is actually the same thing.
In the first usage of the word, you find that the components you get are exactly equal to the $x$ and $y$ co-ordinates of the vector $u$. You have found the components of $u$ in the $x$ direction and in the $y$ direction, but to save breath, we don't say this in full.
In the second usage, it is perhaps not intuitively clear how the $x$ and $y$ components of $u$ combine to give the $v$ component of $u$. The dot product formula gives this. In fact, if you changed your co-ordinate system from units of $x$ and $y$ into units of $v$ and $v_perp$, where $v_perp$ is perpendicular to $v$, then you'd get $$u=av+bv_perptag1$$ for some numbers $a,b$. Then $a$ is the component of $u$ in the direction of $v$. In this new co-ordinate system, (called a basis), one can write $u=(a,b)$ as a short-hand for the expression $(1)$. Then, this looks exactly like the first usage of the word components.
answered Apr 2 at 15:46
John DoeJohn Doe
12.2k11340
$endgroup$
It is actually the same thing.
In the first usage of the word, you find that the components you get are exactly equal to the $x$ and $y$ co-ordinates of the vector $u$. You have found the components of $u$ in the $x$ direction and in the $y$ direction, but to save breath, we don't say this in full.
In the second usage, it is perhaps not intuitively clear how the $x$ and $y$ components of $u$ combine to give the $v$ component of $u$. The dot product formula gives this. In fact, if you changed your co-ordinate system from units of $x$ and $y$ into units of $v$ and $v_perp$, where $v_perp$ is perpendicular to $v$, then you'd get $$u=av+bv_perptag1$$ for some numbers $a,b$. Then $a$ is the component of $u$ in the direction of $v$. In this new co-ordinate system, (called a basis), one can write $u=(a,b)$ as a short-hand for the expression $(1)$. Then, this looks exactly like the first usage of the word components.
answered Apr 2 at 15:46
John DoeJohn Doe
12.2k11340
answered Apr 2 at 15:46
John DoeJohn Doe
12.2k11340
answered Apr 2 at 15:46
John DoeJohn Doe
12.2k11340
answered Apr 2 at 15:46
John DoeJohn Doe
12.2k11340
12.2k11340
add a comment |
add a comment |
$begingroup$
In general, the "component" of a vector in a certain direction is to what extent that vector is pointing in that direction.
So the usual notation $vecv=(x,y)$ says the vector is pointing "$x$ units" in the direction of the $x$-axis, and "$y$ units" in the direction of the $y$-axis. Notice that $x$ and $y$ are also the projection of $vecv$ onto the usual axes. So $vecv=(proj_(1,0)(vecv) , , ,proj_(0,1)(vecv))$
Similarly in your examples, the component of $vecv$ in the direction of some vector $vecu$ is the projection $proj_vecu(vecv)$
answered Apr 2 at 15:45
NazimJNazimJ
890110
$endgroup$
add a comment |
$begingroup$
In general, the "component" of a vector in a certain direction is to what extent that vector is pointing in that direction.
So the usual notation $vecv=(x,y)$ says the vector is pointing "$x$ units" in the direction of the $x$-axis, and "$y$ units" in the direction of the $y$-axis. Notice that $x$ and $y$ are also the projection of $vecv$ onto the usual axes. So $vecv=(proj_(1,0)(vecv) , , ,proj_(0,1)(vecv))$
Similarly in your examples, the component of $vecv$ in the direction of some vector $vecu$ is the projection $proj_vecu(vecv)$
answered Apr 2 at 15:45
NazimJNazimJ
890110
$endgroup$
add a comment |
$begingroup$
In general, the "component" of a vector in a certain direction is to what extent that vector is pointing in that direction.
So the usual notation $vecv=(x,y)$ says the vector is pointing "$x$ units" in the direction of the $x$-axis, and "$y$ units" in the direction of the $y$-axis. Notice that $x$ and $y$ are also the projection of $vecv$ onto the usual axes. So $vecv=(proj_(1,0)(vecv) , , ,proj_(0,1)(vecv))$
Similarly in your examples, the component of $vecv$ in the direction of some vector $vecu$ is the projection $proj_vecu(vecv)$
answered Apr 2 at 15:45
NazimJNazimJ
890110
$endgroup$
In general, the "component" of a vector in a certain direction is to what extent that vector is pointing in that direction.
So the usual notation $vecv=(x,y)$ says the vector is pointing "$x$ units" in the direction of the $x$-axis, and "$y$ units" in the direction of the $y$-axis. Notice that $x$ and $y$ are also the projection of $vecv$ onto the usual axes. So $vecv=(proj_(1,0)(vecv) , , ,proj_(0,1)(vecv))$
Similarly in your examples, the component of $vecv$ in the direction of some vector $vecu$ is the projection $proj_vecu(vecv)$
answered Apr 2 at 15:45
NazimJNazimJ
890110
answered Apr 2 at 15:45
NazimJNazimJ
890110
answered Apr 2 at 15:45
NazimJNazimJ
890110
answered Apr 2 at 15:45
NazimJNazimJ
890110
890110
add a comment |
add a comment |
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$begingroup$
You are right that there is variation in how the notion of components is applied to vector spaces. An introductory class in linear algebra will present the example of Cartesian coordinates and "change of basis" matrices, which effectively give the tools for translating one system of components to another.
$endgroup$
– hardmath
Apr 2 at 15:39