Mixing
Is there an “inverse” of the dot-product?
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The dot product of $[a_1,a_2]$ and $[b_1, b_2]$ is $a_1b_1 + a_2b_2$ (and so on for bigger vectors). What I'm wondering is if there's any definition of a function s.t. the "invdot" product of $[a_1,a_2]$ and $[b_1, b_2]$ is $frac{a_1}{b_1} + frac{a_2}{b_2}$, and if so, are there any other properties/derivations of this function from the dot product? I can see it's the same as taking the reciprocal of all the components of $b$ and then doing the dot product on that, but I can't understand how I'd do that in my specific use case.
For context, I'm trying to solve a problem where I want to algebraically manipulate $a_1b_1 + a_2b_2$ to $frac{a_1}{b_1} + frac{a_2}{b_2}$, and think that this might help if it's a well known operation.
linear-algebra algebra-precalculus vectors
asked Jan 23 at 22:46
Robin AldabanxRobin Aldabanx
1,324415
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add a comment |
$begingroup$
The dot product of $[a_1,a_2]$ and $[b_1, b_2]$ is $a_1b_1 + a_2b_2$ (and so on for bigger vectors). What I'm wondering is if there's any definition of a function s.t. the "invdot" product of $[a_1,a_2]$ and $[b_1, b_2]$ is $frac{a_1}{b_1} + frac{a_2}{b_2}$, and if so, are there any other properties/derivations of this function from the dot product? I can see it's the same as taking the reciprocal of all the components of $b$ and then doing the dot product on that, but I can't understand how I'd do that in my specific use case.
For context, I'm trying to solve a problem where I want to algebraically manipulate $a_1b_1 + a_2b_2$ to $frac{a_1}{b_1} + frac{a_2}{b_2}$, and think that this might help if it's a well known operation.
linear-algebra algebra-precalculus vectors
asked Jan 23 at 22:46
Robin AldabanxRobin Aldabanx
1,324415
$endgroup$
2
$begingroup$
Note that the dot product sends two vectors to a scalar, the inverse of that won't make much sense (there are infinitely many such mappings).
$endgroup$
– lightxbulb
Jan 23 at 22:54
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The dot product is not injective, so there cannot be an inverse. However, maybe you used the term "inverse" by mistake, since you are talking about reciprocals.
$endgroup$
– Ben W
Jan 23 at 22:59
$begingroup$
It's "inverse" as in "opposite", not "reverse". It's not the common meaning in mathematics, but it's pedantic to require everyone here to adhere to such standards. Particularly when they're experimenting.
$endgroup$
– Arthur
Jan 23 at 23:03
$begingroup$
Yep, +Arthur's got it - though I see where you guys are coming too.
$endgroup$
– Robin Aldabanx
Jan 23 at 23:07
add a comment |
$begingroup$
The dot product of $[a_1,a_2]$ and $[b_1, b_2]$ is $a_1b_1 + a_2b_2$ (and so on for bigger vectors). What I'm wondering is if there's any definition of a function s.t. the "invdot" product of $[a_1,a_2]$ and $[b_1, b_2]$ is $frac{a_1}{b_1} + frac{a_2}{b_2}$, and if so, are there any other properties/derivations of this function from the dot product? I can see it's the same as taking the reciprocal of all the components of $b$ and then doing the dot product on that, but I can't understand how I'd do that in my specific use case.
For context, I'm trying to solve a problem where I want to algebraically manipulate $a_1b_1 + a_2b_2$ to $frac{a_1}{b_1} + frac{a_2}{b_2}$, and think that this might help if it's a well known operation.
linear-algebra algebra-precalculus vectors
asked Jan 23 at 22:46
Robin AldabanxRobin Aldabanx
1,324415
$endgroup$
The dot product of $[a_1,a_2]$ and $[b_1, b_2]$ is $a_1b_1 + a_2b_2$ (and so on for bigger vectors). What I'm wondering is if there's any definition of a function s.t. the "invdot" product of $[a_1,a_2]$ and $[b_1, b_2]$ is $frac{a_1}{b_1} + frac{a_2}{b_2}$, and if so, are there any other properties/derivations of this function from the dot product? I can see it's the same as taking the reciprocal of all the components of $b$ and then doing the dot product on that, but I can't understand how I'd do that in my specific use case.
For context, I'm trying to solve a problem where I want to algebraically manipulate $a_1b_1 + a_2b_2$ to $frac{a_1}{b_1} + frac{a_2}{b_2}$, and think that this might help if it's a well known operation.
linear-algebra algebra-precalculus vectors
linear-algebra algebra-precalculus vectors
asked Jan 23 at 22:46
Robin AldabanxRobin Aldabanx
1,324415
asked Jan 23 at 22:46
Robin AldabanxRobin Aldabanx
1,324415
asked Jan 23 at 22:46
Robin AldabanxRobin Aldabanx
1,324415
asked Jan 23 at 22:46
Robin AldabanxRobin Aldabanx
1,324415
asked Jan 23 at 22:46
Robin AldabanxRobin Aldabanx
1,324415
1,324415
2
$begingroup$
Note that the dot product sends two vectors to a scalar, the inverse of that won't make much sense (there are infinitely many such mappings).
$endgroup$
– lightxbulb
Jan 23 at 22:54
$begingroup$
The dot product is not injective, so there cannot be an inverse. However, maybe you used the term "inverse" by mistake, since you are talking about reciprocals.
$endgroup$
– Ben W
Jan 23 at 22:59
$begingroup$
It's "inverse" as in "opposite", not "reverse". It's not the common meaning in mathematics, but it's pedantic to require everyone here to adhere to such standards. Particularly when they're experimenting.
$endgroup$
– Arthur
Jan 23 at 23:03
$begingroup$
Yep, +Arthur's got it - though I see where you guys are coming too.
$endgroup$
– Robin Aldabanx
Jan 23 at 23:07
add a comment |
2
$begingroup$
Note that the dot product sends two vectors to a scalar, the inverse of that won't make much sense (there are infinitely many such mappings).
$endgroup$
– lightxbulb
Jan 23 at 22:54
$begingroup$
The dot product is not injective, so there cannot be an inverse. However, maybe you used the term "inverse" by mistake, since you are talking about reciprocals.
$endgroup$
– Ben W
Jan 23 at 22:59
$begingroup$
It's "inverse" as in "opposite", not "reverse". It's not the common meaning in mathematics, but it's pedantic to require everyone here to adhere to such standards. Particularly when they're experimenting.
$endgroup$
– Arthur
Jan 23 at 23:03
$begingroup$
Yep, +Arthur's got it - though I see where you guys are coming too.
$endgroup$
– Robin Aldabanx
Jan 23 at 23:07
2
2
$begingroup$
Note that the dot product sends two vectors to a scalar, the inverse of that won't make much sense (there are infinitely many such mappings).
$endgroup$
– lightxbulb
Jan 23 at 22:54
$begingroup$
Note that the dot product sends two vectors to a scalar, the inverse of that won't make much sense (there are infinitely many such mappings).
$endgroup$
– lightxbulb
Jan 23 at 22:54
$begingroup$
The dot product is not injective, so there cannot be an inverse. However, maybe you used the term "inverse" by mistake, since you are talking about reciprocals.
$endgroup$
– Ben W
Jan 23 at 22:59
$begingroup$
The dot product is not injective, so there cannot be an inverse. However, maybe you used the term "inverse" by mistake, since you are talking about reciprocals.
$endgroup$
– Ben W
Jan 23 at 22:59
$begingroup$
It's "inverse" as in "opposite", not "reverse". It's not the common meaning in mathematics, but it's pedantic to require everyone here to adhere to such standards. Particularly when they're experimenting.
$endgroup$
– Arthur
Jan 23 at 23:03
$begingroup$
It's "inverse" as in "opposite", not "reverse". It's not the common meaning in mathematics, but it's pedantic to require everyone here to adhere to such standards. Particularly when they're experimenting.
$endgroup$
– Arthur
Jan 23 at 23:03
$begingroup$
Yep, +Arthur's got it - though I see where you guys are coming too.
$endgroup$
– Robin Aldabanx
Jan 23 at 23:07
$begingroup$
Yep, +Arthur's got it - though I see where you guys are coming too.
$endgroup$
– Robin Aldabanx
Jan 23 at 23:07
add a comment |
1 Answer
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Probably not: the dot product has a geometric meaning (the product of the lengths of the vectors times the cosine of the angle between them): it's a scalar, i.e. a number that does not change under orthogonal coordinate transformations ("it's the same number in every coordinate system").
Your operation (it is not an "inverse" of the dot product, so let's not call it that) takes two vectors and produces a number, but the number is not invariant under coordinate transformations (e.g. it becomes infinite when you take a coordinate system where $vec{b}$ is along the x-axis): there is no geometric meaning that can be assigned to it, because it is not the same in every coordinate system.
Note that e.g. the x-coordinate of a vector is such a number: it is not the same in every coordinate system. It can nevertheless be useful. So there might be a use of your number somewhere, but if so, I don't know about it.
answered Jan 23 at 23:13
NickDNickD
1,1921512
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add a comment |
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1 Answer
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$begingroup$
Probably not: the dot product has a geometric meaning (the product of the lengths of the vectors times the cosine of the angle between them): it's a scalar, i.e. a number that does not change under orthogonal coordinate transformations ("it's the same number in every coordinate system").
Your operation (it is not an "inverse" of the dot product, so let's not call it that) takes two vectors and produces a number, but the number is not invariant under coordinate transformations (e.g. it becomes infinite when you take a coordinate system where $vec{b}$ is along the x-axis): there is no geometric meaning that can be assigned to it, because it is not the same in every coordinate system.
Note that e.g. the x-coordinate of a vector is such a number: it is not the same in every coordinate system. It can nevertheless be useful. So there might be a use of your number somewhere, but if so, I don't know about it.
answered Jan 23 at 23:13
NickDNickD
1,1921512
$endgroup$
add a comment |
$begingroup$
Probably not: the dot product has a geometric meaning (the product of the lengths of the vectors times the cosine of the angle between them): it's a scalar, i.e. a number that does not change under orthogonal coordinate transformations ("it's the same number in every coordinate system").
Your operation (it is not an "inverse" of the dot product, so let's not call it that) takes two vectors and produces a number, but the number is not invariant under coordinate transformations (e.g. it becomes infinite when you take a coordinate system where $vec{b}$ is along the x-axis): there is no geometric meaning that can be assigned to it, because it is not the same in every coordinate system.
Note that e.g. the x-coordinate of a vector is such a number: it is not the same in every coordinate system. It can nevertheless be useful. So there might be a use of your number somewhere, but if so, I don't know about it.
answered Jan 23 at 23:13
NickDNickD
1,1921512
$endgroup$
add a comment |
$begingroup$
Probably not: the dot product has a geometric meaning (the product of the lengths of the vectors times the cosine of the angle between them): it's a scalar, i.e. a number that does not change under orthogonal coordinate transformations ("it's the same number in every coordinate system").
Your operation (it is not an "inverse" of the dot product, so let's not call it that) takes two vectors and produces a number, but the number is not invariant under coordinate transformations (e.g. it becomes infinite when you take a coordinate system where $vec{b}$ is along the x-axis): there is no geometric meaning that can be assigned to it, because it is not the same in every coordinate system.
Note that e.g. the x-coordinate of a vector is such a number: it is not the same in every coordinate system. It can nevertheless be useful. So there might be a use of your number somewhere, but if so, I don't know about it.
answered Jan 23 at 23:13
NickDNickD
1,1921512
$endgroup$
Probably not: the dot product has a geometric meaning (the product of the lengths of the vectors times the cosine of the angle between them): it's a scalar, i.e. a number that does not change under orthogonal coordinate transformations ("it's the same number in every coordinate system").
Your operation (it is not an "inverse" of the dot product, so let's not call it that) takes two vectors and produces a number, but the number is not invariant under coordinate transformations (e.g. it becomes infinite when you take a coordinate system where $vec{b}$ is along the x-axis): there is no geometric meaning that can be assigned to it, because it is not the same in every coordinate system.
Note that e.g. the x-coordinate of a vector is such a number: it is not the same in every coordinate system. It can nevertheless be useful. So there might be a use of your number somewhere, but if so, I don't know about it.
answered Jan 23 at 23:13
NickDNickD
1,1921512
answered Jan 23 at 23:13
NickDNickD
1,1921512
answered Jan 23 at 23:13
NickDNickD
1,1921512
answered Jan 23 at 23:13
NickDNickD
1,1921512
1,1921512
add a comment |
add a comment |
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$begingroup$
Note that the dot product sends two vectors to a scalar, the inverse of that won't make much sense (there are infinitely many such mappings).
$endgroup$
– lightxbulb
Jan 23 at 22:54
$begingroup$
The dot product is not injective, so there cannot be an inverse. However, maybe you used the term "inverse" by mistake, since you are talking about reciprocals.
$endgroup$
– Ben W
Jan 23 at 22:59
$begingroup$
It's "inverse" as in "opposite", not "reverse". It's not the common meaning in mathematics, but it's pedantic to require everyone here to adhere to such standards. Particularly when they're experimenting.
$endgroup$
– Arthur
Jan 23 at 23:03
$begingroup$
Yep, +Arthur's got it - though I see where you guys are coming too.
$endgroup$
– Robin Aldabanx
Jan 23 at 23:07