With regard to distributive law of inner product in vector algebraWhy is the inner product between...
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With regard to distributive law of inner product in vector algebra
Why is the inner product between divergence-free current $vec{J}$, and a gradient field$nabla varphi$ zero?How is the MHD magnetic field time evolution equation transformed to the vector potential time evolution equation?Probable mistake in the derivation of the vector form of Biot-Savart's LawWhat are the different definitions of torsion-free connections in GR?Inner product of k-covariant-tensors and inner product of k-formsComplex angle found in inner productIs it valid to construct a work tensor from force and displacement?Linear operators and the inner productWhy does this line integral give the wrong sign?Energy of continious charge distribution
$begingroup$
Consider the equality
begin{align*}
&vec{a}cdotvec{b}+c=0 \
implies &vec{a}cdot(vec{b}+frac{vec{a}}{vec{a}cdotvec{a}}c)=0.
end{align*}
If the above equation is valid for any $vec{a} $, can we say the following equation is valid?
begin{align*}
&vec{b}+frac{vec{a}}{vec{a}cdotvec{a}}c=0\
implies & vec{b}=-frac{c}{|vec{a}cdotvec{a}|}vec{a}
end{align*}
I feel something (in particular, with regard to sign) is wrong.
Can the above process be justified?
If not, let me know the reason.
vectors tensor-calculus vector-fields
asked 8 hours ago
SOQEHSOQEH
257
$endgroup$
|
show 3 more comments
$begingroup$
Consider the equality
begin{align*}
&vec{a}cdotvec{b}+c=0 \
implies &vec{a}cdot(vec{b}+frac{vec{a}}{vec{a}cdotvec{a}}c)=0.
end{align*}
If the above equation is valid for any $vec{a} $, can we say the following equation is valid?
begin{align*}
&vec{b}+frac{vec{a}}{vec{a}cdotvec{a}}c=0\
implies & vec{b}=-frac{c}{|vec{a}cdotvec{a}|}vec{a}
end{align*}
I feel something (in particular, with regard to sign) is wrong.
Can the above process be justified?
If not, let me know the reason.
vectors tensor-calculus vector-fields
asked 8 hours ago
SOQEHSOQEH
257
$endgroup$
$begingroup$
This seems to be a question for Mathematics SE.
$endgroup$
– Steeven
8 hours ago
$begingroup$
It's not valid for $vec{a}=vec{0}$ because $vec{a}cdotvec{a}=0$.
$endgroup$
– TheAverageHijano
8 hours ago
$begingroup$
@TheAverageHijano Thanks. Right! If $vec{a}$ is not zero vector, is it valid?
$endgroup$
– SOQEH
7 hours ago
$begingroup$
It doesn't matter if $vec{a} = vec{0}$ if it's true for every vector $vec{a}$. The fact that it in particular holds for $vec{a} = vec{0}$ is of no consequence. At least if we begin at the second equality.
$endgroup$
– InertialObserver
7 hours ago
$begingroup$
@InertialObserver, I didn't say that it holds for $vec{a}=vec{0}$, I said that it does not hold for that case.
$endgroup$
– TheAverageHijano
7 hours ago
|
show 3 more comments
$begingroup$
Consider the equality
begin{align*}
&vec{a}cdotvec{b}+c=0 \
implies &vec{a}cdot(vec{b}+frac{vec{a}}{vec{a}cdotvec{a}}c)=0.
end{align*}
If the above equation is valid for any $vec{a} $, can we say the following equation is valid?
begin{align*}
&vec{b}+frac{vec{a}}{vec{a}cdotvec{a}}c=0\
implies & vec{b}=-frac{c}{|vec{a}cdotvec{a}|}vec{a}
end{align*}
I feel something (in particular, with regard to sign) is wrong.
Can the above process be justified?
If not, let me know the reason.
vectors tensor-calculus vector-fields
asked 8 hours ago
SOQEHSOQEH
257
$endgroup$
Consider the equality
begin{align*}
&vec{a}cdotvec{b}+c=0 \
implies &vec{a}cdot(vec{b}+frac{vec{a}}{vec{a}cdotvec{a}}c)=0.
end{align*}
If the above equation is valid for any $vec{a} $, can we say the following equation is valid?
begin{align*}
&vec{b}+frac{vec{a}}{vec{a}cdotvec{a}}c=0\
implies & vec{b}=-frac{c}{|vec{a}cdotvec{a}|}vec{a}
end{align*}
I feel something (in particular, with regard to sign) is wrong.
Can the above process be justified?
If not, let me know the reason.
vectors tensor-calculus vector-fields
vectors tensor-calculus vector-fields
asked 8 hours ago
SOQEHSOQEH
257
asked 8 hours ago
SOQEHSOQEH
257
edited 3 hours ago
SOQEH
asked 8 hours ago
SOQEHSOQEH
257
asked 8 hours ago
SOQEHSOQEH
257
asked 8 hours ago
SOQEHSOQEH
257
257
$begingroup$
This seems to be a question for Mathematics SE.
$endgroup$
– Steeven
8 hours ago
$begingroup$
It's not valid for $vec{a}=vec{0}$ because $vec{a}cdotvec{a}=0$.
$endgroup$
– TheAverageHijano
8 hours ago
$begingroup$
@TheAverageHijano Thanks. Right! If $vec{a}$ is not zero vector, is it valid?
$endgroup$
– SOQEH
7 hours ago
$begingroup$
It doesn't matter if $vec{a} = vec{0}$ if it's true for every vector $vec{a}$. The fact that it in particular holds for $vec{a} = vec{0}$ is of no consequence. At least if we begin at the second equality.
$endgroup$
– InertialObserver
7 hours ago
$begingroup$
@InertialObserver, I didn't say that it holds for $vec{a}=vec{0}$, I said that it does not hold for that case.
$endgroup$
– TheAverageHijano
7 hours ago
|
show 3 more comments
$begingroup$
This seems to be a question for Mathematics SE.
$endgroup$
– Steeven
8 hours ago
$begingroup$
It's not valid for $vec{a}=vec{0}$ because $vec{a}cdotvec{a}=0$.
$endgroup$
– TheAverageHijano
8 hours ago
$begingroup$
@TheAverageHijano Thanks. Right! If $vec{a}$ is not zero vector, is it valid?
$endgroup$
– SOQEH
7 hours ago
$begingroup$
It doesn't matter if $vec{a} = vec{0}$ if it's true for every vector $vec{a}$. The fact that it in particular holds for $vec{a} = vec{0}$ is of no consequence. At least if we begin at the second equality.
$endgroup$
– InertialObserver
7 hours ago
$begingroup$
@InertialObserver, I didn't say that it holds for $vec{a}=vec{0}$, I said that it does not hold for that case.
$endgroup$
– TheAverageHijano
7 hours ago
$begingroup$
This seems to be a question for Mathematics SE.
$endgroup$
– Steeven
8 hours ago
$begingroup$
This seems to be a question for Mathematics SE.
$endgroup$
– Steeven
8 hours ago
$begingroup$
It's not valid for $vec{a}=vec{0}$ because $vec{a}cdotvec{a}=0$.
$endgroup$
– TheAverageHijano
8 hours ago
$begingroup$
It's not valid for $vec{a}=vec{0}$ because $vec{a}cdotvec{a}=0$.
$endgroup$
– TheAverageHijano
8 hours ago
$begingroup$
@TheAverageHijano Thanks. Right! If $vec{a}$ is not zero vector, is it valid?
$endgroup$
– SOQEH
7 hours ago
$begingroup$
@TheAverageHijano Thanks. Right! If $vec{a}$ is not zero vector, is it valid?
$endgroup$
– SOQEH
7 hours ago
$begingroup$
It doesn't matter if $vec{a} = vec{0}$ if it's true for every vector $vec{a}$. The fact that it in particular holds for $vec{a} = vec{0}$ is of no consequence. At least if we begin at the second equality.
$endgroup$
– InertialObserver
7 hours ago
$begingroup$
It doesn't matter if $vec{a} = vec{0}$ if it's true for every vector $vec{a}$. The fact that it in particular holds for $vec{a} = vec{0}$ is of no consequence. At least if we begin at the second equality.
$endgroup$
– InertialObserver
7 hours ago
$begingroup$
@InertialObserver, I didn't say that it holds for $vec{a}=vec{0}$, I said that it does not hold for that case.
$endgroup$
– TheAverageHijano
7 hours ago
$begingroup$
@InertialObserver, I didn't say that it holds for $vec{a}=vec{0}$, I said that it does not hold for that case.
$endgroup$
– TheAverageHijano
7 hours ago
|
show 3 more comments
2 Answers
2
active
oldest
votes
$begingroup$
The answer is clearly wrong, because it says that $vec b$ is proportional to $vec a$, i.e. they must point in the same direction. But if you add to any such $vec b_0$ a vector $vec{b'}$ which is perpendicular to $vec a$, then the product $vec a . (vec b_0+vec{b'})$ is the same so your first equation still holds for any $vec b=vec b_0+vec b'$. This equation tells you the component of $vec b$ along $vec a$, but the component of $vec b$ perpendicular to $vec a$ is completely arbitrary.
The error comes in the step (which you rightly have a question mark against) that your second equation is true for all $vec a$, so the bracket must be zero. This works for
$a f(b,a)=0 forall a implies f(b,a)=0$ but not for $vec a.vec f=0 forall vec a implies vec f=0$. You cannot create 3 equations from 1 equation.
The glitch in @InertialObserver's proof is that their $vec b$ is a different $vec b$ for $i=1,2,3$.
answered 6 hours ago
RogerJBarlowRogerJBarlow
3,180518
$endgroup$
$begingroup$
Yea, now that I think about it.. you’re right.. that’s what I get for answering questions at 2 am
$endgroup$
– InertialObserver
6 hours ago
$begingroup$
Unfortunately my answer has been accepted so I can’t delete it to defer to this answer
$endgroup$
– InertialObserver
6 hours ago
$begingroup$
It's a very subtle point. Easy to go adrift. Specially at 2am.
$endgroup$
– RogerJBarlow
5 hours ago
$begingroup$
@RogerJBarlow Thank you for your answer. I want to write the equation in a form of $vec{b}= something $. I'd appreciate it if you could tell me what to do.
$endgroup$
– SOQEH
5 hours ago
$begingroup$
Something like $vec b = -{c vec a over |a^2|} + vec d times vec a$, where $vec d$ is any arbitrary vector. The scalar triple product formula ensures $vec a .(vec d times vec a)=0$
$endgroup$
– RogerJBarlow
4 hours ago
|
show 1 more comment
$begingroup$
Yes, the implication is true given that $vec{a}neq vec{0}$.
Proof: Suppose that the equality
$$ quad vec{a}cdot(vec{b}+frac{vec{a}}{vec{a}cdotvec{a}}c)=0 $$
holds for any $vec{a} neq vec{0}$.
Since we can choose any vector $vec{a}$, choose $vec{a} = vec{e}_i $ where $vec{e}_i$ is a unit basis vector along the axis $x_i$ (in $mathbb{R}^3$ these would be $hat{x}, hat{y}, hat{z}$). Then we have that
$$
vec{e}_icdot(vec{b}+frac{vec{a}}{vec{a}cdotvec{a}}c) = b_i + c frac{a_i}{|vec{a}|^2} = 0
$$
for each component $i$. Or written in vector notation
$$
vec{b} + cfrac{vec{a}}{|vec{a}|^2} = 0.
$$
Note that I have used the notation $vec{a}cdotvec{a} = |vec{a}|^2$.
answered 7 hours ago
InertialObserverInertialObserver
3,0301027
$endgroup$
$begingroup$
But the norm of $vec{a}=vec{0}$ is 0, and you are dividing by $|vec{a}|²$, which is not defined. In fact, if you take the limit of $vec{a}=(epsilon,0,0)$ when $epsilon$ tends to zero, $a_1/|vec{a}|²$ tends to infinity...
$endgroup$
– TheAverageHijano
7 hours ago
$begingroup$
It is obvious the OP's equality does not hold for any $vec a ne vec 0$. (if it holds for some $vec a$, it obviously doesn't also hold for $2vec a$ unless $c = 0$) So this entire argument is wrong - you can prove anything starting from a false premise.
$endgroup$
– alephzero
4 hours ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The answer is clearly wrong, because it says that $vec b$ is proportional to $vec a$, i.e. they must point in the same direction. But if you add to any such $vec b_0$ a vector $vec{b'}$ which is perpendicular to $vec a$, then the product $vec a . (vec b_0+vec{b'})$ is the same so your first equation still holds for any $vec b=vec b_0+vec b'$. This equation tells you the component of $vec b$ along $vec a$, but the component of $vec b$ perpendicular to $vec a$ is completely arbitrary.
The error comes in the step (which you rightly have a question mark against) that your second equation is true for all $vec a$, so the bracket must be zero. This works for
$a f(b,a)=0 forall a implies f(b,a)=0$ but not for $vec a.vec f=0 forall vec a implies vec f=0$. You cannot create 3 equations from 1 equation.
The glitch in @InertialObserver's proof is that their $vec b$ is a different $vec b$ for $i=1,2,3$.
answered 6 hours ago
RogerJBarlowRogerJBarlow
3,180518
$endgroup$
$begingroup$
Yea, now that I think about it.. you’re right.. that’s what I get for answering questions at 2 am
$endgroup$
– InertialObserver
6 hours ago
$begingroup$
Unfortunately my answer has been accepted so I can’t delete it to defer to this answer
$endgroup$
– InertialObserver
6 hours ago
$begingroup$
It's a very subtle point. Easy to go adrift. Specially at 2am.
$endgroup$
– RogerJBarlow
5 hours ago
$begingroup$
@RogerJBarlow Thank you for your answer. I want to write the equation in a form of $vec{b}= something $. I'd appreciate it if you could tell me what to do.
$endgroup$
– SOQEH
5 hours ago
$begingroup$
Something like $vec b = -{c vec a over |a^2|} + vec d times vec a$, where $vec d$ is any arbitrary vector. The scalar triple product formula ensures $vec a .(vec d times vec a)=0$
$endgroup$
– RogerJBarlow
4 hours ago
|
show 1 more comment
$begingroup$
The answer is clearly wrong, because it says that $vec b$ is proportional to $vec a$, i.e. they must point in the same direction. But if you add to any such $vec b_0$ a vector $vec{b'}$ which is perpendicular to $vec a$, then the product $vec a . (vec b_0+vec{b'})$ is the same so your first equation still holds for any $vec b=vec b_0+vec b'$. This equation tells you the component of $vec b$ along $vec a$, but the component of $vec b$ perpendicular to $vec a$ is completely arbitrary.
The error comes in the step (which you rightly have a question mark against) that your second equation is true for all $vec a$, so the bracket must be zero. This works for
$a f(b,a)=0 forall a implies f(b,a)=0$ but not for $vec a.vec f=0 forall vec a implies vec f=0$. You cannot create 3 equations from 1 equation.
The glitch in @InertialObserver's proof is that their $vec b$ is a different $vec b$ for $i=1,2,3$.
answered 6 hours ago
RogerJBarlowRogerJBarlow
3,180518
$endgroup$
$begingroup$
Yea, now that I think about it.. you’re right.. that’s what I get for answering questions at 2 am
$endgroup$
– InertialObserver
6 hours ago
$begingroup$
Unfortunately my answer has been accepted so I can’t delete it to defer to this answer
$endgroup$
– InertialObserver
6 hours ago
$begingroup$
It's a very subtle point. Easy to go adrift. Specially at 2am.
$endgroup$
– RogerJBarlow
5 hours ago
$begingroup$
@RogerJBarlow Thank you for your answer. I want to write the equation in a form of $vec{b}= something $. I'd appreciate it if you could tell me what to do.
$endgroup$
– SOQEH
5 hours ago
$begingroup$
Something like $vec b = -{c vec a over |a^2|} + vec d times vec a$, where $vec d$ is any arbitrary vector. The scalar triple product formula ensures $vec a .(vec d times vec a)=0$
$endgroup$
– RogerJBarlow
4 hours ago
|
show 1 more comment
$begingroup$
The answer is clearly wrong, because it says that $vec b$ is proportional to $vec a$, i.e. they must point in the same direction. But if you add to any such $vec b_0$ a vector $vec{b'}$ which is perpendicular to $vec a$, then the product $vec a . (vec b_0+vec{b'})$ is the same so your first equation still holds for any $vec b=vec b_0+vec b'$. This equation tells you the component of $vec b$ along $vec a$, but the component of $vec b$ perpendicular to $vec a$ is completely arbitrary.
The error comes in the step (which you rightly have a question mark against) that your second equation is true for all $vec a$, so the bracket must be zero. This works for
$a f(b,a)=0 forall a implies f(b,a)=0$ but not for $vec a.vec f=0 forall vec a implies vec f=0$. You cannot create 3 equations from 1 equation.
The glitch in @InertialObserver's proof is that their $vec b$ is a different $vec b$ for $i=1,2,3$.
answered 6 hours ago
RogerJBarlowRogerJBarlow
3,180518
$endgroup$
The answer is clearly wrong, because it says that $vec b$ is proportional to $vec a$, i.e. they must point in the same direction. But if you add to any such $vec b_0$ a vector $vec{b'}$ which is perpendicular to $vec a$, then the product $vec a . (vec b_0+vec{b'})$ is the same so your first equation still holds for any $vec b=vec b_0+vec b'$. This equation tells you the component of $vec b$ along $vec a$, but the component of $vec b$ perpendicular to $vec a$ is completely arbitrary.
The error comes in the step (which you rightly have a question mark against) that your second equation is true for all $vec a$, so the bracket must be zero. This works for
$a f(b,a)=0 forall a implies f(b,a)=0$ but not for $vec a.vec f=0 forall vec a implies vec f=0$. You cannot create 3 equations from 1 equation.
The glitch in @InertialObserver's proof is that their $vec b$ is a different $vec b$ for $i=1,2,3$.
answered 6 hours ago
RogerJBarlowRogerJBarlow
3,180518
answered 6 hours ago
RogerJBarlowRogerJBarlow
3,180518
answered 6 hours ago
RogerJBarlowRogerJBarlow
3,180518
answered 6 hours ago
RogerJBarlowRogerJBarlow
3,180518
3,180518
$begingroup$
Yea, now that I think about it.. you’re right.. that’s what I get for answering questions at 2 am
$endgroup$
– InertialObserver
6 hours ago
$begingroup$
Unfortunately my answer has been accepted so I can’t delete it to defer to this answer
$endgroup$
– InertialObserver
6 hours ago
$begingroup$
It's a very subtle point. Easy to go adrift. Specially at 2am.
$endgroup$
– RogerJBarlow
5 hours ago
$begingroup$
@RogerJBarlow Thank you for your answer. I want to write the equation in a form of $vec{b}= something $. I'd appreciate it if you could tell me what to do.
$endgroup$
– SOQEH
5 hours ago
$begingroup$
Something like $vec b = -{c vec a over |a^2|} + vec d times vec a$, where $vec d$ is any arbitrary vector. The scalar triple product formula ensures $vec a .(vec d times vec a)=0$
$endgroup$
– RogerJBarlow
4 hours ago
|
show 1 more comment
$begingroup$
Yea, now that I think about it.. you’re right.. that’s what I get for answering questions at 2 am
$endgroup$
– InertialObserver
6 hours ago
$begingroup$
Unfortunately my answer has been accepted so I can’t delete it to defer to this answer
$endgroup$
– InertialObserver
6 hours ago
$begingroup$
It's a very subtle point. Easy to go adrift. Specially at 2am.
$endgroup$
– RogerJBarlow
5 hours ago
$begingroup$
@RogerJBarlow Thank you for your answer. I want to write the equation in a form of $vec{b}= something $. I'd appreciate it if you could tell me what to do.
$endgroup$
– SOQEH
5 hours ago
$begingroup$
Something like $vec b = -{c vec a over |a^2|} + vec d times vec a$, where $vec d$ is any arbitrary vector. The scalar triple product formula ensures $vec a .(vec d times vec a)=0$
$endgroup$
– RogerJBarlow
4 hours ago
$begingroup$
Yea, now that I think about it.. you’re right.. that’s what I get for answering questions at 2 am
$endgroup$
– InertialObserver
6 hours ago
$begingroup$
Yea, now that I think about it.. you’re right.. that’s what I get for answering questions at 2 am
$endgroup$
– InertialObserver
6 hours ago
$begingroup$
Unfortunately my answer has been accepted so I can’t delete it to defer to this answer
$endgroup$
– InertialObserver
6 hours ago
$begingroup$
Unfortunately my answer has been accepted so I can’t delete it to defer to this answer
$endgroup$
– InertialObserver
6 hours ago
$begingroup$
It's a very subtle point. Easy to go adrift. Specially at 2am.
$endgroup$
– RogerJBarlow
5 hours ago
$begingroup$
It's a very subtle point. Easy to go adrift. Specially at 2am.
$endgroup$
– RogerJBarlow
5 hours ago
$begingroup$
@RogerJBarlow Thank you for your answer. I want to write the equation in a form of $vec{b}= something $. I'd appreciate it if you could tell me what to do.
$endgroup$
– SOQEH
5 hours ago
$begingroup$
@RogerJBarlow Thank you for your answer. I want to write the equation in a form of $vec{b}= something $. I'd appreciate it if you could tell me what to do.
$endgroup$
– SOQEH
5 hours ago
$begingroup$
Something like $vec b = -{c vec a over |a^2|} + vec d times vec a$, where $vec d$ is any arbitrary vector. The scalar triple product formula ensures $vec a .(vec d times vec a)=0$
$endgroup$
– RogerJBarlow
4 hours ago
$begingroup$
Something like $vec b = -{c vec a over |a^2|} + vec d times vec a$, where $vec d$ is any arbitrary vector. The scalar triple product formula ensures $vec a .(vec d times vec a)=0$
$endgroup$
– RogerJBarlow
4 hours ago
|
show 1 more comment
$begingroup$
Yes, the implication is true given that $vec{a}neq vec{0}$.
Proof: Suppose that the equality
$$ quad vec{a}cdot(vec{b}+frac{vec{a}}{vec{a}cdotvec{a}}c)=0 $$
holds for any $vec{a} neq vec{0}$.
Since we can choose any vector $vec{a}$, choose $vec{a} = vec{e}_i $ where $vec{e}_i$ is a unit basis vector along the axis $x_i$ (in $mathbb{R}^3$ these would be $hat{x}, hat{y}, hat{z}$). Then we have that
$$
vec{e}_icdot(vec{b}+frac{vec{a}}{vec{a}cdotvec{a}}c) = b_i + c frac{a_i}{|vec{a}|^2} = 0
$$
for each component $i$. Or written in vector notation
$$
vec{b} + cfrac{vec{a}}{|vec{a}|^2} = 0.
$$
Note that I have used the notation $vec{a}cdotvec{a} = |vec{a}|^2$.
answered 7 hours ago
InertialObserverInertialObserver
3,0301027
$endgroup$
$begingroup$
But the norm of $vec{a}=vec{0}$ is 0, and you are dividing by $|vec{a}|²$, which is not defined. In fact, if you take the limit of $vec{a}=(epsilon,0,0)$ when $epsilon$ tends to zero, $a_1/|vec{a}|²$ tends to infinity...
$endgroup$
– TheAverageHijano
7 hours ago
$begingroup$
It is obvious the OP's equality does not hold for any $vec a ne vec 0$. (if it holds for some $vec a$, it obviously doesn't also hold for $2vec a$ unless $c = 0$) So this entire argument is wrong - you can prove anything starting from a false premise.
$endgroup$
– alephzero
4 hours ago
add a comment |
$begingroup$
Yes, the implication is true given that $vec{a}neq vec{0}$.
Proof: Suppose that the equality
$$ quad vec{a}cdot(vec{b}+frac{vec{a}}{vec{a}cdotvec{a}}c)=0 $$
holds for any $vec{a} neq vec{0}$.
Since we can choose any vector $vec{a}$, choose $vec{a} = vec{e}_i $ where $vec{e}_i$ is a unit basis vector along the axis $x_i$ (in $mathbb{R}^3$ these would be $hat{x}, hat{y}, hat{z}$). Then we have that
$$
vec{e}_icdot(vec{b}+frac{vec{a}}{vec{a}cdotvec{a}}c) = b_i + c frac{a_i}{|vec{a}|^2} = 0
$$
for each component $i$. Or written in vector notation
$$
vec{b} + cfrac{vec{a}}{|vec{a}|^2} = 0.
$$
Note that I have used the notation $vec{a}cdotvec{a} = |vec{a}|^2$.
answered 7 hours ago
InertialObserverInertialObserver
3,0301027
$endgroup$
$begingroup$
But the norm of $vec{a}=vec{0}$ is 0, and you are dividing by $|vec{a}|²$, which is not defined. In fact, if you take the limit of $vec{a}=(epsilon,0,0)$ when $epsilon$ tends to zero, $a_1/|vec{a}|²$ tends to infinity...
$endgroup$
– TheAverageHijano
7 hours ago
$begingroup$
It is obvious the OP's equality does not hold for any $vec a ne vec 0$. (if it holds for some $vec a$, it obviously doesn't also hold for $2vec a$ unless $c = 0$) So this entire argument is wrong - you can prove anything starting from a false premise.
$endgroup$
– alephzero
4 hours ago
add a comment |
$begingroup$
Yes, the implication is true given that $vec{a}neq vec{0}$.
Proof: Suppose that the equality
$$ quad vec{a}cdot(vec{b}+frac{vec{a}}{vec{a}cdotvec{a}}c)=0 $$
holds for any $vec{a} neq vec{0}$.
Since we can choose any vector $vec{a}$, choose $vec{a} = vec{e}_i $ where $vec{e}_i$ is a unit basis vector along the axis $x_i$ (in $mathbb{R}^3$ these would be $hat{x}, hat{y}, hat{z}$). Then we have that
$$
vec{e}_icdot(vec{b}+frac{vec{a}}{vec{a}cdotvec{a}}c) = b_i + c frac{a_i}{|vec{a}|^2} = 0
$$
for each component $i$. Or written in vector notation
$$
vec{b} + cfrac{vec{a}}{|vec{a}|^2} = 0.
$$
Note that I have used the notation $vec{a}cdotvec{a} = |vec{a}|^2$.
answered 7 hours ago
InertialObserverInertialObserver
3,0301027
$endgroup$
Yes, the implication is true given that $vec{a}neq vec{0}$.
Proof: Suppose that the equality
$$ quad vec{a}cdot(vec{b}+frac{vec{a}}{vec{a}cdotvec{a}}c)=0 $$
holds for any $vec{a} neq vec{0}$.
Since we can choose any vector $vec{a}$, choose $vec{a} = vec{e}_i $ where $vec{e}_i$ is a unit basis vector along the axis $x_i$ (in $mathbb{R}^3$ these would be $hat{x}, hat{y}, hat{z}$). Then we have that
$$
vec{e}_icdot(vec{b}+frac{vec{a}}{vec{a}cdotvec{a}}c) = b_i + c frac{a_i}{|vec{a}|^2} = 0
$$
for each component $i$. Or written in vector notation
$$
vec{b} + cfrac{vec{a}}{|vec{a}|^2} = 0.
$$
Note that I have used the notation $vec{a}cdotvec{a} = |vec{a}|^2$.
answered 7 hours ago
InertialObserverInertialObserver
3,0301027
edited 7 hours ago
answered 7 hours ago
InertialObserverInertialObserver
3,0301027
answered 7 hours ago
InertialObserverInertialObserver
3,0301027
answered 7 hours ago
InertialObserverInertialObserver
3,0301027
3,0301027
$begingroup$
But the norm of $vec{a}=vec{0}$ is 0, and you are dividing by $|vec{a}|²$, which is not defined. In fact, if you take the limit of $vec{a}=(epsilon,0,0)$ when $epsilon$ tends to zero, $a_1/|vec{a}|²$ tends to infinity...
$endgroup$
– TheAverageHijano
7 hours ago
$begingroup$
It is obvious the OP's equality does not hold for any $vec a ne vec 0$. (if it holds for some $vec a$, it obviously doesn't also hold for $2vec a$ unless $c = 0$) So this entire argument is wrong - you can prove anything starting from a false premise.
$endgroup$
– alephzero
4 hours ago
add a comment |
$begingroup$
But the norm of $vec{a}=vec{0}$ is 0, and you are dividing by $|vec{a}|²$, which is not defined. In fact, if you take the limit of $vec{a}=(epsilon,0,0)$ when $epsilon$ tends to zero, $a_1/|vec{a}|²$ tends to infinity...
$endgroup$
– TheAverageHijano
7 hours ago
$begingroup$
It is obvious the OP's equality does not hold for any $vec a ne vec 0$. (if it holds for some $vec a$, it obviously doesn't also hold for $2vec a$ unless $c = 0$) So this entire argument is wrong - you can prove anything starting from a false premise.
$endgroup$
– alephzero
4 hours ago
$begingroup$
But the norm of $vec{a}=vec{0}$ is 0, and you are dividing by $|vec{a}|²$, which is not defined. In fact, if you take the limit of $vec{a}=(epsilon,0,0)$ when $epsilon$ tends to zero, $a_1/|vec{a}|²$ tends to infinity...
$endgroup$
– TheAverageHijano
7 hours ago
$begingroup$
But the norm of $vec{a}=vec{0}$ is 0, and you are dividing by $|vec{a}|²$, which is not defined. In fact, if you take the limit of $vec{a}=(epsilon,0,0)$ when $epsilon$ tends to zero, $a_1/|vec{a}|²$ tends to infinity...
$endgroup$
– TheAverageHijano
7 hours ago
$begingroup$
It is obvious the OP's equality does not hold for any $vec a ne vec 0$. (if it holds for some $vec a$, it obviously doesn't also hold for $2vec a$ unless $c = 0$) So this entire argument is wrong - you can prove anything starting from a false premise.
$endgroup$
– alephzero
4 hours ago
$begingroup$
It is obvious the OP's equality does not hold for any $vec a ne vec 0$. (if it holds for some $vec a$, it obviously doesn't also hold for $2vec a$ unless $c = 0$) So this entire argument is wrong - you can prove anything starting from a false premise.
$endgroup$
– alephzero
4 hours ago
add a comment |
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$begingroup$
This seems to be a question for Mathematics SE.
$endgroup$
– Steeven
8 hours ago
$begingroup$
It's not valid for $vec{a}=vec{0}$ because $vec{a}cdotvec{a}=0$.
$endgroup$
– TheAverageHijano
8 hours ago
$begingroup$
@TheAverageHijano Thanks. Right! If $vec{a}$ is not zero vector, is it valid?
$endgroup$
– SOQEH
7 hours ago
$begingroup$
It doesn't matter if $vec{a} = vec{0}$ if it's true for every vector $vec{a}$. The fact that it in particular holds for $vec{a} = vec{0}$ is of no consequence. At least if we begin at the second equality.
$endgroup$
– InertialObserver
7 hours ago
$begingroup$
@InertialObserver, I didn't say that it holds for $vec{a}=vec{0}$, I said that it does not hold for that case.
$endgroup$
– TheAverageHijano
7 hours ago