Mixing
proof for Dot product and projection.
Clash Royale CLAN TAG#URR8PPP
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dot product can be defined like this.
The dot product a · b is equal to the signed length of the projection of b onto any line parallel to a, multiplied by the length of a.
a.b = (length of projection of b on a) * |a|
How is projection related with dot product ?. And any proof for this formula.
linear-algebra vectors
asked Aug 1 at 13:38
Atul
1106
add a comment |Â
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0
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dot product can be defined like this.
The dot product a · b is equal to the signed length of the projection of b onto any line parallel to a, multiplied by the length of a.
a.b = (length of projection of b on a) * |a|
How is projection related with dot product ?. And any proof for this formula.
linear-algebra vectors
asked Aug 1 at 13:38
Atul
1106
Hint: Lie the vectors on a plane and use trigonometry to express the components.
– ja72
Aug 1 at 18:24
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
dot product can be defined like this.
The dot product a · b is equal to the signed length of the projection of b onto any line parallel to a, multiplied by the length of a.
a.b = (length of projection of b on a) * |a|
How is projection related with dot product ?. And any proof for this formula.
linear-algebra vectors
asked Aug 1 at 13:38
Atul
1106
dot product can be defined like this.
The dot product a · b is equal to the signed length of the projection of b onto any line parallel to a, multiplied by the length of a.
a.b = (length of projection of b on a) * |a|
How is projection related with dot product ?. And any proof for this formula.
linear-algebra vectors
asked Aug 1 at 13:38
Atul
1106
asked Aug 1 at 13:38
Atul
1106
asked Aug 1 at 13:38
Atul
1106
asked Aug 1 at 13:38
Atul
1106
1106
Hint: Lie the vectors on a plane and use trigonometry to express the components.
– ja72
Aug 1 at 18:24
add a comment |Â
Hint: Lie the vectors on a plane and use trigonometry to express the components.
– ja72
Aug 1 at 18:24
Hint: Lie the vectors on a plane and use trigonometry to express the components.
– ja72
Aug 1 at 18:24
Hint: Lie the vectors on a plane and use trigonometry to express the components.
– ja72
Aug 1 at 18:24
add a comment |Â
2 Answers
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The orthogonal projection of $b$ onto the line generated by $a$ is a vector $ra$. such that $b-ra$ is orthogonal to $a$. Then $0=(b-ra)cdot a=bcdot a-r(acdot a)$.
Therefore $acdot b=r|a|^2$. Since the projection is $ra$, its length is $|ra|=|r||a|$.
So, $bcdot a=r|a||a|=sign(r)|ra||a|$.
In your formula I guess you need to interpret "legnth of the projection" as "signed length of the projection", which is the $r|a|$.
For example, if $a=(1,0)$, then the projections of $b_1=(1,1)$ and $b_2=(-1,1)$ onto the line generated by $a$, which is the $x$-axis, have the same length. But $acdot b_1=1$, while $acdot b_2=-1$.
answered Aug 1 at 13:46
byanalogy
1
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up vote
0
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The projection $P_a(b)$ of $b$ on $a$ is in general defined by
$$ |P_a(b)-b| = min_lambda | lambda a - b |. $$
Now if you have a scalar product on your vector space, you can square the norms and get for the right side
$$ |lambda-b|^2 = lambda^2|a|^2 - 2 lambda <a,b> + |b|^2. $$
which is optimal for
$$ 2 lambda |a|^2 = 2 <a,b> $$
or $$lambda = frac<a,b>. $$
Hence, we can calculate the projection then by
$$ P_a(b) = lambda a = <fraca, b> fraca. $$
So, if you have a scalar product you can easily calculate the projection, which should be your desired relation.
Now to your formula: The length of the projection is
$$ |P_a(b)| = |lambda||a|= left|frac<a, b>right| |a| $$
which is
$$ |P_a(b)||a| = |<a, b>|. $$
answered Aug 1 at 20:33
til
694
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The orthogonal projection of $b$ onto the line generated by $a$ is a vector $ra$. such that $b-ra$ is orthogonal to $a$. Then $0=(b-ra)cdot a=bcdot a-r(acdot a)$.
Therefore $acdot b=r|a|^2$. Since the projection is $ra$, its length is $|ra|=|r||a|$.
So, $bcdot a=r|a||a|=sign(r)|ra||a|$.
In your formula I guess you need to interpret "legnth of the projection" as "signed length of the projection", which is the $r|a|$.
For example, if $a=(1,0)$, then the projections of $b_1=(1,1)$ and $b_2=(-1,1)$ onto the line generated by $a$, which is the $x$-axis, have the same length. But $acdot b_1=1$, while $acdot b_2=-1$.
answered Aug 1 at 13:46
byanalogy
1
add a comment |Â
up vote
0
down vote
The orthogonal projection of $b$ onto the line generated by $a$ is a vector $ra$. such that $b-ra$ is orthogonal to $a$. Then $0=(b-ra)cdot a=bcdot a-r(acdot a)$.
Therefore $acdot b=r|a|^2$. Since the projection is $ra$, its length is $|ra|=|r||a|$.
So, $bcdot a=r|a||a|=sign(r)|ra||a|$.
In your formula I guess you need to interpret "legnth of the projection" as "signed length of the projection", which is the $r|a|$.
For example, if $a=(1,0)$, then the projections of $b_1=(1,1)$ and $b_2=(-1,1)$ onto the line generated by $a$, which is the $x$-axis, have the same length. But $acdot b_1=1$, while $acdot b_2=-1$.
answered Aug 1 at 13:46
byanalogy
1
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The orthogonal projection of $b$ onto the line generated by $a$ is a vector $ra$. such that $b-ra$ is orthogonal to $a$. Then $0=(b-ra)cdot a=bcdot a-r(acdot a)$.
Therefore $acdot b=r|a|^2$. Since the projection is $ra$, its length is $|ra|=|r||a|$.
So, $bcdot a=r|a||a|=sign(r)|ra||a|$.
In your formula I guess you need to interpret "legnth of the projection" as "signed length of the projection", which is the $r|a|$.
For example, if $a=(1,0)$, then the projections of $b_1=(1,1)$ and $b_2=(-1,1)$ onto the line generated by $a$, which is the $x$-axis, have the same length. But $acdot b_1=1$, while $acdot b_2=-1$.
answered Aug 1 at 13:46
byanalogy
1
The orthogonal projection of $b$ onto the line generated by $a$ is a vector $ra$. such that $b-ra$ is orthogonal to $a$. Then $0=(b-ra)cdot a=bcdot a-r(acdot a)$.
Therefore $acdot b=r|a|^2$. Since the projection is $ra$, its length is $|ra|=|r||a|$.
So, $bcdot a=r|a||a|=sign(r)|ra||a|$.
In your formula I guess you need to interpret "legnth of the projection" as "signed length of the projection", which is the $r|a|$.
For example, if $a=(1,0)$, then the projections of $b_1=(1,1)$ and $b_2=(-1,1)$ onto the line generated by $a$, which is the $x$-axis, have the same length. But $acdot b_1=1$, while $acdot b_2=-1$.
answered Aug 1 at 13:46
byanalogy
1
answered Aug 1 at 13:46
byanalogy
1
answered Aug 1 at 13:46
byanalogy
1
answered Aug 1 at 13:46
byanalogy
1
1
add a comment |Â
add a comment |Â
up vote
0
down vote
The projection $P_a(b)$ of $b$ on $a$ is in general defined by
$$ |P_a(b)-b| = min_lambda | lambda a - b |. $$
Now if you have a scalar product on your vector space, you can square the norms and get for the right side
$$ |lambda-b|^2 = lambda^2|a|^2 - 2 lambda <a,b> + |b|^2. $$
which is optimal for
$$ 2 lambda |a|^2 = 2 <a,b> $$
or $$lambda = frac<a,b>. $$
Hence, we can calculate the projection then by
$$ P_a(b) = lambda a = <fraca, b> fraca. $$
So, if you have a scalar product you can easily calculate the projection, which should be your desired relation.
Now to your formula: The length of the projection is
$$ |P_a(b)| = |lambda||a|= left|frac<a, b>right| |a| $$
which is
$$ |P_a(b)||a| = |<a, b>|. $$
answered Aug 1 at 20:33
til
694
add a comment |Â
up vote
0
down vote
The projection $P_a(b)$ of $b$ on $a$ is in general defined by
$$ |P_a(b)-b| = min_lambda | lambda a - b |. $$
Now if you have a scalar product on your vector space, you can square the norms and get for the right side
$$ |lambda-b|^2 = lambda^2|a|^2 - 2 lambda <a,b> + |b|^2. $$
which is optimal for
$$ 2 lambda |a|^2 = 2 <a,b> $$
or $$lambda = frac<a,b>. $$
Hence, we can calculate the projection then by
$$ P_a(b) = lambda a = <fraca, b> fraca. $$
So, if you have a scalar product you can easily calculate the projection, which should be your desired relation.
Now to your formula: The length of the projection is
$$ |P_a(b)| = |lambda||a|= left|frac<a, b>right| |a| $$
which is
$$ |P_a(b)||a| = |<a, b>|. $$
answered Aug 1 at 20:33
til
694
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The projection $P_a(b)$ of $b$ on $a$ is in general defined by
$$ |P_a(b)-b| = min_lambda | lambda a - b |. $$
Now if you have a scalar product on your vector space, you can square the norms and get for the right side
$$ |lambda-b|^2 = lambda^2|a|^2 - 2 lambda <a,b> + |b|^2. $$
which is optimal for
$$ 2 lambda |a|^2 = 2 <a,b> $$
or $$lambda = frac<a,b>. $$
Hence, we can calculate the projection then by
$$ P_a(b) = lambda a = <fraca, b> fraca. $$
So, if you have a scalar product you can easily calculate the projection, which should be your desired relation.
Now to your formula: The length of the projection is
$$ |P_a(b)| = |lambda||a|= left|frac<a, b>right| |a| $$
which is
$$ |P_a(b)||a| = |<a, b>|. $$
answered Aug 1 at 20:33
til
694
The projection $P_a(b)$ of $b$ on $a$ is in general defined by
$$ |P_a(b)-b| = min_lambda | lambda a - b |. $$
Now if you have a scalar product on your vector space, you can square the norms and get for the right side
$$ |lambda-b|^2 = lambda^2|a|^2 - 2 lambda <a,b> + |b|^2. $$
which is optimal for
$$ 2 lambda |a|^2 = 2 <a,b> $$
or $$lambda = frac<a,b>. $$
Hence, we can calculate the projection then by
$$ P_a(b) = lambda a = <fraca, b> fraca. $$
So, if you have a scalar product you can easily calculate the projection, which should be your desired relation.
Now to your formula: The length of the projection is
$$ |P_a(b)| = |lambda||a|= left|frac<a, b>right| |a| $$
which is
$$ |P_a(b)||a| = |<a, b>|. $$
answered Aug 1 at 20:33
til
694
answered Aug 1 at 20:33
til
694
answered Aug 1 at 20:33
til
694
answered Aug 1 at 20:33
til
694
694
add a comment |Â
add a comment |Â
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Hint: Lie the vectors on a plane and use trigonometry to express the components.
– ja72
Aug 1 at 18:24