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What is the Euclidean inner product intuition? [duplicate]
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This question already has an answer here:
What does the dot product of two vectors represent?
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The Euclidean inner product $<x,y>$ of the vectors $x , y in mathbbR^n$ is defined by:
$langle x,yrangle = x_1y_1 + x_2y_2 + x_3y_3 + ... + x_ny_n$
I am unable to find the intuition behind this? Why do we need this inner product? What happens to the vectors under equation and what does the scalar answer tell us?
P.S. I have already seareched it, most the articles take it into geometric terms and introduce $costheta$. Can anybody help me visualize thing without angular stuff?
Edit: Question addressed here uses the angle to explain geometric dot product. I am wondering if there exist some explanation for dot product in Vector Space.
vector-spaces
edited Jul 24 at 2:50
max_zorn
3,15151028
asked Jul 24 at 1:22
Wasiq Noor
634
marked as duplicate by Lee Mosher, Travis, JavaMan, Parcly Taxel, user223391 Jul 29 at 22:01
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |Â
up vote
5
down vote
favorite
This question already has an answer here:
What does the dot product of two vectors represent?
7 answers
The Euclidean inner product $<x,y>$ of the vectors $x , y in mathbbR^n$ is defined by:
$langle x,yrangle = x_1y_1 + x_2y_2 + x_3y_3 + ... + x_ny_n$
I am unable to find the intuition behind this? Why do we need this inner product? What happens to the vectors under equation and what does the scalar answer tell us?
P.S. I have already seareched it, most the articles take it into geometric terms and introduce $costheta$. Can anybody help me visualize thing without angular stuff?
Edit: Question addressed here uses the angle to explain geometric dot product. I am wondering if there exist some explanation for dot product in Vector Space.
vector-spaces
edited Jul 24 at 2:50
max_zorn
3,15151028
asked Jul 24 at 1:22
Wasiq Noor
634
marked as duplicate by Lee Mosher, Travis, JavaMan, Parcly Taxel, user223391 Jul 29 at 22:01
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
2
Well, the intuition one must have of the inner product precisely is this "angular stuff", to my opinion.
– Suzet
Jul 24 at 1:24
1
Small $LaTeX$ note: "langle" and "rangle" look nicer than the greater than/less than symbol: They typeset as $langle -, - rangle$.
– JavaMan
Jul 24 at 2:07
I'm answering this here rather than calling it a duplicate because it explicitly asks for explanations that are not "angular." I think @LeeMosher's answer to the other question actually is such an answer but the others mostly are not.
– David K
Jul 24 at 2:14
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
This question already has an answer here:
What does the dot product of two vectors represent?
7 answers
The Euclidean inner product $<x,y>$ of the vectors $x , y in mathbbR^n$ is defined by:
$langle x,yrangle = x_1y_1 + x_2y_2 + x_3y_3 + ... + x_ny_n$
I am unable to find the intuition behind this? Why do we need this inner product? What happens to the vectors under equation and what does the scalar answer tell us?
P.S. I have already seareched it, most the articles take it into geometric terms and introduce $costheta$. Can anybody help me visualize thing without angular stuff?
Edit: Question addressed here uses the angle to explain geometric dot product. I am wondering if there exist some explanation for dot product in Vector Space.
vector-spaces
edited Jul 24 at 2:50
max_zorn
3,15151028
asked Jul 24 at 1:22
Wasiq Noor
634
This question already has an answer here:
What does the dot product of two vectors represent?
7 answers
The Euclidean inner product $<x,y>$ of the vectors $x , y in mathbbR^n$ is defined by:
$langle x,yrangle = x_1y_1 + x_2y_2 + x_3y_3 + ... + x_ny_n$
I am unable to find the intuition behind this? Why do we need this inner product? What happens to the vectors under equation and what does the scalar answer tell us?
P.S. I have already seareched it, most the articles take it into geometric terms and introduce $costheta$. Can anybody help me visualize thing without angular stuff?
Edit: Question addressed here uses the angle to explain geometric dot product. I am wondering if there exist some explanation for dot product in Vector Space.
This question already has an answer here:
What does the dot product of two vectors represent?
7 answers
vector-spaces
edited Jul 24 at 2:50
max_zorn
3,15151028
asked Jul 24 at 1:22
Wasiq Noor
634
edited Jul 24 at 2:50
max_zorn
3,15151028
edited Jul 24 at 2:50
max_zorn
3,15151028
edited Jul 24 at 2:50
max_zorn
3,15151028
3,15151028
asked Jul 24 at 1:22
Wasiq Noor
634
asked Jul 24 at 1:22
Wasiq Noor
634
asked Jul 24 at 1:22
Wasiq Noor
634
634
marked as duplicate by Lee Mosher, Travis, JavaMan, Parcly Taxel, user223391 Jul 29 at 22:01
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Lee Mosher, Travis, JavaMan, Parcly Taxel, user223391 Jul 29 at 22:01
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
2
Well, the intuition one must have of the inner product precisely is this "angular stuff", to my opinion.
– Suzet
Jul 24 at 1:24
1
Small $LaTeX$ note: "langle" and "rangle" look nicer than the greater than/less than symbol: They typeset as $langle -, - rangle$.
– JavaMan
Jul 24 at 2:07
I'm answering this here rather than calling it a duplicate because it explicitly asks for explanations that are not "angular." I think @LeeMosher's answer to the other question actually is such an answer but the others mostly are not.
– David K
Jul 24 at 2:14
add a comment |Â
2
Well, the intuition one must have of the inner product precisely is this "angular stuff", to my opinion.
– Suzet
Jul 24 at 1:24
1
Small $LaTeX$ note: "langle" and "rangle" look nicer than the greater than/less than symbol: They typeset as $langle -, - rangle$.
– JavaMan
Jul 24 at 2:07
I'm answering this here rather than calling it a duplicate because it explicitly asks for explanations that are not "angular." I think @LeeMosher's answer to the other question actually is such an answer but the others mostly are not.
– David K
Jul 24 at 2:14
2
2
Well, the intuition one must have of the inner product precisely is this "angular stuff", to my opinion.
– Suzet
Jul 24 at 1:24
Well, the intuition one must have of the inner product precisely is this "angular stuff", to my opinion.
– Suzet
Jul 24 at 1:24
1
1
Small $LaTeX$ note: "langle" and "rangle" look nicer than the greater than/less than symbol: They typeset as $langle -, - rangle$.
– JavaMan
Jul 24 at 2:07
Small $LaTeX$ note: "langle" and "rangle" look nicer than the greater than/less than symbol: They typeset as $langle -, - rangle$.
– JavaMan
Jul 24 at 2:07
I'm answering this here rather than calling it a duplicate because it explicitly asks for explanations that are not "angular." I think @LeeMosher's answer to the other question actually is such an answer but the others mostly are not.
– David K
Jul 24 at 2:14
I'm answering this here rather than calling it a duplicate because it explicitly asks for explanations that are not "angular." I think @LeeMosher's answer to the other question actually is such an answer but the others mostly are not.
– David K
Jul 24 at 2:14
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
1
down vote
You are probably familiar with the following formula involving
a vector $x = (x_1, ldots, x_n) in mathbb R^n$
and $n$ constants $a_1, ldots, a_n$ that are not all zero:
$$
a_1 x_1 + a_2 x_2 + cdots + a_n x_n = 0.
$$
This formula describes an $(n-1)$-dimensional hyperplane in $mathbb R^n.$
Let's call that hyperplane $A.$ That is, any vector $x$ that satisfies the formula lies in the hyperplane $A.$
If we define a vector $a = (a_1, ldots, a_n),$
then another way to write the formula is
$$ langle x, arangle = 0. $$
Now if we choose an arbitrary vector $x in mathbb R^n,$
it may happen that $langle x, arangle neq 0.$
If we have two such vectors, let's say $x'$ and $x'',$
such that $langle x', arangle > 0$ and $langle x'', arangle > 0,$
then $x'$ and $x''$ will be on the same side of the hyperplane $A.$
But if $langle x', arangle < 0 < langle x'', arangle$ then the vectors are on opposite sides.
For a given vector $x,$ suppose you find a vector $y$ such that
$x - langle x, arangle y$ is in the hyperplane $A.$
Then $x' - langle x', arangle y$ also is in the hyperplane $A$
for any other vector $x'.$
That is, $langle x, arangle$ tells you how many times the length of $y$
you have to travel in the direction of $y$ to get from $x$ to the hyperplane,
or in other words, $langle x', arangle$ is a kind of measurement of the distance of $x$ from the hyperplane (measured in some particular units in some particular direction).
All of this works without any "angles" (unless you consider "parallel" to be a "zero angle").
You don't even need
$(x_1, ldots, x_n)$ to be coordinates over an orthonormal basis,
although if the basis is orthonormal then other nice results follow.
For example,
$$sqrtlangle x, xrangle = sqrtx_1^2 + x_2^2 + cdots + x_n^2,$$
which is the length of $x$ (according to the Pythagorean Theorem)
if the basis of $(x_1, ldots, x_n)$ is orthonormal.
Admittedly, to speak of "normals" one must have a concept of things being perpendicular, which starts to sound like we're dealing with angles again.
But they're right angles, which are especially simple to work with.
answered Jul 24 at 2:09
David K
48.2k340107
add a comment |Â
up vote
0
down vote
$<x,y>$ is positive if and only if the angle between $x$ and $y$ is less than 90 degrees. $<x,y>$ is sort of like a measure of how pointed in the same direction $x$ and $y$ are.
The geometry of least squares problems and separating hyperplanes is useful for gaining intuition about inner products.
answered Jul 24 at 1:29
ted
446312
add a comment |Â
up vote
0
down vote
The intuition is gathered from the following, given by the Law of Cosines
$$| a -b|^2 = | a|^2 + | b|^2 - 2| a|^2 | b|^2cos(theta) $$
$$ | a-b|^2 = (a-b)^2cdot(a-b)^2 $$
$$ | a-b|^2 = acdot a - acdot b - bcdot a + b cdot b $$
useful to note that $a cdot a = | a| $
$$ | a-b|^2 = |a | - acdot b - bcdot a + | b|$$
further more we have $ a cdot b = bcdot a $
$$ | a-b|^2 = |a | - 2 acdot b + | b|$$
so we have
$$| a -b|^2 = | a|^2 + | b|^2 - 2| a|^2 | b|^2cos(theta) $$
$$ | a|^2 - 2 a cdot b + | b|^2 = |a|^2 + |b|^2 - 2|a| |b| cos(theta) - 2 a cdot b = -2 |a | |b| cos(theta) $$
yielding finally
$$a cdot b = | a | | b| cos(theta) $$
Now suppose that
$$ |a| = |b| =1 $$
$$ a cdot b = 1 cdot 1 cos(theta) = cos(theta)$$
ok when is this negative. We have a unit circle here for vectors
$$ a cdot b = cos( theta) < 0 implies fracnpi2 < theta < frac3npi2 $$
that is we are in one part of the circle. However, it goes on forever.
Note from lamar
answered Jul 24 at 2:17
RHowe
1,010815
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
You are probably familiar with the following formula involving
a vector $x = (x_1, ldots, x_n) in mathbb R^n$
and $n$ constants $a_1, ldots, a_n$ that are not all zero:
$$
a_1 x_1 + a_2 x_2 + cdots + a_n x_n = 0.
$$
This formula describes an $(n-1)$-dimensional hyperplane in $mathbb R^n.$
Let's call that hyperplane $A.$ That is, any vector $x$ that satisfies the formula lies in the hyperplane $A.$
If we define a vector $a = (a_1, ldots, a_n),$
then another way to write the formula is
$$ langle x, arangle = 0. $$
Now if we choose an arbitrary vector $x in mathbb R^n,$
it may happen that $langle x, arangle neq 0.$
If we have two such vectors, let's say $x'$ and $x'',$
such that $langle x', arangle > 0$ and $langle x'', arangle > 0,$
then $x'$ and $x''$ will be on the same side of the hyperplane $A.$
But if $langle x', arangle < 0 < langle x'', arangle$ then the vectors are on opposite sides.
For a given vector $x,$ suppose you find a vector $y$ such that
$x - langle x, arangle y$ is in the hyperplane $A.$
Then $x' - langle x', arangle y$ also is in the hyperplane $A$
for any other vector $x'.$
That is, $langle x, arangle$ tells you how many times the length of $y$
you have to travel in the direction of $y$ to get from $x$ to the hyperplane,
or in other words, $langle x', arangle$ is a kind of measurement of the distance of $x$ from the hyperplane (measured in some particular units in some particular direction).
All of this works without any "angles" (unless you consider "parallel" to be a "zero angle").
You don't even need
$(x_1, ldots, x_n)$ to be coordinates over an orthonormal basis,
although if the basis is orthonormal then other nice results follow.
For example,
$$sqrtlangle x, xrangle = sqrtx_1^2 + x_2^2 + cdots + x_n^2,$$
which is the length of $x$ (according to the Pythagorean Theorem)
if the basis of $(x_1, ldots, x_n)$ is orthonormal.
Admittedly, to speak of "normals" one must have a concept of things being perpendicular, which starts to sound like we're dealing with angles again.
But they're right angles, which are especially simple to work with.
answered Jul 24 at 2:09
David K
48.2k340107
add a comment |Â
up vote
1
down vote
You are probably familiar with the following formula involving
a vector $x = (x_1, ldots, x_n) in mathbb R^n$
and $n$ constants $a_1, ldots, a_n$ that are not all zero:
$$
a_1 x_1 + a_2 x_2 + cdots + a_n x_n = 0.
$$
This formula describes an $(n-1)$-dimensional hyperplane in $mathbb R^n.$
Let's call that hyperplane $A.$ That is, any vector $x$ that satisfies the formula lies in the hyperplane $A.$
If we define a vector $a = (a_1, ldots, a_n),$
then another way to write the formula is
$$ langle x, arangle = 0. $$
Now if we choose an arbitrary vector $x in mathbb R^n,$
it may happen that $langle x, arangle neq 0.$
If we have two such vectors, let's say $x'$ and $x'',$
such that $langle x', arangle > 0$ and $langle x'', arangle > 0,$
then $x'$ and $x''$ will be on the same side of the hyperplane $A.$
But if $langle x', arangle < 0 < langle x'', arangle$ then the vectors are on opposite sides.
For a given vector $x,$ suppose you find a vector $y$ such that
$x - langle x, arangle y$ is in the hyperplane $A.$
Then $x' - langle x', arangle y$ also is in the hyperplane $A$
for any other vector $x'.$
That is, $langle x, arangle$ tells you how many times the length of $y$
you have to travel in the direction of $y$ to get from $x$ to the hyperplane,
or in other words, $langle x', arangle$ is a kind of measurement of the distance of $x$ from the hyperplane (measured in some particular units in some particular direction).
All of this works without any "angles" (unless you consider "parallel" to be a "zero angle").
You don't even need
$(x_1, ldots, x_n)$ to be coordinates over an orthonormal basis,
although if the basis is orthonormal then other nice results follow.
For example,
$$sqrtlangle x, xrangle = sqrtx_1^2 + x_2^2 + cdots + x_n^2,$$
which is the length of $x$ (according to the Pythagorean Theorem)
if the basis of $(x_1, ldots, x_n)$ is orthonormal.
Admittedly, to speak of "normals" one must have a concept of things being perpendicular, which starts to sound like we're dealing with angles again.
But they're right angles, which are especially simple to work with.
answered Jul 24 at 2:09
David K
48.2k340107
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You are probably familiar with the following formula involving
a vector $x = (x_1, ldots, x_n) in mathbb R^n$
and $n$ constants $a_1, ldots, a_n$ that are not all zero:
$$
a_1 x_1 + a_2 x_2 + cdots + a_n x_n = 0.
$$
This formula describes an $(n-1)$-dimensional hyperplane in $mathbb R^n.$
Let's call that hyperplane $A.$ That is, any vector $x$ that satisfies the formula lies in the hyperplane $A.$
If we define a vector $a = (a_1, ldots, a_n),$
then another way to write the formula is
$$ langle x, arangle = 0. $$
Now if we choose an arbitrary vector $x in mathbb R^n,$
it may happen that $langle x, arangle neq 0.$
If we have two such vectors, let's say $x'$ and $x'',$
such that $langle x', arangle > 0$ and $langle x'', arangle > 0,$
then $x'$ and $x''$ will be on the same side of the hyperplane $A.$
But if $langle x', arangle < 0 < langle x'', arangle$ then the vectors are on opposite sides.
For a given vector $x,$ suppose you find a vector $y$ such that
$x - langle x, arangle y$ is in the hyperplane $A.$
Then $x' - langle x', arangle y$ also is in the hyperplane $A$
for any other vector $x'.$
That is, $langle x, arangle$ tells you how many times the length of $y$
you have to travel in the direction of $y$ to get from $x$ to the hyperplane,
or in other words, $langle x', arangle$ is a kind of measurement of the distance of $x$ from the hyperplane (measured in some particular units in some particular direction).
All of this works without any "angles" (unless you consider "parallel" to be a "zero angle").
You don't even need
$(x_1, ldots, x_n)$ to be coordinates over an orthonormal basis,
although if the basis is orthonormal then other nice results follow.
For example,
$$sqrtlangle x, xrangle = sqrtx_1^2 + x_2^2 + cdots + x_n^2,$$
which is the length of $x$ (according to the Pythagorean Theorem)
if the basis of $(x_1, ldots, x_n)$ is orthonormal.
Admittedly, to speak of "normals" one must have a concept of things being perpendicular, which starts to sound like we're dealing with angles again.
But they're right angles, which are especially simple to work with.
answered Jul 24 at 2:09
David K
48.2k340107
You are probably familiar with the following formula involving
a vector $x = (x_1, ldots, x_n) in mathbb R^n$
and $n$ constants $a_1, ldots, a_n$ that are not all zero:
$$
a_1 x_1 + a_2 x_2 + cdots + a_n x_n = 0.
$$
This formula describes an $(n-1)$-dimensional hyperplane in $mathbb R^n.$
Let's call that hyperplane $A.$ That is, any vector $x$ that satisfies the formula lies in the hyperplane $A.$
If we define a vector $a = (a_1, ldots, a_n),$
then another way to write the formula is
$$ langle x, arangle = 0. $$
Now if we choose an arbitrary vector $x in mathbb R^n,$
it may happen that $langle x, arangle neq 0.$
If we have two such vectors, let's say $x'$ and $x'',$
such that $langle x', arangle > 0$ and $langle x'', arangle > 0,$
then $x'$ and $x''$ will be on the same side of the hyperplane $A.$
But if $langle x', arangle < 0 < langle x'', arangle$ then the vectors are on opposite sides.
For a given vector $x,$ suppose you find a vector $y$ such that
$x - langle x, arangle y$ is in the hyperplane $A.$
Then $x' - langle x', arangle y$ also is in the hyperplane $A$
for any other vector $x'.$
That is, $langle x, arangle$ tells you how many times the length of $y$
you have to travel in the direction of $y$ to get from $x$ to the hyperplane,
or in other words, $langle x', arangle$ is a kind of measurement of the distance of $x$ from the hyperplane (measured in some particular units in some particular direction).
All of this works without any "angles" (unless you consider "parallel" to be a "zero angle").
You don't even need
$(x_1, ldots, x_n)$ to be coordinates over an orthonormal basis,
although if the basis is orthonormal then other nice results follow.
For example,
$$sqrtlangle x, xrangle = sqrtx_1^2 + x_2^2 + cdots + x_n^2,$$
which is the length of $x$ (according to the Pythagorean Theorem)
if the basis of $(x_1, ldots, x_n)$ is orthonormal.
Admittedly, to speak of "normals" one must have a concept of things being perpendicular, which starts to sound like we're dealing with angles again.
But they're right angles, which are especially simple to work with.
answered Jul 24 at 2:09
David K
48.2k340107
answered Jul 24 at 2:09
David K
48.2k340107
answered Jul 24 at 2:09
David K
48.2k340107
answered Jul 24 at 2:09
David K
48.2k340107
48.2k340107
add a comment |Â
add a comment |Â
up vote
0
down vote
$<x,y>$ is positive if and only if the angle between $x$ and $y$ is less than 90 degrees. $<x,y>$ is sort of like a measure of how pointed in the same direction $x$ and $y$ are.
The geometry of least squares problems and separating hyperplanes is useful for gaining intuition about inner products.
answered Jul 24 at 1:29
ted
446312
add a comment |Â
up vote
0
down vote
$<x,y>$ is positive if and only if the angle between $x$ and $y$ is less than 90 degrees. $<x,y>$ is sort of like a measure of how pointed in the same direction $x$ and $y$ are.
The geometry of least squares problems and separating hyperplanes is useful for gaining intuition about inner products.
answered Jul 24 at 1:29
ted
446312
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$<x,y>$ is positive if and only if the angle between $x$ and $y$ is less than 90 degrees. $<x,y>$ is sort of like a measure of how pointed in the same direction $x$ and $y$ are.
The geometry of least squares problems and separating hyperplanes is useful for gaining intuition about inner products.
answered Jul 24 at 1:29
ted
446312
$<x,y>$ is positive if and only if the angle between $x$ and $y$ is less than 90 degrees. $<x,y>$ is sort of like a measure of how pointed in the same direction $x$ and $y$ are.
The geometry of least squares problems and separating hyperplanes is useful for gaining intuition about inner products.
answered Jul 24 at 1:29
ted
446312
answered Jul 24 at 1:29
ted
446312
answered Jul 24 at 1:29
ted
446312
answered Jul 24 at 1:29
ted
446312
446312
add a comment |Â
add a comment |Â
up vote
0
down vote
The intuition is gathered from the following, given by the Law of Cosines
$$| a -b|^2 = | a|^2 + | b|^2 - 2| a|^2 | b|^2cos(theta) $$
$$ | a-b|^2 = (a-b)^2cdot(a-b)^2 $$
$$ | a-b|^2 = acdot a - acdot b - bcdot a + b cdot b $$
useful to note that $a cdot a = | a| $
$$ | a-b|^2 = |a | - acdot b - bcdot a + | b|$$
further more we have $ a cdot b = bcdot a $
$$ | a-b|^2 = |a | - 2 acdot b + | b|$$
so we have
$$| a -b|^2 = | a|^2 + | b|^2 - 2| a|^2 | b|^2cos(theta) $$
$$ | a|^2 - 2 a cdot b + | b|^2 = |a|^2 + |b|^2 - 2|a| |b| cos(theta) - 2 a cdot b = -2 |a | |b| cos(theta) $$
yielding finally
$$a cdot b = | a | | b| cos(theta) $$
Now suppose that
$$ |a| = |b| =1 $$
$$ a cdot b = 1 cdot 1 cos(theta) = cos(theta)$$
ok when is this negative. We have a unit circle here for vectors
$$ a cdot b = cos( theta) < 0 implies fracnpi2 < theta < frac3npi2 $$
that is we are in one part of the circle. However, it goes on forever.
Note from lamar
answered Jul 24 at 2:17
RHowe
1,010815
add a comment |Â
up vote
0
down vote
The intuition is gathered from the following, given by the Law of Cosines
$$| a -b|^2 = | a|^2 + | b|^2 - 2| a|^2 | b|^2cos(theta) $$
$$ | a-b|^2 = (a-b)^2cdot(a-b)^2 $$
$$ | a-b|^2 = acdot a - acdot b - bcdot a + b cdot b $$
useful to note that $a cdot a = | a| $
$$ | a-b|^2 = |a | - acdot b - bcdot a + | b|$$
further more we have $ a cdot b = bcdot a $
$$ | a-b|^2 = |a | - 2 acdot b + | b|$$
so we have
$$| a -b|^2 = | a|^2 + | b|^2 - 2| a|^2 | b|^2cos(theta) $$
$$ | a|^2 - 2 a cdot b + | b|^2 = |a|^2 + |b|^2 - 2|a| |b| cos(theta) - 2 a cdot b = -2 |a | |b| cos(theta) $$
yielding finally
$$a cdot b = | a | | b| cos(theta) $$
Now suppose that
$$ |a| = |b| =1 $$
$$ a cdot b = 1 cdot 1 cos(theta) = cos(theta)$$
ok when is this negative. We have a unit circle here for vectors
$$ a cdot b = cos( theta) < 0 implies fracnpi2 < theta < frac3npi2 $$
that is we are in one part of the circle. However, it goes on forever.
Note from lamar
answered Jul 24 at 2:17
RHowe
1,010815
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The intuition is gathered from the following, given by the Law of Cosines
$$| a -b|^2 = | a|^2 + | b|^2 - 2| a|^2 | b|^2cos(theta) $$
$$ | a-b|^2 = (a-b)^2cdot(a-b)^2 $$
$$ | a-b|^2 = acdot a - acdot b - bcdot a + b cdot b $$
useful to note that $a cdot a = | a| $
$$ | a-b|^2 = |a | - acdot b - bcdot a + | b|$$
further more we have $ a cdot b = bcdot a $
$$ | a-b|^2 = |a | - 2 acdot b + | b|$$
so we have
$$| a -b|^2 = | a|^2 + | b|^2 - 2| a|^2 | b|^2cos(theta) $$
$$ | a|^2 - 2 a cdot b + | b|^2 = |a|^2 + |b|^2 - 2|a| |b| cos(theta) - 2 a cdot b = -2 |a | |b| cos(theta) $$
yielding finally
$$a cdot b = | a | | b| cos(theta) $$
Now suppose that
$$ |a| = |b| =1 $$
$$ a cdot b = 1 cdot 1 cos(theta) = cos(theta)$$
ok when is this negative. We have a unit circle here for vectors
$$ a cdot b = cos( theta) < 0 implies fracnpi2 < theta < frac3npi2 $$
that is we are in one part of the circle. However, it goes on forever.
Note from lamar
answered Jul 24 at 2:17
RHowe
1,010815
The intuition is gathered from the following, given by the Law of Cosines
$$| a -b|^2 = | a|^2 + | b|^2 - 2| a|^2 | b|^2cos(theta) $$
$$ | a-b|^2 = (a-b)^2cdot(a-b)^2 $$
$$ | a-b|^2 = acdot a - acdot b - bcdot a + b cdot b $$
useful to note that $a cdot a = | a| $
$$ | a-b|^2 = |a | - acdot b - bcdot a + | b|$$
further more we have $ a cdot b = bcdot a $
$$ | a-b|^2 = |a | - 2 acdot b + | b|$$
so we have
$$| a -b|^2 = | a|^2 + | b|^2 - 2| a|^2 | b|^2cos(theta) $$
$$ | a|^2 - 2 a cdot b + | b|^2 = |a|^2 + |b|^2 - 2|a| |b| cos(theta) - 2 a cdot b = -2 |a | |b| cos(theta) $$
yielding finally
$$a cdot b = | a | | b| cos(theta) $$
Now suppose that
$$ |a| = |b| =1 $$
$$ a cdot b = 1 cdot 1 cos(theta) = cos(theta)$$
ok when is this negative. We have a unit circle here for vectors
$$ a cdot b = cos( theta) < 0 implies fracnpi2 < theta < frac3npi2 $$
that is we are in one part of the circle. However, it goes on forever.
Note from lamar
answered Jul 24 at 2:17
RHowe
1,010815
answered Jul 24 at 2:17
RHowe
1,010815
answered Jul 24 at 2:17
RHowe
1,010815
answered Jul 24 at 2:17
RHowe
1,010815
1,010815
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2
Well, the intuition one must have of the inner product precisely is this "angular stuff", to my opinion.
– Suzet
Jul 24 at 1:24
1
Small $LaTeX$ note: "langle" and "rangle" look nicer than the greater than/less than symbol: They typeset as $langle -, - rangle$.
– JavaMan
Jul 24 at 2:07
I'm answering this here rather than calling it a duplicate because it explicitly asks for explanations that are not "angular." I think @LeeMosher's answer to the other question actually is such an answer but the others mostly are not.
– David K
Jul 24 at 2:14