Mixing
Prove the distributive property of the dot product using its geometric definition?
$begingroup$
The geometric definition of the dot product:
$$mathbf{a} cdot mathbf{b} = a cdot b cos thetaqquadforallmathbf{a} , mathbf{b} in mathbb{R} ^n.$$
The distributive property of the dot product:
$$mathbf{a} cdot ( mathbf{b} + mathbf{c} ) = mathbf{a} cdot mathbf{b} + mathbf{a} cdot mathbf{c} qquadforallmathbf{a} , mathbf{b},mathbf{c} in mathbb{R} ^n.$$
I don't understand how the concepts 'length' and 'angle' make sense in $n geq 4$. They seem to be defined using the dot product. But we're defining the dot product with them so that's a circular definition. Maybe the geometric definition isn't valid for those higher dimensions.
Anyway, I want to prove the distributive property of the dot product using its geometric definition for all $n$ or at least for $n = 3$ using basic things like geometry, trigonometric identities etc.
I've proven it for $n = 2$ by defining the vectors $hat{mathbf{x}}$ and $hat{mathbf{y}}$ as vectors that are of unit length and are perpendicular to each other. Then I assumed that all vectors in $mathbb{R} ^2$ can be written as a linear combination of these unit vectors since you can reach anywhere on a plane by going forwards/backwards and left/right. And then I rewrote the scalars of the linear combinations in polar coordinates. And finally by using trigonometric identities the distributive property can then be demonstrated to be true.
I'm not sure how to prove the case $n = 3$ since the geometry is more complicated.
And I'm not even sure if geometry works for $n geq 4$ as I've mentioned above.
This would be a lot simpler using the algebraic definition of the dot product but the proof that the two definitions are equivalent on Wikipedia requires the distributive property.
linear-algebra geometry
edited Sep 13 '18 at 6:45
joriki
171k10188349
asked Mar 31 '15 at 23:47
m009m009
212
$endgroup$
add a comment |
$begingroup$
The geometric definition of the dot product:
$$mathbf{a} cdot mathbf{b} = a cdot b cos thetaqquadforallmathbf{a} , mathbf{b} in mathbb{R} ^n.$$
The distributive property of the dot product:
$$mathbf{a} cdot ( mathbf{b} + mathbf{c} ) = mathbf{a} cdot mathbf{b} + mathbf{a} cdot mathbf{c} qquadforallmathbf{a} , mathbf{b},mathbf{c} in mathbb{R} ^n.$$
I don't understand how the concepts 'length' and 'angle' make sense in $n geq 4$. They seem to be defined using the dot product. But we're defining the dot product with them so that's a circular definition. Maybe the geometric definition isn't valid for those higher dimensions.
Anyway, I want to prove the distributive property of the dot product using its geometric definition for all $n$ or at least for $n = 3$ using basic things like geometry, trigonometric identities etc.
I've proven it for $n = 2$ by defining the vectors $hat{mathbf{x}}$ and $hat{mathbf{y}}$ as vectors that are of unit length and are perpendicular to each other. Then I assumed that all vectors in $mathbb{R} ^2$ can be written as a linear combination of these unit vectors since you can reach anywhere on a plane by going forwards/backwards and left/right. And then I rewrote the scalars of the linear combinations in polar coordinates. And finally by using trigonometric identities the distributive property can then be demonstrated to be true.
I'm not sure how to prove the case $n = 3$ since the geometry is more complicated.
And I'm not even sure if geometry works for $n geq 4$ as I've mentioned above.
This would be a lot simpler using the algebraic definition of the dot product but the proof that the two definitions are equivalent on Wikipedia requires the distributive property.
linear-algebra geometry
edited Sep 13 '18 at 6:45
joriki
171k10188349
asked Mar 31 '15 at 23:47
m009m009
212
$endgroup$
add a comment |
$begingroup$
The geometric definition of the dot product:
$$mathbf{a} cdot mathbf{b} = a cdot b cos thetaqquadforallmathbf{a} , mathbf{b} in mathbb{R} ^n.$$
The distributive property of the dot product:
$$mathbf{a} cdot ( mathbf{b} + mathbf{c} ) = mathbf{a} cdot mathbf{b} + mathbf{a} cdot mathbf{c} qquadforallmathbf{a} , mathbf{b},mathbf{c} in mathbb{R} ^n.$$
I don't understand how the concepts 'length' and 'angle' make sense in $n geq 4$. They seem to be defined using the dot product. But we're defining the dot product with them so that's a circular definition. Maybe the geometric definition isn't valid for those higher dimensions.
Anyway, I want to prove the distributive property of the dot product using its geometric definition for all $n$ or at least for $n = 3$ using basic things like geometry, trigonometric identities etc.
I've proven it for $n = 2$ by defining the vectors $hat{mathbf{x}}$ and $hat{mathbf{y}}$ as vectors that are of unit length and are perpendicular to each other. Then I assumed that all vectors in $mathbb{R} ^2$ can be written as a linear combination of these unit vectors since you can reach anywhere on a plane by going forwards/backwards and left/right. And then I rewrote the scalars of the linear combinations in polar coordinates. And finally by using trigonometric identities the distributive property can then be demonstrated to be true.
I'm not sure how to prove the case $n = 3$ since the geometry is more complicated.
And I'm not even sure if geometry works for $n geq 4$ as I've mentioned above.
This would be a lot simpler using the algebraic definition of the dot product but the proof that the two definitions are equivalent on Wikipedia requires the distributive property.
linear-algebra geometry
edited Sep 13 '18 at 6:45
joriki
171k10188349
asked Mar 31 '15 at 23:47
m009m009
212
$endgroup$
The geometric definition of the dot product:
$$mathbf{a} cdot mathbf{b} = a cdot b cos thetaqquadforallmathbf{a} , mathbf{b} in mathbb{R} ^n.$$
The distributive property of the dot product:
$$mathbf{a} cdot ( mathbf{b} + mathbf{c} ) = mathbf{a} cdot mathbf{b} + mathbf{a} cdot mathbf{c} qquadforallmathbf{a} , mathbf{b},mathbf{c} in mathbb{R} ^n.$$
I don't understand how the concepts 'length' and 'angle' make sense in $n geq 4$. They seem to be defined using the dot product. But we're defining the dot product with them so that's a circular definition. Maybe the geometric definition isn't valid for those higher dimensions.
Anyway, I want to prove the distributive property of the dot product using its geometric definition for all $n$ or at least for $n = 3$ using basic things like geometry, trigonometric identities etc.
I've proven it for $n = 2$ by defining the vectors $hat{mathbf{x}}$ and $hat{mathbf{y}}$ as vectors that are of unit length and are perpendicular to each other. Then I assumed that all vectors in $mathbb{R} ^2$ can be written as a linear combination of these unit vectors since you can reach anywhere on a plane by going forwards/backwards and left/right. And then I rewrote the scalars of the linear combinations in polar coordinates. And finally by using trigonometric identities the distributive property can then be demonstrated to be true.
I'm not sure how to prove the case $n = 3$ since the geometry is more complicated.
And I'm not even sure if geometry works for $n geq 4$ as I've mentioned above.
This would be a lot simpler using the algebraic definition of the dot product but the proof that the two definitions are equivalent on Wikipedia requires the distributive property.
linear-algebra geometry
linear-algebra geometry
edited Sep 13 '18 at 6:45
joriki
171k10188349
asked Mar 31 '15 at 23:47
m009m009
212
edited Sep 13 '18 at 6:45
joriki
171k10188349
asked Mar 31 '15 at 23:47
m009m009
212
edited Sep 13 '18 at 6:45
joriki
171k10188349
edited Sep 13 '18 at 6:45
joriki
171k10188349
edited Sep 13 '18 at 6:45
joriki
171k10188349
171k10188349
asked Mar 31 '15 at 23:47
m009m009
212
asked Mar 31 '15 at 23:47
m009m009
212
asked Mar 31 '15 at 23:47
m009m009
212
212
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
The geometric definition gives us the dot product as the magnitude of a multiplied by the scalar projection of b onto a. This is given for any a,b in n-space.
$textbf{a}⋅textbf{b}=a⋅b cosθ_{a} = a⋅b_{a}$
The dot product of a with b+c is just the magnitude of a times the scalar projection of b+c onto a. But that can be broken up into components, after which normal distribution takes over.
$textbf{a}⋅(textbf{b}+textbf{c})=a⋅(b+c)_{a}=a⋅(b_{a}+c_{a})=a⋅b_{a}+a⋅c_{a}=textbf{a}⋅textbf{b}+textbf{a}⋅textbf{c}$
answered Apr 1 '15 at 0:37
zahbazzahbaz
8,43921938
$endgroup$
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
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oldest
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active
oldest
votes
$begingroup$
The geometric definition gives us the dot product as the magnitude of a multiplied by the scalar projection of b onto a. This is given for any a,b in n-space.
$textbf{a}⋅textbf{b}=a⋅b cosθ_{a} = a⋅b_{a}$
The dot product of a with b+c is just the magnitude of a times the scalar projection of b+c onto a. But that can be broken up into components, after which normal distribution takes over.
$textbf{a}⋅(textbf{b}+textbf{c})=a⋅(b+c)_{a}=a⋅(b_{a}+c_{a})=a⋅b_{a}+a⋅c_{a}=textbf{a}⋅textbf{b}+textbf{a}⋅textbf{c}$
answered Apr 1 '15 at 0:37
zahbazzahbaz
8,43921938
$endgroup$
add a comment |
$begingroup$
The geometric definition gives us the dot product as the magnitude of a multiplied by the scalar projection of b onto a. This is given for any a,b in n-space.
$textbf{a}⋅textbf{b}=a⋅b cosθ_{a} = a⋅b_{a}$
The dot product of a with b+c is just the magnitude of a times the scalar projection of b+c onto a. But that can be broken up into components, after which normal distribution takes over.
$textbf{a}⋅(textbf{b}+textbf{c})=a⋅(b+c)_{a}=a⋅(b_{a}+c_{a})=a⋅b_{a}+a⋅c_{a}=textbf{a}⋅textbf{b}+textbf{a}⋅textbf{c}$
answered Apr 1 '15 at 0:37
zahbazzahbaz
8,43921938
$endgroup$
add a comment |
$begingroup$
The geometric definition gives us the dot product as the magnitude of a multiplied by the scalar projection of b onto a. This is given for any a,b in n-space.
$textbf{a}⋅textbf{b}=a⋅b cosθ_{a} = a⋅b_{a}$
The dot product of a with b+c is just the magnitude of a times the scalar projection of b+c onto a. But that can be broken up into components, after which normal distribution takes over.
$textbf{a}⋅(textbf{b}+textbf{c})=a⋅(b+c)_{a}=a⋅(b_{a}+c_{a})=a⋅b_{a}+a⋅c_{a}=textbf{a}⋅textbf{b}+textbf{a}⋅textbf{c}$
answered Apr 1 '15 at 0:37
zahbazzahbaz
8,43921938
$endgroup$
The geometric definition gives us the dot product as the magnitude of a multiplied by the scalar projection of b onto a. This is given for any a,b in n-space.
$textbf{a}⋅textbf{b}=a⋅b cosθ_{a} = a⋅b_{a}$
The dot product of a with b+c is just the magnitude of a times the scalar projection of b+c onto a. But that can be broken up into components, after which normal distribution takes over.
$textbf{a}⋅(textbf{b}+textbf{c})=a⋅(b+c)_{a}=a⋅(b_{a}+c_{a})=a⋅b_{a}+a⋅c_{a}=textbf{a}⋅textbf{b}+textbf{a}⋅textbf{c}$
answered Apr 1 '15 at 0:37
zahbazzahbaz
8,43921938
answered Apr 1 '15 at 0:37
zahbazzahbaz
8,43921938
answered Apr 1 '15 at 0:37
zahbazzahbaz
8,43921938
answered Apr 1 '15 at 0:37
zahbazzahbaz
8,43921938
8,43921938
add a comment |
add a comment |
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