Find the distance between $u$ and $v$
$begingroup$
Find the distance between $u$ and $v$ Where $u = [1,2,3]$ and $v = [-1,0,1]$
Am I calculating the norm of both vectors and substracting them together?
Thank you very much
linear-algebra eigenvalues-eigenvectors
edited Dec 2 '18 at 13:25
Key Flex
7,83961232
asked Dec 1 '18 at 23:41
AlexAlex
173
$endgroup$
add a comment |
$begingroup$
Find the distance between $u$ and $v$ Where $u = [1,2,3]$ and $v = [-1,0,1]$
Am I calculating the norm of both vectors and substracting them together?
Thank you very much
linear-algebra eigenvalues-eigenvectors
edited Dec 2 '18 at 13:25
Key Flex
7,83961232
asked Dec 1 '18 at 23:41
AlexAlex
173
$endgroup$
add a comment |
$begingroup$
Find the distance between $u$ and $v$ Where $u = [1,2,3]$ and $v = [-1,0,1]$
Am I calculating the norm of both vectors and substracting them together?
Thank you very much
linear-algebra eigenvalues-eigenvectors
edited Dec 2 '18 at 13:25
Key Flex
7,83961232
asked Dec 1 '18 at 23:41
AlexAlex
173
$endgroup$
Find the distance between $u$ and $v$ Where $u = [1,2,3]$ and $v = [-1,0,1]$
Am I calculating the norm of both vectors and substracting them together?
Thank you very much
linear-algebra eigenvalues-eigenvectors
linear-algebra eigenvalues-eigenvectors
edited Dec 2 '18 at 13:25
Key Flex
7,83961232
asked Dec 1 '18 at 23:41
AlexAlex
173
edited Dec 2 '18 at 13:25
Key Flex
7,83961232
asked Dec 1 '18 at 23:41
AlexAlex
173
edited Dec 2 '18 at 13:25
Key Flex
7,83961232
edited Dec 2 '18 at 13:25
Key Flex
7,83961232
edited Dec 2 '18 at 13:25
Key Flex
7,83961232
7,83961232
asked Dec 1 '18 at 23:41
AlexAlex
173
asked Dec 1 '18 at 23:41
AlexAlex
173
asked Dec 1 '18 at 23:41
AlexAlex
173
173
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Try it with the example $[0,0,0]$ and $[1,0,0]$.
The norms are $0$ and $1$ and the distance now can either be $-1$ or $1$ (problematic).
Reversing the order of your operations leads to calculating the norm of either $[1,0,0]$ or $[-1,0,0]$ which both is $1$.
edited Dec 2 '18 at 7:23
Tianlalu
3,08621038
answered Dec 1 '18 at 23:46
DennisDennis
412
$endgroup$
add a comment |
$begingroup$
Given $u=[1,2,3]$ and $v=[-1,0,1]$
The distance between $u$ and $v$ is
$$d(u,v)=sqrt{(1-(-1))^2+(2-0)^2+(3-1)^2}$$
$$d(u,v)=sqrt{(2)^2+(2)^2+(2)^2}=sqrt{12}=2sqrt{3}$$
answered Dec 1 '18 at 23:49
Key FlexKey Flex
7,83961232
$endgroup$
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
$begingroup$
Try it with the example $[0,0,0]$ and $[1,0,0]$.
The norms are $0$ and $1$ and the distance now can either be $-1$ or $1$ (problematic).
Reversing the order of your operations leads to calculating the norm of either $[1,0,0]$ or $[-1,0,0]$ which both is $1$.
edited Dec 2 '18 at 7:23
Tianlalu
3,08621038
answered Dec 1 '18 at 23:46
DennisDennis
412
$endgroup$
add a comment |
$begingroup$
Try it with the example $[0,0,0]$ and $[1,0,0]$.
The norms are $0$ and $1$ and the distance now can either be $-1$ or $1$ (problematic).
Reversing the order of your operations leads to calculating the norm of either $[1,0,0]$ or $[-1,0,0]$ which both is $1$.
edited Dec 2 '18 at 7:23
Tianlalu
3,08621038
answered Dec 1 '18 at 23:46
DennisDennis
412
$endgroup$
add a comment |
$begingroup$
Try it with the example $[0,0,0]$ and $[1,0,0]$.
The norms are $0$ and $1$ and the distance now can either be $-1$ or $1$ (problematic).
Reversing the order of your operations leads to calculating the norm of either $[1,0,0]$ or $[-1,0,0]$ which both is $1$.
edited Dec 2 '18 at 7:23
Tianlalu
3,08621038
answered Dec 1 '18 at 23:46
DennisDennis
412
$endgroup$
Try it with the example $[0,0,0]$ and $[1,0,0]$.
The norms are $0$ and $1$ and the distance now can either be $-1$ or $1$ (problematic).
Reversing the order of your operations leads to calculating the norm of either $[1,0,0]$ or $[-1,0,0]$ which both is $1$.
edited Dec 2 '18 at 7:23
Tianlalu
3,08621038
answered Dec 1 '18 at 23:46
DennisDennis
412
edited Dec 2 '18 at 7:23
Tianlalu
3,08621038
edited Dec 2 '18 at 7:23
Tianlalu
3,08621038
edited Dec 2 '18 at 7:23
Tianlalu
3,08621038
3,08621038
answered Dec 1 '18 at 23:46
DennisDennis
412
answered Dec 1 '18 at 23:46
DennisDennis
412
answered Dec 1 '18 at 23:46
DennisDennis
412
412
add a comment |
add a comment |
$begingroup$
Given $u=[1,2,3]$ and $v=[-1,0,1]$
The distance between $u$ and $v$ is
$$d(u,v)=sqrt{(1-(-1))^2+(2-0)^2+(3-1)^2}$$
$$d(u,v)=sqrt{(2)^2+(2)^2+(2)^2}=sqrt{12}=2sqrt{3}$$
answered Dec 1 '18 at 23:49
Key FlexKey Flex
7,83961232
$endgroup$
add a comment |
$begingroup$
Given $u=[1,2,3]$ and $v=[-1,0,1]$
The distance between $u$ and $v$ is
$$d(u,v)=sqrt{(1-(-1))^2+(2-0)^2+(3-1)^2}$$
$$d(u,v)=sqrt{(2)^2+(2)^2+(2)^2}=sqrt{12}=2sqrt{3}$$
answered Dec 1 '18 at 23:49
Key FlexKey Flex
7,83961232
$endgroup$
add a comment |
$begingroup$
Given $u=[1,2,3]$ and $v=[-1,0,1]$
The distance between $u$ and $v$ is
$$d(u,v)=sqrt{(1-(-1))^2+(2-0)^2+(3-1)^2}$$
$$d(u,v)=sqrt{(2)^2+(2)^2+(2)^2}=sqrt{12}=2sqrt{3}$$
answered Dec 1 '18 at 23:49
Key FlexKey Flex
7,83961232
$endgroup$
Given $u=[1,2,3]$ and $v=[-1,0,1]$
The distance between $u$ and $v$ is
$$d(u,v)=sqrt{(1-(-1))^2+(2-0)^2+(3-1)^2}$$
$$d(u,v)=sqrt{(2)^2+(2)^2+(2)^2}=sqrt{12}=2sqrt{3}$$
answered Dec 1 '18 at 23:49
Key FlexKey Flex
7,83961232
answered Dec 1 '18 at 23:49
Key FlexKey Flex
7,83961232
answered Dec 1 '18 at 23:49
Key FlexKey Flex
7,83961232
answered Dec 1 '18 at 23:49
Key FlexKey Flex
7,83961232
7,83961232
add a comment |
add a comment |
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