Scalar product of two points coordinates
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In one of the algorithms I encountered an formula to calculate a scalar number from 2 points. Given Point 1 has coordinates $(x1,y1)$ and Point 2 has coordinates $(x2, y2)$ the formula was following
$sqrt{(y1-x1)^2 + (y2-x2)^2}$
If the $x1$ is equal to $y1$ and $x2$ is equal to $y2$ then the result is $0$. On the other hand the result grows when coordinates are far from the diagonal.
Does this formula has any name or does it serve any well known purpose e.g. in gaming?
linear-algebra euclidean-geometry
asked Jan 10 at 21:21
nosalannosalan
1061
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add a comment |
$begingroup$
In one of the algorithms I encountered an formula to calculate a scalar number from 2 points. Given Point 1 has coordinates $(x1,y1)$ and Point 2 has coordinates $(x2, y2)$ the formula was following
$sqrt{(y1-x1)^2 + (y2-x2)^2}$
If the $x1$ is equal to $y1$ and $x2$ is equal to $y2$ then the result is $0$. On the other hand the result grows when coordinates are far from the diagonal.
Does this formula has any name or does it serve any well known purpose e.g. in gaming?
linear-algebra euclidean-geometry
asked Jan 10 at 21:21
nosalannosalan
1061
$endgroup$
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Pythagorean theorem
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– caverac
Jan 10 at 21:25
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Nope. Check the coordinates.... Oh, and this has nothing to do with "scalar product."
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– David G. Stork
Jan 10 at 21:26
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@caverac But Pytagorean hypothenuse whould be $sqrt{(x1-x2)^2+(y1-y2)^2}$
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– nosalan
Jan 10 at 21:36
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You're right, I misread the variables. Are you sure this is the right expression?
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– caverac
Jan 10 at 23:51
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Some context for this expression would be helpful.
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– amd
Jan 11 at 0:50
add a comment |
$begingroup$
In one of the algorithms I encountered an formula to calculate a scalar number from 2 points. Given Point 1 has coordinates $(x1,y1)$ and Point 2 has coordinates $(x2, y2)$ the formula was following
$sqrt{(y1-x1)^2 + (y2-x2)^2}$
If the $x1$ is equal to $y1$ and $x2$ is equal to $y2$ then the result is $0$. On the other hand the result grows when coordinates are far from the diagonal.
Does this formula has any name or does it serve any well known purpose e.g. in gaming?
linear-algebra euclidean-geometry
asked Jan 10 at 21:21
nosalannosalan
1061
$endgroup$
In one of the algorithms I encountered an formula to calculate a scalar number from 2 points. Given Point 1 has coordinates $(x1,y1)$ and Point 2 has coordinates $(x2, y2)$ the formula was following
$sqrt{(y1-x1)^2 + (y2-x2)^2}$
If the $x1$ is equal to $y1$ and $x2$ is equal to $y2$ then the result is $0$. On the other hand the result grows when coordinates are far from the diagonal.
Does this formula has any name or does it serve any well known purpose e.g. in gaming?
linear-algebra euclidean-geometry
linear-algebra euclidean-geometry
asked Jan 10 at 21:21
nosalannosalan
1061
asked Jan 10 at 21:21
nosalannosalan
1061
asked Jan 10 at 21:21
nosalannosalan
1061
asked Jan 10 at 21:21
nosalannosalan
1061
asked Jan 10 at 21:21
nosalannosalan
1061
1061
$begingroup$
Pythagorean theorem
$endgroup$
– caverac
Jan 10 at 21:25
$begingroup$
Nope. Check the coordinates.... Oh, and this has nothing to do with "scalar product."
$endgroup$
– David G. Stork
Jan 10 at 21:26
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@caverac But Pytagorean hypothenuse whould be $sqrt{(x1-x2)^2+(y1-y2)^2}$
$endgroup$
– nosalan
Jan 10 at 21:36
$begingroup$
You're right, I misread the variables. Are you sure this is the right expression?
$endgroup$
– caverac
Jan 10 at 23:51
$begingroup$
Some context for this expression would be helpful.
$endgroup$
– amd
Jan 11 at 0:50
add a comment |
$begingroup$
Pythagorean theorem
$endgroup$
– caverac
Jan 10 at 21:25
$begingroup$
Nope. Check the coordinates.... Oh, and this has nothing to do with "scalar product."
$endgroup$
– David G. Stork
Jan 10 at 21:26
$begingroup$
@caverac But Pytagorean hypothenuse whould be $sqrt{(x1-x2)^2+(y1-y2)^2}$
$endgroup$
– nosalan
Jan 10 at 21:36
$begingroup$
You're right, I misread the variables. Are you sure this is the right expression?
$endgroup$
– caverac
Jan 10 at 23:51
$begingroup$
Some context for this expression would be helpful.
$endgroup$
– amd
Jan 11 at 0:50
$begingroup$
Pythagorean theorem
$endgroup$
– caverac
Jan 10 at 21:25
$begingroup$
Pythagorean theorem
$endgroup$
– caverac
Jan 10 at 21:25
$begingroup$
Nope. Check the coordinates.... Oh, and this has nothing to do with "scalar product."
$endgroup$
– David G. Stork
Jan 10 at 21:26
$begingroup$
Nope. Check the coordinates.... Oh, and this has nothing to do with "scalar product."
$endgroup$
– David G. Stork
Jan 10 at 21:26
$begingroup$
@caverac But Pytagorean hypothenuse whould be $sqrt{(x1-x2)^2+(y1-y2)^2}$
$endgroup$
– nosalan
Jan 10 at 21:36
$begingroup$
@caverac But Pytagorean hypothenuse whould be $sqrt{(x1-x2)^2+(y1-y2)^2}$
$endgroup$
– nosalan
Jan 10 at 21:36
$begingroup$
You're right, I misread the variables. Are you sure this is the right expression?
$endgroup$
– caverac
Jan 10 at 23:51
$begingroup$
You're right, I misread the variables. Are you sure this is the right expression?
$endgroup$
– caverac
Jan 10 at 23:51
$begingroup$
Some context for this expression would be helpful.
$endgroup$
– amd
Jan 11 at 0:50
$begingroup$
Some context for this expression would be helpful.
$endgroup$
– amd
Jan 11 at 0:50
add a comment |
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$begingroup$
Pythagorean theorem
$endgroup$
– caverac
Jan 10 at 21:25
$begingroup$
Nope. Check the coordinates.... Oh, and this has nothing to do with "scalar product."
$endgroup$
– David G. Stork
Jan 10 at 21:26
$begingroup$
@caverac But Pytagorean hypothenuse whould be $sqrt{(x1-x2)^2+(y1-y2)^2}$
$endgroup$
– nosalan
Jan 10 at 21:36
$begingroup$
You're right, I misread the variables. Are you sure this is the right expression?
$endgroup$
– caverac
Jan 10 at 23:51
$begingroup$
Some context for this expression would be helpful.
$endgroup$
– amd
Jan 11 at 0:50