What's the geometric interpretation of this “vector cross product”?Geometric Interpretation of Jacobi identity for cross producttesting parallelity/perpendicularity of two 3D vectors with lengths close to zero using dot productFind the geometric interpretation of general solutionGeometric interpretation of the Dot ProductHow to define the inverse of a vector?Geometric interpretation of the quadruple vector product.Geometric interpretation of cross & dot product relationCovariant Contravariant Dot product and LengthMagnitude of vector cross productComputing an area of a parallelogram given the area of its planar projections
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What's the geometric interpretation of this “vector cross product”?
Geometric Interpretation of Jacobi identity for cross producttesting parallelity/perpendicularity of two 3D vectors with lengths close to zero using dot productFind the geometric interpretation of general solutionGeometric interpretation of the Dot ProductHow to define the inverse of a vector?Geometric interpretation of the quadruple vector product.Geometric interpretation of cross & dot product relationCovariant Contravariant Dot product and LengthMagnitude of vector cross productComputing an area of a parallelogram given the area of its planar projections
$begingroup$
This answer on StackOverflow answers a question about intersection of two segments. Right at the beginning, it introduces a “vector cross product”.
Define the 2-dimensional vector cross product $vec v times vec w$ to be $v_x w_y − v_y w_x$.
However, this doesn't seem to be a regular cross product (nor does it even produce a vector). I realize that the formula is a simple determinant of the two vectors, but I cannot understand its meaning or relation to the rest of the post.
Does it have a meaning (motivation) or is it just a “lucky-guess” operation in order to transform the equation $vec p + tvec r = vec q + u vec s$ into a solvable state? In other words, how does this operation (intuitively) relate to the described algorithm?
vectors analytic-geometry motivation geometric-interpretation
asked Mar 29 at 11:44
Lazar LjubenovićLazar Ljubenović
1,3471029
$endgroup$
add a comment |
$begingroup$
This answer on StackOverflow answers a question about intersection of two segments. Right at the beginning, it introduces a “vector cross product”.
Define the 2-dimensional vector cross product $vec v times vec w$ to be $v_x w_y − v_y w_x$.
However, this doesn't seem to be a regular cross product (nor does it even produce a vector). I realize that the formula is a simple determinant of the two vectors, but I cannot understand its meaning or relation to the rest of the post.
Does it have a meaning (motivation) or is it just a “lucky-guess” operation in order to transform the equation $vec p + tvec r = vec q + u vec s$ into a solvable state? In other words, how does this operation (intuitively) relate to the described algorithm?
vectors analytic-geometry motivation geometric-interpretation
asked Mar 29 at 11:44
Lazar LjubenovićLazar Ljubenović
1,3471029
$endgroup$
add a comment |
$begingroup$
This answer on StackOverflow answers a question about intersection of two segments. Right at the beginning, it introduces a “vector cross product”.
Define the 2-dimensional vector cross product $vec v times vec w$ to be $v_x w_y − v_y w_x$.
However, this doesn't seem to be a regular cross product (nor does it even produce a vector). I realize that the formula is a simple determinant of the two vectors, but I cannot understand its meaning or relation to the rest of the post.
Does it have a meaning (motivation) or is it just a “lucky-guess” operation in order to transform the equation $vec p + tvec r = vec q + u vec s$ into a solvable state? In other words, how does this operation (intuitively) relate to the described algorithm?
vectors analytic-geometry motivation geometric-interpretation
asked Mar 29 at 11:44
Lazar LjubenovićLazar Ljubenović
1,3471029
$endgroup$
This answer on StackOverflow answers a question about intersection of two segments. Right at the beginning, it introduces a “vector cross product”.
Define the 2-dimensional vector cross product $vec v times vec w$ to be $v_x w_y − v_y w_x$.
However, this doesn't seem to be a regular cross product (nor does it even produce a vector). I realize that the formula is a simple determinant of the two vectors, but I cannot understand its meaning or relation to the rest of the post.
Does it have a meaning (motivation) or is it just a “lucky-guess” operation in order to transform the equation $vec p + tvec r = vec q + u vec s$ into a solvable state? In other words, how does this operation (intuitively) relate to the described algorithm?
vectors analytic-geometry motivation geometric-interpretation
vectors analytic-geometry motivation geometric-interpretation
asked Mar 29 at 11:44
Lazar LjubenovićLazar Ljubenović
1,3471029
asked Mar 29 at 11:44
Lazar LjubenovićLazar Ljubenović
1,3471029
asked Mar 29 at 11:44
Lazar LjubenovićLazar Ljubenović
1,3471029
asked Mar 29 at 11:44
Lazar LjubenovićLazar Ljubenović
1,3471029
asked Mar 29 at 11:44
Lazar LjubenovićLazar Ljubenović
1,3471029
1,3471029
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Geometrically it gives the (signed) area of the parallelogram defined by the two vectors.
If you multiply by the appropriate unit normal to the plane you get the normal three dimensional cross product. You don't get a vector in the plane though.
If you try to define a "cross product" in four dimensions, you might appreciate that the familiar situation in three dimensions is a happy coincidence which trips up people who try to generalise in the wrong way.
answered Mar 29 at 12:02
Mark BennetMark Bennet
81.9k984183
$endgroup$
$begingroup$
Isn't that true for the regular cross product definition? This one is different.
$endgroup$
– Lazar Ljubenović
Mar 30 at 9:19
$begingroup$
Ah, apparently it holds for both. Sorry. +1
$endgroup$
– Lazar Ljubenović
Mar 30 at 9:32
add a comment |
$begingroup$
Apparently, this is known as perp product; however, the only online reference I could find after a (rather quick) search is this geomalgorithms page. It's also on Wolfram, where it mentions that Hill introduced it in 1994, in a chapter in “Graphic Gems IV”.
He firstly defines perp operator (perpendicular operator) which gives a rotation of a vector by 90 degeres:
$$ vec v^perp = (v_x, v_y)^perp = (-v_y, v_x). $$
Then a perp product (perpendicular product) is defined, denoted as an infix operator $perp$:
$$ vec v perp vec w := vec v ^ perp cdot vec w = v_x w_y - v_y w_x. $$
The idea of using the perp product in the algorithm seems to come from the following property: $$vec v perp vec w = 0 Leftrightarrow text$vec v$ and $vec w$ are collinear. $$
At the end of the algorithm, the denominators are $vec r perp vec s$, which is then discussed for being zero, implying collinearity of the vectors and parallelity of the segments.
Either way, a similar but much more intuitive discussion of the problem with the same idea of using parametric equations and perp product is given at this page called “Intersections of Lines and Planes” on geomalgorithms. It also gives more details in the algorithm at the bottom, considering cases where one or both segments are degenerated into a single point.
answered Mar 30 at 9:37
Lazar LjubenovićLazar Ljubenović
1,3471029
$endgroup$
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
$begingroup$
Geometrically it gives the (signed) area of the parallelogram defined by the two vectors.
If you multiply by the appropriate unit normal to the plane you get the normal three dimensional cross product. You don't get a vector in the plane though.
If you try to define a "cross product" in four dimensions, you might appreciate that the familiar situation in three dimensions is a happy coincidence which trips up people who try to generalise in the wrong way.
answered Mar 29 at 12:02
Mark BennetMark Bennet
81.9k984183
$endgroup$
$begingroup$
Isn't that true for the regular cross product definition? This one is different.
$endgroup$
– Lazar Ljubenović
Mar 30 at 9:19
$begingroup$
Ah, apparently it holds for both. Sorry. +1
$endgroup$
– Lazar Ljubenović
Mar 30 at 9:32
add a comment |
$begingroup$
Geometrically it gives the (signed) area of the parallelogram defined by the two vectors.
If you multiply by the appropriate unit normal to the plane you get the normal three dimensional cross product. You don't get a vector in the plane though.
If you try to define a "cross product" in four dimensions, you might appreciate that the familiar situation in three dimensions is a happy coincidence which trips up people who try to generalise in the wrong way.
answered Mar 29 at 12:02
Mark BennetMark Bennet
81.9k984183
$endgroup$
$begingroup$
Isn't that true for the regular cross product definition? This one is different.
$endgroup$
– Lazar Ljubenović
Mar 30 at 9:19
$begingroup$
Ah, apparently it holds for both. Sorry. +1
$endgroup$
– Lazar Ljubenović
Mar 30 at 9:32
add a comment |
$begingroup$
Geometrically it gives the (signed) area of the parallelogram defined by the two vectors.
If you multiply by the appropriate unit normal to the plane you get the normal three dimensional cross product. You don't get a vector in the plane though.
If you try to define a "cross product" in four dimensions, you might appreciate that the familiar situation in three dimensions is a happy coincidence which trips up people who try to generalise in the wrong way.
answered Mar 29 at 12:02
Mark BennetMark Bennet
81.9k984183
$endgroup$
Geometrically it gives the (signed) area of the parallelogram defined by the two vectors.
If you multiply by the appropriate unit normal to the plane you get the normal three dimensional cross product. You don't get a vector in the plane though.
If you try to define a "cross product" in four dimensions, you might appreciate that the familiar situation in three dimensions is a happy coincidence which trips up people who try to generalise in the wrong way.
answered Mar 29 at 12:02
Mark BennetMark Bennet
81.9k984183
answered Mar 29 at 12:02
Mark BennetMark Bennet
81.9k984183
answered Mar 29 at 12:02
Mark BennetMark Bennet
81.9k984183
answered Mar 29 at 12:02
Mark BennetMark Bennet
81.9k984183
81.9k984183
$begingroup$
Isn't that true for the regular cross product definition? This one is different.
$endgroup$
– Lazar Ljubenović
Mar 30 at 9:19
$begingroup$
Ah, apparently it holds for both. Sorry. +1
$endgroup$
– Lazar Ljubenović
Mar 30 at 9:32
add a comment |
$begingroup$
Isn't that true for the regular cross product definition? This one is different.
$endgroup$
– Lazar Ljubenović
Mar 30 at 9:19
$begingroup$
Ah, apparently it holds for both. Sorry. +1
$endgroup$
– Lazar Ljubenović
Mar 30 at 9:32
$begingroup$
Isn't that true for the regular cross product definition? This one is different.
$endgroup$
– Lazar Ljubenović
Mar 30 at 9:19
$begingroup$
Isn't that true for the regular cross product definition? This one is different.
$endgroup$
– Lazar Ljubenović
Mar 30 at 9:19
$begingroup$
Ah, apparently it holds for both. Sorry. +1
$endgroup$
– Lazar Ljubenović
Mar 30 at 9:32
$begingroup$
Ah, apparently it holds for both. Sorry. +1
$endgroup$
– Lazar Ljubenović
Mar 30 at 9:32
add a comment |
$begingroup$
Apparently, this is known as perp product; however, the only online reference I could find after a (rather quick) search is this geomalgorithms page. It's also on Wolfram, where it mentions that Hill introduced it in 1994, in a chapter in “Graphic Gems IV”.
He firstly defines perp operator (perpendicular operator) which gives a rotation of a vector by 90 degeres:
$$ vec v^perp = (v_x, v_y)^perp = (-v_y, v_x). $$
Then a perp product (perpendicular product) is defined, denoted as an infix operator $perp$:
$$ vec v perp vec w := vec v ^ perp cdot vec w = v_x w_y - v_y w_x. $$
The idea of using the perp product in the algorithm seems to come from the following property: $$vec v perp vec w = 0 Leftrightarrow text$vec v$ and $vec w$ are collinear. $$
At the end of the algorithm, the denominators are $vec r perp vec s$, which is then discussed for being zero, implying collinearity of the vectors and parallelity of the segments.
Either way, a similar but much more intuitive discussion of the problem with the same idea of using parametric equations and perp product is given at this page called “Intersections of Lines and Planes” on geomalgorithms. It also gives more details in the algorithm at the bottom, considering cases where one or both segments are degenerated into a single point.
answered Mar 30 at 9:37
Lazar LjubenovićLazar Ljubenović
1,3471029
$endgroup$
add a comment |
$begingroup$
Apparently, this is known as perp product; however, the only online reference I could find after a (rather quick) search is this geomalgorithms page. It's also on Wolfram, where it mentions that Hill introduced it in 1994, in a chapter in “Graphic Gems IV”.
He firstly defines perp operator (perpendicular operator) which gives a rotation of a vector by 90 degeres:
$$ vec v^perp = (v_x, v_y)^perp = (-v_y, v_x). $$
Then a perp product (perpendicular product) is defined, denoted as an infix operator $perp$:
$$ vec v perp vec w := vec v ^ perp cdot vec w = v_x w_y - v_y w_x. $$
The idea of using the perp product in the algorithm seems to come from the following property: $$vec v perp vec w = 0 Leftrightarrow text$vec v$ and $vec w$ are collinear. $$
At the end of the algorithm, the denominators are $vec r perp vec s$, which is then discussed for being zero, implying collinearity of the vectors and parallelity of the segments.
Either way, a similar but much more intuitive discussion of the problem with the same idea of using parametric equations and perp product is given at this page called “Intersections of Lines and Planes” on geomalgorithms. It also gives more details in the algorithm at the bottom, considering cases where one or both segments are degenerated into a single point.
answered Mar 30 at 9:37
Lazar LjubenovićLazar Ljubenović
1,3471029
$endgroup$
add a comment |
$begingroup$
Apparently, this is known as perp product; however, the only online reference I could find after a (rather quick) search is this geomalgorithms page. It's also on Wolfram, where it mentions that Hill introduced it in 1994, in a chapter in “Graphic Gems IV”.
He firstly defines perp operator (perpendicular operator) which gives a rotation of a vector by 90 degeres:
$$ vec v^perp = (v_x, v_y)^perp = (-v_y, v_x). $$
Then a perp product (perpendicular product) is defined, denoted as an infix operator $perp$:
$$ vec v perp vec w := vec v ^ perp cdot vec w = v_x w_y - v_y w_x. $$
The idea of using the perp product in the algorithm seems to come from the following property: $$vec v perp vec w = 0 Leftrightarrow text$vec v$ and $vec w$ are collinear. $$
At the end of the algorithm, the denominators are $vec r perp vec s$, which is then discussed for being zero, implying collinearity of the vectors and parallelity of the segments.
Either way, a similar but much more intuitive discussion of the problem with the same idea of using parametric equations and perp product is given at this page called “Intersections of Lines and Planes” on geomalgorithms. It also gives more details in the algorithm at the bottom, considering cases where one or both segments are degenerated into a single point.
answered Mar 30 at 9:37
Lazar LjubenovićLazar Ljubenović
1,3471029
$endgroup$
Apparently, this is known as perp product; however, the only online reference I could find after a (rather quick) search is this geomalgorithms page. It's also on Wolfram, where it mentions that Hill introduced it in 1994, in a chapter in “Graphic Gems IV”.
He firstly defines perp operator (perpendicular operator) which gives a rotation of a vector by 90 degeres:
$$ vec v^perp = (v_x, v_y)^perp = (-v_y, v_x). $$
Then a perp product (perpendicular product) is defined, denoted as an infix operator $perp$:
$$ vec v perp vec w := vec v ^ perp cdot vec w = v_x w_y - v_y w_x. $$
The idea of using the perp product in the algorithm seems to come from the following property: $$vec v perp vec w = 0 Leftrightarrow text$vec v$ and $vec w$ are collinear. $$
At the end of the algorithm, the denominators are $vec r perp vec s$, which is then discussed for being zero, implying collinearity of the vectors and parallelity of the segments.
Either way, a similar but much more intuitive discussion of the problem with the same idea of using parametric equations and perp product is given at this page called “Intersections of Lines and Planes” on geomalgorithms. It also gives more details in the algorithm at the bottom, considering cases where one or both segments are degenerated into a single point.
answered Mar 30 at 9:37
Lazar LjubenovićLazar Ljubenović
1,3471029
edited Apr 1 at 16:33
answered Mar 30 at 9:37
Lazar LjubenovićLazar Ljubenović
1,3471029
answered Mar 30 at 9:37
Lazar LjubenovićLazar Ljubenović
1,3471029
answered Mar 30 at 9:37
Lazar LjubenovićLazar Ljubenović
1,3471029
1,3471029
add a comment |
add a comment |
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