Visualizing the area described by the dot product?












1












$begingroup$


Since the dot product of two vectors is an area (if your vectors have units of meters, then the dot product would be in m$^2$), I was wondering if there is a good way to visualize that area. The wedge product in 2D has an easy visualization, the directed area of the parallelogram formed by the vectors in question. For the dot product you can put an arbitrarily placed and oriented $x_1$ by $x_2$ rectangle adjacent to a $y_1$ by $y_2$ rectangle for $[x_1,y_1] cdot [x_2, y_2]$, but that doesn't show how the area changes as your rotate or scale the vectors, or the relationship to the original vectors.



Thanks!






geometry vectors area geometric-algebras







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$endgroup$








  • 1




    $begingroup$
    See this diagram of the Law of Cosines, and note that, say, $$overrightarrow{CB}cdotoverrightarrow{CA} = a b cos C$$ Two of the rectangles shown represent the indicated area.
    $endgroup$
    – Blue
    Nov 28 '18 at 6:16








  • 1




    $begingroup$
    This might be helpful-math.stackexchange.com/questions/805954/…
    $endgroup$
    – Thomas Shelby
    Nov 28 '18 at 6:17






  • 2




    $begingroup$
    The dot product of two vectors has units of area, but personally I have never encountered a situation where it was useful to visualize it as an area of some geometrical shape.
    $endgroup$
    – Rahul
    Nov 28 '18 at 6:53










  • $begingroup$
    Do you see the rectangle?
    $endgroup$
    – Moti
    Nov 28 '18 at 7:11
















1












$begingroup$


Since the dot product of two vectors is an area (if your vectors have units of meters, then the dot product would be in m$^2$), I was wondering if there is a good way to visualize that area. The wedge product in 2D has an easy visualization, the directed area of the parallelogram formed by the vectors in question. For the dot product you can put an arbitrarily placed and oriented $x_1$ by $x_2$ rectangle adjacent to a $y_1$ by $y_2$ rectangle for $[x_1,y_1] cdot [x_2, y_2]$, but that doesn't show how the area changes as your rotate or scale the vectors, or the relationship to the original vectors.



Thanks!






geometry vectors area geometric-algebras







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    See this diagram of the Law of Cosines, and note that, say, $$overrightarrow{CB}cdotoverrightarrow{CA} = a b cos C$$ Two of the rectangles shown represent the indicated area.
    $endgroup$
    – Blue
    Nov 28 '18 at 6:16








  • 1




    $begingroup$
    This might be helpful-math.stackexchange.com/questions/805954/…
    $endgroup$
    – Thomas Shelby
    Nov 28 '18 at 6:17






  • 2




    $begingroup$
    The dot product of two vectors has units of area, but personally I have never encountered a situation where it was useful to visualize it as an area of some geometrical shape.
    $endgroup$
    – Rahul
    Nov 28 '18 at 6:53










  • $begingroup$
    Do you see the rectangle?
    $endgroup$
    – Moti
    Nov 28 '18 at 7:11














1












1








1





$begingroup$


Since the dot product of two vectors is an area (if your vectors have units of meters, then the dot product would be in m$^2$), I was wondering if there is a good way to visualize that area. The wedge product in 2D has an easy visualization, the directed area of the parallelogram formed by the vectors in question. For the dot product you can put an arbitrarily placed and oriented $x_1$ by $x_2$ rectangle adjacent to a $y_1$ by $y_2$ rectangle for $[x_1,y_1] cdot [x_2, y_2]$, but that doesn't show how the area changes as your rotate or scale the vectors, or the relationship to the original vectors.



Thanks!






geometry vectors area geometric-algebras







share|cite|improve this question











$endgroup$




Since the dot product of two vectors is an area (if your vectors have units of meters, then the dot product would be in m$^2$), I was wondering if there is a good way to visualize that area. The wedge product in 2D has an easy visualization, the directed area of the parallelogram formed by the vectors in question. For the dot product you can put an arbitrarily placed and oriented $x_1$ by $x_2$ rectangle adjacent to a $y_1$ by $y_2$ rectangle for $[x_1,y_1] cdot [x_2, y_2]$, but that doesn't show how the area changes as your rotate or scale the vectors, or the relationship to the original vectors.



Thanks!






geometry vectors area geometric-algebras




geometry vectors area geometric-algebras






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 28 '18 at 6:15







Greg Buchholz

















asked Nov 28 '18 at 6:10









Greg BuchholzGreg Buchholz

1533




1533








  • 1




    $begingroup$
    See this diagram of the Law of Cosines, and note that, say, $$overrightarrow{CB}cdotoverrightarrow{CA} = a b cos C$$ Two of the rectangles shown represent the indicated area.
    $endgroup$
    – Blue
    Nov 28 '18 at 6:16








  • 1




    $begingroup$
    This might be helpful-math.stackexchange.com/questions/805954/…
    $endgroup$
    – Thomas Shelby
    Nov 28 '18 at 6:17






  • 2




    $begingroup$
    The dot product of two vectors has units of area, but personally I have never encountered a situation where it was useful to visualize it as an area of some geometrical shape.
    $endgroup$
    – Rahul
    Nov 28 '18 at 6:53










  • $begingroup$
    Do you see the rectangle?
    $endgroup$
    – Moti
    Nov 28 '18 at 7:11














  • 1




    $begingroup$
    See this diagram of the Law of Cosines, and note that, say, $$overrightarrow{CB}cdotoverrightarrow{CA} = a b cos C$$ Two of the rectangles shown represent the indicated area.
    $endgroup$
    – Blue
    Nov 28 '18 at 6:16








  • 1




    $begingroup$
    This might be helpful-math.stackexchange.com/questions/805954/…
    $endgroup$
    – Thomas Shelby
    Nov 28 '18 at 6:17






  • 2




    $begingroup$
    The dot product of two vectors has units of area, but personally I have never encountered a situation where it was useful to visualize it as an area of some geometrical shape.
    $endgroup$
    – Rahul
    Nov 28 '18 at 6:53










  • $begingroup$
    Do you see the rectangle?
    $endgroup$
    – Moti
    Nov 28 '18 at 7:11








1




1




$begingroup$
See this diagram of the Law of Cosines, and note that, say, $$overrightarrow{CB}cdotoverrightarrow{CA} = a b cos C$$ Two of the rectangles shown represent the indicated area.
$endgroup$
– Blue
Nov 28 '18 at 6:16






$begingroup$
See this diagram of the Law of Cosines, and note that, say, $$overrightarrow{CB}cdotoverrightarrow{CA} = a b cos C$$ Two of the rectangles shown represent the indicated area.
$endgroup$
– Blue
Nov 28 '18 at 6:16






1




1




$begingroup$
This might be helpful-math.stackexchange.com/questions/805954/…
$endgroup$
– Thomas Shelby
Nov 28 '18 at 6:17




$begingroup$
This might be helpful-math.stackexchange.com/questions/805954/…
$endgroup$
– Thomas Shelby
Nov 28 '18 at 6:17




2




2




$begingroup$
The dot product of two vectors has units of area, but personally I have never encountered a situation where it was useful to visualize it as an area of some geometrical shape.
$endgroup$
– Rahul
Nov 28 '18 at 6:53




$begingroup$
The dot product of two vectors has units of area, but personally I have never encountered a situation where it was useful to visualize it as an area of some geometrical shape.
$endgroup$
– Rahul
Nov 28 '18 at 6:53












$begingroup$
Do you see the rectangle?
$endgroup$
– Moti
Nov 28 '18 at 7:11




$begingroup$
Do you see the rectangle?
$endgroup$
– Moti
Nov 28 '18 at 7:11










1 Answer
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1












$begingroup$

There is a generalized form of the Pythagorean theorem. If $mathbf{v}$ and $mathbf{w}$ are real vectors, then



$$ (mathbf{v} cdot mathbf{w})^2 + lVert mathbf{v} wedge mathbf{w} rVert^2 = lVert mathbf{v}rVert^2 lVert mathbf{w}rVert^2.$$



So you can (kind of) think of the wedge product as how much the parallelogram spanned by $mathbf{v}$ and $mathbf{w}$ succeeds at being a rectangle, and the dot product as the extent to which the parallelogram fails at being a rectangle. Or you could see them as legs of a "right triangle", except the legs have dimensions of area. IDK.



The formula generalizes further; if $mathbf{B}$ is a blade, then



$$ lVert mathbf{v} ⨼ mathbf{B} rVert^2 + lVert mathbf{v} wedge mathbf{B} rVert^2 = lVert mathbf{v}rVert^2 lVert mathbf{B}rVert^2,$$



where the ⨼ denotes the left contraction, a generalization of the dot product (as defined by Leo Dorst in "The inner products of geometric algebra").






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    $begingroup$

    There is a generalized form of the Pythagorean theorem. If $mathbf{v}$ and $mathbf{w}$ are real vectors, then



    $$ (mathbf{v} cdot mathbf{w})^2 + lVert mathbf{v} wedge mathbf{w} rVert^2 = lVert mathbf{v}rVert^2 lVert mathbf{w}rVert^2.$$



    So you can (kind of) think of the wedge product as how much the parallelogram spanned by $mathbf{v}$ and $mathbf{w}$ succeeds at being a rectangle, and the dot product as the extent to which the parallelogram fails at being a rectangle. Or you could see them as legs of a "right triangle", except the legs have dimensions of area. IDK.



    The formula generalizes further; if $mathbf{B}$ is a blade, then



    $$ lVert mathbf{v} ⨼ mathbf{B} rVert^2 + lVert mathbf{v} wedge mathbf{B} rVert^2 = lVert mathbf{v}rVert^2 lVert mathbf{B}rVert^2,$$



    where the ⨼ denotes the left contraction, a generalization of the dot product (as defined by Leo Dorst in "The inner products of geometric algebra").






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      There is a generalized form of the Pythagorean theorem. If $mathbf{v}$ and $mathbf{w}$ are real vectors, then



      $$ (mathbf{v} cdot mathbf{w})^2 + lVert mathbf{v} wedge mathbf{w} rVert^2 = lVert mathbf{v}rVert^2 lVert mathbf{w}rVert^2.$$



      So you can (kind of) think of the wedge product as how much the parallelogram spanned by $mathbf{v}$ and $mathbf{w}$ succeeds at being a rectangle, and the dot product as the extent to which the parallelogram fails at being a rectangle. Or you could see them as legs of a "right triangle", except the legs have dimensions of area. IDK.



      The formula generalizes further; if $mathbf{B}$ is a blade, then



      $$ lVert mathbf{v} ⨼ mathbf{B} rVert^2 + lVert mathbf{v} wedge mathbf{B} rVert^2 = lVert mathbf{v}rVert^2 lVert mathbf{B}rVert^2,$$



      where the ⨼ denotes the left contraction, a generalization of the dot product (as defined by Leo Dorst in "The inner products of geometric algebra").






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        There is a generalized form of the Pythagorean theorem. If $mathbf{v}$ and $mathbf{w}$ are real vectors, then



        $$ (mathbf{v} cdot mathbf{w})^2 + lVert mathbf{v} wedge mathbf{w} rVert^2 = lVert mathbf{v}rVert^2 lVert mathbf{w}rVert^2.$$



        So you can (kind of) think of the wedge product as how much the parallelogram spanned by $mathbf{v}$ and $mathbf{w}$ succeeds at being a rectangle, and the dot product as the extent to which the parallelogram fails at being a rectangle. Or you could see them as legs of a "right triangle", except the legs have dimensions of area. IDK.



        The formula generalizes further; if $mathbf{B}$ is a blade, then



        $$ lVert mathbf{v} ⨼ mathbf{B} rVert^2 + lVert mathbf{v} wedge mathbf{B} rVert^2 = lVert mathbf{v}rVert^2 lVert mathbf{B}rVert^2,$$



        where the ⨼ denotes the left contraction, a generalization of the dot product (as defined by Leo Dorst in "The inner products of geometric algebra").






        share|cite|improve this answer









        $endgroup$



        There is a generalized form of the Pythagorean theorem. If $mathbf{v}$ and $mathbf{w}$ are real vectors, then



        $$ (mathbf{v} cdot mathbf{w})^2 + lVert mathbf{v} wedge mathbf{w} rVert^2 = lVert mathbf{v}rVert^2 lVert mathbf{w}rVert^2.$$



        So you can (kind of) think of the wedge product as how much the parallelogram spanned by $mathbf{v}$ and $mathbf{w}$ succeeds at being a rectangle, and the dot product as the extent to which the parallelogram fails at being a rectangle. Or you could see them as legs of a "right triangle", except the legs have dimensions of area. IDK.



        The formula generalizes further; if $mathbf{B}$ is a blade, then



        $$ lVert mathbf{v} ⨼ mathbf{B} rVert^2 + lVert mathbf{v} wedge mathbf{B} rVert^2 = lVert mathbf{v}rVert^2 lVert mathbf{B}rVert^2,$$



        where the ⨼ denotes the left contraction, a generalization of the dot product (as defined by Leo Dorst in "The inner products of geometric algebra").







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 29 '18 at 11:58









        Nikifuj908Nikifuj908

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