why do we use cosine as the expression of vector dot product?












3












$begingroup$


When we do vector products, we use two different methods.
One is the vector dot product, another is vector cross product.
The equation of the vector dot product is
$$textbf A cdot textbf B =|textbf A| | textbf B| costheta,$$
where $theta$ is the angle between the vectors $textbf A$ and $textbf B$.



Why do we use cosine as the expression?










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migrated from physics.stackexchange.com Feb 15 '15 at 15:54


This question came from our site for active researchers, academics and students of physics.














  • 1




    $begingroup$
    It really doesn't matter, use $cos x$ or $sin(x+pi /2)$ if you like, they are equivalent.
    $endgroup$
    – JamalS
    Feb 15 '15 at 9:01










  • $begingroup$
    possible duplicate of What is the physical meaning of a product of vectors?
    $endgroup$
    – Ali
    Feb 15 '15 at 12:45
















3












$begingroup$


When we do vector products, we use two different methods.
One is the vector dot product, another is vector cross product.
The equation of the vector dot product is
$$textbf A cdot textbf B =|textbf A| | textbf B| costheta,$$
where $theta$ is the angle between the vectors $textbf A$ and $textbf B$.



Why do we use cosine as the expression?










share|cite|improve this question









$endgroup$



migrated from physics.stackexchange.com Feb 15 '15 at 15:54


This question came from our site for active researchers, academics and students of physics.














  • 1




    $begingroup$
    It really doesn't matter, use $cos x$ or $sin(x+pi /2)$ if you like, they are equivalent.
    $endgroup$
    – JamalS
    Feb 15 '15 at 9:01










  • $begingroup$
    possible duplicate of What is the physical meaning of a product of vectors?
    $endgroup$
    – Ali
    Feb 15 '15 at 12:45














3












3








3





$begingroup$


When we do vector products, we use two different methods.
One is the vector dot product, another is vector cross product.
The equation of the vector dot product is
$$textbf A cdot textbf B =|textbf A| | textbf B| costheta,$$
where $theta$ is the angle between the vectors $textbf A$ and $textbf B$.



Why do we use cosine as the expression?










share|cite|improve this question









$endgroup$




When we do vector products, we use two different methods.
One is the vector dot product, another is vector cross product.
The equation of the vector dot product is
$$textbf A cdot textbf B =|textbf A| | textbf B| costheta,$$
where $theta$ is the angle between the vectors $textbf A$ and $textbf B$.



Why do we use cosine as the expression?







vectors






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share|cite|improve this question











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share|cite|improve this question










asked Feb 15 '15 at 6:24







Shams Tarek











migrated from physics.stackexchange.com Feb 15 '15 at 15:54


This question came from our site for active researchers, academics and students of physics.









migrated from physics.stackexchange.com Feb 15 '15 at 15:54


This question came from our site for active researchers, academics and students of physics.










  • 1




    $begingroup$
    It really doesn't matter, use $cos x$ or $sin(x+pi /2)$ if you like, they are equivalent.
    $endgroup$
    – JamalS
    Feb 15 '15 at 9:01










  • $begingroup$
    possible duplicate of What is the physical meaning of a product of vectors?
    $endgroup$
    – Ali
    Feb 15 '15 at 12:45














  • 1




    $begingroup$
    It really doesn't matter, use $cos x$ or $sin(x+pi /2)$ if you like, they are equivalent.
    $endgroup$
    – JamalS
    Feb 15 '15 at 9:01










  • $begingroup$
    possible duplicate of What is the physical meaning of a product of vectors?
    $endgroup$
    – Ali
    Feb 15 '15 at 12:45








1




1




$begingroup$
It really doesn't matter, use $cos x$ or $sin(x+pi /2)$ if you like, they are equivalent.
$endgroup$
– JamalS
Feb 15 '15 at 9:01




$begingroup$
It really doesn't matter, use $cos x$ or $sin(x+pi /2)$ if you like, they are equivalent.
$endgroup$
– JamalS
Feb 15 '15 at 9:01












$begingroup$
possible duplicate of What is the physical meaning of a product of vectors?
$endgroup$
– Ali
Feb 15 '15 at 12:45




$begingroup$
possible duplicate of What is the physical meaning of a product of vectors?
$endgroup$
– Ali
Feb 15 '15 at 12:45










7 Answers
7






active

oldest

votes


















10












$begingroup$

enter image description here



The dot product of two vectors $A$ and $B$ is just the product of the magnitude of one vector with the scalar projection of the other one on itself. Hence the $cos$ term. Also, note that the $cos$ function is greater for smaller angles, and lesser for larger ones, just like the length of the projection. (both are the same thing, actually.)






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    The dot product is defined in that way.
    Note that $costheta$ is a suitable function; since by the Schwarz inequality:
    $$|mathbf{A} cdot mathbf{B}| leq |mathbf{A}| |mathbf{B}|$$
    and thus the dot product ranges continuosly between -1 and 1, as $cos theta$ for $theta in [0,pi] $.






    share|cite|improve this answer











    $endgroup$





















      2












      $begingroup$

      Let $vec{mathbf a} = (x_1, y_1) = (a cos alpha, a sin alpha)$



      Let $vec{mathbf b} = (x_2, y_2) = (b cos beta, b sin beta)$



      Then $theta = beta - alpha$



      By definition,



      begin{align}
      vec{mathbf a} circ vec{mathbf b}
      &= x_1x_2 + y_1y_2 \
      &= ab(cos alpha cos beta + sin alpha sin beta) \
      &= ab cos(beta - alpha) \
      &= ab cos theta
      end{align}






      share|cite|improve this answer











      $endgroup$





















        0












        $begingroup$

        This is entirely determined by what we consider a rotation in the plane.



        Let $u, v$ be two unit vectors. Let the angle between them be $theta$, and we can naturally write $v$ as



        $$v = u cos theta + u_perp sin theta$$



        where $u_perp$ is a unit vector perpendicular to $u$. Then clearly, the dot product is $u cdot v = cos theta$.



        But, if you're not in a Euclidean plane anymore, this relationship no longer holds. For example, in a Lorentzian space, instead of cosine and sine, we get hyperbolic functions instead:



        $$v = u cosh theta + u_perp sinh theta$$



        And the dot product is $u cdot v = cosh theta$. A physicist should recognize that this $theta$ is the "rapidity", and that the form of $v$ given here is exactly that of any Lorentz boost.



        So the reason we use sine and cosine in Euclidean space is because they are dictated by the use of sine and cosine in rotations. In other spaces, with different rotation operators, you use the functions that are associated with those rotations instead.






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          What does this have to do with rotations? This just looks like decomposition of $v$ into $v_{|}$ and $v_{bot}$ parts.
          $endgroup$
          – got it--thanks
          Feb 22 '15 at 20:22






        • 1




          $begingroup$
          Whether you use cosine or hyperbolic cosine depends on what the rotation operations are, on which set of vectors is the set of unit vectors.
          $endgroup$
          – Muphrid
          Feb 22 '15 at 21:17



















        0












        $begingroup$

        In dot product we use cos theta because in this type of product
        1.) One vector is the projection over the other.
        2.) The distance is covered along one axis or in the direction of force and there is no need of perpendicular axis or sin theta.
        In cross product the angle between must be greater than 0 and less than 180 degree it is max at 90 degree. let take the example of torque if the angle between applied force and moment arm is 90 degree than torque will be max.
        That's why we use cos theta for dot product and sin theta for cross product.






        share|cite|improve this answer









        $endgroup$





















          0












          $begingroup$

          Because here cosine is responsible for doing any type of work e.g work done cosine is use . Here one axis is use for work done I.e x -axis and for x -axis we are using cosine






          share|cite|improve this answer









          $endgroup$





















            -1












            $begingroup$

            It is actually the definition of the dot product of two vectors.






            share|cite|improve this answer









            $endgroup$









            • 7




              $begingroup$
              The point of this question is to ask why one would want to express the dot product as something involving the cosine, not what the definition is.
              $endgroup$
              – Danu
              Feb 15 '15 at 11:31






            • 4




              $begingroup$
              Besides, it is more commonly seen the other way around - the angle $theta$ is defined by this formula, while the dot product is defined by a symmetric matrix.
              $endgroup$
              – ACuriousMind
              Feb 15 '15 at 13:57












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            7 Answers
            7






            active

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            7 Answers
            7






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            10












            $begingroup$

            enter image description here



            The dot product of two vectors $A$ and $B$ is just the product of the magnitude of one vector with the scalar projection of the other one on itself. Hence the $cos$ term. Also, note that the $cos$ function is greater for smaller angles, and lesser for larger ones, just like the length of the projection. (both are the same thing, actually.)






            share|cite|improve this answer









            $endgroup$


















              10












              $begingroup$

              enter image description here



              The dot product of two vectors $A$ and $B$ is just the product of the magnitude of one vector with the scalar projection of the other one on itself. Hence the $cos$ term. Also, note that the $cos$ function is greater for smaller angles, and lesser for larger ones, just like the length of the projection. (both are the same thing, actually.)






              share|cite|improve this answer









              $endgroup$
















                10












                10








                10





                $begingroup$

                enter image description here



                The dot product of two vectors $A$ and $B$ is just the product of the magnitude of one vector with the scalar projection of the other one on itself. Hence the $cos$ term. Also, note that the $cos$ function is greater for smaller angles, and lesser for larger ones, just like the length of the projection. (both are the same thing, actually.)






                share|cite|improve this answer









                $endgroup$



                enter image description here



                The dot product of two vectors $A$ and $B$ is just the product of the magnitude of one vector with the scalar projection of the other one on itself. Hence the $cos$ term. Also, note that the $cos$ function is greater for smaller angles, and lesser for larger ones, just like the length of the projection. (both are the same thing, actually.)







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Feb 15 '15 at 6:52









                Hritik NarayanHritik Narayan

                382313




                382313























                    2












                    $begingroup$

                    The dot product is defined in that way.
                    Note that $costheta$ is a suitable function; since by the Schwarz inequality:
                    $$|mathbf{A} cdot mathbf{B}| leq |mathbf{A}| |mathbf{B}|$$
                    and thus the dot product ranges continuosly between -1 and 1, as $cos theta$ for $theta in [0,pi] $.






                    share|cite|improve this answer











                    $endgroup$


















                      2












                      $begingroup$

                      The dot product is defined in that way.
                      Note that $costheta$ is a suitable function; since by the Schwarz inequality:
                      $$|mathbf{A} cdot mathbf{B}| leq |mathbf{A}| |mathbf{B}|$$
                      and thus the dot product ranges continuosly between -1 and 1, as $cos theta$ for $theta in [0,pi] $.






                      share|cite|improve this answer











                      $endgroup$
















                        2












                        2








                        2





                        $begingroup$

                        The dot product is defined in that way.
                        Note that $costheta$ is a suitable function; since by the Schwarz inequality:
                        $$|mathbf{A} cdot mathbf{B}| leq |mathbf{A}| |mathbf{B}|$$
                        and thus the dot product ranges continuosly between -1 and 1, as $cos theta$ for $theta in [0,pi] $.






                        share|cite|improve this answer











                        $endgroup$



                        The dot product is defined in that way.
                        Note that $costheta$ is a suitable function; since by the Schwarz inequality:
                        $$|mathbf{A} cdot mathbf{B}| leq |mathbf{A}| |mathbf{B}|$$
                        and thus the dot product ranges continuosly between -1 and 1, as $cos theta$ for $theta in [0,pi] $.







                        share|cite|improve this answer














                        share|cite|improve this answer



                        share|cite|improve this answer








                        edited Jul 29 '16 at 6:28









                        rubik

                        6,77632662




                        6,77632662










                        answered Feb 15 '15 at 8:46









                        NNecNNec

                        31717




                        31717























                            2












                            $begingroup$

                            Let $vec{mathbf a} = (x_1, y_1) = (a cos alpha, a sin alpha)$



                            Let $vec{mathbf b} = (x_2, y_2) = (b cos beta, b sin beta)$



                            Then $theta = beta - alpha$



                            By definition,



                            begin{align}
                            vec{mathbf a} circ vec{mathbf b}
                            &= x_1x_2 + y_1y_2 \
                            &= ab(cos alpha cos beta + sin alpha sin beta) \
                            &= ab cos(beta - alpha) \
                            &= ab cos theta
                            end{align}






                            share|cite|improve this answer











                            $endgroup$


















                              2












                              $begingroup$

                              Let $vec{mathbf a} = (x_1, y_1) = (a cos alpha, a sin alpha)$



                              Let $vec{mathbf b} = (x_2, y_2) = (b cos beta, b sin beta)$



                              Then $theta = beta - alpha$



                              By definition,



                              begin{align}
                              vec{mathbf a} circ vec{mathbf b}
                              &= x_1x_2 + y_1y_2 \
                              &= ab(cos alpha cos beta + sin alpha sin beta) \
                              &= ab cos(beta - alpha) \
                              &= ab cos theta
                              end{align}






                              share|cite|improve this answer











                              $endgroup$
















                                2












                                2








                                2





                                $begingroup$

                                Let $vec{mathbf a} = (x_1, y_1) = (a cos alpha, a sin alpha)$



                                Let $vec{mathbf b} = (x_2, y_2) = (b cos beta, b sin beta)$



                                Then $theta = beta - alpha$



                                By definition,



                                begin{align}
                                vec{mathbf a} circ vec{mathbf b}
                                &= x_1x_2 + y_1y_2 \
                                &= ab(cos alpha cos beta + sin alpha sin beta) \
                                &= ab cos(beta - alpha) \
                                &= ab cos theta
                                end{align}






                                share|cite|improve this answer











                                $endgroup$



                                Let $vec{mathbf a} = (x_1, y_1) = (a cos alpha, a sin alpha)$



                                Let $vec{mathbf b} = (x_2, y_2) = (b cos beta, b sin beta)$



                                Then $theta = beta - alpha$



                                By definition,



                                begin{align}
                                vec{mathbf a} circ vec{mathbf b}
                                &= x_1x_2 + y_1y_2 \
                                &= ab(cos alpha cos beta + sin alpha sin beta) \
                                &= ab cos(beta - alpha) \
                                &= ab cos theta
                                end{align}







                                share|cite|improve this answer














                                share|cite|improve this answer



                                share|cite|improve this answer








                                edited Jan 16 at 6:15

























                                answered Jan 15 at 23:00









                                steven gregorysteven gregory

                                18.5k32459




                                18.5k32459























                                    0












                                    $begingroup$

                                    This is entirely determined by what we consider a rotation in the plane.



                                    Let $u, v$ be two unit vectors. Let the angle between them be $theta$, and we can naturally write $v$ as



                                    $$v = u cos theta + u_perp sin theta$$



                                    where $u_perp$ is a unit vector perpendicular to $u$. Then clearly, the dot product is $u cdot v = cos theta$.



                                    But, if you're not in a Euclidean plane anymore, this relationship no longer holds. For example, in a Lorentzian space, instead of cosine and sine, we get hyperbolic functions instead:



                                    $$v = u cosh theta + u_perp sinh theta$$



                                    And the dot product is $u cdot v = cosh theta$. A physicist should recognize that this $theta$ is the "rapidity", and that the form of $v$ given here is exactly that of any Lorentz boost.



                                    So the reason we use sine and cosine in Euclidean space is because they are dictated by the use of sine and cosine in rotations. In other spaces, with different rotation operators, you use the functions that are associated with those rotations instead.






                                    share|cite|improve this answer









                                    $endgroup$













                                    • $begingroup$
                                      What does this have to do with rotations? This just looks like decomposition of $v$ into $v_{|}$ and $v_{bot}$ parts.
                                      $endgroup$
                                      – got it--thanks
                                      Feb 22 '15 at 20:22






                                    • 1




                                      $begingroup$
                                      Whether you use cosine or hyperbolic cosine depends on what the rotation operations are, on which set of vectors is the set of unit vectors.
                                      $endgroup$
                                      – Muphrid
                                      Feb 22 '15 at 21:17
















                                    0












                                    $begingroup$

                                    This is entirely determined by what we consider a rotation in the plane.



                                    Let $u, v$ be two unit vectors. Let the angle between them be $theta$, and we can naturally write $v$ as



                                    $$v = u cos theta + u_perp sin theta$$



                                    where $u_perp$ is a unit vector perpendicular to $u$. Then clearly, the dot product is $u cdot v = cos theta$.



                                    But, if you're not in a Euclidean plane anymore, this relationship no longer holds. For example, in a Lorentzian space, instead of cosine and sine, we get hyperbolic functions instead:



                                    $$v = u cosh theta + u_perp sinh theta$$



                                    And the dot product is $u cdot v = cosh theta$. A physicist should recognize that this $theta$ is the "rapidity", and that the form of $v$ given here is exactly that of any Lorentz boost.



                                    So the reason we use sine and cosine in Euclidean space is because they are dictated by the use of sine and cosine in rotations. In other spaces, with different rotation operators, you use the functions that are associated with those rotations instead.






                                    share|cite|improve this answer









                                    $endgroup$













                                    • $begingroup$
                                      What does this have to do with rotations? This just looks like decomposition of $v$ into $v_{|}$ and $v_{bot}$ parts.
                                      $endgroup$
                                      – got it--thanks
                                      Feb 22 '15 at 20:22






                                    • 1




                                      $begingroup$
                                      Whether you use cosine or hyperbolic cosine depends on what the rotation operations are, on which set of vectors is the set of unit vectors.
                                      $endgroup$
                                      – Muphrid
                                      Feb 22 '15 at 21:17














                                    0












                                    0








                                    0





                                    $begingroup$

                                    This is entirely determined by what we consider a rotation in the plane.



                                    Let $u, v$ be two unit vectors. Let the angle between them be $theta$, and we can naturally write $v$ as



                                    $$v = u cos theta + u_perp sin theta$$



                                    where $u_perp$ is a unit vector perpendicular to $u$. Then clearly, the dot product is $u cdot v = cos theta$.



                                    But, if you're not in a Euclidean plane anymore, this relationship no longer holds. For example, in a Lorentzian space, instead of cosine and sine, we get hyperbolic functions instead:



                                    $$v = u cosh theta + u_perp sinh theta$$



                                    And the dot product is $u cdot v = cosh theta$. A physicist should recognize that this $theta$ is the "rapidity", and that the form of $v$ given here is exactly that of any Lorentz boost.



                                    So the reason we use sine and cosine in Euclidean space is because they are dictated by the use of sine and cosine in rotations. In other spaces, with different rotation operators, you use the functions that are associated with those rotations instead.






                                    share|cite|improve this answer









                                    $endgroup$



                                    This is entirely determined by what we consider a rotation in the plane.



                                    Let $u, v$ be two unit vectors. Let the angle between them be $theta$, and we can naturally write $v$ as



                                    $$v = u cos theta + u_perp sin theta$$



                                    where $u_perp$ is a unit vector perpendicular to $u$. Then clearly, the dot product is $u cdot v = cos theta$.



                                    But, if you're not in a Euclidean plane anymore, this relationship no longer holds. For example, in a Lorentzian space, instead of cosine and sine, we get hyperbolic functions instead:



                                    $$v = u cosh theta + u_perp sinh theta$$



                                    And the dot product is $u cdot v = cosh theta$. A physicist should recognize that this $theta$ is the "rapidity", and that the form of $v$ given here is exactly that of any Lorentz boost.



                                    So the reason we use sine and cosine in Euclidean space is because they are dictated by the use of sine and cosine in rotations. In other spaces, with different rotation operators, you use the functions that are associated with those rotations instead.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Feb 15 '15 at 16:06









                                    MuphridMuphrid

                                    15.9k11543




                                    15.9k11543












                                    • $begingroup$
                                      What does this have to do with rotations? This just looks like decomposition of $v$ into $v_{|}$ and $v_{bot}$ parts.
                                      $endgroup$
                                      – got it--thanks
                                      Feb 22 '15 at 20:22






                                    • 1




                                      $begingroup$
                                      Whether you use cosine or hyperbolic cosine depends on what the rotation operations are, on which set of vectors is the set of unit vectors.
                                      $endgroup$
                                      – Muphrid
                                      Feb 22 '15 at 21:17


















                                    • $begingroup$
                                      What does this have to do with rotations? This just looks like decomposition of $v$ into $v_{|}$ and $v_{bot}$ parts.
                                      $endgroup$
                                      – got it--thanks
                                      Feb 22 '15 at 20:22






                                    • 1




                                      $begingroup$
                                      Whether you use cosine or hyperbolic cosine depends on what the rotation operations are, on which set of vectors is the set of unit vectors.
                                      $endgroup$
                                      – Muphrid
                                      Feb 22 '15 at 21:17
















                                    $begingroup$
                                    What does this have to do with rotations? This just looks like decomposition of $v$ into $v_{|}$ and $v_{bot}$ parts.
                                    $endgroup$
                                    – got it--thanks
                                    Feb 22 '15 at 20:22




                                    $begingroup$
                                    What does this have to do with rotations? This just looks like decomposition of $v$ into $v_{|}$ and $v_{bot}$ parts.
                                    $endgroup$
                                    – got it--thanks
                                    Feb 22 '15 at 20:22




                                    1




                                    1




                                    $begingroup$
                                    Whether you use cosine or hyperbolic cosine depends on what the rotation operations are, on which set of vectors is the set of unit vectors.
                                    $endgroup$
                                    – Muphrid
                                    Feb 22 '15 at 21:17




                                    $begingroup$
                                    Whether you use cosine or hyperbolic cosine depends on what the rotation operations are, on which set of vectors is the set of unit vectors.
                                    $endgroup$
                                    – Muphrid
                                    Feb 22 '15 at 21:17











                                    0












                                    $begingroup$

                                    In dot product we use cos theta because in this type of product
                                    1.) One vector is the projection over the other.
                                    2.) The distance is covered along one axis or in the direction of force and there is no need of perpendicular axis or sin theta.
                                    In cross product the angle between must be greater than 0 and less than 180 degree it is max at 90 degree. let take the example of torque if the angle between applied force and moment arm is 90 degree than torque will be max.
                                    That's why we use cos theta for dot product and sin theta for cross product.






                                    share|cite|improve this answer









                                    $endgroup$


















                                      0












                                      $begingroup$

                                      In dot product we use cos theta because in this type of product
                                      1.) One vector is the projection over the other.
                                      2.) The distance is covered along one axis or in the direction of force and there is no need of perpendicular axis or sin theta.
                                      In cross product the angle between must be greater than 0 and less than 180 degree it is max at 90 degree. let take the example of torque if the angle between applied force and moment arm is 90 degree than torque will be max.
                                      That's why we use cos theta for dot product and sin theta for cross product.






                                      share|cite|improve this answer









                                      $endgroup$
















                                        0












                                        0








                                        0





                                        $begingroup$

                                        In dot product we use cos theta because in this type of product
                                        1.) One vector is the projection over the other.
                                        2.) The distance is covered along one axis or in the direction of force and there is no need of perpendicular axis or sin theta.
                                        In cross product the angle between must be greater than 0 and less than 180 degree it is max at 90 degree. let take the example of torque if the angle between applied force and moment arm is 90 degree than torque will be max.
                                        That's why we use cos theta for dot product and sin theta for cross product.






                                        share|cite|improve this answer









                                        $endgroup$



                                        In dot product we use cos theta because in this type of product
                                        1.) One vector is the projection over the other.
                                        2.) The distance is covered along one axis or in the direction of force and there is no need of perpendicular axis or sin theta.
                                        In cross product the angle between must be greater than 0 and less than 180 degree it is max at 90 degree. let take the example of torque if the angle between applied force and moment arm is 90 degree than torque will be max.
                                        That's why we use cos theta for dot product and sin theta for cross product.







                                        share|cite|improve this answer












                                        share|cite|improve this answer



                                        share|cite|improve this answer










                                        answered Jul 29 '16 at 5:56









                                        muhammad mubeenmuhammad mubeen

                                        1




                                        1























                                            0












                                            $begingroup$

                                            Because here cosine is responsible for doing any type of work e.g work done cosine is use . Here one axis is use for work done I.e x -axis and for x -axis we are using cosine






                                            share|cite|improve this answer









                                            $endgroup$


















                                              0












                                              $begingroup$

                                              Because here cosine is responsible for doing any type of work e.g work done cosine is use . Here one axis is use for work done I.e x -axis and for x -axis we are using cosine






                                              share|cite|improve this answer









                                              $endgroup$
















                                                0












                                                0








                                                0





                                                $begingroup$

                                                Because here cosine is responsible for doing any type of work e.g work done cosine is use . Here one axis is use for work done I.e x -axis and for x -axis we are using cosine






                                                share|cite|improve this answer









                                                $endgroup$



                                                Because here cosine is responsible for doing any type of work e.g work done cosine is use . Here one axis is use for work done I.e x -axis and for x -axis we are using cosine







                                                share|cite|improve this answer












                                                share|cite|improve this answer



                                                share|cite|improve this answer










                                                answered May 25 '18 at 8:52









                                                Bibi AlamBibi Alam

                                                1




                                                1























                                                    -1












                                                    $begingroup$

                                                    It is actually the definition of the dot product of two vectors.






                                                    share|cite|improve this answer









                                                    $endgroup$









                                                    • 7




                                                      $begingroup$
                                                      The point of this question is to ask why one would want to express the dot product as something involving the cosine, not what the definition is.
                                                      $endgroup$
                                                      – Danu
                                                      Feb 15 '15 at 11:31






                                                    • 4




                                                      $begingroup$
                                                      Besides, it is more commonly seen the other way around - the angle $theta$ is defined by this formula, while the dot product is defined by a symmetric matrix.
                                                      $endgroup$
                                                      – ACuriousMind
                                                      Feb 15 '15 at 13:57
















                                                    -1












                                                    $begingroup$

                                                    It is actually the definition of the dot product of two vectors.






                                                    share|cite|improve this answer









                                                    $endgroup$









                                                    • 7




                                                      $begingroup$
                                                      The point of this question is to ask why one would want to express the dot product as something involving the cosine, not what the definition is.
                                                      $endgroup$
                                                      – Danu
                                                      Feb 15 '15 at 11:31






                                                    • 4




                                                      $begingroup$
                                                      Besides, it is more commonly seen the other way around - the angle $theta$ is defined by this formula, while the dot product is defined by a symmetric matrix.
                                                      $endgroup$
                                                      – ACuriousMind
                                                      Feb 15 '15 at 13:57














                                                    -1












                                                    -1








                                                    -1





                                                    $begingroup$

                                                    It is actually the definition of the dot product of two vectors.






                                                    share|cite|improve this answer









                                                    $endgroup$



                                                    It is actually the definition of the dot product of two vectors.







                                                    share|cite|improve this answer












                                                    share|cite|improve this answer



                                                    share|cite|improve this answer










                                                    answered Feb 15 '15 at 8:38









                                                    omehoqueomehoque

                                                    204312




                                                    204312








                                                    • 7




                                                      $begingroup$
                                                      The point of this question is to ask why one would want to express the dot product as something involving the cosine, not what the definition is.
                                                      $endgroup$
                                                      – Danu
                                                      Feb 15 '15 at 11:31






                                                    • 4




                                                      $begingroup$
                                                      Besides, it is more commonly seen the other way around - the angle $theta$ is defined by this formula, while the dot product is defined by a symmetric matrix.
                                                      $endgroup$
                                                      – ACuriousMind
                                                      Feb 15 '15 at 13:57














                                                    • 7




                                                      $begingroup$
                                                      The point of this question is to ask why one would want to express the dot product as something involving the cosine, not what the definition is.
                                                      $endgroup$
                                                      – Danu
                                                      Feb 15 '15 at 11:31






                                                    • 4




                                                      $begingroup$
                                                      Besides, it is more commonly seen the other way around - the angle $theta$ is defined by this formula, while the dot product is defined by a symmetric matrix.
                                                      $endgroup$
                                                      – ACuriousMind
                                                      Feb 15 '15 at 13:57








                                                    7




                                                    7




                                                    $begingroup$
                                                    The point of this question is to ask why one would want to express the dot product as something involving the cosine, not what the definition is.
                                                    $endgroup$
                                                    – Danu
                                                    Feb 15 '15 at 11:31




                                                    $begingroup$
                                                    The point of this question is to ask why one would want to express the dot product as something involving the cosine, not what the definition is.
                                                    $endgroup$
                                                    – Danu
                                                    Feb 15 '15 at 11:31




                                                    4




                                                    4




                                                    $begingroup$
                                                    Besides, it is more commonly seen the other way around - the angle $theta$ is defined by this formula, while the dot product is defined by a symmetric matrix.
                                                    $endgroup$
                                                    – ACuriousMind
                                                    Feb 15 '15 at 13:57




                                                    $begingroup$
                                                    Besides, it is more commonly seen the other way around - the angle $theta$ is defined by this formula, while the dot product is defined by a symmetric matrix.
                                                    $endgroup$
                                                    – ACuriousMind
                                                    Feb 15 '15 at 13:57


















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