Scalar product of vectors with different basis?

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Let's say I have this scalar product with the basis in both Cartesian and cylindrical coordinates.
$$ê_ycdotê_rho$$



Do I need to convert one of them to the other basis, e.g. convert $ê_rho$ to $cosphiê_x+sinphiê_y$ and then perform the scalar product $(0,1,0)cdot(cosphi, sinphi,0)=(0,sinphi,0)$?



Or how does it work?







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    up vote
    0
    down vote

    favorite












    Let's say I have this scalar product with the basis in both Cartesian and cylindrical coordinates.
    $$ê_ycdotê_rho$$



    Do I need to convert one of them to the other basis, e.g. convert $ê_rho$ to $cosphiê_x+sinphiê_y$ and then perform the scalar product $(0,1,0)cdot(cosphi, sinphi,0)=(0,sinphi,0)$?



    Or how does it work?







    share|cite|improve this question






















      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Let's say I have this scalar product with the basis in both Cartesian and cylindrical coordinates.
      $$ê_ycdotê_rho$$



      Do I need to convert one of them to the other basis, e.g. convert $ê_rho$ to $cosphiê_x+sinphiê_y$ and then perform the scalar product $(0,1,0)cdot(cosphi, sinphi,0)=(0,sinphi,0)$?



      Or how does it work?







      share|cite|improve this question












      Let's say I have this scalar product with the basis in both Cartesian and cylindrical coordinates.
      $$ê_ycdotê_rho$$



      Do I need to convert one of them to the other basis, e.g. convert $ê_rho$ to $cosphiê_x+sinphiê_y$ and then perform the scalar product $(0,1,0)cdot(cosphi, sinphi,0)=(0,sinphi,0)$?



      Or how does it work?









      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Aug 28 at 13:14









      Defindun

      296




      296




















          1 Answer
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          The scalar product must be a scalar, so it is just $sin phi$. You can see this by converting one vector from cylindrical to cartesian co-ordinates or by using the co-ordinate free definition of the scalar product which is



          $overrightarrowa cdot overrightarrowb = |overrightarrowa||overrightarrowb| cos theta$



          where $theta$ is the angle between vectors $overrightarrowa$ and $overrightarrowb$. In you example $|overrightarrowa|=|overrightarrowb=1$ and $theta = fracpi2-phi$.






          share|cite|improve this answer




















          • Yes, of course, it's just sin(phi). Scalar product! Okay, thank you, so to convert is the answer (or use the coordinate free definition of scalar product). If I was to convert to cartesian to cylindrical the answer would always be the same I guess?
            – Defindun
            Aug 28 at 15:06










          • Yes, but the formula for scalar product in cylindrical co-ordinates becomes more complex. It is $(r_1, theta_1, z_1) cdot (r_2, theta_2, z_2) = r_1r_2 cos(theta_1-theta_2) + z_1z_2$
            – gandalf61
            Aug 28 at 15:34











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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          0
          down vote



          accepted










          The scalar product must be a scalar, so it is just $sin phi$. You can see this by converting one vector from cylindrical to cartesian co-ordinates or by using the co-ordinate free definition of the scalar product which is



          $overrightarrowa cdot overrightarrowb = |overrightarrowa||overrightarrowb| cos theta$



          where $theta$ is the angle between vectors $overrightarrowa$ and $overrightarrowb$. In you example $|overrightarrowa|=|overrightarrowb=1$ and $theta = fracpi2-phi$.






          share|cite|improve this answer




















          • Yes, of course, it's just sin(phi). Scalar product! Okay, thank you, so to convert is the answer (or use the coordinate free definition of scalar product). If I was to convert to cartesian to cylindrical the answer would always be the same I guess?
            – Defindun
            Aug 28 at 15:06










          • Yes, but the formula for scalar product in cylindrical co-ordinates becomes more complex. It is $(r_1, theta_1, z_1) cdot (r_2, theta_2, z_2) = r_1r_2 cos(theta_1-theta_2) + z_1z_2$
            – gandalf61
            Aug 28 at 15:34















          up vote
          0
          down vote



          accepted










          The scalar product must be a scalar, so it is just $sin phi$. You can see this by converting one vector from cylindrical to cartesian co-ordinates or by using the co-ordinate free definition of the scalar product which is



          $overrightarrowa cdot overrightarrowb = |overrightarrowa||overrightarrowb| cos theta$



          where $theta$ is the angle between vectors $overrightarrowa$ and $overrightarrowb$. In you example $|overrightarrowa|=|overrightarrowb=1$ and $theta = fracpi2-phi$.






          share|cite|improve this answer




















          • Yes, of course, it's just sin(phi). Scalar product! Okay, thank you, so to convert is the answer (or use the coordinate free definition of scalar product). If I was to convert to cartesian to cylindrical the answer would always be the same I guess?
            – Defindun
            Aug 28 at 15:06










          • Yes, but the formula for scalar product in cylindrical co-ordinates becomes more complex. It is $(r_1, theta_1, z_1) cdot (r_2, theta_2, z_2) = r_1r_2 cos(theta_1-theta_2) + z_1z_2$
            – gandalf61
            Aug 28 at 15:34













          up vote
          0
          down vote



          accepted







          up vote
          0
          down vote



          accepted






          The scalar product must be a scalar, so it is just $sin phi$. You can see this by converting one vector from cylindrical to cartesian co-ordinates or by using the co-ordinate free definition of the scalar product which is



          $overrightarrowa cdot overrightarrowb = |overrightarrowa||overrightarrowb| cos theta$



          where $theta$ is the angle between vectors $overrightarrowa$ and $overrightarrowb$. In you example $|overrightarrowa|=|overrightarrowb=1$ and $theta = fracpi2-phi$.






          share|cite|improve this answer












          The scalar product must be a scalar, so it is just $sin phi$. You can see this by converting one vector from cylindrical to cartesian co-ordinates or by using the co-ordinate free definition of the scalar product which is



          $overrightarrowa cdot overrightarrowb = |overrightarrowa||overrightarrowb| cos theta$



          where $theta$ is the angle between vectors $overrightarrowa$ and $overrightarrowb$. In you example $|overrightarrowa|=|overrightarrowb=1$ and $theta = fracpi2-phi$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 28 at 14:14









          gandalf61

          6,061522




          6,061522











          • Yes, of course, it's just sin(phi). Scalar product! Okay, thank you, so to convert is the answer (or use the coordinate free definition of scalar product). If I was to convert to cartesian to cylindrical the answer would always be the same I guess?
            – Defindun
            Aug 28 at 15:06










          • Yes, but the formula for scalar product in cylindrical co-ordinates becomes more complex. It is $(r_1, theta_1, z_1) cdot (r_2, theta_2, z_2) = r_1r_2 cos(theta_1-theta_2) + z_1z_2$
            – gandalf61
            Aug 28 at 15:34

















          • Yes, of course, it's just sin(phi). Scalar product! Okay, thank you, so to convert is the answer (or use the coordinate free definition of scalar product). If I was to convert to cartesian to cylindrical the answer would always be the same I guess?
            – Defindun
            Aug 28 at 15:06










          • Yes, but the formula for scalar product in cylindrical co-ordinates becomes more complex. It is $(r_1, theta_1, z_1) cdot (r_2, theta_2, z_2) = r_1r_2 cos(theta_1-theta_2) + z_1z_2$
            – gandalf61
            Aug 28 at 15:34
















          Yes, of course, it's just sin(phi). Scalar product! Okay, thank you, so to convert is the answer (or use the coordinate free definition of scalar product). If I was to convert to cartesian to cylindrical the answer would always be the same I guess?
          – Defindun
          Aug 28 at 15:06




          Yes, of course, it's just sin(phi). Scalar product! Okay, thank you, so to convert is the answer (or use the coordinate free definition of scalar product). If I was to convert to cartesian to cylindrical the answer would always be the same I guess?
          – Defindun
          Aug 28 at 15:06












          Yes, but the formula for scalar product in cylindrical co-ordinates becomes more complex. It is $(r_1, theta_1, z_1) cdot (r_2, theta_2, z_2) = r_1r_2 cos(theta_1-theta_2) + z_1z_2$
          – gandalf61
          Aug 28 at 15:34





          Yes, but the formula for scalar product in cylindrical co-ordinates becomes more complex. It is $(r_1, theta_1, z_1) cdot (r_2, theta_2, z_2) = r_1r_2 cos(theta_1-theta_2) + z_1z_2$
          – gandalf61
          Aug 28 at 15:34


















           

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