Signed angle in plane












0














What is the formula to compute the signed angle between two vectors $u, vinmathbb{R}^2$ where positive angle is equivalent to a counter-clockwise rotation on the plane



In other words I would like to create a function $angle(u, v)$ where



$$angle(u, v) = -angle(v, u)$$
$$angle(u, v) = -angle(u, -v)$$
$$angle(u, v) = angle(-u, -v)$$
$$angle(u, v) = pi + angle(-u, v)$$



and I want that $angle(e_1, e_2)=frac{pi}{2}$










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    0














    What is the formula to compute the signed angle between two vectors $u, vinmathbb{R}^2$ where positive angle is equivalent to a counter-clockwise rotation on the plane



    In other words I would like to create a function $angle(u, v)$ where



    $$angle(u, v) = -angle(v, u)$$
    $$angle(u, v) = -angle(u, -v)$$
    $$angle(u, v) = angle(-u, -v)$$
    $$angle(u, v) = pi + angle(-u, v)$$



    and I want that $angle(e_1, e_2)=frac{pi}{2}$










    share|cite|improve this question



























      0












      0








      0







      What is the formula to compute the signed angle between two vectors $u, vinmathbb{R}^2$ where positive angle is equivalent to a counter-clockwise rotation on the plane



      In other words I would like to create a function $angle(u, v)$ where



      $$angle(u, v) = -angle(v, u)$$
      $$angle(u, v) = -angle(u, -v)$$
      $$angle(u, v) = angle(-u, -v)$$
      $$angle(u, v) = pi + angle(-u, v)$$



      and I want that $angle(e_1, e_2)=frac{pi}{2}$










      share|cite|improve this question















      What is the formula to compute the signed angle between two vectors $u, vinmathbb{R}^2$ where positive angle is equivalent to a counter-clockwise rotation on the plane



      In other words I would like to create a function $angle(u, v)$ where



      $$angle(u, v) = -angle(v, u)$$
      $$angle(u, v) = -angle(u, -v)$$
      $$angle(u, v) = angle(-u, -v)$$
      $$angle(u, v) = pi + angle(-u, v)$$



      and I want that $angle(e_1, e_2)=frac{pi}{2}$







      linear-algebra angle






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      edited Nov 30 at 14:03









      Andrei

      11k21025




      11k21025










      asked Nov 30 at 13:33









      gota

      404315




      404315






















          1 Answer
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          active

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          3














          Just a small observation first, you will have some minor inconsistencies when the angle is either $pi$ or $0$.



          To calculate the angle, you can use the $sin$ and $cos$ function, at the same time. Using both will allow you to uniquely define the angle between $-pi$ and $pi$. The cosine of the angle is given by the scalar product:$$ucdot v=|u||v|costheta$$In $mathbb R^2$ you can write the scalar product as $ucdot v=u_xv_x+u_yv_y$.



          You get the sine of the angle using cross product $$utimes v=|u||v|sintheta$$
          In $mathbb R^2$ you can write the cross product as $utimes v=u_xv_y-u_yv_x$.



          Note that the ratio of the cross product and scalar product is the tangent of the angle. But the range of the $arctan$ function is from $-pi/2$ to $pi/2$. Note however that there is a function $mathrm{arctan2}$ or $mathrm{atan2}$ in some programming languages that take the sine and cosine and give you the correct angle. See for example the definition on wikipedia
          Then $$angle(u,v)=mathrm{arctan2}(utimes v,ucdot v)$$






          share|cite|improve this answer





















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            1 Answer
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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3














            Just a small observation first, you will have some minor inconsistencies when the angle is either $pi$ or $0$.



            To calculate the angle, you can use the $sin$ and $cos$ function, at the same time. Using both will allow you to uniquely define the angle between $-pi$ and $pi$. The cosine of the angle is given by the scalar product:$$ucdot v=|u||v|costheta$$In $mathbb R^2$ you can write the scalar product as $ucdot v=u_xv_x+u_yv_y$.



            You get the sine of the angle using cross product $$utimes v=|u||v|sintheta$$
            In $mathbb R^2$ you can write the cross product as $utimes v=u_xv_y-u_yv_x$.



            Note that the ratio of the cross product and scalar product is the tangent of the angle. But the range of the $arctan$ function is from $-pi/2$ to $pi/2$. Note however that there is a function $mathrm{arctan2}$ or $mathrm{atan2}$ in some programming languages that take the sine and cosine and give you the correct angle. See for example the definition on wikipedia
            Then $$angle(u,v)=mathrm{arctan2}(utimes v,ucdot v)$$






            share|cite|improve this answer


























              3














              Just a small observation first, you will have some minor inconsistencies when the angle is either $pi$ or $0$.



              To calculate the angle, you can use the $sin$ and $cos$ function, at the same time. Using both will allow you to uniquely define the angle between $-pi$ and $pi$. The cosine of the angle is given by the scalar product:$$ucdot v=|u||v|costheta$$In $mathbb R^2$ you can write the scalar product as $ucdot v=u_xv_x+u_yv_y$.



              You get the sine of the angle using cross product $$utimes v=|u||v|sintheta$$
              In $mathbb R^2$ you can write the cross product as $utimes v=u_xv_y-u_yv_x$.



              Note that the ratio of the cross product and scalar product is the tangent of the angle. But the range of the $arctan$ function is from $-pi/2$ to $pi/2$. Note however that there is a function $mathrm{arctan2}$ or $mathrm{atan2}$ in some programming languages that take the sine and cosine and give you the correct angle. See for example the definition on wikipedia
              Then $$angle(u,v)=mathrm{arctan2}(utimes v,ucdot v)$$






              share|cite|improve this answer
























                3












                3








                3






                Just a small observation first, you will have some minor inconsistencies when the angle is either $pi$ or $0$.



                To calculate the angle, you can use the $sin$ and $cos$ function, at the same time. Using both will allow you to uniquely define the angle between $-pi$ and $pi$. The cosine of the angle is given by the scalar product:$$ucdot v=|u||v|costheta$$In $mathbb R^2$ you can write the scalar product as $ucdot v=u_xv_x+u_yv_y$.



                You get the sine of the angle using cross product $$utimes v=|u||v|sintheta$$
                In $mathbb R^2$ you can write the cross product as $utimes v=u_xv_y-u_yv_x$.



                Note that the ratio of the cross product and scalar product is the tangent of the angle. But the range of the $arctan$ function is from $-pi/2$ to $pi/2$. Note however that there is a function $mathrm{arctan2}$ or $mathrm{atan2}$ in some programming languages that take the sine and cosine and give you the correct angle. See for example the definition on wikipedia
                Then $$angle(u,v)=mathrm{arctan2}(utimes v,ucdot v)$$






                share|cite|improve this answer












                Just a small observation first, you will have some minor inconsistencies when the angle is either $pi$ or $0$.



                To calculate the angle, you can use the $sin$ and $cos$ function, at the same time. Using both will allow you to uniquely define the angle between $-pi$ and $pi$. The cosine of the angle is given by the scalar product:$$ucdot v=|u||v|costheta$$In $mathbb R^2$ you can write the scalar product as $ucdot v=u_xv_x+u_yv_y$.



                You get the sine of the angle using cross product $$utimes v=|u||v|sintheta$$
                In $mathbb R^2$ you can write the cross product as $utimes v=u_xv_y-u_yv_x$.



                Note that the ratio of the cross product and scalar product is the tangent of the angle. But the range of the $arctan$ function is from $-pi/2$ to $pi/2$. Note however that there is a function $mathrm{arctan2}$ or $mathrm{atan2}$ in some programming languages that take the sine and cosine and give you the correct angle. See for example the definition on wikipedia
                Then $$angle(u,v)=mathrm{arctan2}(utimes v,ucdot v)$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 30 at 14:22









                Andrei

                11k21025




                11k21025






























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