Why do we divide by $ sqrt{A^2 + B^2} $ to convert a function from the standard form to the normal form?












3












$begingroup$


To convert an equation in the standard form, $ Ax + By + C = 0 $ to the normal form $ cos omega x + sin omega y = p $, we first divide by $ sqrt{A^2 + B^2} $. Why?



For example, to convert the equation $ sqrt 3 x + y - 8 = 0 $, we divide by $ sqrt{A^2 + B^2} = 2 $ to get $ frac{sqrt 3}{2} x + frac 12 y = 4 $ and then solve for $ cos omega = frac{sqrt 3}{2} $ and $ sin omega = frac 12 $to find the value of $ omega $.



I understand that we can't possibly convert this directly, since there is no $ omega $ for which $ cos omega = sqrt 3 $--but why do we specifically divide by $ sqrt{A^2 + B^2}?$ Why not any other number to bring the values of $ cos omega $ and $ sin omega $ between 0 and 1 and then solve it?






linear-algebra algebra-precalculus







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  • 3




    $begingroup$
    Because $cos^2 omega + sin^2 omega = 1$
    $endgroup$
    – Martin R
    Jan 13 at 16:59
















3












$begingroup$


To convert an equation in the standard form, $ Ax + By + C = 0 $ to the normal form $ cos omega x + sin omega y = p $, we first divide by $ sqrt{A^2 + B^2} $. Why?



For example, to convert the equation $ sqrt 3 x + y - 8 = 0 $, we divide by $ sqrt{A^2 + B^2} = 2 $ to get $ frac{sqrt 3}{2} x + frac 12 y = 4 $ and then solve for $ cos omega = frac{sqrt 3}{2} $ and $ sin omega = frac 12 $to find the value of $ omega $.



I understand that we can't possibly convert this directly, since there is no $ omega $ for which $ cos omega = sqrt 3 $--but why do we specifically divide by $ sqrt{A^2 + B^2}?$ Why not any other number to bring the values of $ cos omega $ and $ sin omega $ between 0 and 1 and then solve it?






linear-algebra algebra-precalculus







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Because $cos^2 omega + sin^2 omega = 1$
    $endgroup$
    – Martin R
    Jan 13 at 16:59














3












3








3





$begingroup$


To convert an equation in the standard form, $ Ax + By + C = 0 $ to the normal form $ cos omega x + sin omega y = p $, we first divide by $ sqrt{A^2 + B^2} $. Why?



For example, to convert the equation $ sqrt 3 x + y - 8 = 0 $, we divide by $ sqrt{A^2 + B^2} = 2 $ to get $ frac{sqrt 3}{2} x + frac 12 y = 4 $ and then solve for $ cos omega = frac{sqrt 3}{2} $ and $ sin omega = frac 12 $to find the value of $ omega $.



I understand that we can't possibly convert this directly, since there is no $ omega $ for which $ cos omega = sqrt 3 $--but why do we specifically divide by $ sqrt{A^2 + B^2}?$ Why not any other number to bring the values of $ cos omega $ and $ sin omega $ between 0 and 1 and then solve it?






linear-algebra algebra-precalculus







share|cite|improve this question











$endgroup$




To convert an equation in the standard form, $ Ax + By + C = 0 $ to the normal form $ cos omega x + sin omega y = p $, we first divide by $ sqrt{A^2 + B^2} $. Why?



For example, to convert the equation $ sqrt 3 x + y - 8 = 0 $, we divide by $ sqrt{A^2 + B^2} = 2 $ to get $ frac{sqrt 3}{2} x + frac 12 y = 4 $ and then solve for $ cos omega = frac{sqrt 3}{2} $ and $ sin omega = frac 12 $to find the value of $ omega $.



I understand that we can't possibly convert this directly, since there is no $ omega $ for which $ cos omega = sqrt 3 $--but why do we specifically divide by $ sqrt{A^2 + B^2}?$ Why not any other number to bring the values of $ cos omega $ and $ sin omega $ between 0 and 1 and then solve it?






linear-algebra algebra-precalculus




linear-algebra algebra-precalculus






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edited Jan 13 at 17:43









J. W. Tanner

3,4601320




3,4601320










asked Jan 13 at 16:57









WorldGovWorldGov

324211




324211








  • 3




    $begingroup$
    Because $cos^2 omega + sin^2 omega = 1$
    $endgroup$
    – Martin R
    Jan 13 at 16:59














  • 3




    $begingroup$
    Because $cos^2 omega + sin^2 omega = 1$
    $endgroup$
    – Martin R
    Jan 13 at 16:59








3




3




$begingroup$
Because $cos^2 omega + sin^2 omega = 1$
$endgroup$
– Martin R
Jan 13 at 16:59




$begingroup$
Because $cos^2 omega + sin^2 omega = 1$
$endgroup$
– Martin R
Jan 13 at 16:59










3 Answers
3






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4












$begingroup$

Given any two numbers $A$ and $B$ (not both equal zero), the numbers $a=A/sqrt{A^2+B^2}$ and $b=B/sqrt{A^2+B^2}$ satisfy $a^2+b^2=1$, so $a$ and $b$ are, respectively, the cosine and the sine of some angle. You are just finding a unit normal vector out of the given normal vector $(A,B)$ in the equation of the line. The same can be done in any number of dimensions: you can transform a given non-zero normal vector into a unit normal vector.






share|cite|improve this answer











$endgroup$





















    3












    $begingroup$

    We can solve $(cosomega,sinomega)=(a,b)$ if and only if $(a,b)$ is a unit vector (i.e. the point $(a,b)$ is on the unit circle). The original vector $(A,B)$ is not unit, so we need to normalize it, that is, to find another vector of the length one that has the same direction. It is done by dividing the vector by its length
    $$
    (a,b)=frac{(A,B)}{|(A,B)|}=frac{(A,B)}{sqrt{A^2+B^2}}.
    $$






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      Because the most important property of $sin$ and $cos$ must be preserved ($sin^2x+cos^2 x=1$).






      share|cite|improve this answer









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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        4












        $begingroup$

        Given any two numbers $A$ and $B$ (not both equal zero), the numbers $a=A/sqrt{A^2+B^2}$ and $b=B/sqrt{A^2+B^2}$ satisfy $a^2+b^2=1$, so $a$ and $b$ are, respectively, the cosine and the sine of some angle. You are just finding a unit normal vector out of the given normal vector $(A,B)$ in the equation of the line. The same can be done in any number of dimensions: you can transform a given non-zero normal vector into a unit normal vector.






        share|cite|improve this answer











        $endgroup$


















          4












          $begingroup$

          Given any two numbers $A$ and $B$ (not both equal zero), the numbers $a=A/sqrt{A^2+B^2}$ and $b=B/sqrt{A^2+B^2}$ satisfy $a^2+b^2=1$, so $a$ and $b$ are, respectively, the cosine and the sine of some angle. You are just finding a unit normal vector out of the given normal vector $(A,B)$ in the equation of the line. The same can be done in any number of dimensions: you can transform a given non-zero normal vector into a unit normal vector.






          share|cite|improve this answer











          $endgroup$
















            4












            4








            4





            $begingroup$

            Given any two numbers $A$ and $B$ (not both equal zero), the numbers $a=A/sqrt{A^2+B^2}$ and $b=B/sqrt{A^2+B^2}$ satisfy $a^2+b^2=1$, so $a$ and $b$ are, respectively, the cosine and the sine of some angle. You are just finding a unit normal vector out of the given normal vector $(A,B)$ in the equation of the line. The same can be done in any number of dimensions: you can transform a given non-zero normal vector into a unit normal vector.






            share|cite|improve this answer











            $endgroup$



            Given any two numbers $A$ and $B$ (not both equal zero), the numbers $a=A/sqrt{A^2+B^2}$ and $b=B/sqrt{A^2+B^2}$ satisfy $a^2+b^2=1$, so $a$ and $b$ are, respectively, the cosine and the sine of some angle. You are just finding a unit normal vector out of the given normal vector $(A,B)$ in the equation of the line. The same can be done in any number of dimensions: you can transform a given non-zero normal vector into a unit normal vector.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 13 at 18:59

























            answered Jan 13 at 17:09









            GReyesGReyes

            2,22815




            2,22815























                3












                $begingroup$

                We can solve $(cosomega,sinomega)=(a,b)$ if and only if $(a,b)$ is a unit vector (i.e. the point $(a,b)$ is on the unit circle). The original vector $(A,B)$ is not unit, so we need to normalize it, that is, to find another vector of the length one that has the same direction. It is done by dividing the vector by its length
                $$
                (a,b)=frac{(A,B)}{|(A,B)|}=frac{(A,B)}{sqrt{A^2+B^2}}.
                $$






                share|cite|improve this answer









                $endgroup$


















                  3












                  $begingroup$

                  We can solve $(cosomega,sinomega)=(a,b)$ if and only if $(a,b)$ is a unit vector (i.e. the point $(a,b)$ is on the unit circle). The original vector $(A,B)$ is not unit, so we need to normalize it, that is, to find another vector of the length one that has the same direction. It is done by dividing the vector by its length
                  $$
                  (a,b)=frac{(A,B)}{|(A,B)|}=frac{(A,B)}{sqrt{A^2+B^2}}.
                  $$






                  share|cite|improve this answer









                  $endgroup$
















                    3












                    3








                    3





                    $begingroup$

                    We can solve $(cosomega,sinomega)=(a,b)$ if and only if $(a,b)$ is a unit vector (i.e. the point $(a,b)$ is on the unit circle). The original vector $(A,B)$ is not unit, so we need to normalize it, that is, to find another vector of the length one that has the same direction. It is done by dividing the vector by its length
                    $$
                    (a,b)=frac{(A,B)}{|(A,B)|}=frac{(A,B)}{sqrt{A^2+B^2}}.
                    $$






                    share|cite|improve this answer









                    $endgroup$



                    We can solve $(cosomega,sinomega)=(a,b)$ if and only if $(a,b)$ is a unit vector (i.e. the point $(a,b)$ is on the unit circle). The original vector $(A,B)$ is not unit, so we need to normalize it, that is, to find another vector of the length one that has the same direction. It is done by dividing the vector by its length
                    $$
                    (a,b)=frac{(A,B)}{|(A,B)|}=frac{(A,B)}{sqrt{A^2+B^2}}.
                    $$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 13 at 17:07









                    A.Γ.A.Γ.

                    22.9k32656




                    22.9k32656























                        1












                        $begingroup$

                        Because the most important property of $sin$ and $cos$ must be preserved ($sin^2x+cos^2 x=1$).






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          Because the most important property of $sin$ and $cos$ must be preserved ($sin^2x+cos^2 x=1$).






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            Because the most important property of $sin$ and $cos$ must be preserved ($sin^2x+cos^2 x=1$).






                            share|cite|improve this answer









                            $endgroup$



                            Because the most important property of $sin$ and $cos$ must be preserved ($sin^2x+cos^2 x=1$).







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 13 at 17:12









                            Mostafa AyazMostafa Ayaz

                            16.7k3939




                            16.7k3939






























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