Angle between vector and axis












0












$begingroup$


I am trying to make some data work for Malus Law with 3 polarizers (don't worry, my question is about geometry, not physics), and I think I should know this but I'm loosing my mind at this point.



Let's say we have a vector $v$ in two dimensions making an angle $theta$ with the $Y$ axis, then what's the expresion for the angle with the $X$ axis?



I was thinking about adding something to $theta$, but I believe that doesn't work for a domain $[0,2pi]$.






geometry trigonometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    Is this in two dimensions or three?
    $endgroup$
    – amd
    Nov 26 '18 at 2:58










  • $begingroup$
    Two dimensions, I will edit my question.
    $endgroup$
    – IchVerloren
    Nov 26 '18 at 3:05










  • $begingroup$
    If you want the angle to lie in a certain range, you’ll almost certainly have to perform some normalization after the arithmetic.
    $endgroup$
    – amd
    Nov 26 '18 at 3:08
















0












$begingroup$


I am trying to make some data work for Malus Law with 3 polarizers (don't worry, my question is about geometry, not physics), and I think I should know this but I'm loosing my mind at this point.



Let's say we have a vector $v$ in two dimensions making an angle $theta$ with the $Y$ axis, then what's the expresion for the angle with the $X$ axis?



I was thinking about adding something to $theta$, but I believe that doesn't work for a domain $[0,2pi]$.






geometry trigonometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    Is this in two dimensions or three?
    $endgroup$
    – amd
    Nov 26 '18 at 2:58










  • $begingroup$
    Two dimensions, I will edit my question.
    $endgroup$
    – IchVerloren
    Nov 26 '18 at 3:05










  • $begingroup$
    If you want the angle to lie in a certain range, you’ll almost certainly have to perform some normalization after the arithmetic.
    $endgroup$
    – amd
    Nov 26 '18 at 3:08














0












0








0





$begingroup$


I am trying to make some data work for Malus Law with 3 polarizers (don't worry, my question is about geometry, not physics), and I think I should know this but I'm loosing my mind at this point.



Let's say we have a vector $v$ in two dimensions making an angle $theta$ with the $Y$ axis, then what's the expresion for the angle with the $X$ axis?



I was thinking about adding something to $theta$, but I believe that doesn't work for a domain $[0,2pi]$.






geometry trigonometry







share|cite|improve this question











$endgroup$




I am trying to make some data work for Malus Law with 3 polarizers (don't worry, my question is about geometry, not physics), and I think I should know this but I'm loosing my mind at this point.



Let's say we have a vector $v$ in two dimensions making an angle $theta$ with the $Y$ axis, then what's the expresion for the angle with the $X$ axis?



I was thinking about adding something to $theta$, but I believe that doesn't work for a domain $[0,2pi]$.






geometry trigonometry




geometry trigonometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 26 '18 at 3:06







IchVerloren

















asked Nov 26 '18 at 2:57









IchVerlorenIchVerloren

969




969












  • $begingroup$
    Is this in two dimensions or three?
    $endgroup$
    – amd
    Nov 26 '18 at 2:58










  • $begingroup$
    Two dimensions, I will edit my question.
    $endgroup$
    – IchVerloren
    Nov 26 '18 at 3:05










  • $begingroup$
    If you want the angle to lie in a certain range, you’ll almost certainly have to perform some normalization after the arithmetic.
    $endgroup$
    – amd
    Nov 26 '18 at 3:08


















  • $begingroup$
    Is this in two dimensions or three?
    $endgroup$
    – amd
    Nov 26 '18 at 2:58










  • $begingroup$
    Two dimensions, I will edit my question.
    $endgroup$
    – IchVerloren
    Nov 26 '18 at 3:05










  • $begingroup$
    If you want the angle to lie in a certain range, you’ll almost certainly have to perform some normalization after the arithmetic.
    $endgroup$
    – amd
    Nov 26 '18 at 3:08
















$begingroup$
Is this in two dimensions or three?
$endgroup$
– amd
Nov 26 '18 at 2:58




$begingroup$
Is this in two dimensions or three?
$endgroup$
– amd
Nov 26 '18 at 2:58












$begingroup$
Two dimensions, I will edit my question.
$endgroup$
– IchVerloren
Nov 26 '18 at 3:05




$begingroup$
Two dimensions, I will edit my question.
$endgroup$
– IchVerloren
Nov 26 '18 at 3:05












$begingroup$
If you want the angle to lie in a certain range, you’ll almost certainly have to perform some normalization after the arithmetic.
$endgroup$
– amd
Nov 26 '18 at 3:08




$begingroup$
If you want the angle to lie in a certain range, you’ll almost certainly have to perform some normalization after the arithmetic.
$endgroup$
– amd
Nov 26 '18 at 3:08










1 Answer
1






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0












$begingroup$

Do this using coordinates:



$x = ||v||sin(theta)$



$y = ||v||cos(theta)$



Then your angle with the x Axis is $phi = atan2(y, x)$ where atan2 is the arc tangent corrected for the quadrant by considering the sign of x and y.






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









    0












    $begingroup$

    Do this using coordinates:



    $x = ||v||sin(theta)$



    $y = ||v||cos(theta)$



    Then your angle with the x Axis is $phi = atan2(y, x)$ where atan2 is the arc tangent corrected for the quadrant by considering the sign of x and y.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Do this using coordinates:



      $x = ||v||sin(theta)$



      $y = ||v||cos(theta)$



      Then your angle with the x Axis is $phi = atan2(y, x)$ where atan2 is the arc tangent corrected for the quadrant by considering the sign of x and y.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Do this using coordinates:



        $x = ||v||sin(theta)$



        $y = ||v||cos(theta)$



        Then your angle with the x Axis is $phi = atan2(y, x)$ where atan2 is the arc tangent corrected for the quadrant by considering the sign of x and y.






        share|cite|improve this answer









        $endgroup$



        Do this using coordinates:



        $x = ||v||sin(theta)$



        $y = ||v||cos(theta)$



        Then your angle with the x Axis is $phi = atan2(y, x)$ where atan2 is the arc tangent corrected for the quadrant by considering the sign of x and y.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 26 '18 at 3:39









        Michael StachowskyMichael Stachowsky

        1,250417




        1,250417






























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