Angle between vector and axis

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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 $[0,2pi]$.

edited Nov 26 at 3:06

asked Nov 26 at 2:57

IchVerloren

849

Is this in two dimensions or three?
– amd
Nov 26 at 2:58

Two dimensions, I will edit my question.
– IchVerloren
Nov 26 at 3:05

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

add a comment |

up vote
0
down vote

favorite

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 $[0,2pi]$.

edited Nov 26 at 3:06

asked Nov 26 at 2:57

IchVerloren

849

Is this in two dimensions or three?
– amd
Nov 26 at 2:58

Two dimensions, I will edit my question.
– IchVerloren
Nov 26 at 3:05

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

add a comment |

up vote
0
down vote

favorite

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 $[0,2pi]$.

edited Nov 26 at 3:06

asked Nov 26 at 2:57

IchVerloren

849

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

edited Nov 26 at 3:06

asked Nov 26 at 2:57

IchVerloren

849

edited Nov 26 at 3:06

asked Nov 26 at 2:57

IchVerloren

849

edited Nov 26 at 3:06

asked Nov 26 at 2:57

IchVerloren

849

asked Nov 26 at 2:57

IchVerloren

849

asked Nov 26 at 2:57

IchVerloren

849

Is this in two dimensions or three?
– amd
Nov 26 at 2:58

Two dimensions, I will edit my question.
– IchVerloren
Nov 26 at 3:05

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

add a comment |

Is this in two dimensions or three?
– amd
Nov 26 at 2:58

Two dimensions, I will edit my question.
– IchVerloren
Nov 26 at 3:05

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

Is this in two dimensions or three?
– amd
Nov 26 at 2:58

Two dimensions, I will edit my question.
– IchVerloren
Nov 26 at 3:05

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

add a comment |

1 Answer
1

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up vote
0
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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.

answered Nov 26 at 3:39

Michael Stachowsky

1,260417

add a comment |

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

up vote
0
down vote

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.

answered Nov 26 at 3:39

Michael Stachowsky

1,260417

add a comment |

up vote
0
down vote

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.

answered Nov 26 at 3:39

Michael Stachowsky

1,260417

add a comment |

up vote
0
down vote

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.

answered Nov 26 at 3:39

Michael Stachowsky

1,260417

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.

answered Nov 26 at 3:39

Michael Stachowsky

1,260417

answered Nov 26 at 3:39

Michael Stachowsky

1,260417

answered Nov 26 at 3:39

Michael Stachowsky

1,260417

answered Nov 26 at 3:39

Michael Stachowsky

1,260417

add a comment |

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