Angle between vector and axis
$begingroup$
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
asked Nov 26 '18 at 2:57
IchVerlorenIchVerloren
969
$endgroup$
add a comment |
$begingroup$
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
asked Nov 26 '18 at 2:57
IchVerlorenIchVerloren
969
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
asked Nov 26 '18 at 2:57
IchVerlorenIchVerloren
969
$endgroup$
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
asked Nov 26 '18 at 2:57
IchVerlorenIchVerloren
969
asked Nov 26 '18 at 2:57
IchVerlorenIchVerloren
969
edited Nov 26 '18 at 3:06
IchVerloren
asked Nov 26 '18 at 2:57
IchVerlorenIchVerloren
969
asked Nov 26 '18 at 2:57
IchVerlorenIchVerloren
969
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Nov 26 '18 at 3:39
Michael StachowskyMichael Stachowsky
1,250417
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3013757%2fangle-between-vector-and-axis%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered Nov 26 '18 at 3:39
Michael StachowskyMichael Stachowsky
1,250417
$endgroup$
add a comment |
$begingroup$
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.
answered Nov 26 '18 at 3:39
Michael StachowskyMichael Stachowsky
1,250417
$endgroup$
add a comment |
$begingroup$
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.
answered Nov 26 '18 at 3:39
Michael StachowskyMichael Stachowsky
1,250417
$endgroup$
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.
answered Nov 26 '18 at 3:39
Michael StachowskyMichael Stachowsky
1,250417
answered Nov 26 '18 at 3:39
Michael StachowskyMichael Stachowsky
1,250417
answered Nov 26 '18 at 3:39
Michael StachowskyMichael Stachowsky
1,250417
answered Nov 26 '18 at 3:39
Michael StachowskyMichael Stachowsky
1,250417
1,250417
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3013757%2fangle-between-vector-and-axis%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
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