Compare a constant with the angle between 2 vectors without using Trigonometric functions
I have a constant X that is the value of an angle. I want to check if the angle formed by 2 given vectors is less than the value X. I know it's possible to calculate the angle using the arccos of the dot product of the vectors divided by the product of the magnitude of the vectors, however I want to do this without applying trigonometric functions to the vectors. Using trigonometric functions on the constant X is fine.
My thoughts were that the dot product of the vectors divided by the product of the magnitude of the vectors gives me the cosine of that angle so if I do the cosine of the angle X maybe I can compare these 2 cosines. Is this viable?
Out of curiosity, is this possible without using trigonometric functions at all?
trigonometry vector-spaces
asked Nov 30 at 19:17
Dozed12
255
add a comment |
I have a constant X that is the value of an angle. I want to check if the angle formed by 2 given vectors is less than the value X. I know it's possible to calculate the angle using the arccos of the dot product of the vectors divided by the product of the magnitude of the vectors, however I want to do this without applying trigonometric functions to the vectors. Using trigonometric functions on the constant X is fine.
My thoughts were that the dot product of the vectors divided by the product of the magnitude of the vectors gives me the cosine of that angle so if I do the cosine of the angle X maybe I can compare these 2 cosines. Is this viable?
Out of curiosity, is this possible without using trigonometric functions at all?
trigonometry vector-spaces
asked Nov 30 at 19:17
Dozed12
255
Certainly, but how is this not using trigonometric functions? You’re still computing the cosine of the angle that you’re testing against.
– amd
Nov 30 at 20:11
"...without applying trigonometric functions to the vectors. Using them on the constant X is fine." Sorry, maybe I should have made it clear. I'm fine with using trigonometric functions on the angle X.
– Dozed12
Nov 30 at 20:41
add a comment |
I have a constant X that is the value of an angle. I want to check if the angle formed by 2 given vectors is less than the value X. I know it's possible to calculate the angle using the arccos of the dot product of the vectors divided by the product of the magnitude of the vectors, however I want to do this without applying trigonometric functions to the vectors. Using trigonometric functions on the constant X is fine.
My thoughts were that the dot product of the vectors divided by the product of the magnitude of the vectors gives me the cosine of that angle so if I do the cosine of the angle X maybe I can compare these 2 cosines. Is this viable?
Out of curiosity, is this possible without using trigonometric functions at all?
trigonometry vector-spaces
asked Nov 30 at 19:17
Dozed12
255
I have a constant X that is the value of an angle. I want to check if the angle formed by 2 given vectors is less than the value X. I know it's possible to calculate the angle using the arccos of the dot product of the vectors divided by the product of the magnitude of the vectors, however I want to do this without applying trigonometric functions to the vectors. Using trigonometric functions on the constant X is fine.
My thoughts were that the dot product of the vectors divided by the product of the magnitude of the vectors gives me the cosine of that angle so if I do the cosine of the angle X maybe I can compare these 2 cosines. Is this viable?
Out of curiosity, is this possible without using trigonometric functions at all?
trigonometry vector-spaces
trigonometry vector-spaces
asked Nov 30 at 19:17
Dozed12
255
asked Nov 30 at 19:17
Dozed12
255
edited Nov 30 at 20:41
asked Nov 30 at 19:17
Dozed12
255
asked Nov 30 at 19:17
Dozed12
255
asked Nov 30 at 19:17
Dozed12
255
255
Certainly, but how is this not using trigonometric functions? You’re still computing the cosine of the angle that you’re testing against.
– amd
Nov 30 at 20:11
"...without applying trigonometric functions to the vectors. Using them on the constant X is fine." Sorry, maybe I should have made it clear. I'm fine with using trigonometric functions on the angle X.
– Dozed12
Nov 30 at 20:41
add a comment |
Certainly, but how is this not using trigonometric functions? You’re still computing the cosine of the angle that you’re testing against.
– amd
Nov 30 at 20:11
"...without applying trigonometric functions to the vectors. Using them on the constant X is fine." Sorry, maybe I should have made it clear. I'm fine with using trigonometric functions on the angle X.
– Dozed12
Nov 30 at 20:41
Certainly, but how is this not using trigonometric functions? You’re still computing the cosine of the angle that you’re testing against.
– amd
Nov 30 at 20:11
Certainly, but how is this not using trigonometric functions? You’re still computing the cosine of the angle that you’re testing against.
– amd
Nov 30 at 20:11
"...without applying trigonometric functions to the vectors. Using them on the constant X is fine." Sorry, maybe I should have made it clear. I'm fine with using trigonometric functions on the angle X.
– Dozed12
Nov 30 at 20:41
"...without applying trigonometric functions to the vectors. Using them on the constant X is fine." Sorry, maybe I should have made it clear. I'm fine with using trigonometric functions on the angle X.
– Dozed12
Nov 30 at 20:41
add a comment |
1 Answer
1
active
oldest
votes
Solution involving complex numbers
Assume the given vectors $vec{u},vec{v};$ are represented by the numbers $z_
{u}=a+ib,;z_{v}=c+id.;$
Then the oriented angle $theta$ from $vec{u}$ to $vec{v}$ is $$arg frac {c+id}{a+ib}.$$ Compute $$frac {c+id}{a+ib}cdot frac{a-ib}{a-ib}=frac{ac+bd}{a^2+b^2}+ifrac{ad-bc}{a^2+b^2}$$
*Assume $vec{u}not perpvec{v}.;$ From the signs of the real and imaginary parts, one can locate $theta$ into $left(-frac pi2, frac pi2right); text{or}; left(frac pi2, frac {3pi}{2}right).$
Further, $$frac{text{imaginary part}}{text{real part}}=frac{ad-bc}{ac+bd}=tantheta.$$
To finish, it suffices to locate the given angle $X$ into one of the above intervals and use the monotony of $;tan;$ on this interval.
Note *
Denote $mathcal{R}=left{frac{(2k-1)pi}{2}, k =0,1 right}.;$ If $thetain mathcal{R}; text{or}; X in mathcal{R},;$ we have a trivial case easy to solve.
answered Nov 30 at 22:09
user376343
2,7982822
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%2f3020524%2fcompare-a-constant-with-the-angle-between-2-vectors-without-using-trigonometric%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
Solution involving complex numbers
Assume the given vectors $vec{u},vec{v};$ are represented by the numbers $z_
{u}=a+ib,;z_{v}=c+id.;$
Then the oriented angle $theta$ from $vec{u}$ to $vec{v}$ is $$arg frac {c+id}{a+ib}.$$ Compute $$frac {c+id}{a+ib}cdot frac{a-ib}{a-ib}=frac{ac+bd}{a^2+b^2}+ifrac{ad-bc}{a^2+b^2}$$
*Assume $vec{u}not perpvec{v}.;$ From the signs of the real and imaginary parts, one can locate $theta$ into $left(-frac pi2, frac pi2right); text{or}; left(frac pi2, frac {3pi}{2}right).$
Further, $$frac{text{imaginary part}}{text{real part}}=frac{ad-bc}{ac+bd}=tantheta.$$
To finish, it suffices to locate the given angle $X$ into one of the above intervals and use the monotony of $;tan;$ on this interval.
Note *
Denote $mathcal{R}=left{frac{(2k-1)pi}{2}, k =0,1 right}.;$ If $thetain mathcal{R}; text{or}; X in mathcal{R},;$ we have a trivial case easy to solve.
answered Nov 30 at 22:09
user376343
2,7982822
add a comment |
Solution involving complex numbers
Assume the given vectors $vec{u},vec{v};$ are represented by the numbers $z_
{u}=a+ib,;z_{v}=c+id.;$
Then the oriented angle $theta$ from $vec{u}$ to $vec{v}$ is $$arg frac {c+id}{a+ib}.$$ Compute $$frac {c+id}{a+ib}cdot frac{a-ib}{a-ib}=frac{ac+bd}{a^2+b^2}+ifrac{ad-bc}{a^2+b^2}$$
*Assume $vec{u}not perpvec{v}.;$ From the signs of the real and imaginary parts, one can locate $theta$ into $left(-frac pi2, frac pi2right); text{or}; left(frac pi2, frac {3pi}{2}right).$
Further, $$frac{text{imaginary part}}{text{real part}}=frac{ad-bc}{ac+bd}=tantheta.$$
To finish, it suffices to locate the given angle $X$ into one of the above intervals and use the monotony of $;tan;$ on this interval.
Note *
Denote $mathcal{R}=left{frac{(2k-1)pi}{2}, k =0,1 right}.;$ If $thetain mathcal{R}; text{or}; X in mathcal{R},;$ we have a trivial case easy to solve.
answered Nov 30 at 22:09
user376343
2,7982822
add a comment |
Solution involving complex numbers
Assume the given vectors $vec{u},vec{v};$ are represented by the numbers $z_
{u}=a+ib,;z_{v}=c+id.;$
Then the oriented angle $theta$ from $vec{u}$ to $vec{v}$ is $$arg frac {c+id}{a+ib}.$$ Compute $$frac {c+id}{a+ib}cdot frac{a-ib}{a-ib}=frac{ac+bd}{a^2+b^2}+ifrac{ad-bc}{a^2+b^2}$$
*Assume $vec{u}not perpvec{v}.;$ From the signs of the real and imaginary parts, one can locate $theta$ into $left(-frac pi2, frac pi2right); text{or}; left(frac pi2, frac {3pi}{2}right).$
Further, $$frac{text{imaginary part}}{text{real part}}=frac{ad-bc}{ac+bd}=tantheta.$$
To finish, it suffices to locate the given angle $X$ into one of the above intervals and use the monotony of $;tan;$ on this interval.
Note *
Denote $mathcal{R}=left{frac{(2k-1)pi}{2}, k =0,1 right}.;$ If $thetain mathcal{R}; text{or}; X in mathcal{R},;$ we have a trivial case easy to solve.
answered Nov 30 at 22:09
user376343
2,7982822
Solution involving complex numbers
Assume the given vectors $vec{u},vec{v};$ are represented by the numbers $z_
{u}=a+ib,;z_{v}=c+id.;$
Then the oriented angle $theta$ from $vec{u}$ to $vec{v}$ is $$arg frac {c+id}{a+ib}.$$ Compute $$frac {c+id}{a+ib}cdot frac{a-ib}{a-ib}=frac{ac+bd}{a^2+b^2}+ifrac{ad-bc}{a^2+b^2}$$
*Assume $vec{u}not perpvec{v}.;$ From the signs of the real and imaginary parts, one can locate $theta$ into $left(-frac pi2, frac pi2right); text{or}; left(frac pi2, frac {3pi}{2}right).$
Further, $$frac{text{imaginary part}}{text{real part}}=frac{ad-bc}{ac+bd}=tantheta.$$
To finish, it suffices to locate the given angle $X$ into one of the above intervals and use the monotony of $;tan;$ on this interval.
Note *
Denote $mathcal{R}=left{frac{(2k-1)pi}{2}, k =0,1 right}.;$ If $thetain mathcal{R}; text{or}; X in mathcal{R},;$ we have a trivial case easy to solve.
answered Nov 30 at 22:09
user376343
2,7982822
answered Nov 30 at 22:09
user376343
2,7982822
answered Nov 30 at 22:09
user376343
2,7982822
answered Nov 30 at 22:09
user376343
2,7982822
2,7982822
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3020524%2fcompare-a-constant-with-the-angle-between-2-vectors-without-using-trigonometric%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
Certainly, but how is this not using trigonometric functions? You’re still computing the cosine of the angle that you’re testing against.
– amd
Nov 30 at 20:11
"...without applying trigonometric functions to the vectors. Using them on the constant X is fine." Sorry, maybe I should have made it clear. I'm fine with using trigonometric functions on the angle X.
– Dozed12
Nov 30 at 20:41