Finding sign of an angle without calculating the angle itself












0














Say we have an angle between vectors $P_1$ and $P_2$ in $mathbb R^2$ whose vertex, $O$, is at the origin. I would like to know the sign of the smallest signed angle between $P_1$ and $P_2$ such that $overrightarrow{O{P_1}}$ is rotated to become overlaid on $overrightarrow{O{P_2}}$. It is too computationally expensive for me to use trigonometric functions to calculate the angle itself and then find the sign. Is there a good way to do this?






linear-algebra geometry rotations angle







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  • Do you mean find the sine or the sign?
    – Seth
    Nov 25 '18 at 23:04










  • The sign as in positive or negative
    – rbjacob
    Nov 25 '18 at 23:06
















0














Say we have an angle between vectors $P_1$ and $P_2$ in $mathbb R^2$ whose vertex, $O$, is at the origin. I would like to know the sign of the smallest signed angle between $P_1$ and $P_2$ such that $overrightarrow{O{P_1}}$ is rotated to become overlaid on $overrightarrow{O{P_2}}$. It is too computationally expensive for me to use trigonometric functions to calculate the angle itself and then find the sign. Is there a good way to do this?






linear-algebra geometry rotations angle







share|cite|improve this question
























  • Do you mean find the sine or the sign?
    – Seth
    Nov 25 '18 at 23:04










  • The sign as in positive or negative
    – rbjacob
    Nov 25 '18 at 23:06














0












0








0







Say we have an angle between vectors $P_1$ and $P_2$ in $mathbb R^2$ whose vertex, $O$, is at the origin. I would like to know the sign of the smallest signed angle between $P_1$ and $P_2$ such that $overrightarrow{O{P_1}}$ is rotated to become overlaid on $overrightarrow{O{P_2}}$. It is too computationally expensive for me to use trigonometric functions to calculate the angle itself and then find the sign. Is there a good way to do this?






linear-algebra geometry rotations angle







share|cite|improve this question















Say we have an angle between vectors $P_1$ and $P_2$ in $mathbb R^2$ whose vertex, $O$, is at the origin. I would like to know the sign of the smallest signed angle between $P_1$ and $P_2$ such that $overrightarrow{O{P_1}}$ is rotated to become overlaid on $overrightarrow{O{P_2}}$. It is too computationally expensive for me to use trigonometric functions to calculate the angle itself and then find the sign. Is there a good way to do this?






linear-algebra geometry rotations angle




linear-algebra geometry rotations angle






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share|cite|improve this question













share|cite|improve this question




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edited Nov 26 '18 at 1:21









KReiser

9,31211435




9,31211435










asked Nov 25 '18 at 22:54









rbjacob

223




223












  • Do you mean find the sine or the sign?
    – Seth
    Nov 25 '18 at 23:04










  • The sign as in positive or negative
    – rbjacob
    Nov 25 '18 at 23:06


















  • Do you mean find the sine or the sign?
    – Seth
    Nov 25 '18 at 23:04










  • The sign as in positive or negative
    – rbjacob
    Nov 25 '18 at 23:06
















Do you mean find the sine or the sign?
– Seth
Nov 25 '18 at 23:04




Do you mean find the sine or the sign?
– Seth
Nov 25 '18 at 23:04












The sign as in positive or negative
– rbjacob
Nov 25 '18 at 23:06




The sign as in positive or negative
– rbjacob
Nov 25 '18 at 23:06










1 Answer
1






active

oldest

votes


















2














Check the sign of the coefficient of $hat{k}$ of the cross product of both vectors.



For example, let $P_1=(5,0)$, $P_2=(-3,6)$ , $overrightarrow{OP_1}times overrightarrow{OP_2}=30hat{k}$, so it's a counterclockwise smallest turn. Switching both vectors will result in a negative coefficient for $hat{k}$, indicating a clockwise smallest turn.



Basically for $P_1(a,b), P_2(c,d)$, and turning from $overrightarrow{OP_1}$ to $overrightarrow{OP_2}$, you test sign of $ad-bc$ for the direction of smallest turn (positive for counterclockwise). In the case the result is $0$, either way is the same.






share|cite|improve this answer























  • Thanks, perfect
    – rbjacob
    Nov 25 '18 at 23:19






  • 1




    In other words, examine $det[P_1,P_2]$.
    – amd
    Nov 25 '18 at 23:56










  • @amd Thanks for the insight!
    – Lance
    Nov 26 '18 at 0:38











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














Check the sign of the coefficient of $hat{k}$ of the cross product of both vectors.



For example, let $P_1=(5,0)$, $P_2=(-3,6)$ , $overrightarrow{OP_1}times overrightarrow{OP_2}=30hat{k}$, so it's a counterclockwise smallest turn. Switching both vectors will result in a negative coefficient for $hat{k}$, indicating a clockwise smallest turn.



Basically for $P_1(a,b), P_2(c,d)$, and turning from $overrightarrow{OP_1}$ to $overrightarrow{OP_2}$, you test sign of $ad-bc$ for the direction of smallest turn (positive for counterclockwise). In the case the result is $0$, either way is the same.






share|cite|improve this answer























  • Thanks, perfect
    – rbjacob
    Nov 25 '18 at 23:19






  • 1




    In other words, examine $det[P_1,P_2]$.
    – amd
    Nov 25 '18 at 23:56










  • @amd Thanks for the insight!
    – Lance
    Nov 26 '18 at 0:38
















2














Check the sign of the coefficient of $hat{k}$ of the cross product of both vectors.



For example, let $P_1=(5,0)$, $P_2=(-3,6)$ , $overrightarrow{OP_1}times overrightarrow{OP_2}=30hat{k}$, so it's a counterclockwise smallest turn. Switching both vectors will result in a negative coefficient for $hat{k}$, indicating a clockwise smallest turn.



Basically for $P_1(a,b), P_2(c,d)$, and turning from $overrightarrow{OP_1}$ to $overrightarrow{OP_2}$, you test sign of $ad-bc$ for the direction of smallest turn (positive for counterclockwise). In the case the result is $0$, either way is the same.






share|cite|improve this answer























  • Thanks, perfect
    – rbjacob
    Nov 25 '18 at 23:19






  • 1




    In other words, examine $det[P_1,P_2]$.
    – amd
    Nov 25 '18 at 23:56










  • @amd Thanks for the insight!
    – Lance
    Nov 26 '18 at 0:38














2












2








2






Check the sign of the coefficient of $hat{k}$ of the cross product of both vectors.



For example, let $P_1=(5,0)$, $P_2=(-3,6)$ , $overrightarrow{OP_1}times overrightarrow{OP_2}=30hat{k}$, so it's a counterclockwise smallest turn. Switching both vectors will result in a negative coefficient for $hat{k}$, indicating a clockwise smallest turn.



Basically for $P_1(a,b), P_2(c,d)$, and turning from $overrightarrow{OP_1}$ to $overrightarrow{OP_2}$, you test sign of $ad-bc$ for the direction of smallest turn (positive for counterclockwise). In the case the result is $0$, either way is the same.






share|cite|improve this answer














Check the sign of the coefficient of $hat{k}$ of the cross product of both vectors.



For example, let $P_1=(5,0)$, $P_2=(-3,6)$ , $overrightarrow{OP_1}times overrightarrow{OP_2}=30hat{k}$, so it's a counterclockwise smallest turn. Switching both vectors will result in a negative coefficient for $hat{k}$, indicating a clockwise smallest turn.



Basically for $P_1(a,b), P_2(c,d)$, and turning from $overrightarrow{OP_1}$ to $overrightarrow{OP_2}$, you test sign of $ad-bc$ for the direction of smallest turn (positive for counterclockwise). In the case the result is $0$, either way is the same.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 25 '18 at 23:16

























answered Nov 25 '18 at 23:11









Lance

61229




61229












  • Thanks, perfect
    – rbjacob
    Nov 25 '18 at 23:19






  • 1




    In other words, examine $det[P_1,P_2]$.
    – amd
    Nov 25 '18 at 23:56










  • @amd Thanks for the insight!
    – Lance
    Nov 26 '18 at 0:38


















  • Thanks, perfect
    – rbjacob
    Nov 25 '18 at 23:19






  • 1




    In other words, examine $det[P_1,P_2]$.
    – amd
    Nov 25 '18 at 23:56










  • @amd Thanks for the insight!
    – Lance
    Nov 26 '18 at 0:38
















Thanks, perfect
– rbjacob
Nov 25 '18 at 23:19




Thanks, perfect
– rbjacob
Nov 25 '18 at 23:19




1




1




In other words, examine $det[P_1,P_2]$.
– amd
Nov 25 '18 at 23:56




In other words, examine $det[P_1,P_2]$.
– amd
Nov 25 '18 at 23:56












@amd Thanks for the insight!
– Lance
Nov 26 '18 at 0:38




@amd Thanks for the insight!
– Lance
Nov 26 '18 at 0:38


















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