Atypical way to find angle between unit vectors: $theta = 2 sin^-1left(frac12left|hatA-hatBright|right)$

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At my work, I have come across code with the following way of calculating the angle between two vectors.




$$theta = 2 sin^-1left(frac12left|hatA-hatB right|right)$$




(Note the physics convention: $hatv$ indicates the normalization of $v$; ie, $hatv:=v/|v|$).



I've spent some time, but I can't think of how this was derived using typical methodologies (law of sines, law of cosines, dot product, cross product). It is pretty different. So my questions is,




How could this have been derived?











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  • $begingroup$
    @DavidK Thank you. I have addressed that.
    $endgroup$
    – WolcottR
    Jan 22 at 15:46















6












$begingroup$


At my work, I have come across code with the following way of calculating the angle between two vectors.




$$theta = 2 sin^-1left(frac12left|hatA-hatB right|right)$$




(Note the physics convention: $hatv$ indicates the normalization of $v$; ie, $hatv:=v/|v|$).



I've spent some time, but I can't think of how this was derived using typical methodologies (law of sines, law of cosines, dot product, cross product). It is pretty different. So my questions is,




How could this have been derived?











share|cite|improve this question











$endgroup$











  • $begingroup$
    @DavidK Thank you. I have addressed that.
    $endgroup$
    – WolcottR
    Jan 22 at 15:46













6












6








6





$begingroup$


At my work, I have come across code with the following way of calculating the angle between two vectors.




$$theta = 2 sin^-1left(frac12left|hatA-hatB right|right)$$




(Note the physics convention: $hatv$ indicates the normalization of $v$; ie, $hatv:=v/|v|$).



I've spent some time, but I can't think of how this was derived using typical methodologies (law of sines, law of cosines, dot product, cross product). It is pretty different. So my questions is,




How could this have been derived?











share|cite|improve this question











$endgroup$




At my work, I have come across code with the following way of calculating the angle between two vectors.




$$theta = 2 sin^-1left(frac12left|hatA-hatB right|right)$$




(Note the physics convention: $hatv$ indicates the normalization of $v$; ie, $hatv:=v/|v|$).



I've spent some time, but I can't think of how this was derived using typical methodologies (law of sines, law of cosines, dot product, cross product). It is pretty different. So my questions is,




How could this have been derived?








linear-algebra trigonometry vectors






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share|cite|improve this question













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edited Jan 22 at 16:42









Blue

48.4k870154




48.4k870154










asked Jan 21 at 21:49









WolcottRWolcottR

455




455











  • $begingroup$
    @DavidK Thank you. I have addressed that.
    $endgroup$
    – WolcottR
    Jan 22 at 15:46
















  • $begingroup$
    @DavidK Thank you. I have addressed that.
    $endgroup$
    – WolcottR
    Jan 22 at 15:46















$begingroup$
@DavidK Thank you. I have addressed that.
$endgroup$
– WolcottR
Jan 22 at 15:46




$begingroup$
@DavidK Thank you. I have addressed that.
$endgroup$
– WolcottR
Jan 22 at 15:46










1 Answer
1






active

oldest

votes


















13












$begingroup$

This works when $A$ and $B$ are unit vectors.



You have
$$
|A-B|^2=(A-B)cdot(A-B)=|A|^2+|B|^2-2,Acdot B=2-2costheta=4sin^2tfractheta2.
$$

Solving,
$$
theta=2arcsintfrac2.
$$






share|cite|improve this answer









$endgroup$








  • 1




    $begingroup$
    A general comment: note the hats on the vectors, which often indicates normalization.
    $endgroup$
    – Will R
    Jan 21 at 21:56







  • 1




    $begingroup$
    Good to know! I haven't seen that in the last 30 years of doing math.
    $endgroup$
    – Martin Argerami
    Jan 21 at 21:57






  • 1




    $begingroup$
    It's fairly common in physics, I think. Some people denote the standard unit vectors in $mathbbR^3$ with $i,j,k,$ and some of those people put hats on them, to emphasize that they are unit vectors.
    $endgroup$
    – Will R
    Jan 21 at 22:02






  • 3




    $begingroup$
    I have been using the hats on $i,j,k$ for many years, but I never saw the notation being used for other vectors.
    $endgroup$
    – Martin Argerami
    Jan 21 at 22:16






  • 2




    $begingroup$
    @WillR It's indeed common in physics. We often go one step further too, and write stuff like $vecv=vhatv$. Makes for fairly compact notation.
    $endgroup$
    – Anyon
    Jan 22 at 2:47










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









13












$begingroup$

This works when $A$ and $B$ are unit vectors.



You have
$$
|A-B|^2=(A-B)cdot(A-B)=|A|^2+|B|^2-2,Acdot B=2-2costheta=4sin^2tfractheta2.
$$

Solving,
$$
theta=2arcsintfrac2.
$$






share|cite|improve this answer









$endgroup$








  • 1




    $begingroup$
    A general comment: note the hats on the vectors, which often indicates normalization.
    $endgroup$
    – Will R
    Jan 21 at 21:56







  • 1




    $begingroup$
    Good to know! I haven't seen that in the last 30 years of doing math.
    $endgroup$
    – Martin Argerami
    Jan 21 at 21:57






  • 1




    $begingroup$
    It's fairly common in physics, I think. Some people denote the standard unit vectors in $mathbbR^3$ with $i,j,k,$ and some of those people put hats on them, to emphasize that they are unit vectors.
    $endgroup$
    – Will R
    Jan 21 at 22:02






  • 3




    $begingroup$
    I have been using the hats on $i,j,k$ for many years, but I never saw the notation being used for other vectors.
    $endgroup$
    – Martin Argerami
    Jan 21 at 22:16






  • 2




    $begingroup$
    @WillR It's indeed common in physics. We often go one step further too, and write stuff like $vecv=vhatv$. Makes for fairly compact notation.
    $endgroup$
    – Anyon
    Jan 22 at 2:47















13












$begingroup$

This works when $A$ and $B$ are unit vectors.



You have
$$
|A-B|^2=(A-B)cdot(A-B)=|A|^2+|B|^2-2,Acdot B=2-2costheta=4sin^2tfractheta2.
$$

Solving,
$$
theta=2arcsintfrac2.
$$






share|cite|improve this answer









$endgroup$








  • 1




    $begingroup$
    A general comment: note the hats on the vectors, which often indicates normalization.
    $endgroup$
    – Will R
    Jan 21 at 21:56







  • 1




    $begingroup$
    Good to know! I haven't seen that in the last 30 years of doing math.
    $endgroup$
    – Martin Argerami
    Jan 21 at 21:57






  • 1




    $begingroup$
    It's fairly common in physics, I think. Some people denote the standard unit vectors in $mathbbR^3$ with $i,j,k,$ and some of those people put hats on them, to emphasize that they are unit vectors.
    $endgroup$
    – Will R
    Jan 21 at 22:02






  • 3




    $begingroup$
    I have been using the hats on $i,j,k$ for many years, but I never saw the notation being used for other vectors.
    $endgroup$
    – Martin Argerami
    Jan 21 at 22:16






  • 2




    $begingroup$
    @WillR It's indeed common in physics. We often go one step further too, and write stuff like $vecv=vhatv$. Makes for fairly compact notation.
    $endgroup$
    – Anyon
    Jan 22 at 2:47













13












13








13





$begingroup$

This works when $A$ and $B$ are unit vectors.



You have
$$
|A-B|^2=(A-B)cdot(A-B)=|A|^2+|B|^2-2,Acdot B=2-2costheta=4sin^2tfractheta2.
$$

Solving,
$$
theta=2arcsintfrac2.
$$






share|cite|improve this answer









$endgroup$



This works when $A$ and $B$ are unit vectors.



You have
$$
|A-B|^2=(A-B)cdot(A-B)=|A|^2+|B|^2-2,Acdot B=2-2costheta=4sin^2tfractheta2.
$$

Solving,
$$
theta=2arcsintfrac2.
$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 21 at 21:54









Martin ArgeramiMartin Argerami

127k1182181




127k1182181







  • 1




    $begingroup$
    A general comment: note the hats on the vectors, which often indicates normalization.
    $endgroup$
    – Will R
    Jan 21 at 21:56







  • 1




    $begingroup$
    Good to know! I haven't seen that in the last 30 years of doing math.
    $endgroup$
    – Martin Argerami
    Jan 21 at 21:57






  • 1




    $begingroup$
    It's fairly common in physics, I think. Some people denote the standard unit vectors in $mathbbR^3$ with $i,j,k,$ and some of those people put hats on them, to emphasize that they are unit vectors.
    $endgroup$
    – Will R
    Jan 21 at 22:02






  • 3




    $begingroup$
    I have been using the hats on $i,j,k$ for many years, but I never saw the notation being used for other vectors.
    $endgroup$
    – Martin Argerami
    Jan 21 at 22:16






  • 2




    $begingroup$
    @WillR It's indeed common in physics. We often go one step further too, and write stuff like $vecv=vhatv$. Makes for fairly compact notation.
    $endgroup$
    – Anyon
    Jan 22 at 2:47












  • 1




    $begingroup$
    A general comment: note the hats on the vectors, which often indicates normalization.
    $endgroup$
    – Will R
    Jan 21 at 21:56







  • 1




    $begingroup$
    Good to know! I haven't seen that in the last 30 years of doing math.
    $endgroup$
    – Martin Argerami
    Jan 21 at 21:57






  • 1




    $begingroup$
    It's fairly common in physics, I think. Some people denote the standard unit vectors in $mathbbR^3$ with $i,j,k,$ and some of those people put hats on them, to emphasize that they are unit vectors.
    $endgroup$
    – Will R
    Jan 21 at 22:02






  • 3




    $begingroup$
    I have been using the hats on $i,j,k$ for many years, but I never saw the notation being used for other vectors.
    $endgroup$
    – Martin Argerami
    Jan 21 at 22:16






  • 2




    $begingroup$
    @WillR It's indeed common in physics. We often go one step further too, and write stuff like $vecv=vhatv$. Makes for fairly compact notation.
    $endgroup$
    – Anyon
    Jan 22 at 2:47







1




1




$begingroup$
A general comment: note the hats on the vectors, which often indicates normalization.
$endgroup$
– Will R
Jan 21 at 21:56





$begingroup$
A general comment: note the hats on the vectors, which often indicates normalization.
$endgroup$
– Will R
Jan 21 at 21:56





1




1




$begingroup$
Good to know! I haven't seen that in the last 30 years of doing math.
$endgroup$
– Martin Argerami
Jan 21 at 21:57




$begingroup$
Good to know! I haven't seen that in the last 30 years of doing math.
$endgroup$
– Martin Argerami
Jan 21 at 21:57




1




1




$begingroup$
It's fairly common in physics, I think. Some people denote the standard unit vectors in $mathbbR^3$ with $i,j,k,$ and some of those people put hats on them, to emphasize that they are unit vectors.
$endgroup$
– Will R
Jan 21 at 22:02




$begingroup$
It's fairly common in physics, I think. Some people denote the standard unit vectors in $mathbbR^3$ with $i,j,k,$ and some of those people put hats on them, to emphasize that they are unit vectors.
$endgroup$
– Will R
Jan 21 at 22:02




3




3




$begingroup$
I have been using the hats on $i,j,k$ for many years, but I never saw the notation being used for other vectors.
$endgroup$
– Martin Argerami
Jan 21 at 22:16




$begingroup$
I have been using the hats on $i,j,k$ for many years, but I never saw the notation being used for other vectors.
$endgroup$
– Martin Argerami
Jan 21 at 22:16




2




2




$begingroup$
@WillR It's indeed common in physics. We often go one step further too, and write stuff like $vecv=vhatv$. Makes for fairly compact notation.
$endgroup$
– Anyon
Jan 22 at 2:47




$begingroup$
@WillR It's indeed common in physics. We often go one step further too, and write stuff like $vecv=vhatv$. Makes for fairly compact notation.
$endgroup$
– Anyon
Jan 22 at 2:47

















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