Atypical way to find angle between vectors












1












$begingroup$


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



$theta = 2 sin^{-1}(|hat{A}-hat{B} | frac{1}{2})$



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







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I would write $2 sin^{-1}(frac12 | hat{A} - hat{B} |).$ The way you wrote it, it almost looks like you wanted to write $2 sin^{-1}(| hat{A} - hat{B} |^{1/2})$ but made a mistake in the typesetting.
    $endgroup$
    – David K
    1 hour ago
















1












$begingroup$


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



$theta = 2 sin^{-1}(|hat{A}-hat{B} | frac{1}{2})$



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







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I would write $2 sin^{-1}(frac12 | hat{A} - hat{B} |).$ The way you wrote it, it almost looks like you wanted to write $2 sin^{-1}(| hat{A} - hat{B} |^{1/2})$ but made a mistake in the typesetting.
    $endgroup$
    – David K
    1 hour ago














1












1








1





$begingroup$


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



$theta = 2 sin^{-1}(|hat{A}-hat{B} | frac{1}{2})$



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







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.



$theta = 2 sin^{-1}(|hat{A}-hat{B} | frac{1}{2})$



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




linear-algebra trigonometry vectors






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 6 hours ago









Martin Argerami

125k1179178




125k1179178










asked 7 hours ago









WolcottRWolcottR

202




202








  • 1




    $begingroup$
    I would write $2 sin^{-1}(frac12 | hat{A} - hat{B} |).$ The way you wrote it, it almost looks like you wanted to write $2 sin^{-1}(| hat{A} - hat{B} |^{1/2})$ but made a mistake in the typesetting.
    $endgroup$
    – David K
    1 hour ago














  • 1




    $begingroup$
    I would write $2 sin^{-1}(frac12 | hat{A} - hat{B} |).$ The way you wrote it, it almost looks like you wanted to write $2 sin^{-1}(| hat{A} - hat{B} |^{1/2})$ but made a mistake in the typesetting.
    $endgroup$
    – David K
    1 hour ago








1




1




$begingroup$
I would write $2 sin^{-1}(frac12 | hat{A} - hat{B} |).$ The way you wrote it, it almost looks like you wanted to write $2 sin^{-1}(| hat{A} - hat{B} |^{1/2})$ but made a mistake in the typesetting.
$endgroup$
– David K
1 hour ago




$begingroup$
I would write $2 sin^{-1}(frac12 | hat{A} - hat{B} |).$ The way you wrote it, it almost looks like you wanted to write $2 sin^{-1}(| hat{A} - hat{B} |^{1/2})$ but made a mistake in the typesetting.
$endgroup$
– David K
1 hour ago










1 Answer
1






active

oldest

votes


















7












$begingroup$

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,
$$
theta=2arcsintfrac{|A-B|}2.
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    A general comment: note the hats on the vectors, which often indicates normalization.
    $endgroup$
    – Will R
    6 hours ago












  • $begingroup$
    Good to know! I haven't seen that in the last 30 years of doing math.
    $endgroup$
    – Martin Argerami
    6 hours ago






  • 1




    $begingroup$
    It's fairly common in physics, I think. Some people denote the standard unit vectors in $mathbb{R}^{3}$ with $i,j,k,$ and some of those people put hats on them, to emphasize that they are unit vectors.
    $endgroup$
    – Will R
    6 hours ago






  • 2




    $begingroup$
    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
    6 hours ago










  • $begingroup$
    Thank you! I was thinking there was going to be a square/root step in there somewhere
    $endgroup$
    – WolcottR
    6 hours ago











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









7












$begingroup$

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,
$$
theta=2arcsintfrac{|A-B|}2.
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    A general comment: note the hats on the vectors, which often indicates normalization.
    $endgroup$
    – Will R
    6 hours ago












  • $begingroup$
    Good to know! I haven't seen that in the last 30 years of doing math.
    $endgroup$
    – Martin Argerami
    6 hours ago






  • 1




    $begingroup$
    It's fairly common in physics, I think. Some people denote the standard unit vectors in $mathbb{R}^{3}$ with $i,j,k,$ and some of those people put hats on them, to emphasize that they are unit vectors.
    $endgroup$
    – Will R
    6 hours ago






  • 2




    $begingroup$
    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
    6 hours ago










  • $begingroup$
    Thank you! I was thinking there was going to be a square/root step in there somewhere
    $endgroup$
    – WolcottR
    6 hours ago
















7












$begingroup$

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,
$$
theta=2arcsintfrac{|A-B|}2.
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    A general comment: note the hats on the vectors, which often indicates normalization.
    $endgroup$
    – Will R
    6 hours ago












  • $begingroup$
    Good to know! I haven't seen that in the last 30 years of doing math.
    $endgroup$
    – Martin Argerami
    6 hours ago






  • 1




    $begingroup$
    It's fairly common in physics, I think. Some people denote the standard unit vectors in $mathbb{R}^{3}$ with $i,j,k,$ and some of those people put hats on them, to emphasize that they are unit vectors.
    $endgroup$
    – Will R
    6 hours ago






  • 2




    $begingroup$
    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
    6 hours ago










  • $begingroup$
    Thank you! I was thinking there was going to be a square/root step in there somewhere
    $endgroup$
    – WolcottR
    6 hours ago














7












7








7





$begingroup$

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,
$$
theta=2arcsintfrac{|A-B|}2.
$$






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,
$$
theta=2arcsintfrac{|A-B|}2.
$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 6 hours ago









Martin ArgeramiMartin Argerami

125k1179178




125k1179178












  • $begingroup$
    A general comment: note the hats on the vectors, which often indicates normalization.
    $endgroup$
    – Will R
    6 hours ago












  • $begingroup$
    Good to know! I haven't seen that in the last 30 years of doing math.
    $endgroup$
    – Martin Argerami
    6 hours ago






  • 1




    $begingroup$
    It's fairly common in physics, I think. Some people denote the standard unit vectors in $mathbb{R}^{3}$ with $i,j,k,$ and some of those people put hats on them, to emphasize that they are unit vectors.
    $endgroup$
    – Will R
    6 hours ago






  • 2




    $begingroup$
    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
    6 hours ago










  • $begingroup$
    Thank you! I was thinking there was going to be a square/root step in there somewhere
    $endgroup$
    – WolcottR
    6 hours ago


















  • $begingroup$
    A general comment: note the hats on the vectors, which often indicates normalization.
    $endgroup$
    – Will R
    6 hours ago












  • $begingroup$
    Good to know! I haven't seen that in the last 30 years of doing math.
    $endgroup$
    – Martin Argerami
    6 hours ago






  • 1




    $begingroup$
    It's fairly common in physics, I think. Some people denote the standard unit vectors in $mathbb{R}^{3}$ with $i,j,k,$ and some of those people put hats on them, to emphasize that they are unit vectors.
    $endgroup$
    – Will R
    6 hours ago






  • 2




    $begingroup$
    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
    6 hours ago










  • $begingroup$
    Thank you! I was thinking there was going to be a square/root step in there somewhere
    $endgroup$
    – WolcottR
    6 hours ago
















$begingroup$
A general comment: note the hats on the vectors, which often indicates normalization.
$endgroup$
– Will R
6 hours ago






$begingroup$
A general comment: note the hats on the vectors, which often indicates normalization.
$endgroup$
– Will R
6 hours ago














$begingroup$
Good to know! I haven't seen that in the last 30 years of doing math.
$endgroup$
– Martin Argerami
6 hours ago




$begingroup$
Good to know! I haven't seen that in the last 30 years of doing math.
$endgroup$
– Martin Argerami
6 hours ago




1




1




$begingroup$
It's fairly common in physics, I think. Some people denote the standard unit vectors in $mathbb{R}^{3}$ with $i,j,k,$ and some of those people put hats on them, to emphasize that they are unit vectors.
$endgroup$
– Will R
6 hours ago




$begingroup$
It's fairly common in physics, I think. Some people denote the standard unit vectors in $mathbb{R}^{3}$ with $i,j,k,$ and some of those people put hats on them, to emphasize that they are unit vectors.
$endgroup$
– Will R
6 hours ago




2




2




$begingroup$
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
6 hours ago




$begingroup$
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
6 hours ago












$begingroup$
Thank you! I was thinking there was going to be a square/root step in there somewhere
$endgroup$
– WolcottR
6 hours ago




$begingroup$
Thank you! I was thinking there was going to be a square/root step in there somewhere
$endgroup$
– WolcottR
6 hours ago


















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