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












    $begingroup$


    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












      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 4 hours ago









      Martin Argerami

      125k1179178




      125k1179178










      asked 4 hours ago









      WolcottRWolcottR

      202




      202






















          1 Answer
          1






          active

          oldest

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          6












          $begingroup$

          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
            4 hours ago












          • $begingroup$
            Good to know! I haven't seen that in the last 30 years of doing math.
            $endgroup$
            – Martin Argerami
            4 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
            4 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
            4 hours ago










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











          Your Answer





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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          6












          $begingroup$

          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
            4 hours ago












          • $begingroup$
            Good to know! I haven't seen that in the last 30 years of doing math.
            $endgroup$
            – Martin Argerami
            4 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
            4 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
            4 hours ago










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
















          6












          $begingroup$

          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
            4 hours ago












          • $begingroup$
            Good to know! I haven't seen that in the last 30 years of doing math.
            $endgroup$
            – Martin Argerami
            4 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
            4 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
            4 hours ago










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














          6












          6








          6





          $begingroup$

          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 4 hours ago









          Martin ArgeramiMartin Argerami

          125k1179178




          125k1179178












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












          • $begingroup$
            Good to know! I haven't seen that in the last 30 years of doing math.
            $endgroup$
            – Martin Argerami
            4 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
            4 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
            4 hours ago










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


















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












          • $begingroup$
            Good to know! I haven't seen that in the last 30 years of doing math.
            $endgroup$
            – Martin Argerami
            4 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
            4 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
            4 hours ago










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
















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






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














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




          $begingroup$
          Good to know! I haven't seen that in the last 30 years of doing math.
          $endgroup$
          – Martin Argerami
          4 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
          4 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
          4 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
          4 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
          4 hours ago












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




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


















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