How to compute angle between vectors a and b if |a|=|b|=|a+b|












0














How to compute angle between vectors a and b if |a|=|b|=|a+b|. This is everything I have. I start with formula $cos(alpha)=frac{ab}{|a| |b|}$.



Thank you for your help.






vectors







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    Have you tried drawing a diagram?
    – Mark Bennet
    2 days ago










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0














How to compute angle between vectors a and b if |a|=|b|=|a+b|. This is everything I have. I start with formula $cos(alpha)=frac{ab}{|a| |b|}$.



Thank you for your help.






vectors







share|cite|improve this question


















  • 1




    Have you tried drawing a diagram?
    – Mark Bennet
    2 days ago










  • No, but I will.
    – J.Doe
    2 days ago














0












0








0







How to compute angle between vectors a and b if |a|=|b|=|a+b|. This is everything I have. I start with formula $cos(alpha)=frac{ab}{|a| |b|}$.



Thank you for your help.






vectors







share|cite|improve this question













How to compute angle between vectors a and b if |a|=|b|=|a+b|. This is everything I have. I start with formula $cos(alpha)=frac{ab}{|a| |b|}$.



Thank you for your help.






vectors




vectors






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share|cite|improve this question











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asked 2 days ago









J.DoeJ.Doe

539




539








  • 1




    Have you tried drawing a diagram?
    – Mark Bennet
    2 days ago










  • No, but I will.
    – J.Doe
    2 days ago














  • 1




    Have you tried drawing a diagram?
    – Mark Bennet
    2 days ago










  • No, but I will.
    – J.Doe
    2 days ago








1




1




Have you tried drawing a diagram?
– Mark Bennet
2 days ago




Have you tried drawing a diagram?
– Mark Bennet
2 days ago












No, but I will.
– J.Doe
2 days ago




No, but I will.
– J.Doe
2 days ago










2 Answers
2






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oldest

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Sometimes a figure is worth a thousand words:



enter image description here






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  • 60°. Thank you!
    – J.Doe
    2 days ago



















3














Since $|a+b|^2=langle a+b,a+b rangle = |a|^2 + 2langle a, b rangle + |b|^2$, it follows that $2langle a,b rangle=-|a|^2$. Hence $cos(alpha)= -frac{1}{2}$.






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    2 Answers
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    active

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3














    Sometimes a figure is worth a thousand words:



    enter image description here






    share|cite|improve this answer





















    • 60°. Thank you!
      – J.Doe
      2 days ago
















    3














    Sometimes a figure is worth a thousand words:



    enter image description here






    share|cite|improve this answer





















    • 60°. Thank you!
      – J.Doe
      2 days ago














    3












    3








    3






    Sometimes a figure is worth a thousand words:



    enter image description here






    share|cite|improve this answer












    Sometimes a figure is worth a thousand words:



    enter image description here







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 2 days ago









    David G. StorkDavid G. Stork

    10k21332




    10k21332












    • 60°. Thank you!
      – J.Doe
      2 days ago


















    • 60°. Thank you!
      – J.Doe
      2 days ago
















    60°. Thank you!
    – J.Doe
    2 days ago




    60°. Thank you!
    – J.Doe
    2 days ago











    3














    Since $|a+b|^2=langle a+b,a+b rangle = |a|^2 + 2langle a, b rangle + |b|^2$, it follows that $2langle a,b rangle=-|a|^2$. Hence $cos(alpha)= -frac{1}{2}$.






    share|cite|improve this answer


























      3














      Since $|a+b|^2=langle a+b,a+b rangle = |a|^2 + 2langle a, b rangle + |b|^2$, it follows that $2langle a,b rangle=-|a|^2$. Hence $cos(alpha)= -frac{1}{2}$.






      share|cite|improve this answer
























        3












        3








        3






        Since $|a+b|^2=langle a+b,a+b rangle = |a|^2 + 2langle a, b rangle + |b|^2$, it follows that $2langle a,b rangle=-|a|^2$. Hence $cos(alpha)= -frac{1}{2}$.






        share|cite|improve this answer












        Since $|a+b|^2=langle a+b,a+b rangle = |a|^2 + 2langle a, b rangle + |b|^2$, it follows that $2langle a,b rangle=-|a|^2$. Hence $cos(alpha)= -frac{1}{2}$.







        share|cite|improve this answer












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        answered 2 days ago









        MirceaMircea

        1736




        1736






























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