Maximum angle between vectors












4














Consider 3 vectors $textbf{v},textbf{v}',textbf{u}$ related by



$$textbf{v}=textbf{u}+textbf{v}'$$



Let $theta$ be the angle between $textbf{v}$ and $textbf{u}$ and let $phi$ be the angle between $textbf{v}$ and $-textbf{v}'$.



For what angle $theta$ is the angle $phi$ a maximum? The magnitudes $v,u$ are given and you can express the angle in terms of them.



I started by doing the dot product of the equation with itself getting



$$v^{2}=vucostheta+vv'cosphi=u^{2}+v^{'2}+2uv'cos (pi/2-phi-theta)$$



I thought I would then find the derivative of this expression with respect to $theta$ and set $dphi/dtheta=0$. That gives



$$-vusintheta=2uv'sin(pi/2-phi-theta)$$ or



$$vsintheta=2v'sin(phi+theta-pi/2)=-2v'cos(phi+theta)$$



which seems overdetermined.






calculus vectors







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  • 1




    Lenghts are given?
    – greedoid
    2 days ago










  • On geometric grounds, if $|u|<|v'|$, then $theta$ should be $0$ or $pi$ such that $v$ and $v'$ are opposite ...
    – Henning Makholm
    2 days ago






  • 2




    If all three magnitudes are given, then both $theta$ and $phi$ can be computed by standard triangle solving. There's no choice and no maxima or minima.
    – Henning Makholm
    2 days ago












  • @HenningMakholm Only v,u are given (not v'). I edited the question.
    – user2175783
    2 days ago
















4














Consider 3 vectors $textbf{v},textbf{v}',textbf{u}$ related by



$$textbf{v}=textbf{u}+textbf{v}'$$



Let $theta$ be the angle between $textbf{v}$ and $textbf{u}$ and let $phi$ be the angle between $textbf{v}$ and $-textbf{v}'$.



For what angle $theta$ is the angle $phi$ a maximum? The magnitudes $v,u$ are given and you can express the angle in terms of them.



I started by doing the dot product of the equation with itself getting



$$v^{2}=vucostheta+vv'cosphi=u^{2}+v^{'2}+2uv'cos (pi/2-phi-theta)$$



I thought I would then find the derivative of this expression with respect to $theta$ and set $dphi/dtheta=0$. That gives



$$-vusintheta=2uv'sin(pi/2-phi-theta)$$ or



$$vsintheta=2v'sin(phi+theta-pi/2)=-2v'cos(phi+theta)$$



which seems overdetermined.






calculus vectors







share|cite|improve this question




















  • 1




    Lenghts are given?
    – greedoid
    2 days ago










  • On geometric grounds, if $|u|<|v'|$, then $theta$ should be $0$ or $pi$ such that $v$ and $v'$ are opposite ...
    – Henning Makholm
    2 days ago






  • 2




    If all three magnitudes are given, then both $theta$ and $phi$ can be computed by standard triangle solving. There's no choice and no maxima or minima.
    – Henning Makholm
    2 days ago












  • @HenningMakholm Only v,u are given (not v'). I edited the question.
    – user2175783
    2 days ago














4












4








4


2





Consider 3 vectors $textbf{v},textbf{v}',textbf{u}$ related by



$$textbf{v}=textbf{u}+textbf{v}'$$



Let $theta$ be the angle between $textbf{v}$ and $textbf{u}$ and let $phi$ be the angle between $textbf{v}$ and $-textbf{v}'$.



For what angle $theta$ is the angle $phi$ a maximum? The magnitudes $v,u$ are given and you can express the angle in terms of them.



I started by doing the dot product of the equation with itself getting



$$v^{2}=vucostheta+vv'cosphi=u^{2}+v^{'2}+2uv'cos (pi/2-phi-theta)$$



I thought I would then find the derivative of this expression with respect to $theta$ and set $dphi/dtheta=0$. That gives



$$-vusintheta=2uv'sin(pi/2-phi-theta)$$ or



$$vsintheta=2v'sin(phi+theta-pi/2)=-2v'cos(phi+theta)$$



which seems overdetermined.






calculus vectors







share|cite|improve this question















Consider 3 vectors $textbf{v},textbf{v}',textbf{u}$ related by



$$textbf{v}=textbf{u}+textbf{v}'$$



Let $theta$ be the angle between $textbf{v}$ and $textbf{u}$ and let $phi$ be the angle between $textbf{v}$ and $-textbf{v}'$.



For what angle $theta$ is the angle $phi$ a maximum? The magnitudes $v,u$ are given and you can express the angle in terms of them.



I started by doing the dot product of the equation with itself getting



$$v^{2}=vucostheta+vv'cosphi=u^{2}+v^{'2}+2uv'cos (pi/2-phi-theta)$$



I thought I would then find the derivative of this expression with respect to $theta$ and set $dphi/dtheta=0$. That gives



$$-vusintheta=2uv'sin(pi/2-phi-theta)$$ or



$$vsintheta=2v'sin(phi+theta-pi/2)=-2v'cos(phi+theta)$$



which seems overdetermined.






calculus vectors




calculus vectors






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago

























asked 2 days ago









user2175783

1575




1575








  • 1




    Lenghts are given?
    – greedoid
    2 days ago










  • On geometric grounds, if $|u|<|v'|$, then $theta$ should be $0$ or $pi$ such that $v$ and $v'$ are opposite ...
    – Henning Makholm
    2 days ago






  • 2




    If all three magnitudes are given, then both $theta$ and $phi$ can be computed by standard triangle solving. There's no choice and no maxima or minima.
    – Henning Makholm
    2 days ago












  • @HenningMakholm Only v,u are given (not v'). I edited the question.
    – user2175783
    2 days ago














  • 1




    Lenghts are given?
    – greedoid
    2 days ago










  • On geometric grounds, if $|u|<|v'|$, then $theta$ should be $0$ or $pi$ such that $v$ and $v'$ are opposite ...
    – Henning Makholm
    2 days ago






  • 2




    If all three magnitudes are given, then both $theta$ and $phi$ can be computed by standard triangle solving. There's no choice and no maxima or minima.
    – Henning Makholm
    2 days ago












  • @HenningMakholm Only v,u are given (not v'). I edited the question.
    – user2175783
    2 days ago








1




1




Lenghts are given?
– greedoid
2 days ago




Lenghts are given?
– greedoid
2 days ago












On geometric grounds, if $|u|<|v'|$, then $theta$ should be $0$ or $pi$ such that $v$ and $v'$ are opposite ...
– Henning Makholm
2 days ago




On geometric grounds, if $|u|<|v'|$, then $theta$ should be $0$ or $pi$ such that $v$ and $v'$ are opposite ...
– Henning Makholm
2 days ago




2




2




If all three magnitudes are given, then both $theta$ and $phi$ can be computed by standard triangle solving. There's no choice and no maxima or minima.
– Henning Makholm
2 days ago






If all three magnitudes are given, then both $theta$ and $phi$ can be computed by standard triangle solving. There's no choice and no maxima or minima.
– Henning Makholm
2 days ago














@HenningMakholm Only v,u are given (not v'). I edited the question.
– user2175783
2 days ago




@HenningMakholm Only v,u are given (not v'). I edited the question.
– user2175783
2 days ago










2 Answers
2






active

oldest

votes


















2














Unless you're working under peculiar constraints, doing this algebraically is almost certainly a tactical error. Instead, draw a diagram:



      A

     / 

   u/   v

  |/     |

  B------>C

      v'


The angles are $phi$ at C and $theta$ at A.



$phi$ will certainly be a maximum if we can make it $pi$ which happens if $|u|>|v|$ and $theta=0$.



Then consider the case $|u|<|v|$. If we have already decided on the positions of A and C, the possible positions of B is a circle with center A. Among those, the ones that maximize the angle at C are the ones where BC is tangent to the circle -- which are the ones where the angle is B is right!



Basic trigonometry then gives us $|u|=|v|costheta$.






share|cite|improve this answer





















  • "Among those, the ones that maximize the angle at C are the ones where BC is tangent to the circle". Why is that?
    – user35202
    2 days ago






  • 1




    @user35202: Geometrically, the tangent to the circle through C separates the rays that intersect the circle (which are the ones that represent possible $phi$s) from those that don't intersect the circle (and therefor cannot be possible $phi$s).
    – Henning Makholm
    2 days ago





















2














$$
vec v = vec u -(-vec v') = vec v = vec u -vec w
$$



so now



$$
vec ucdotvec v = ||vec u||^2-vec ucdotvec w
$$



or



$$
||vec u||||vec v||costheta=||vec u|| - ||vec u||||vec w||cosphi
$$



and then assuming that $vec ucdotvec w ne 0$



$$
phi = arccosleft(a+bcosthetaright)
$$



and now deriving



$$
frac{dphi}{dtheta} = frac{b sin (theta )}{sqrt{1-(a+b cos (theta ))^2}} = 0
$$



which gives $theta = 0 + kpi$






share|cite|improve this answer























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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2














    Unless you're working under peculiar constraints, doing this algebraically is almost certainly a tactical error. Instead, draw a diagram:



          A
    
     / 
    
   u/   v
    
  |/     |
    
  B------>C
    
      v'


    The angles are $phi$ at C and $theta$ at A.



    $phi$ will certainly be a maximum if we can make it $pi$ which happens if $|u|>|v|$ and $theta=0$.



    Then consider the case $|u|<|v|$. If we have already decided on the positions of A and C, the possible positions of B is a circle with center A. Among those, the ones that maximize the angle at C are the ones where BC is tangent to the circle -- which are the ones where the angle is B is right!



    Basic trigonometry then gives us $|u|=|v|costheta$.






    share|cite|improve this answer





















    • "Among those, the ones that maximize the angle at C are the ones where BC is tangent to the circle". Why is that?
      – user35202
      2 days ago






    • 1




      @user35202: Geometrically, the tangent to the circle through C separates the rays that intersect the circle (which are the ones that represent possible $phi$s) from those that don't intersect the circle (and therefor cannot be possible $phi$s).
      – Henning Makholm
      2 days ago


















    2














    Unless you're working under peculiar constraints, doing this algebraically is almost certainly a tactical error. Instead, draw a diagram:



          A
    
     / 
    
   u/   v
    
  |/     |
    
  B------>C
    
      v'


    The angles are $phi$ at C and $theta$ at A.



    $phi$ will certainly be a maximum if we can make it $pi$ which happens if $|u|>|v|$ and $theta=0$.



    Then consider the case $|u|<|v|$. If we have already decided on the positions of A and C, the possible positions of B is a circle with center A. Among those, the ones that maximize the angle at C are the ones where BC is tangent to the circle -- which are the ones where the angle is B is right!



    Basic trigonometry then gives us $|u|=|v|costheta$.






    share|cite|improve this answer





















    • "Among those, the ones that maximize the angle at C are the ones where BC is tangent to the circle". Why is that?
      – user35202
      2 days ago






    • 1




      @user35202: Geometrically, the tangent to the circle through C separates the rays that intersect the circle (which are the ones that represent possible $phi$s) from those that don't intersect the circle (and therefor cannot be possible $phi$s).
      – Henning Makholm
      2 days ago
















    2












    2








    2






    Unless you're working under peculiar constraints, doing this algebraically is almost certainly a tactical error. Instead, draw a diagram:



          A
    
     / 
    
   u/   v
    
  |/     |
    
  B------>C
    
      v'


    The angles are $phi$ at C and $theta$ at A.



    $phi$ will certainly be a maximum if we can make it $pi$ which happens if $|u|>|v|$ and $theta=0$.



    Then consider the case $|u|<|v|$. If we have already decided on the positions of A and C, the possible positions of B is a circle with center A. Among those, the ones that maximize the angle at C are the ones where BC is tangent to the circle -- which are the ones where the angle is B is right!



    Basic trigonometry then gives us $|u|=|v|costheta$.






    share|cite|improve this answer












    Unless you're working under peculiar constraints, doing this algebraically is almost certainly a tactical error. Instead, draw a diagram:



          A
    
     / 
    
   u/   v
    
  |/     |
    
  B------>C
    
      v'


    The angles are $phi$ at C and $theta$ at A.



    $phi$ will certainly be a maximum if we can make it $pi$ which happens if $|u|>|v|$ and $theta=0$.



    Then consider the case $|u|<|v|$. If we have already decided on the positions of A and C, the possible positions of B is a circle with center A. Among those, the ones that maximize the angle at C are the ones where BC is tangent to the circle -- which are the ones where the angle is B is right!



    Basic trigonometry then gives us $|u|=|v|costheta$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 2 days ago









    Henning Makholm

    237k16302537




    237k16302537












    • "Among those, the ones that maximize the angle at C are the ones where BC is tangent to the circle". Why is that?
      – user35202
      2 days ago






    • 1




      @user35202: Geometrically, the tangent to the circle through C separates the rays that intersect the circle (which are the ones that represent possible $phi$s) from those that don't intersect the circle (and therefor cannot be possible $phi$s).
      – Henning Makholm
      2 days ago




















    • "Among those, the ones that maximize the angle at C are the ones where BC is tangent to the circle". Why is that?
      – user35202
      2 days ago






    • 1




      @user35202: Geometrically, the tangent to the circle through C separates the rays that intersect the circle (which are the ones that represent possible $phi$s) from those that don't intersect the circle (and therefor cannot be possible $phi$s).
      – Henning Makholm
      2 days ago


















    "Among those, the ones that maximize the angle at C are the ones where BC is tangent to the circle". Why is that?
    – user35202
    2 days ago




    "Among those, the ones that maximize the angle at C are the ones where BC is tangent to the circle". Why is that?
    – user35202
    2 days ago




    1




    1




    @user35202: Geometrically, the tangent to the circle through C separates the rays that intersect the circle (which are the ones that represent possible $phi$s) from those that don't intersect the circle (and therefor cannot be possible $phi$s).
    – Henning Makholm
    2 days ago






    @user35202: Geometrically, the tangent to the circle through C separates the rays that intersect the circle (which are the ones that represent possible $phi$s) from those that don't intersect the circle (and therefor cannot be possible $phi$s).
    – Henning Makholm
    2 days ago













    2














    $$
    vec v = vec u -(-vec v') = vec v = vec u -vec w
    $$



    so now



    $$
    vec ucdotvec v = ||vec u||^2-vec ucdotvec w
    $$



    or



    $$
    ||vec u||||vec v||costheta=||vec u|| - ||vec u||||vec w||cosphi
    $$



    and then assuming that $vec ucdotvec w ne 0$



    $$
    phi = arccosleft(a+bcosthetaright)
    $$



    and now deriving



    $$
    frac{dphi}{dtheta} = frac{b sin (theta )}{sqrt{1-(a+b cos (theta ))^2}} = 0
    $$



    which gives $theta = 0 + kpi$






    share|cite|improve this answer




























      2














      $$
      vec v = vec u -(-vec v') = vec v = vec u -vec w
      $$



      so now



      $$
      vec ucdotvec v = ||vec u||^2-vec ucdotvec w
      $$



      or



      $$
      ||vec u||||vec v||costheta=||vec u|| - ||vec u||||vec w||cosphi
      $$



      and then assuming that $vec ucdotvec w ne 0$



      $$
      phi = arccosleft(a+bcosthetaright)
      $$



      and now deriving



      $$
      frac{dphi}{dtheta} = frac{b sin (theta )}{sqrt{1-(a+b cos (theta ))^2}} = 0
      $$



      which gives $theta = 0 + kpi$






      share|cite|improve this answer


























        2












        2








        2






        $$
        vec v = vec u -(-vec v') = vec v = vec u -vec w
        $$



        so now



        $$
        vec ucdotvec v = ||vec u||^2-vec ucdotvec w
        $$



        or



        $$
        ||vec u||||vec v||costheta=||vec u|| - ||vec u||||vec w||cosphi
        $$



        and then assuming that $vec ucdotvec w ne 0$



        $$
        phi = arccosleft(a+bcosthetaright)
        $$



        and now deriving



        $$
        frac{dphi}{dtheta} = frac{b sin (theta )}{sqrt{1-(a+b cos (theta ))^2}} = 0
        $$



        which gives $theta = 0 + kpi$






        share|cite|improve this answer














        $$
        vec v = vec u -(-vec v') = vec v = vec u -vec w
        $$



        so now



        $$
        vec ucdotvec v = ||vec u||^2-vec ucdotvec w
        $$



        or



        $$
        ||vec u||||vec v||costheta=||vec u|| - ||vec u||||vec w||cosphi
        $$



        and then assuming that $vec ucdotvec w ne 0$



        $$
        phi = arccosleft(a+bcosthetaright)
        $$



        and now deriving



        $$
        frac{dphi}{dtheta} = frac{b sin (theta )}{sqrt{1-(a+b cos (theta ))^2}} = 0
        $$



        which gives $theta = 0 + kpi$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 2 days ago

























        answered 2 days ago









        Cesareo

        8,1633516




        8,1633516






























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