If $B$ is added to $A$, under what condition does the resultant vector have a magnitude equal to $A+B$?












0












$begingroup$


If $B$ is added to $A$, under what condition does the resultant vector have a magnitude equal to $A+B$? Under what conditions is the resultant vector equal to zero?



My Attempt:
Let $theta $ be the angle between $vec {A}$ and $vec {B}$. Then,



$$R=A+B$$
$$sqrt {A^2+B^2+2A.Bcos theta}=A+B$$
$$A^2+B^2+2A.Bcos theta=A^2+B^2+2A.B$$
$$cos theta =1$$
$$cos theta = cos 0$$
$$theta =0°$$
The resultant have a magnitude $A+B$ when the angle between the vectors is $0°$. How do I solve the second part of the question.?






vectors







share|cite|improve this question









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  • 3




    $begingroup$
    First point looks good. For the second one, all you need is $,A+B=0 iff A = -B,$.
    $endgroup$
    – dxiv
    Sep 13 '17 at 2:31












  • $begingroup$
    @dxiv, And how do I get that? Please elaborate.
    $endgroup$
    – blue_eyed_...
    Sep 13 '17 at 2:39










  • $begingroup$
    First $|A|=|B|$, then $costheta=-1$
    $endgroup$
    – Andrei
    Sep 13 '17 at 2:46








  • 1




    $begingroup$
    @blue_eyed_... how do I get that? It's part of the definition of a vector space that every vector $V$ has an additive inverse $-V$ such that $V + (-V) = 0,$. Then, if $A+B=0$ you just add $(-B)$ to both sides of the equality and get $A=-B,$.
    $endgroup$
    – dxiv
    Sep 13 '17 at 3:18


















0












$begingroup$


If $B$ is added to $A$, under what condition does the resultant vector have a magnitude equal to $A+B$? Under what conditions is the resultant vector equal to zero?



My Attempt:
Let $theta $ be the angle between $vec {A}$ and $vec {B}$. Then,



$$R=A+B$$
$$sqrt {A^2+B^2+2A.Bcos theta}=A+B$$
$$A^2+B^2+2A.Bcos theta=A^2+B^2+2A.B$$
$$cos theta =1$$
$$cos theta = cos 0$$
$$theta =0°$$
The resultant have a magnitude $A+B$ when the angle between the vectors is $0°$. How do I solve the second part of the question.?






vectors







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    First point looks good. For the second one, all you need is $,A+B=0 iff A = -B,$.
    $endgroup$
    – dxiv
    Sep 13 '17 at 2:31












  • $begingroup$
    @dxiv, And how do I get that? Please elaborate.
    $endgroup$
    – blue_eyed_...
    Sep 13 '17 at 2:39










  • $begingroup$
    First $|A|=|B|$, then $costheta=-1$
    $endgroup$
    – Andrei
    Sep 13 '17 at 2:46








  • 1




    $begingroup$
    @blue_eyed_... how do I get that? It's part of the definition of a vector space that every vector $V$ has an additive inverse $-V$ such that $V + (-V) = 0,$. Then, if $A+B=0$ you just add $(-B)$ to both sides of the equality and get $A=-B,$.
    $endgroup$
    – dxiv
    Sep 13 '17 at 3:18
















0












0








0





$begingroup$


If $B$ is added to $A$, under what condition does the resultant vector have a magnitude equal to $A+B$? Under what conditions is the resultant vector equal to zero?



My Attempt:
Let $theta $ be the angle between $vec {A}$ and $vec {B}$. Then,



$$R=A+B$$
$$sqrt {A^2+B^2+2A.Bcos theta}=A+B$$
$$A^2+B^2+2A.Bcos theta=A^2+B^2+2A.B$$
$$cos theta =1$$
$$cos theta = cos 0$$
$$theta =0°$$
The resultant have a magnitude $A+B$ when the angle between the vectors is $0°$. How do I solve the second part of the question.?






vectors







share|cite|improve this question









$endgroup$




If $B$ is added to $A$, under what condition does the resultant vector have a magnitude equal to $A+B$? Under what conditions is the resultant vector equal to zero?



My Attempt:
Let $theta $ be the angle between $vec {A}$ and $vec {B}$. Then,



$$R=A+B$$
$$sqrt {A^2+B^2+2A.Bcos theta}=A+B$$
$$A^2+B^2+2A.Bcos theta=A^2+B^2+2A.B$$
$$cos theta =1$$
$$cos theta = cos 0$$
$$theta =0°$$
The resultant have a magnitude $A+B$ when the angle between the vectors is $0°$. How do I solve the second part of the question.?






vectors




vectors






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Sep 13 '17 at 2:25









blue_eyed_...blue_eyed_...

3,26721750




3,26721750








  • 3




    $begingroup$
    First point looks good. For the second one, all you need is $,A+B=0 iff A = -B,$.
    $endgroup$
    – dxiv
    Sep 13 '17 at 2:31












  • $begingroup$
    @dxiv, And how do I get that? Please elaborate.
    $endgroup$
    – blue_eyed_...
    Sep 13 '17 at 2:39










  • $begingroup$
    First $|A|=|B|$, then $costheta=-1$
    $endgroup$
    – Andrei
    Sep 13 '17 at 2:46








  • 1




    $begingroup$
    @blue_eyed_... how do I get that? It's part of the definition of a vector space that every vector $V$ has an additive inverse $-V$ such that $V + (-V) = 0,$. Then, if $A+B=0$ you just add $(-B)$ to both sides of the equality and get $A=-B,$.
    $endgroup$
    – dxiv
    Sep 13 '17 at 3:18
















  • 3




    $begingroup$
    First point looks good. For the second one, all you need is $,A+B=0 iff A = -B,$.
    $endgroup$
    – dxiv
    Sep 13 '17 at 2:31












  • $begingroup$
    @dxiv, And how do I get that? Please elaborate.
    $endgroup$
    – blue_eyed_...
    Sep 13 '17 at 2:39










  • $begingroup$
    First $|A|=|B|$, then $costheta=-1$
    $endgroup$
    – Andrei
    Sep 13 '17 at 2:46








  • 1




    $begingroup$
    @blue_eyed_... how do I get that? It's part of the definition of a vector space that every vector $V$ has an additive inverse $-V$ such that $V + (-V) = 0,$. Then, if $A+B=0$ you just add $(-B)$ to both sides of the equality and get $A=-B,$.
    $endgroup$
    – dxiv
    Sep 13 '17 at 3:18










3




3




$begingroup$
First point looks good. For the second one, all you need is $,A+B=0 iff A = -B,$.
$endgroup$
– dxiv
Sep 13 '17 at 2:31






$begingroup$
First point looks good. For the second one, all you need is $,A+B=0 iff A = -B,$.
$endgroup$
– dxiv
Sep 13 '17 at 2:31














$begingroup$
@dxiv, And how do I get that? Please elaborate.
$endgroup$
– blue_eyed_...
Sep 13 '17 at 2:39




$begingroup$
@dxiv, And how do I get that? Please elaborate.
$endgroup$
– blue_eyed_...
Sep 13 '17 at 2:39












$begingroup$
First $|A|=|B|$, then $costheta=-1$
$endgroup$
– Andrei
Sep 13 '17 at 2:46






$begingroup$
First $|A|=|B|$, then $costheta=-1$
$endgroup$
– Andrei
Sep 13 '17 at 2:46






1




1




$begingroup$
@blue_eyed_... how do I get that? It's part of the definition of a vector space that every vector $V$ has an additive inverse $-V$ such that $V + (-V) = 0,$. Then, if $A+B=0$ you just add $(-B)$ to both sides of the equality and get $A=-B,$.
$endgroup$
– dxiv
Sep 13 '17 at 3:18






$begingroup$
@blue_eyed_... how do I get that? It's part of the definition of a vector space that every vector $V$ has an additive inverse $-V$ such that $V + (-V) = 0,$. Then, if $A+B=0$ you just add $(-B)$ to both sides of the equality and get $A=-B,$.
$endgroup$
– dxiv
Sep 13 '17 at 3:18












1 Answer
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$begingroup$

Though the obvious fact that $vec B=-vec A$, we can infer this from your previous expression:



$$A^2+B^2+2ABcostheta=0$$



If $A=0$, then trivially $B^2=0 to B=0$, and viceversa, and the angle can be any value.



If $A,Bgt0$, this leads to:
$$
costheta=-{A^2+B^2 over 2AB}
$$



Knowing that $cos(cdot)$ must be bounded in [-1,1]:
$$
-1le {A^2+B^2 over 2AB}le1\
-2ABle A^2+B^2le2AB\
-4ABle (A-B)^2le0\
$$
which can only be true for $A=B$, and the angle in this case is $
costheta=-{2over2}=-1 to theta=pi$






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    0












    $begingroup$

    Though the obvious fact that $vec B=-vec A$, we can infer this from your previous expression:



    $$A^2+B^2+2ABcostheta=0$$



    If $A=0$, then trivially $B^2=0 to B=0$, and viceversa, and the angle can be any value.



    If $A,Bgt0$, this leads to:
    $$
    costheta=-{A^2+B^2 over 2AB}
    $$



    Knowing that $cos(cdot)$ must be bounded in [-1,1]:
    $$
    -1le {A^2+B^2 over 2AB}le1\
    -2ABle A^2+B^2le2AB\
    -4ABle (A-B)^2le0\
    $$
    which can only be true for $A=B$, and the angle in this case is $
    costheta=-{2over2}=-1 to theta=pi$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Though the obvious fact that $vec B=-vec A$, we can infer this from your previous expression:



      $$A^2+B^2+2ABcostheta=0$$



      If $A=0$, then trivially $B^2=0 to B=0$, and viceversa, and the angle can be any value.



      If $A,Bgt0$, this leads to:
      $$
      costheta=-{A^2+B^2 over 2AB}
      $$



      Knowing that $cos(cdot)$ must be bounded in [-1,1]:
      $$
      -1le {A^2+B^2 over 2AB}le1\
      -2ABle A^2+B^2le2AB\
      -4ABle (A-B)^2le0\
      $$
      which can only be true for $A=B$, and the angle in this case is $
      costheta=-{2over2}=-1 to theta=pi$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Though the obvious fact that $vec B=-vec A$, we can infer this from your previous expression:



        $$A^2+B^2+2ABcostheta=0$$



        If $A=0$, then trivially $B^2=0 to B=0$, and viceversa, and the angle can be any value.



        If $A,Bgt0$, this leads to:
        $$
        costheta=-{A^2+B^2 over 2AB}
        $$



        Knowing that $cos(cdot)$ must be bounded in [-1,1]:
        $$
        -1le {A^2+B^2 over 2AB}le1\
        -2ABle A^2+B^2le2AB\
        -4ABle (A-B)^2le0\
        $$
        which can only be true for $A=B$, and the angle in this case is $
        costheta=-{2over2}=-1 to theta=pi$






        share|cite|improve this answer









        $endgroup$



        Though the obvious fact that $vec B=-vec A$, we can infer this from your previous expression:



        $$A^2+B^2+2ABcostheta=0$$



        If $A=0$, then trivially $B^2=0 to B=0$, and viceversa, and the angle can be any value.



        If $A,Bgt0$, this leads to:
        $$
        costheta=-{A^2+B^2 over 2AB}
        $$



        Knowing that $cos(cdot)$ must be bounded in [-1,1]:
        $$
        -1le {A^2+B^2 over 2AB}le1\
        -2ABle A^2+B^2le2AB\
        -4ABle (A-B)^2le0\
        $$
        which can only be true for $A=B$, and the angle in this case is $
        costheta=-{2over2}=-1 to theta=pi$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Sep 13 '17 at 3:51









        BrethloszeBrethlosze

        2,149316




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