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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,

$$R=A+B$$ $$\sqrt {A^2+B^2+2A.B\cos \theta}=A+B$$ $$A^2+B^2+2A.B\cos \theta=A^2+B^2+2A.B$$ $$\cos \theta =1$$ $$\cos \theta = \cos 0$$ $$\theta =0°$$ 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.?

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    $\begingroup$ 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. $\endgroup$ – pi-π Sep 13 '17 at 2:39
  • $\begingroup$ First $|A|=|B|$, then $\cos\theta=-1$ $\endgroup$ – Andrei Sep 13 '17 at 2:46
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    $\begingroup$ @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
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Though the obvious fact that $\vec B=-\vec A$, we can infer this from your previous expression:

$$A^2+B^2+2AB\cos\theta=0$$

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

If $A,B\gt0$, this leads to: $$ \cos\theta=-{A^2+B^2 \over 2AB} $$

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

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