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.?
Best Answer
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$