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If the magnitude of the resultant of two equal vectors is equal to that of either vector, find the angle between them.
My Attempt: Let $\vec {a}$ and $\vec {b}$ be two vectors such that $\|\vec {a}\|=\|\vec {b}\|$
Magnitude of Resultant: $$=\sqrt {a^2+b^2+2ab\cos \theta}$$ $$=\sqrt {2a^2+2ab\cos \theta}$$
How do I proceed further?
From $\|a+b\|^2=\|a\|^2$ and $\|a\|=\|b\|$ we have $$\|a\|^2=\|a+b\|^2=\|a\|^2+\|b\|^2+2\langle a,b\rangle=2\|a\|^2+2\langle a,b\rangle,$$ so $\langle a,b\rangle=-\|a\|^2/2$. Hence the cosine of the angle between $a$ and $b$ using $\|a\|=\|b\|$ again is $$\frac{\langle a,b\rangle}{\|a\|\|b\|}=\frac{-1}{2}.$$
The resultant vector $\mathbf{OC}$ is the diagonal of the rhombus whose adjacent sides are $\mathbf{OA}, \mathbf{OB}$. Since its length is equal to either, we have an equilateral triangle $\mathbf{OBC}$. Thus the angle between the vectors is $120^\circ$
Let,$R$ be the resultant of the sum of the two vectors $\vec A$ and $\vec B$ such that $|\vec A|=|\vec B|$.So,$$R=\sqrt{|\vec A|^2+|\vec B|^2+2|\vec A||\vec B|\cos\theta}$$
where $\theta$ is the angle between $\vec A$ and $\vec B$.So,
$$R=\sqrt{2|\vec A|^2+2|\vec A|^2\cos\theta}$$.
Now magnitude of resultant $R$ is equal to either of the vectors $\vec A$ and $\vec B$.So,
$$|A|=\sqrt{2|\vec A|^2+2|\vec A|^2\cos\theta}$$ $$\implies|\vec A|^2=2|\vec A|^2+2|\vec A|^2\cos\theta$$ $$\implies\frac{-|\vec A|^2}{2|\vec A|^2}=\cos \theta$$ $$\implies\theta=\cos^{-1}-\frac{1}{2}$$.
Hope this helps!!
