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?
Best Answer
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 $$\frac{\langle a,b\rangle}{\|a\|\|b\|}=\frac{-1}{2}.$$