Two vectors $$\vec{A}$$ and $$\vec{B}$$ have equal magnitudes. If magnitude of $$\vec{A} + \vec{B}$$ is equal to two times the magnitude of $$\vec{A} - \vec{B}$$, then the angle between $$\vec{A}$$ and $$\vec{B}$$ will be

We are given that $$|\vec{A}| = |\vec{B}| = A$$ (say), and $$|\vec{A} + \vec{B}| = 2|\vec{A} - \vec{B}|$$. Let $$\theta$$ be the angle between $$\vec{A}$$ and $$\vec{B}$$. Then $$|\vec{A} + \vec{B}|^2 = A^2 + A^2 + 2A^2\cos\theta = 2A^2(1 + \cos\theta)$$ and $$|\vec{A} - \vec{B}|^2 = A^2 + A^2 - 2A^2\cos\theta = 2A^2(1 - \cos\theta)$$.

Squaring both sides of $$|\vec{A} + \vec{B}| = 2|\vec{A} - \vec{B}|$$ gives $$2A^2(1 + \cos\theta) = 4 \times 2A^2(1 - \cos\theta)$$, so $$1 + \cos\theta = 4(1 - \cos\theta)$$ which simplifies to $$1 + \cos\theta = 4 - 4\cos\theta$$, hence $$5\cos\theta = 3$$ and $$\cos\theta = \frac{3}{5}$$.

Therefore $$\theta = \cos^{-1}\left(\frac{3}{5}\right)$$. The correct answer is Option A: $$\cos^{-1}\left(\frac{3}{5}\right)$$.