Two vectors → X and → Y have equal magnitude. The magnitude of ( →...

Given $X = Y$

$\sqrt{{X}^{2} + {Y}^{2} - 2 \times Ycos\theta}$

$= n\sqrt{{X}^{2} + {Y}^{2} + 2 \times Ycos\theta}$

Square both sides

$2{X}^{2} ( 1 - cos\theta ) = {n}^{2} \cdot 2{X}^{2} ( 1 + cos\theta )$

$1 - cos\theta = {n}^{2} + {n}^{2}cos\theta$

$cos\theta = \dfrac{1 - {n}^{2}}{1 + {n}^{2}}$

$\theta = \cos^{- 1}\left[\dfrac{{n}^{2} - 1}{- {n}^{2} - 1}\right]$