# Question #9451d

Given that the vectors have same length, it is meant that they have same magnitude. Let this be a and angle between them is $\theta$. By the problem the resultant of the two being twice of either i.e. 2a, we can write
${\left(2 a\right)}^{2} = {a}^{2} + {a}^{2} + 2 \times a \times a \cos \theta$
$\implies \cos \theta = 1 \implies \theta = {0}^{\circ}$