d
(d)\(|\,\overrightarrow A \times \overrightarrow B |\, = \sqrt 3 (\overrightarrow A .\overrightarrow B )\)

\(AB\sin \theta = \sqrt 3 AB\cos \theta  \Rightarrow \)\(\tan \theta = \sqrt 3 \) \(⇒\) \(\theta = 60^\circ \)

Now \(|\overrightarrow R |\, = \,|\overrightarrow A + \overrightarrow B |\, = \sqrt {{A^2} + {B^2} + 2AB\cos \theta } \)

\( = \sqrt {{A^2} + {B^2} + 2AB\left( {\frac{1}{2}} \right)} \)

\( = {({A^2} + {B^2} + AB)^{1/2}}\)