JEE Main 2025 — Complex Number Question with Solution

$\begin{aligned} & O A=\left|z_1\right|=\sqrt{3+8}=\sqrt{11} \\ & \text { and } O B=\frac{1}{\sqrt{3}}\left|z_1\right|=\sqrt{\frac{11}{3}} \\ & A B^2=O A^2+O B^2-2 \cdot O A \cdot O B \cos \frac{\pi}{6} \\ & \quad=11+\frac{11}{3}-2 \cdot \frac{11}{\sqrt{3}} \cdot \frac{\sqrt{3}}{2} \\ & \therefore \quad A B=\sqrt{\frac{11}{3}} \\ & \therefore \quad \text { Area of } \triangle A B D=\frac{1}{2} \cdot O A \cdot O B \cdot \sin \frac{\pi}{6} \\ & \quad=\frac{11}{4 \sqrt{3}} \text { sq. units } \end{aligned}$ Here $OB=AB$ and $\angle A=\frac{2\pi }{3}$ $\therefore △ABD$ is an obtuse angled isosceles triangle.