JEE Main 2025 Jan 28 Shift 1, Mathematics Q18: Rotation And Geometric Applications

Compute side lengths from the origin.

$|z_1{|}^2=(\sqrt{3}{)}^2+(2\sqrt{2}{)}^2=3+8=11$, so $|OA|=|z_1|=\sqrt{11}$.

$|z_2|=\frac{|z_1|}{\sqrt{3}}=\sqrt{\frac{11}{3}}$, so $|OB|=\sqrt{11/3}$.

Angle $\angle AOB=\arg (z_2)-\arg (z_1)=\frac{\pi }{6}$.

By the law of cosines:

$|AB{|}^2=|OA{|}^2+|OB{|}^2-2|OA||OB|\cos (\pi /6)=11+\frac{11}{3}-2\sqrt{11}\sqrt{11/3}\cdot \frac{\sqrt{3}}{2}$

$=\frac{44}{3}-\frac{11\sqrt{3}}{\sqrt{3}}=\frac{44}{3}-11=\frac{11}{3}$.

So $|AB|=\sqrt{11/3}=|OB|$. The triangle is isosceles with $|OB|=|AB|$, and the apex (equal sides meet) is at $B$. The angles opposite the equal sides are equal: $\angle BAO=\angle BOA=\frac{\pi }{6}$, and the remaining angle at $B$ is $\pi -\frac{\pi }{3}=\frac{2\pi }{3}$, which is obtuse.

$\therefore △ABO$ is an obtuse-angled isosceles triangle. Area $=\frac{1}{2}|OA||OB|\sin \frac{\pi }{6}=\frac{1}{2}\sqrt{11}\cdot \sqrt{11/3}\cdot \frac{1}{2}=\frac{11}{4\sqrt{3}}$ (which is neither $\frac{11}{\sqrt{3}}$ nor $\frac{11}{4}$).

Option (4) is the correct statement.