Let O be the origin, the point A be z_1= 3 +2 2 i , the point B(z_2) be such that 3 |z_2|=|z_1| and (z_2)= (z_1)+(π)/(6) . Then

Correct answer: B. ABO is an obtuse angled isosceles triangle aligned & O A=|z_1|= 3+8 = 11 & and O B= 1 3 |z_1|= (11)/(3) & A B^2=O A^2+O B^2-2 · O A · O B (π)/(6) & =11+(11)/(3)-2 · 11 3 · 3 2 & A B= (11)/(3) & Area of A B D=(1)/(2) · O A · O B · (π)/(6) & = 11 4 3 sq. units aligned Here O B=A B and A=(2 π)/(3) A B D is an obtuse angled isosceles triangle.

JEE Main 202528 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 28 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Complex Number

JEE Main 2025 · 28 January · Morning Shift · Mathematics · Complex Number

Let O be the origin, the point A be z_1= 3 +2 2 i , the point...

Let

O

be the origin, the point

A

z_1=\sqrt{3}+2 \sqrt{2} i

, the point $

B\left(z_2\right)

be such that

\sqrt{3}\left|z_2\right|=\left|z_1\right| and

\arg \left(z_2\right)=\arg \left(z_1\right)+\frac{\pi}{6}$. Then

Options

Aarea of triangle ABO is $\frac{11}{\sqrt{3}}$
BABO is an obtuse angled isosceles triangle
Carea of triangle ABO is $\frac{11}{4}$
DABO is a scalene triangle

Correct answer

B. ABO is an obtuse angled isosceles triangle

Step-by-step solution

\begin{aligned} & O A=\left|z_1\right|=\sqrt{3+8}=\sqrt{11} \\ & \text{\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 O B=A B and

\angle A=\frac{2 \pi}{3} \therefore \quad \triangle A B D

is an obtuse angled isosceles triangle.

Practice Complex Number on Quantrex Academy →

More from Complex Number