Let O be the origin, the point A be z_1=\sqrt{3}+2 \sqrt{2} i the point B (z_2) be such that \sqrt{3}{|z_2 |}=|z_1 |} and \arg (z_2) =\arg (z_1) +\frac{\pi}{6}. Then [28 Jan 2025 Shift 1]

Correct Answer is : ABO is an obtuse angled isosceles triangle

A B O is an obtuse angled isosceles triangle

area of triangle A B O is {11}/{\sqrt{3}}

area of triangle A B O is \frac{11}{4}

JEE Main 28 Jan 2025 Shift 1 Paper

Section: Mathematics

Question : 7 of 75

Marks: +1, -0

Let O be the origin, the point A be z1=√3+2√2i the point B(z2) be such that √3|z2|=|z1| and arg(z2)=arg(z1)+‌

. Then

[28 Jan 2025 Shift 1]

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