JEE Mains · Maths · STD 11 - 4.1 complex nubers
Let \(O\) be the origin, the point \(A\) be \(z_1=\sqrt{3}+2 \sqrt{2} i\), 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
- A area of triangle ABO is \(\frac{11}{\sqrt{3}}\)
- B ABO is an obtuse angled isosceles triangle
- C area of triangle ABO is \(\frac{11}{4}\)
- D ABO is a scalene triangle
Answer & Solution
Correct Answer
(B) ABO is an obtuse angled isosceles triangle
Step-by-step Solution
Detailed explanation
\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…
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