Let O be the origin, the point A be , the point be such that and . Then [2025]

Question

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 $\arg (z_{2}) = \arg (z_{1}) + \frac{π}{6}$ . Then [2025]

Smart Achievers · Accepted Answer

(2)

We have, $z_{1} = \sqrt{3} + 2 \sqrt{2} i$ ; $\sqrt{3} | z_{2} | = | z_{1} |$ and $\arg (z_{2}) = \arg (z_{1}) + \frac{π}{6}$

$\Rightarrow \arg (z_{2}) - \arg (z_{1}) = \frac{π}{6}$

$∴ z_{2} = \frac{| z_{2} |}{| z_{1} |} \cdot z_{1} e^{i (π / 6)}$

$= \frac{1}{\sqrt{3}} [\frac{(\sqrt{3} + 2 \sqrt{2} i) (\sqrt{3} + i)}{2}]$

$\Rightarrow z_{2} = \frac{1}{2 \sqrt{3}} [3 - 2 \sqrt{2} + i (2 \sqrt{6} + \sqrt{3})]$

Now, $z_{1} - z_{2} = \sqrt{3} + 2 \sqrt{2} i - \frac{(3 - 2 \sqrt{2} + i) (2 \sqrt{6} + \sqrt{3})}{2 \sqrt{3}}$

$= \frac{6 + 4 \sqrt{6} i - 3 + 2 \sqrt{2} - 2 \sqrt{6} i - i \sqrt{3}}{2 \sqrt{3}}$

$= \frac{3 + 2 \sqrt{2} + i (2 \sqrt{6} - \sqrt{3})}{2 \sqrt{3}}$ $∴$ $| z_{1} - z_{2} |$ $=$ $| z_{2} |$

$\Rightarrow$ $△$ ABO is isosceles with angles $\frac{π}{6}, \frac{π}{6}$ and $\frac{2 π}{3}$ .

$∴$ Area of $△$ ABO = $\frac{1}{2} \sqrt{11} \times \frac{\sqrt{11}}{\sqrt{3}} \sin \frac{π}{6} = \frac{11}{4 \sqrt{3}}$ .

Solution

Q.
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 $\arg (z_{2}) = \arg (z_{1}) + \frac{π}{6}$ . Then [2025]

1 area of triangle ABO is $\frac{11}{\sqrt{3}}$

2 ABO is an obtuse angled isosceles triangle

3 ABO is a scalene triangle

4 area of triangle ABO is $\frac{11}{4}$

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