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MHT CET · Maths · Complex Number

The area of the triangle whose vertices are \(i, \omega\) and \(\omega^2\) is (Where \(\omega\) is a complex cube root of unity other than \(1, \mathrm{i}\) is an imaginary number) ______ sq.units

  1. A \(\frac{3 \sqrt{3}}{4}\)
  2. B \(\frac{\sqrt{3}}{2}\)
  3. C \(\frac{3 \sqrt{3}}{2}\)
  4. D \(\frac{\sqrt{3}}{4}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{\sqrt{3}}{4}\)

Step-by-step Solution

Detailed explanation

Vertices: \(z_1 = (0, 1)\), \(z_2 = (-\frac{1}{2}, \frac{\sqrt{3}}{2})\), \(z_3 = (-\frac{1}{2}, -\frac{\sqrt{3}}{2})\) Area \( = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|\)