Let $A = \frac{2}{\sqrt{3}} e^{- i \frac{π}{6}}, B = \frac{2}{\sqrt{3}} e^{i \frac{π}{2}}, C = \frac{2}{\sqrt{3}} e^{- \frac{5 π}{6}}$ be three points forming a triangle ABC in the Gaussian plane,then triangle ABC is

equilateral

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isosceles

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scalene

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None of these

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Solution

The correct option is A equilateral
$A = \frac{2}{\sqrt{3}} (\frac{\sqrt{3} - i}{2})$
$\Rightarrow A = \frac{\sqrt{3} - i}{\sqrt{3}}$
$\Rightarrow A = \frac{3 - \sqrt{3} i}{3}$ ...(i)
$B = \frac{2}{\sqrt{3}} [i]$
$= \frac{2 \sqrt{3} i}{3}$ ...(ii)
$C = \frac{2}{\sqrt{3}} [\frac{- \sqrt{3} - i}{2}]$
$= \frac{- \sqrt{3} - i}{\sqrt{3}}$
$= \frac{- 3 - i \sqrt{3}}{3}$ ...(ii)
Now
$A - B = 3 (\frac{1 - \sqrt{3} i}{3})$
$B - C = 3 (\frac{1 + \sqrt{3} i}{3})$
$C - A = - 2$
Thus, $| A - B | = | B - C | = | C - A | = 2$
Therefore, the triangle is equiangular/equilateral.
Hence, option 'A' is correct.

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Similar questions

A = \frac{2}{\sqrt{3}} e^{- i \frac{π}{6}}

B = \frac{2}{\sqrt{3}} e^{i \frac{π}{2}}

C = \frac{2}{\sqrt{3}} e^{- \frac{i 5 π}{6}}

\frac{2}{\sqrt{3}} e^{- i \frac{π}{6}}

\frac{2}{\sqrt{3}} e^{i \frac{π}{2}}

\frac{2}{\sqrt{3}} e^{- i \frac{5 π}{6}}

A (\frac{2}{\sqrt{3}} e^{i π / 2}), B (\frac{2}{\sqrt{3}} e^{- i π / 6})

C (\frac{2}{\sqrt{3}} e^{- i 5 π / 6}) .

P

A P^{2} + B P^{2} + C P^{2} =

A = \frac{2}{3} e^{\frac{i π}{2}}, B = \frac{2}{3} e^{\frac{i π}{6}}, C = \frac{2}{3} e^{\frac{i 5 π}{6}}

△ A B C

△ A B C

A (\frac{2}{\sqrt{3}} e^{i π / 2}), B (\frac{2}{\sqrt{3}} e^{- i π / 6}), C (\frac{2}{\sqrt{3}} e^{- i 5 π / 6})

P

A P^{2} + B P^{2} + C P^{2} =

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