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Let P(e^(itheta1)), Q(e^(itheta2)) and ...

Let `P(e^(itheta_1)), Q(e^(itheta_2)) and R(e^(itheta_3))` be the vertices of a triangle `PQR` in the Argand Plane. Theorthocenter of the triangle `PQR` is

A

`e^(i(theta_(1)+theta_(2)+theta_(3))`

B

`2/3e^(i(theta_(1)+theta_(2)+theta_(3))`

C

`e^(i(theta_(1))+e^(itheta_(2))+e^(itheta_(3))`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
C

We have,
`|e^(itheta_(1))|=|e^(itheta_(2))|=|e^(itheta_(3))|=1`
`rArr OP=OQ=OR=1`, where O is the origin.
`rArr` Origin O is the circumference of `trianglePQR`.
The affix of the centroid is `1/3e^(itheta_(1))+e^(itheta_(2))+e^(itheta_(3))`.
Let z be the affix of the orthocenter. Since centroid divides the segment joining circumference and othocenter in the ratio `1:2`.
`therefore 1/3(e^(itheta_(1))+e^(itheta_(2))+e^(itheta_(3)))=(1 xx z + 2 xx 0)/(1+2) rArr z=e^(itheta_(1)) + e^(itheta_(2))+e^(itheta_(3))`
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