Let Az1,Bz2,Cz3 be the vertices of △ABC such that z3+iωz2=(1+iω)z1 where ω(≠1) is a cube root of unity,t hen Coordinates of orthocenter of △ABC is

Since the side BC makes an angle at orthocenter is complementary angle with AHence,z3−z0z2−z0=eiπ/3=cos⁡π3+isin⁡π/3=12+i32=−ω2=1+ω⇒z3−z0=(1+ω)z2−z0=z2−z0+ωz2−ωz0⇒z3−z2−ωz2=−ωz0⇒ωz0=(1+ω)z2−z3⇒z0=(1+ω)z2−z3ω

Let Az1,Bz2,Cz3 be the vertices of △ABC such that z3+iωz2=(1+iω)z1 where ω(≠1) is a cube root of unity,t hen Coordinates of orthocenter of △ABC is

Since the side BC makes an angle at orthocenter is complementary angle with AHence,z3−z0z2−z0=eiπ/3=cos⁡π3+isin⁡π/3=12+i32=−ω2=1+ω⇒z3−z0=(1+ω)z2−z0=z2−z0+ωz2−ωz0⇒z3−z2−ωz2=−ωz0⇒ωz0=(1+ω)z2−z3⇒z0=(1+ω)z2−z3ω

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

$Let A (z_{1}), B (z_{2}), C (z_{3}) be the vertices of △ A B C such that z_{3} + i ω z_{2} = (1 + i ω) z_{1} where ω (\neq 1) is a cube root of unity, t hen Coordinates of orthocenter of △ A B C is$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{(1 + ω) z_{2} - z_{3}}{ω}$

$\frac{ω (z_{1} - z_{2}) + z_{3}}{ω}$

$(\frac{(z_{1} - z_{2}) + (ω - 4) z_{3}}{ω})$

$\frac{ω (z_{1} - z_{2}) + (ω + 4) z_{3}}{ω}$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} S i n c e t h e s i d e B C m a k e s a n a n g l e a t o r t h o c e n t e r i s c o m p l e m e n t a r y a n g l e w i t h A \\ H e n c e, \\ \frac{z_{3} - z_{0}}{z_{2} - z_{0}} = e^{i π / 3} = \cos \frac{π}{3} + i \sin π / 3 \end{array}$

$\begin{array}{l} = \frac{1}{2} + i \frac{\sqrt{3}}{2} = - ω^{2} = 1 + ω \\ \Rightarrow z_{3} - z_{0} = (1 + ω) (z_{2} - z_{0}) = z_{2} - z_{0} + ω z_{2} - ω z_{0} \\ \Rightarrow z_{3} - z_{2} - ω z_{2} = - ω z_{0} \\ \Rightarrow ω z_{0} = (1 + ω) z_{2} - z_{3} \\ \Rightarrow z_{0} = \frac{(1 + ω) z_{2} - z_{3}}{ω} \end{array}$