Let R , z 1 , z 2 , z 3 be three distinct complex numbers such that | z 1 | = | z 2 | = | z 3 | = 3 & | ( k z 1 + z 2 ) - ( k z 2 + z 3 ) | min = | z 3 - z 2 | | z 3 - z 1 | , k R - { 0 } then 36 =

Question

Let $α \in R, z_{1}, z_{2}, z_{3}$ be three distinct complex numbers such that $| z_{1} |$ $=$ $| z_{2} |$ $=$ $| z_{3} |$ $= 3$ & $| (k z_{1} + z_{2}) - (k z_{2} + z_{3}) |_{\min}$ $= α$ $| z_{3} - z_{2} |$ $| z_{3} - z_{1} |,$ $\forall k \in R - {0}$ then $36 α =____$

Smart Achievers · Accepted Answer

(6)

$| k z_{1} + (1 - k) z_{2} - z_{3} |$ $= height from vertex C .$

$∴$ $=$ $| z_{1} - z_{3} |$ $| z_{2} - z_{3} |$ $= α$

$α = \frac{\sin C}{A B} = \frac{1}{2 R} = \frac{1}{6}$

Solution

Q.
Let $α \in R, z_{1}, z_{2}, z_{3}$ be three distinct complex numbers such that $| z_{1} |$ $=$ $| z_{2} |$ $=$ $| z_{3} |$ $= 3$ & $| (k z_{1} + z_{2}) - (k z_{2} + z_{3}) |_{\min}$ $= α$ $| z_{3} - z_{2} |$ $| z_{3} - z_{1} |,$ $\forall k \in R - {0}$ then $36 α =____$

Ans.
(6)

$| k z_{1} + (1 - k) z_{2} - z_{3} |$ $= height from vertex C .$

$∴$ $Area = \frac{1}{2} (A B) (height) = \frac{1}{2}$ $| z_{1} - z_{3} |$ $| z_{2} - z_{3} |$ $\sin C$

$α = \frac{\sin C}{A B} = \frac{1}{2 R} = \frac{1}{6}$

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Solution

Q. Let α∈R, z1,z2,z3 be three distinct complex numbers such that |z1|=|z2|=|z3|=3 & |(kz1+z2)-(kz2+z3)|min=α|z3-z2||z3-z1|,∀k∈R-{0} then 36α=____

Ans. (6) |kz1+(1-k)z2-z3|=height from vertex C. ∴ Area=12(AB)(height)=12|z1-z3||z2-z3|sinC α=sinCAB=12R=16

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Q.
Let $α \in R, z_{1}, z_{2}, z_{3}$ be three distinct complex numbers such that $| z_{1} |$ $=$ $| z_{2} |$ $=$ $| z_{3} |$ $= 3$ & $| (k z_{1} + z_{2}) - (k z_{2} + z_{3}) |_{\min}$ $= α$ $| z_{3} - z_{2} |$ $| z_{3} - z_{1} |,$ $\forall k \in R - {0}$ then $36 α =____$

Ans.
(6)

$| k z_{1} + (1 - k) z_{2} - z_{3} |$ $= height from vertex C .$

$∴$ $Area = \frac{1}{2} (A B) (height) = \frac{1}{2}$ $| z_{1} - z_{3} |$ $| z_{2} - z_{3} |$ $\sin C$

$α = \frac{\sin C}{A B} = \frac{1}{2 R} = \frac{1}{6}$