The minimum value of the expression B = ∣z∣^2 +∣z−3∣^2 +∣z−6i∣^2 is p. What the value of (p/(10)) .?


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Question Number 131196 by john_santu last updated on 02/Feb/21

$T h e m i n i m u m v a l u e o f t h e$ $e x p r e s s i o n B = ∣ z ∣^{2} + ∣ z - 3 ∣^{2} + ∣ z - 6 i ∣^{2}$ $i s p . W h a t t h e v a l u e o f \frac{p}{10} . ?$
	Answered by liberty last updated on 02/Feb/21
	$z=a+bi → { ((∣z∣^2 =a^2 +b^2 )),((∣z−3∣^2 = (a−3)^2 +b^2 )),((∣z−6i∣^2 = a^2 +b^2 +36−12b)) :} now we find B=3a^2 +3b^2 −6a−12b+45 B=3(a−1)^2 +3(b−2)^2 +30 so the minimum value of B is 30 when { ((a=1)),((b=2)) :} ⇒ z=1+2i then p = 30 and (p/(10)) = ((30)/(10)) = 3.$
	$z = a + b i \to {\begin{cases} ∣ z ∣^{2} = a^{2} + b^{2} \\ ∣ z - 3 ∣^{2} = {(a - 3)}^{2} + b^{2} \\ ∣ z - 6 i ∣^{2} = a^{2} + b^{2} + 36 - 12 b \end{cases}$ $n o w w e f i n d$ $B = 3 a^{2} + 3 b^{2} - 6 a - 12 b + 45$ $B = 3 {(a - 1)}^{2} + 3 {(b - 2)}^{2} + 30$ $s o t h e m i n i m u m v a l u e o f B is 30 when$ ${\begin{cases} a = 1 \\ b = 2 \end{cases} \Rightarrow z = 1 + 2 i t h e n p = 30$ $a n d \frac{p}{10} = \frac{30}{10} = 3 .$
	Commented by john_santu last updated on 02/Feb/21

	$g a v e k u d o s$
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