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Question-127568




Question Number 127568 by Ar Brandon last updated on 30/Dec/20
Commented by bramlexs22 last updated on 30/Dec/20
R_(max)  = (√(∣a^→ ∣^2 +∣b^→ ∣^2 +2∣a^→ ∣∣b^→ ∣))   R_(min)  = (√(∣a^→ ∣^2 +∣b^→ ∣^2 −2∣a^→ ∣∣b^→ ∣))  ⇔ (9/1) = ((∣a^→ ∣^2 +∣b^→ ∣^2 +2∣a^→ ∣∣b^→ ∣)/(∣a^→ ∣^2 +∣b^→ ∣^2 −2∣a^→ ∣∣b^→ ∣))  then ∣a^→ ∣ = 2∣b^→ ∣
$$\mathrm{R}_{\mathrm{max}} \:=\:\sqrt{\mid\overset{\rightarrow} {\mathrm{a}}\mid^{\mathrm{2}} +\mid\overset{\rightarrow} {\mathrm{b}}\mid^{\mathrm{2}} +\mathrm{2}\mid\overset{\rightarrow} {\mathrm{a}}\mid\mid\overset{\rightarrow} {\mathrm{b}}\mid}\: \\ $$$$\mathrm{R}_{\mathrm{min}} \:=\:\sqrt{\mid\overset{\rightarrow} {\mathrm{a}}\mid^{\mathrm{2}} +\mid\overset{\rightarrow} {\mathrm{b}}\mid^{\mathrm{2}} −\mathrm{2}\mid\overset{\rightarrow} {\mathrm{a}}\mid\mid\overset{\rightarrow} {\mathrm{b}}\mid} \\ $$$$\Leftrightarrow\:\frac{\mathrm{9}}{\mathrm{1}}\:=\:\frac{\mid\overset{\rightarrow} {\mathrm{a}}\mid^{\mathrm{2}} +\mid\overset{\rightarrow} {\mathrm{b}}\mid^{\mathrm{2}} +\mathrm{2}\mid\overset{\rightarrow} {\mathrm{a}}\mid\mid\overset{\rightarrow} {\mathrm{b}}\mid}{\mid\overset{\rightarrow} {\mathrm{a}}\mid^{\mathrm{2}} +\mid\overset{\rightarrow} {\mathrm{b}}\mid^{\mathrm{2}} −\mathrm{2}\mid\overset{\rightarrow} {\mathrm{a}}\mid\mid\overset{\rightarrow} {\mathrm{b}}\mid} \\ $$$$\mathrm{then}\:\mid\overset{\rightarrow} {\mathrm{a}}\mid\:=\:\mathrm{2}\mid\overset{\rightarrow} {\mathrm{b}}\mid\: \\ $$

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