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} = \sqrt{∣ \vec{a} ∣^{2} + ∣ \vec{b} ∣^{2} + 2 ∣ \vec{a} ∣∣ \vec{b} ∣}

R_{\min} = \sqrt{∣ \vec{a} ∣^{2} + ∣ \vec{b} ∣^{2} - 2 ∣ \vec{a} ∣∣ \vec{b} ∣}

\Leftrightarrow \frac{9}{1} = \frac{∣ \vec{a} ∣^{2} + ∣ \vec{b} ∣^{2} + 2 ∣ \vec{a} ∣∣ \vec{b} ∣}{∣ \vec{a} ∣^{2} + ∣ \vec{b} ∣^{2} - 2 ∣ \vec{a} ∣∣ \vec{b} ∣}

then ∣ \vec{a} ∣ = 2 ∣ \vec{b} ∣