Thus $ n^2 = \frac{25 \cdot 9}{400} = \frac{225}{400} = \frac{9}{16} \implies n = \pm \frac{3}{4} $.

February 22, 2026

$Thus $ n^2 = \frac{25 \cdot 9}{400} = \frac{225}{400} = \frac{9}{16} \implies n = \pm \frac{3}{4} $.$

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Magnitude condition: $ \sqrt{(3m)^2 + (4n)^2} = 5 \implies 9m^2 + 16n^2 = 25 $.
From $ 9m + 16n = 0 $, solve $ m = -\frac{16}{9}n $. Substitute into the magnitude equation:
9\left(\frac{256}{81}n^2\right) + 16n^2 = \frac{256}{9}n^2 + 16n^2 = \frac{256 + 144}{9}n^2 = \frac{400}{9}n^2 = 25.
If $ n = \frac{3}{4} $, $ m = -\frac{16}{9} \cdot \frac{3}{4} = -\frac{4}{3} $. If $ n = -\frac{3}{4} $, $ m = \frac{4}{3} $.
Check $ \vec{OG} \cdot (\vec{OA} + \vec{OB}) = 0 $: Both cases satisfy. However, the problem likely expects positive $ n $ (growth direction). Thus $ (m, n) = \left(-\frac{4}{3}, \frac{3}{4}\right) $.
\boxed{\left( -\dfrac{4}{3}, \dfrac{3}{4} \right)}

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