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) $.

["Understanding the Vector Equation: $ \vec{OG} \cdot (\vec{OA} + \vec{OB}) = 0 $ — Geometry, Interpretation, and the Expected Solution $ \left(-\frac{4}{3}, \frac{3}{4}\right) $", "When analyzing vector relationships in geometry, equations like $ \vec{OG} \cdot (\vec{OA} + \vec{OB}) = 0 $ reveal meaningful spatial conditions—particularly orthogonality. This article explores the mathematical implications of this dot product equation, its two possible interpretive cases, and why the solution $ (m, n) = \left(-\frac{4}{3}, \frac{3}{4}\right) $ fits expected conventions, especially in growth-oriented contexts.", "---", "### The Equation: Geometry and Meaning", "The equation", "$$\n\vec{OG} \cdot (\vec{OA} + \vec{OB}) = 0\n$$", "states that the vector $ \vec{OG} $ is perpendicular to the sum of vectors $ \vec{OA} $ and $ \vec{OB} $. In vector geometry, this orthogonality imposes a strict relationship between the directions and magnitudes of the involved vectors.", "---", "### Two Possible Interpretive Cases", "At first glance, the dot product equating to zero yields infinitely many solutions satisfying orthogonality. However, typical mathematical problems condition this with physical or directional expectations—such as expecting $ \vec{OG} $ to represent a growth vector (positive direction), typically aligned with $ x > 0 $, $ y > 0 $, especially in growth modeling scenarios.", "#### Case 1: Trivial Orthogonality (Zero Vector)\n$ \vec{OA} + \vec{OB} = \vec{0} $ implies $ \vec{OG} $ is orthogonal to the zero vector—valid for any $ \vec{OG} $, but mathematically degenerate. This case usually excludes nontrivial solutions in application-oriented settings.", "#### Case 2: Growth-Consistent Positive Direction\nMore constructively, the condition $ \vec{OG} \cdot (\vec{OA} + \vec{OB}) = 0 $ under the expectation of positive directional growth constrains the solution to those where $ \vec{OG} $ is not zero but orthogonal to $ \vec{OA} + \vec{OB} $. This preference for “positive” $ m $ and $ n $ (e.g., in $ (m, n) $) reflects realism in modeling contexts—such as growth trajectories constrained to positive quadrant vectors.", "---", "### Deriving the Solution: $ \left(-\frac{4}{3}, \frac{3}{4}\right) $", "Let $ \vec{OA} = \langle a, 0 \rangle $, $ \vec{OB} = \langle b, 0 \rangle $, assuming alignment along the $ x $-axis (common in growth modeling where direction matters). Then:", "$$\n\vec{OA} + \vec{OB} = \langle a + b, 0 \rangle\n$$\n$$\n\vec{OG} = \langle m, n \rangle\n$$", "Dot product:", "$$\n\vec{OG} \cdot (\vec{OA} + \vec{OB}) = \langle m, n \rangle \cdot \langle a + b, 0 \rangle = m(a + b)\n$$", "Setting equal to zero:", "$$\nm(a + b) = 0\n$$", "If $ a + b <br/>\ne 0 $, then $ m = 0 $—but that violates the expectation of positive $ m $. Instead, real-world interpretation (e.g., projective growth) allows homogeneous scaling. Assume $ \vec{OA} $ and $ \vec{OB} $ define a direction not aligned with $ \vec{OG} $’s dominant axis, and seek a balanced, non-zero orthogonal solution.", "Suppose $ \vec{OA} + \vec{OB} $ lies along the $ x $-axis, so $ \vec{OA} = \langle p, 0 \rangle $, $ \vec{OB} = \langle q, 0 \rangle $, $ p, q > 0 $. Orthogonality forces $ \vec{OG} $’s $ x $-component to be zero: $ m = 0 $—but again inconsistent with growth assumptions.", "Hence, generalize: Let $ \vec{OA} + \vec{OB} $ have components $ \vec{v} = (u, v) $, $ u > 0 $ or $ v > 0 $. Then:", "$$\n\vec{OG} \cdot \vec{v} = 0 \Rightarrow m u + n v = 0\n$$", "This linear equation gives a family of solutions. To select a unique pair $ (m, n) $, data or context requires proportionality reflecting relative magnitudes. The expected solution $ \left(-\frac{4}{3}, \frac{3}{4}\right) $ satisfies:", "$$\nm u - \frac{4}{3}u + \frac{3}{4}v = 0\n\Rightarrow m u = \frac{4}{3}u - \frac{3}{4}v\n$$", "But more elegantly, the ratio $ \frac{n}{m} = -\frac{u}{v} $. Choosing $ u = 3 $, $ v = 4 $ yields $ \frac{n}{m} = -\frac{3}{4} \Rightarrow n = -\frac{4}{3}m $. Plug into orthogonality:", "$$\nm(3) + \left(-\frac{4}{3}m\right)(4) = 3m - \frac{16}{3}m = \left(\frac{9 - 16}{3}\right)m = -\frac{7}{3}m <br/>\ne 0\n$$", "Not quite. However, if $ \vec{OA} = \langle 1, 1 \rangle $, $ \vec{OB} = \langle 1, -4 \rangle $, then:", "$$\n\vec{OA} + \vec{OB} = \langle 2, -3 \rangle\n$$\nSolve $ \vec{OG} = \langle m, n \rangle \cdot \langle 2, -3 \rangle = 0 \Rightarrow 2m - 3n = 0 \Rightarrow m = \frac{3}{2}n $", "To satisfy positive growth assumption ($ m > 0 $, $ n > 0 $), pick $ n = \frac{3}{4} \Rightarrow m = \frac{9}{8} $, not matching.", "But suppose the directional vector is $ \vec{v} = \langle 3, -4 \rangle $, so $ \vec{OA} + \vec{OB} \parallel \langle 3, -4 \rangle $. Then any $ \vec{OG} \perp \langle 3, -4 \rangle $ satisfies:", "$$\n3m - 4n = 0 \Rightarrow \frac{m}{n} = \frac{"]