Solution: Since $ \vec{OA} \cdot \vec{OB} = 0 $, they are perpendicular. Let $ \vec{OA} = (3, 0) $, $ \vec{OB} = (0, 4) $. Then $ \vec{OG} = (3m, 4n) $.

["Solution: Using Vector Perpendicularity to Simplify Geometric Problems", "In vector geometry, determining perpendicularity between vectors is fundamental for solving coordinate-based problems efficiently. A key fact is that two vectors are perpendicular if and only if their dot product equals zero. This principle enables swift verification and simplifies complex geometric constructions—especially when working with coordinate planes.", "## Understanding the Perpendicularity Condition", "Given two vectors $ \vec{OA} $ and $ \vec{OB} $, they are perpendicular if:\n$$\n\vec{OA} \cdot \vec{OB} = 0\n$$\nFor $ \vec{OA} = (3, 0) $ and $ \vec{OB} = (0, 4) $, computing the dot product yields:\n$$\n\vec{OA} \cdot \vec{OB} = (3)(0) + (0)(4) = 0\n$$\nThis confirms that $ \vec{OA} \perp \vec{OB} $, forming a right angle at point $ O $.", "## Extending to a New Point $ G $ with Vector $ \vec{OG} = (3m, 4n) $", "Now consider a third point $ G $ defined by vector $ \vec{OG} = (3m, 4n) $. The relationship established above—$ \vec{OA} \cdot \vec{OB} = 0 $—can guide analysis of geometric compatibility involving $ G $. Specifically, we examine whether $ \vec{OG} $, as a linear combination of $ \vec{OA} $ and $ \vec{OB} $, maintains key vector relationships in its direction and magnitude.", "Since $ \vec{OA} = (3, 0) $ spans the horizontal axis and $ \vec{OB} = (0, 4) $ spans the vertical axis, any vector of the form $ \vec{OG} = (3m, 4n) $ lies in the span of $ \vec{OA} $ and $ \vec{OB} $, making it a point in the rectangle formed by these perpendicular axes.", "### Checking if $ G $ Lies on a Perpendicular Line or Subspace", "To assess meaningful geometric alignment, suppose $ G $ must lie on a coordinate-like line perpendicular to $ OG $, or relate to projections along $ OA $ and $ OB $. The equality $ \vec{OA} \cdot \vec{OB} = 0 $ acts as a foundation when computing projections or ensuring orthogonality in derived vectors.", "For instance, since $ \vec{OA} $ is horizontal and $ \vec{OB} $ is vertical, $ \vec{OG} = (3m, 4n) $ lies generically in the plane, but its coefficients $ m $ and $ n $ determine its locus:\n- When $ m = 1, n = 1 $, $ \vec{OG} = (3, 4) $, forming a diagonal in the rectangle.\n- When $ m = 0 $, $ G = (0, 4n) $ lies on $ \vec{OB} $; when $ n = 0 $, $ G = (3m, 0) $ lies on $ \vec{OA} $.", "Thus, $ G $ traces a path across the coordinate grid only when both $ m $ and $ n $ vary within real scalars—never perpendicular to $ \vec{OG} $ itself unless $ m = 0 $ or $ n = 0 $, which collapses $ G $ onto single axes.", "### Solution Summary", "- $ \vec{OA} \cdot \vec{OB} = 0 $ proves $ \vec{OA} \perp \vec{OB} $, forming a right coordinate system.\n- $ \vec{OG} = (3m, 4n) $ lies in this perpendicular framework, spanned by $ \vec{OA} $ and $ \vec{OB} $.\n- For $ G $ to preserve orthogonality in derived products (e.g., $ \vec{OG} \cdot \vec{OA} = 0 $), one must solve $ 3(3m) + 0(4n) = 0 \Rightarrow m = 0 $, or $ \vec{OG} \cdot \vec{OB} = 0 \Rightarrow 0 + 4(4n) = 0 \Rightarrow n = 0 $—i.e., $ G $ must lie on $ \vec{OA} $ or $ \vec{OB} $, not in their orthogonal direction.", "This solution demonstrates how the fundamental dot product condition enables efficient analysis of geometric constraints in coordinate vector systems.", "---", "In essence, recognizing vector perpendicularity via dot products powers precise geometric reasoning—critical when defining and analyzing points like $ G $ within constructed perpendicular frameworks. Leveraging known orthogonal pairs such as $ \vec{OA} $ and $ \vec{OB} $ allows quick validation of alignment, projections, and orthogonality in advanced coordinate-based problems."]