Solution: Compute $ 2\overrightarrow{OA} = \begin{bmatrix} 4 \\ 2 \end{bmatrix} $ and $ 3\overrightarrow{OB} = \begin{bmatrix} -3 \\ 9 \end{bmatrix} $. Subtract:
["Understanding Vector Scaling and Subtraction: Computing $ 2\overrightarrow{OA} $ and $ 3\overrightarrow{OB} $", "In linear algebra and vector mathematics, scaling vectors by constants is a fundamental operation that simplifies analysis in geometry, physics, and computer graphics. This article explores how to compute scaled vectors and demonstrate subtraction—key tools in vector manipulation.", "---", "### Solving $ 2\overrightarrow{OA} = \begin{bmatrix} 4 \ 2 \end{bmatrix} $", "To find vector $ \overrightarrow{OA} $ when $ 2\overrightarrow{OA} = \begin{bmatrix} 4 \ 2 \end{bmatrix} $, divide both sides by 2:", "$$\n\overrightarrow{OA} = \frac{1}{2} \begin{bmatrix} 4 \ 2 \end{bmatrix} = \begin{bmatrix} 2 \ 1 \end{bmatrix}\n$$", "This scalar multiplication scales each component of $ \overrightarrow{OA} $ by $ \frac{1}{2} $, effectively doubling the input vector to match the target value.", "---", "### Solving $ 3\overrightarrow{OB} = \begin{bmatrix} -3 \ 9 \end{bmatrix} $", "To determine $ \overrightarrow{OB} $, divide the given vector by 3:", "$$\n\overrightarrow{OB} = \frac{1}{3} \begin{bmatrix} -3 \ 9 \end{bmatrix} = \begin{bmatrix} -1 \ 3 \end{bmatrix}\n$$", "Here, each component of $ \overrightarrow{OB} $ is obtained by dividing the corresponding component of $ \begin{bmatrix} -3 \ 9 \end{bmatrix} $ by 3. This operation evenly scales vectors, preserving direction while modifying magnitude.", "---", "### Subtracting Vectors: What It Adds Up To", "Vector subtraction extends concepts from basic algebra into multidimensional spaces. Given $ \vec{u} $ and $ \vec{v} $, the operation $ \vec{u} - \vec{v} $ computes the vector pointing from $ \vec{v} $ to $ \vec{u} $, or equivalently, $ -\vec{v} + \vec{u} $.", "For example, subtracting vectors:", "$$\n\overrightarrow{OA} - 3\overrightarrow{OB} = \begin{bmatrix} 2 \ 1 \end{bmatrix} - 3\overrightarrow{OB} = \begin{bmatrix} 2 \ 1 \end{bmatrix} - \begin{bmatrix} -3 \ 3 \end{bmatrix} = \begin{bmatrix} 2 + 3 \ 1 - 3 \end{bmatrix} = \begin{bmatrix} 5 \ -2 \end{bmatrix}\n$$", "Alternatively, using the decomposition:", "$$\n\overrightarrow{OA} - 3\overrightarrow{OB} = \overrightarrow{OA} + (-3\overrightarrow{OB}) = \begin{bmatrix} 2 \ 1 \end{bmatrix} + \begin{bmatrix} -3(-1) \ -3(3) \end{bmatrix} = \begin{bmatrix} 2 + 3 \ 1 - 9 \end{bmatrix} = \begin{bmatrix} 5 \ -2 \end{bmatrix}\n$$", "This result shows how subtracting scaled vectors produces a new vector representing displacement or difference.", "---", "### Why These Operations Matter", "Scaling vectors enables proportional resizing—vital in modeling, simulations, and transformations. Subtracting vectors supports difference calculations, pathfinding, and relative positioning—cornerstones of computational geometry.", "Whether working in 2D spaces or higher dimensions, mastering these operations equips you with powerful tools for vector-based reasoning and applications.", "---", "### Conclusion", "Computing $ 2\overrightarrow{OA} $ and $ 3\overrightarrow{OB} $ reveals how vector scaling transforms vectors efficiently. Subtraction builds upon these scaled vectors to express differences elegantly. Together, these operations form essential building blocks for linear algebra applications in science, engineering, and computer programming.", "---", "Keywords: vector scaling, compute $ 2\overrightarrow{OA} $, $ 3\overrightarrow{OB} $, vector subtraction, linear algebra, mathematical operations, displacement vector, proportional vectors, coordinate geometry", "---", "Explore how these vector computations extend into rotation, projection, and coordinate transformations—key steps in mastering multidimensional mathematics."]
