Solution: The closest point occurs where the vector from $ (5, 7) $ to the line is perpendicular to the direction vector $ \begin{pmatrix} 3 \\ 4 \end{pmatrix} $. Let the point on the line be $ (2 + 3t, -1 + 4t) $. The vector from $ (5, 7) $ to this point is $ \begin{pmatrix} 2 + 3t - 5 \\ -1 + 4t - 7 \end{pmatrix} = \begin{pmatrix} -3 + 3t \\ -8 + 4t \end{pmatrix} $. Dot product with $ \begin{pmatrix} 3 \\ 4 \end{pmatrix} $ must be zero: $ 3(-3 + 3t) + 4(-8 + 4t) = 0 $. Simplify: $ -9 + 9t - 32

["Finding the Closest Point on a Line: A Step-by-Step Geometric Solution", "Determining the closest point on a line to a given point is a fundamental problem in geometry and linear algebra—critical in computer graphics, path planning, and optimization. This article walks through a precise mathematical solution to finding the closest point on a line using vector methods, offering clarity on the core principles behind the procedure.", "---", "### The Problem Setup", "Given a point ( P = (5, 7) ) and a line defined parametrically, our goal is to find the point ( Q ) on the line such that the vector ( \vec{PQ} ) is perpendicular to the direction vector of the line. This perpendicularity condition ensures minimal distance from ( P ) to the line.", "Let the line be parametrized as:", "$$\nQ(t) = (2 + 3t,, -1 + 4t)\n$$", "This comes from a direction vector ( \begin{pmatrix} 3 \ 4 \end{pmatrix} ) and a base point ( (2, -1) ). The vector from ( P = (5, 7) ) to a general point on the line is:", "$$\n\vec{PQ} = \begin{pmatrix} (2 + 3t) - 5 \ (-1 + 4t) - 7 \end{pmatrix} = \begin{pmatrix} -3 + 3t \ -8 + 4t \end{pmatrix}\n$$", "---", "### Enforcing Perpendicularity via the Dot Product", "Since ( \vec{PQ} ) must be perpendicular to the direction vector ( \begin{pmatrix} 3 \ 4 \end{pmatrix} ), their dot product equals zero:", "$$\n\vec{PQ} \cdot \begin{pmatrix} 3 \ 4 \end{pmatrix} = 0\n$$", "Substitute the components:", "$$\n(-3 + 3t)(3) + (-8 + 4t)(4) = 0\n$$", "Compute:", "$$\n-9 + 9t - 32 + 16t = 0\n$$", "$$\n(9t + 16t) + (-9 - 32) = 0\n\Rightarrow 25t - 41 = 0\n\Rightarrow t = \frac{41}{25}\n$$", "---", "### Finding the Exact Coordinates of the Closest Point", "Plug ( t = \frac{41}{25} ) back into the parametric equation:", "- For ( x ):\n ( 2 + 3 \cdot \frac{41}{25} = \frac{50}{25} + \frac{123}{25} = \frac{173}{25} )", "- For ( y ):\n ( -1 + 4 \cdot \frac{41}{25} = -\frac{25}{25} + \frac{164}{25} = \frac{156}{25} )", "Thus, the closest point on the line is:", "$$\n\left( \frac{173}{25},\ \frac{156}{25} \right)\n$$", "---", "### Why This Method Works", "This solution leverages key properties of vectors and dot products. The vector from the external point to the line is perpendicular to the line’s direction when the shortest distance is achieved. By solving the resulting linear equation, we find the exact parameter value that satisfies this orthogonality condition.", "---", "### Real-World Applications", "This approach is widely used in computational geometry:\n- Computer graphics: Computing nearest points on surfaces\n- Robotics: Determining optimal robot paths\n- Machine learning: Finding projections in support vector machines", "Understanding this mathematical foundation enables robust implementation across domains.", "---", "### Summary", "- The closest point on a line to a point occurs where the connecting vector is perpendicular to the line’s direction.\n- Using parametric line equations and dot product conditions, we derive the exact value of the parameter ( t ).\n- Substituting back gives the precise coordinates of the nearest point.\n- This method is efficient, geometrically intuitive, and widely applicable.", "Mastering vector projection principles empowers you to solve not just this problem—but an entire class of geometric optimization challenges."]