Question: A hydrologist models the flow of groundwater in a 3D aquifer using vectors. If the flow vector $\vec{f} = \begin{pmatrix} 4 \\ -2 \\ 5 \end{pmatrix}$ represents the direction and rate of flow, find the point on the line $\vec{r}(t) = \begin{pmatrix} 1 \\ 0 \\ -3 \end{pmatrix} + t\begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}$ that is closest to the point $P = (3, -1, 2)$.

["Understanding the Closest Point on a Line in 3D Using Vector Geometry\nHydrology, Groundwater Flow, and Vector Modeling", "When modeling groundwater flow in a 3D aquifer, hydrologists often use vector mathematics to describe the direction and magnitude of flow through subsurface materials. In this article, we explore how to find the point on a line that is closest to a given point—a common and essential computation in hydrological simulations.", "Suppose the flow of groundwater is modeled by the vector\n$$\n\vec{f} = \begin{pmatrix} 4 \ -2 \ 5 \end{pmatrix},\n$$\nand the line representing the flow path is defined by\n$$\n\vec{r}(t) = \begin{pmatrix} 1 \ 0 \ -3 \end{pmatrix} + t\begin{pmatrix} 2 \ 1 \ -1 \end{pmatrix}.\n$$\nWe are tasked with finding the point on this line that is closest to the point $P = (3, -1, 2)$.", "---", "### Step 1: Understand the Geometry", "The line is defined parametrically as\n$$\n\vec{r}(t) = \vec{r}_0 + t\vec{d},\n$$\nwhere\n- $\vec{r}_0 = \begin{pmatrix} 1 \ 0 \ -3 \end{pmatrix}$ is a point on the line,\n- $\vec{d} = \begin{pmatrix} 2 \ 1 \ -1 \end{pmatrix}$ is the direction vector,\n- $t$ is a real scalar parameter.", "We want to find the value of $t$ such that the vector from $P = (3, -1, 2)$ to a point $\vec{r}(t)$ on the line is perpendicular to the direction vector $\vec{d}$. This condition guarantees that $\vec{r}(t) - \vec{P}$ is orthogonal to the line, meaning it reaches the closest point.", "---", "### Step 2: Define the Vector from $P$ to a General Point on the Line", "Let\n$$\n\vec{v}(t) = \vec{r}(t) - \vec{P} = \left( \vec{r}_0 + t\vec{d} \right) - \vec{P} = \vec{r}_0 - \vec{P} + t\vec{d}.\n$$", "Compute $\vec{r}_0 - \vec{P}$:\n$$\n\vec{r}_0 - \vec{P} = \begin{pmatrix} 1 - 3 \ 0 - (-1) \ -3 - 2 \end{pmatrix} = \begin{pmatrix} -2 \ 1 \ -5 \end{pmatrix}.\n$$", "So,\n$$\n\vec{v}(t) = \begin{pmatrix} -2 \ 1 \ -5 \end{pmatrix} + t\begin{pmatrix} 2 \ 1 \ -1 \end{pmatrix} = \begin{pmatrix} -2 + 2t \ 1 + t \ -5 - t \end{pmatrix}.\n$$", "---", "### Step 3: Enforce Orthogonality (Dot Product = 0)", "For $\vec{v}(t)$ to be perpendicular to the direction vector $\vec{d}$, their dot product must be zero:\n$$\n\vec{v}(t) \cdot \vec{d} = 0.\n$$", "Compute the dot product:\n$$\n\begin{pmatrix} -2 + 2t \ 1 + t \ -5 - t \end{pmatrix} \cdot \begin{pmatrix} 2 \ 1 \ -1 \end{pmatrix} = (-2 + 2t)(2) + (1 + t)(1) + (-5 - t)(-1)\n$$", "Break it down:\n- First component: $(-2 + 2t) \cdot 2 = -4 + 4t$\n- Second component: $(1 + t) \cdot 1 = 1 + t$\n- Third component: $(-5 - t)(-1) = 5 + t$", "Add them:\n$$\n(-4 + 4t) + (1 + t) + (5 + t) = (-4 + 1 + 5) + (4t + t + t) = 2 + 6t\n$$", "Set equal to zero:\n$$\n2 + 6t = 0 \Rightarrow t = -\frac{1}{3}\n$$", "---", "### Step 4: Find the Closest Point", "Substitute $t = -\frac{1}{3}$ into $\vec{r}(t)$:\n$$\n\vec{r}\left(-\frac{1}{3}\right) = \begin{pmatrix} 1 \ 0 \ -3 \end{pmatrix} + \left(-\frac{1}{3}\right)\begin{pmatrix} 2 \ 1 \ -1 \end{pmatrix} = \begin{pmatrix} 1 - \frac{2}{3} \ 0 - \frac{1}{3} \ -3 + \frac{1}{3} \end{pmatrix} = \begin{pmatrix} \frac{1}{3} \ -\frac{1}{3} \ -\frac{8}{3} \end{pmatrix}\n$$", "---", "### Final Answer", "The point on the line $\vec{r}(t)$ closest to $P = (3, -1, 2)$ is\n$$\n\boxed{\left( \frac{1}{3},\ -\frac{1}{3},\ -\frac{8}{3} \right)}.\n$$", "This geometric approach—using vector projection and orthogonality—is fundamental in hydrology for modeling optimal flow paths, contaminant transport, and well placement in 3D aquifer systems.", "---", "Keywords: hydrology, groundwater flow, vector modeling, 3D aquifer, closest point on line, vector dot product, hydrological simulation, (\vec{f} = \begin{pmatrix} 4 \ -2 \ 5 \end{pmatrix}), (\vec{r}(t) = \begin{pmatrix} 1 \ 0 \ -3 \end{pmatrix} + t\begin{pmatrix} 2 \ 1 \ -1 \end{pmatrix}), point $P = (3, -1, 2)$"]