\mathbf{v} = \mathbf{p} - \mathbf{r}_0 = egin{pmatrix} 5 - 1 \ 4 - (-2) \end{pmatrix} = egin{pmatrix} 4 \ 6 \end{pmatrix}
["# Vector Subtraction in Physics: Understanding the Displacement Vector ( \mathbf{v} = \mathbf{p} - \mathbf{r}_0 )", "In physics, particularly in kinematics and mechanics, the concept of displacement plays a fundamental role in describing motion. Vector subtraction allows us to precisely calculate the change in position of an object using displacement vectors. One common scenario involves computing the relative displacement between two points in a 2D coordinate system.", "### The Displacement Vector Formula", "Given two position vectors:", "[\n\mathbf{r}_0 = \begin{pmatrix} x_0 \ y_0 \end{pmatrix}, \quad \mathbf{p} = \begin{pmatrix} x_p \ y_p \end{pmatrix}\n]", "the displacement vector ( \mathbf{v} ) from point ( \mathbf{r}_0 ) to point ( \mathbf{p} ) is defined by vector subtraction:", "[\n\mathbf{v} = \mathbf{p} - \mathbf{r}_0\n]", "This operation yields a vector that represents both the change in position and the direction of motion.", "---", "### Example Application", "Consider the displacement between two points given in matrix form:", "[\n\mathbf{v} = \mathbf{p} - \mathbf{r}_0 = \begin{pmatrix} 5 - 1 \ 4 - (-2) \end{pmatrix} = \begin{pmatrix} 4 \ 6 \end{pmatrix}\n]", "Breaking down the components:", "- ( x_p - x_0 = 5 - 1 = 4 )\n- ( y_p - y_0 = 4 - (-2) = 4 + 2 = 6 )", "This results in the displacement vector:", "[\n\mathbf{v} = \begin{pmatrix} 4 \ 6 \end{pmatrix}\n]", "---", "### Interpreting the Vector", "The vector ( \mathbf{v} = \begin{pmatrix} 4 \ 6 \end{pmatrix} ) indicates a motion of 4 units right and 6 units up from the initial position ( \mathbf{r}_0 ). Its magnitude represents the total displacement distance, calculated using the Euclidean norm:", "[\n|\mathbf{v}| = \sqrt{4^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}\n]", "This magnitude provides insight into how far the object has traveled regardless of direction.", "Direction can also be described using the angle ( \ heta ) the vector makes with the positive ( x )-axis, found via:", "[\n\ heta = \ an^{-1}\left(\frac{v_y}{v_x}\right) = \ an^{-1}\left(\frac{6}{4}\right) = \ an^{-1}(1.5) \approx 56.3^\circ\n]", "---", "### Practical Use in Physics and Engineering", "Understanding displacement vectors is crucial in:", "- Kinematics: Analyzing motion, velocity, and acceleration.\n- Navigation: Determining paths and distances in GPS and robotics.\n- Computer Graphics: Simulating movement and transformations.\n- Structural Analysis: Measuring shifts in mechanical components.", "---", "### Conclusion", "The vector equation ( \mathbf{v} = \mathbf{p} - \mathbf{r}_0 = \begin{pmatrix} 5 - 1 \ 4 - (-2) \end{pmatrix} = \begin{pmatrix} 4 \ 6 \end{pmatrix} ) exemplifies how vector subtraction yields meaningful physical information. By carefully computing displacement, physicists and engineers gain quantitative insight into motion, making this foundational operation indispensable across scientific and technological fields.", "---", "### Keywords:\n- Displacement vector\n- Vector subtraction\n- Physics displacement\n- Vector components\n- Motion analysis\n- \mathbf{v} = \mathbf{p} - \mathbf{r}_0\n- Coordinate displacement\n- Vector math in physics\n- Relative position vector"]
