Problem 16 - Advanced Engineering Mathematics

Question

Find the vector \(P_{1} P_{2}\). Graph \(P_{1} P_{2}\) and its corresponding position vector. \(P_{1}(-2,-1), P_{2}(4,-5)\)

Step-by-Step Solution

Answer

The vector \(P_1 P_2\) is \((6, -4)\), and it can be graphed from \((-2,-1)\) to \((4,-5)\).

Step 1: Identify the Coordinates

First, recognize the coordinates of points \(P_1\) and \(P_2\). \(P_1\) is at \((-2,-1)\) and \(P_2\) is at \((4,-5)\).

Step 2: Use the Vector Formula

To find the vector \(P_1P_2\), use the formula for the vector between two points: \(P_1P_2 = (x_2 - x_1, y_2 - y_1)\). For \(P_1 = (-2,-1)\) and \(P_2 = (4,-5)\), calculate the differences: \[(x_2 - x_1, y_2 - y_1) = (4 - (-2), -5 - (-1)) = (4 + 2, -5 + 1)\].

Step 3: Compute the Vector Components

Calculate each component of the vector based on the differences identified: \((4 + 2, -5 + 1) = (6, -4)\). Therefore, the vector \(P_1P_2\) is \((6, -4)\).

Step 4: Graph the Vector

To graph the vector \(P_1P_2\) starting from \(P_1\) at \((-2,-1)\), move horizontally 6 units to the right and vertically 4 units down. The vector ends at \(P_2\), which is at the coordinates \((4,-5)\). Draw an arrow from \((-2,-1)\) to \((4,-5)\).

Step 5: Create the Position Vector

A position vector is typically drawn from the origin \((0,0)\) to a point. For \(P_1P_2\), interpret the vector as a position vector \((6, -4)\), meaning it starts at the origin \((0,0)\) and ends at \((6,-4)\). This can be drawn similarly with an arrow starting at \((0,0)\) and pointing to \((6,-4)\).

Key Concepts

Vector Operations

Vector operations include addition, subtraction, and scalar multiplication. Understanding how to calculate these is crucial when working with vectors.

In this exercise, the focus is on finding the vector between two points. Using the vector formula, we subtract the coordinates of the initial point from the final point. Here, the coordinates \(P_1 = (-2, -1)\) and \(P_2 = (4, -5)\) were involved.

The calculation \(P_1P_2 = (x_2 - x_1, y_2 - y_1)\) is simple:

• Subtract the x-coordinates: \((4 - (-2)) = 6\)
• Subtract the y-coordinates: \((-5 - (-1)) = -4\)

This results in the vector \(P_1P_2 = (6, -4)\). It represents the direction and magnitude from the point \(P_1\) to \(P_2\), essential for understanding their spatial relationship.

Graphing Vectors

Graphing vectors is a visual method to understand and interpret vector direction and magnitude.

To graph a vector, start at the initial point. Here, for the vector \(P_1P_2\) with coordinates from \((-2,-1)\) to \( (4,-5)\), we use simple steps:

• From the point \(-2,-1\), move 6 units right to reach \(4\) in the x-direction.
• Then, go 4 units down to \(-5\) in the y-direction.

Mark the endpoint and draw an arrow from the starting point to this marked end. This arrow helps to visualize the vector's impact physically.

Additionally, graphing a position vector from the origin to \(6,-4\) further solidifies understanding. It provides a clear view of how much and in which direction the vector moves from the origin.

Coordinate Geometry

Coordinate geometry seamlessly integrates algebra with geometry to analyze points, lines, and shapes in a plane.

With vectors in coordinate geometry, we leverage coordinates to track vectors' properties. Here, we're anchoring calculations to specific points \(P_1 = (-2, -1)\) and \(P_2 = (4, -5)\) in a plane.

The vector \(P_1P_2 = (6, -4)\) derived through subtraction displays the precise translation in the coordinate plane:

• The x-component indicates a move right by 6 units.
• The y-component shows a downward travel by 4 units.

These vector components facilitate understanding transformations and movements in space. Position vectors like \( (6, -4) \) drawn from the origin also bridge concepts between coordinate systems and spatial interpretations, crucial for complex geometrical problems.