Problem 3 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the parametric equations of the line through the given pair of points. \((4,2,3),(6,2,-1)\)

Step-by-Step Solution

Answer

The parametric equations are \(x=4+2t\), \(y=2\), \(z=3-4t\).

Step 1: Find the Direction Vector

To find the parametric equation of a line through two points, we first need the direction vector. This vector can be found by subtracting the coordinates of the first point from the coordinates of the second point. Let the points be \((x_1, y_1, z_1) = (4, 2, 3)\) and \((x_2, y_2, z_2) = (6, 2, -1)\). The direction vector \(\mathbf{d}\) is \((x_2-x_1, y_2-y_1, z_2-z_1) = (6-4, 2-2, -1-3) = (2, 0, -4)\).

Step 2: Write the Parametric Equations

Now, use the direction vector and one of the points to write the parametric equations of the line. Using point \((4, 2, 3)\) and direction vector \((2, 0, -4)\), the parametric equations are: \(x = 4 + 2t\), \(y = 2 + 0t = 2\), \(z = 3 - 4t\), where \(t\) is a parameter.

Key Concepts

Direction Vector

In the world of parametric equations, the direction vector plays a crucial role. Think of it as the arrow pointing the way your line will move across the 3D space. To find this vector, you need to observe the change in each coordinate from one point to the other. In the provided exercise, we are given two points

• \((x_1, y_1, z_1) = (4, 2, 3)\)
• \((x_2, y_2, z_2) = (6, 2, -1)\)

To calculate the direction vector \(\mathbf{d}\), you use the formula: \((x_2 - x_1, y_2 - y_1, z_2 - z_1)\). Plugging in the given numbers, we subtract the coordinates of the first point from the second:

• For \(x\): \(6 - 4 = 2\)
• For \(y\): \(2 - 2 = 0\)
• For \(z\): \(-1 - 3 = -4\)

The result is the direction vector \((2, 0, -4)\). This vector indicates how the line will extend—two steps forward in the \(x\) direction, no change in the \(y\) direction, and four steps back in the \(z\) direction. This vector is integral to forming the parametric equations of the line.

Line Through Points

Forming a line through points in 3D space involves a combination of a given point and the direction vector. First, let's remind ourselves why we need a direction vector and a starting point:

• The starting point gives us the initial position from which the line begins its path through space.
• The direction vector dictates how the line will project from that starting point.

In the exercise, we are developing the parametric equations from the point \((4, 2, 3)\) and the direction vector \((2, 0, -4)\). Using a parameter, usually \(t\), the line is expressed as \(x = x_0 + at\), \(y = y_0 + bt\), \(z = z_0 + ct\), where \((a, b, c)\) are components of the direction vector. So, we write:

• \(x = 4 + 2t\)
• \(y = 2 + 0t = 2\)
• \(z = 3 - 4t\)

The parameter \(t\) is typically understood as a time variable indicating how far along the line you are, which allows the line to extend infinitely through the space in its designated direction.

3D Coordinate Geometry

In 3D coordinate geometry, you’re working with three axes: \(x\), \(y\), and \(z\). Together, they define a space where any point can be represented as \((x, y, z)\), distinctly different from 2D geometry, which only utilizes the \(x\) and \(y\) axes. This third dimension adds depth to lines, vectors, and shapes.
When you work with lines in this space, as demonstrated by the problem-solving strategy with the two points and direction vector, it provides a more comprehensive look into how objects interact within the three dimensions. By using parametric equations, we express lines not just in terms of \(x\) and \(y\) but with the added \(z\) component, opening up applications in real-world scenarios like physics and engineering.

• The \(x\), \(y\), and \(z\) values in parametric equations allow the line to traverse effortlessly through 3D space.
• Each coordinate provides a detailed path that clarifies the positioning and orientation of the line.
• Applications in this space are vast, stretching to fields such as computer graphics, 3D modeling, and vector calculus.

Understanding 3D geometry and its components ensures a solid foundation for more advanced explorations into mathematics and its practical applications.