Problem 7 - Essential Calculus Early Transcendentals

Question

\(6-10=\) Find parametric equations and symmetric equations for the line. The line through the points \(\left(0, \frac{1}{2}, 1\right)\) and \((2,1,-3)\)

Step-by-Step Solution

Answer

The parametric equations are \( x=2t, y=\frac{1}{2} + \frac{1}{2}t, z=1-4t \) and the symmetric equation is \( \frac{x}{2} = 2y-1 = \frac{1-z}{4} \).

Step 1: Find Direction Vector

To find the direction vector of the line, subtract the coordinates of the first point from the corresponding coordinates of the second point. Given points are \( (0, \frac{1}{2}, 1) \) and \( (2, 1, -3) \).Direction vector \( \mathbf{d} \) is:\[ \mathbf{d} = (2-0, 1-\frac{1}{2}, -3-1) = (2, \frac{1}{2}, -4) \].

Step 2: Find Parametric Equations

Using the direction vector \( \mathbf{d} = (2, \frac{1}{2}, -4) \) and one of the points, say \( (0, \frac{1}{2}, 1) \), the parametric equations of the line are:\[ x = 0 + 2t \]\[ y = \frac{1}{2} + \frac{1}{2}t \]\[ z = 1 - 4t \] where \( t \) is the parameter.

Step 3: Find Symmetric Equations

The symmetric equations are found from the parametric form by solving each equation for \( t \): From \( x = 2t \), \( t = \frac{x}{2} \).From \( y = \frac{1}{2} + \frac{1}{2}t \), \( t = 2y - 1 \).From \( z = 1 - 4t \), \( t = \frac{1-z}{4} \).Equating these:\[ \frac{x}{2} = 2y - 1 = \frac{1-z}{4} \].

Key Concepts

Direction Vector

A direction vector is a crucial concept when dealing with lines in three-dimensional space. It acts like a guide or an arrow showing the path along the line. To find the direction vector between two points, you simply subtract the coordinates of the starting point from the corresponding coordinates of the ending point.
Take for example our points:

• Starting point: \((0, \frac{1}{2}, 1)\)
• Ending point: \((2, 1, -3)\)

Subtracting these coordinates gives us the direction vector \((2, \frac{1}{2}, -4)\).
This vector describes the change needed to move from one point to the other, essentially giving the line its orientation in space. Knowing the direction vector is essential for constructing the parametric equation of a line, as it provides the "slope" or direction along which the line extends.

Symmetric Equations

Symmetric equations provide a different perspective for representing lines. Instead of expressing the line with a parameter, these equations set up a condition that each point on the line satisfies.
The symmetric form is derived from the parametric equations. Starting from:

• \( x = 0 + 2t \) gives \( t = \frac{x}{2} \)
• \( y = \frac{1}{2} + \frac{1}{2}t \) gives \( t = 2y - 1 \)
• \( z = 1 - 4t \) gives \( t = \frac{1-z}{4} \)

To form the symmetric equation, we equate these expressions for \( t \): \[ \frac{x}{2} = 2y - 1 = \frac{1-z}{4} \]
This equation links \( x \), \( y \), and \( z \) without the need for a parameter like \( t \), making it often easier to analyze the spatial relationship between points on the line.

Linear Algebra

Linear Algebra is a field of mathematics that studies vectors, vector spaces, and linear transformations. It is a foundational tool for understanding the geometry of lines and planes in space, providing the language and framework for explaining concepts like direction vectors and symmetric equations.
When we delve into vectors in Linear Algebra, we encounter:

• **Vectors:** Direct representations of magnitude and direction
• **Scalar multiplication:** Scaling a vector without changing its direction
• **Vector addition and subtraction:** Creating new vectors through combination

These operations allow us to manipulate vector equations to solve problems. In the context of straight lines, Linear Algebra helps derive parametric and symmetric forms from basic vector operations. Understanding these concepts provides insight into how lines can intersect, how they align in space, and how transformations affect their orientation and position.