Question

In Exercises \(9-12,\) find sets of (a) parametric equations and (b) symmetric equations of the line through the two points. (For each line, write the direction numbers as integers.) $$ (5,-3,-2),\left(-\frac{2}{3}, \frac{2}{3}, 1\right) $$

Short answer

The parametric equations of the line through the given points are \(x = 5 - 17/3t\), \(y = -3 + 11/3t\), and \(z = -2 + 3t\). The symmetric equations are \((3x + 51)/-17 = (3y - 9)/11 = z + 2\).

Step-by-step solution

Determining the Direction Vector

The direction vector of a line passing through the points \(P_1(x_1, y_1, z_1)\) and \(P_2(x_2, y_2, z_2)\) is given by \(\vec{d} = P_2 - P_1\), which can be calculated as \(\vec{d} = (x_2 - x_1, y_2 - y_1, z_2 - z_1)\). In this case, \(P_1 = (5, -3, -2)\) and \(P_2 = (-2/3, 2/3, 1)\), so \(\vec{d} = ((-2/3 - 5), (2/3 - -3), (1 - -2)) = (-17/3, 11/3, 3)\).

Formulating the Parametric Equations

To construct the parametric equations of the line, use the point \(P_1\) and the direction vector \(\vec{d}\). The parametric equations are \(x = x_1 + at\), \(y = y_1 + bt\), and \(z = z_1 + ct\), where \((a, b, c)\) is the direction vector and \(t\) is the parameter. Substituting the known values gives \(x = 5 - 17/3t\), \(y = -3 + 11/3t\), and \(z = -2 + 3t\).

Deriving the Symmetric Equations

To derive the symmetric equations of the line, the parameter \(t\) is eliminated from the parametric equations, which gives \((x - x_1)/a = (y - y_1)/b = (z - z_1)/c\). Substituting the known values results in \((x - 5)/-17/3 = (y + 3)/11/3 = (z + 2)/3\). To state the direction numbers as integers, it is helpful to multiply the fractions by 3, which gives \((3x + 51)/-17 = (3y - 9)/11 = z + 2\).

Key concepts

Understanding the Direction Vector

A direction vector is crucial in defining the course of a straight line in three-dimensional space. It not only points the way but also dictates the steepness of the line's path. In the given exercise, the direction vector, denoted as \( \vec{d} \), elegantly encapsulates the line's orientation.

To find this vector, one simply subtracts the coordinates of the starting point \( P_1 \) from the ending point \( P_2 \) for each dimension—X, Y, and Z, respectively. This act of subtraction is a fundamental operation in vector algebra which directly provides us with the components of the direction vector. In the context of our problem, \( \vec{d} = (-17/3, 11/3, 3) \) is that pivotal guidepost.

In representing movement or change in space, each number within \( \vec{d} \) serves as a 'direction number', informing exactly how much we move in each axis for each step along the line. This set of numbers forms the backbone for constructing both parametric and symmetric equations of the line.

Deciphering Symmetric Equations of a Line

Symmetric equations provide a way to express a line in a balanced, viewable form when it spans across multiple dimensions and, unlike parametric equations, do not require an explicit parameter like \( t \). They are obtained by isolating the parameter from the parametric equations, thus displaying the relationship between the coordinates directly.

For our particular example, the symmetric equations attained are \( (3x + 51)/-17 = (3y - 9)/11 = z + 2 \). This format is beneficial as it shows how each coordinate correlates with the others on the line, maintaining a symmetric form that makes analytical and graphical interpretations more straightforward.

Using integers in the symmetric equations aids clarity and simplicity. It's easier for us to conceptualize and work with whole numbers than fractions, especially when discussing the ratios of change among the coordinates. Thus, small adjustments, such as scaling fractions to obtain integers, significantly enhance the usability of these equations.

Vector Algebra: The Language of Direction and Magnitude

Vector algebra is the mathematical language that describes vectors—entities with both magnitude and direction. This field of mathematics is essential when dealing with parametric equations and geometrical interpretations in spaces of any dimension.

Vector operations, such as the addition, subtraction, and scalar multiplication used in this exercise, not only allow us to calculate direction vectors but are also integral in deriving other vector-related quantities, such as dot products and cross products. These quantities and operations are part of the larger picture, encompassing everything from simple displacement to complex physical simulations. In the context of our exercise, vector algebra enables us to move seamlessly from a pair of points to the equations that define the line that intercepts them, highlighting its foundational role in geometry and physics alike.

The elegance of vector algebra lies in its ability to simplify complex spatial problems into solvable equations, thus providing a cornerstone for fields ranging from computer graphics to engineering, and in our case, navigating students through the world of three-dimensional lines.