Problem 6 - Calculus Early Transcendentals: Pearson New International Edition

Question

Write both the parametric equations and the symmetric equations for the line through the given point parallel to the given vector. $$(-1,3,-6),(-2,0,5)$$

Step-by-Step Solution

Answer

Parametric: \(x = -1 - 2t,\ y = 3,\ z = -6 + 5t\). Symmetric: \(\frac{x + 1}{-2} = \frac{z + 6}{5},\ y = 3\).

Step 1: Identify Point and Direction Vector

The given point is \((-1, 3, -6)\) and the direction vector is \((-2, 0, 5)\). We will use this direction vector for both parametric and symmetric equations.

Step 2: Write Parametric Equations

The parametric equations of a line parallel to a vector \((a, b, c)\) through a point \((x_0, y_0, z_0)\) are:\[ x = x_0 + at, \]\[ y = y_0 + bt, \]\[ z = z_0 + ct \]Substitute the given point and vector:\[ x = -1 - 2t, \]\[ y = 3 + 0t, \]\[ z = -6 + 5t \] This simplifies to:\[ x = -1 - 2t, \]\[ y = 3, \]\[ z = -6 + 5t \]

Step 3: Derive Symmetric Equations

The symmetric equations are derived from the parametric equations by eliminating the parameter \(t\). Set each equation equal to \(t\) and solve:For \(x = -1 - 2t\): \[ t = \frac{x + 1}{-2} \]For \(y = 3\), no \(t\) is involved, meaning it is a constant.For \(z = -6 + 5t\):\[ t = \frac{z + 6}{5} \]Set them equal:\[ \frac{x + 1}{-2} = \frac{z + 6}{5} \]Since \(y\) is constant, the equation for \(y\) is simply \(y = 3\).

Step 4: Write Final Symmetric Equations

The final symmetric equations of the line are:\[ \frac{x + 1}{-2} = \frac{z + 6}{5} \] and \(y = 3\).

Key Concepts

Line Equations

Understanding line equations is fundamental in geometry, and it involves defining a line using mathematical expressions.
A line can be expressed in various forms such as parametric, symmetric, and vector equations depending on what information is provided or required.
For a line given by a point \((x_0, y_0, z_0)\) and direction vector \((a, b, c)\), parametric equations describe how each coordinate of the line depends on a parameter (often denoted as \(t\)). This is expressed as:

• \(x = x_0 + at\)
• \(y = y_0 + bt\)
• \(z = z_0 + ct\)

Lines allow us to explore linear relationships and are the foundation for understanding more complex shapes and concepts in geometry.

Symmetric Equations

Symmetric equations provide a convenient method to describe a line without explicitly using a parameter. They are especially useful in solving problems involving intersection of lines or finding coplanar lines.
To derive symmetric equations, we eliminate the parameter \(t\) from parametric forms. For instance, consider the equations:

• \(x = -1 - 2t\)
• \(y = 3\)
• \(z = -6 + 5t\)

To remove \(t\), solve each equation for \(t\) and equate them, resulting in:

• \(t = \frac{x + 1}{-2}\)
• \(t = \frac{z + 6}{5}\)

Thus, the symmetric equations become:

• \(\frac{x + 1}{-2} = \frac{z + 6}{5}\)
• \(y = 3\)

These equations capture the essence of the line in a more straightforward manner compared to parametrics.

Vector Algebra

Vector algebra is a powerful tool in both geometry and physics, enabling us to succinctly describe lines, planes, and spaces in n-dimensional settings.
A vector has both magnitude and direction, which makes it an ideal representation for quantities like force or velocity.
In the context of line equations, vectors help us characterize the direction of the line.
Given a direction vector \((a, b, c)\), it indicates the direction in which the line extends.

• The direction vector also aids in performing operations like dot product, cross product, and determining angles between vectors in space.
• The vector equation of a line can be formulated as: \( \mathbf{r} = \mathbf{r_0} + t \mathbf{v} \), where \( \mathbf{r_0} \) is the position vector of a point on the line, and \( \mathbf{v} \) is the direction vector.

This algebraic approach provides a robust framework for solving geometric problems.

3D Geometry

Three-dimensional geometry extends concepts of lines and shapes from two dimensions into three, adding depth to our mathematical models.
Working in 3D allows us to describe the world around us with more realism, where every point, line, and object has an x, y, and z coordinate.
Key features include:

• The use of the Cartesian coordinate system to pinpoint locations in space.
• Understanding of planes, which are flat surfaces extending infinitely and can be described using vector and line equations.
• Applications include computer graphics, engineering, and designing physical spaces.

3D geometry, paired with vector algebra, allows for the exploration of complex structures and movements within a spatial context, forming an integral part of both theoretical and applied mathematics.