Problem 24 - Advanced Engineering Mathematics

Question

Find symmetric equations for the line through \((4,-11,-7)\) that is parallel to the line \(x=2+5 t, y=-1+\frac{1}{3} t, z=9-2 t\).

Step-by-Step Solution

Answer

\(\frac{x-4}{5} = \frac{y+11}{1/3} = \frac{z+7}{-2}\)

Step 1: Understand the Given Line Equations

The line is represented by the parametric equations: \(x = 2 + 5t\), \(y = -1 + \frac{1}{3}t\), \(z = 9 - 2t\). The line direction can be determined as a vector: \(\langle 5, \frac{1}{3}, -2 \rangle\).

Step 2: Note the Point and Direction Vector of New Line

The new line must pass through the point \((4,-11,-7)\) and be parallel to the given line. Therefore, it will have the same direction vector \(\langle 5, \frac{1}{3}, -2 \rangle\) as the given line.

Step 3: Write Parametric Equations for the New Line

Using the direction vector \(\langle 5, \frac{1}{3}, -2 \rangle\) and the point \((4,-11,-7)\), the parametric equations are: \(x = 4 + 5s\), \(y = -11 + \frac{1}{3}s\), and \(z = -7 - 2s\), where \(s\) is a parameter.

Step 4: Find Symmetric Equations from Parametric Equations

To convert parametric equations into symmetric form, solve each for \(s\): \(s = \frac{x-4}{5}\), \(s = \frac{y+11}{1/3}\), \(s = \frac{z+7}{-2}\). Thus, the symmetric equations are: \(\frac{x-4}{5} = \frac{y+11}{1/3} = \frac{z+7}{-2}\).

Step 5: Adjust Symmetric Equations for Simplicity

To simplify, multiply \(\frac{y+11}{1/3}\) by 3 to get the symmetric equation: \(\frac{x-4}{5} = 3(y+11) = \frac{z+7}{-2}\). The equations are now: \(\frac{x-4}{5} = \frac{y+11}{1/3} = \frac{z+7}{-2}\).

Key Concepts

Parametric Equations

Parametric equations are a way to express the coordinates of the points that make up a geometric object, such as a line, using one or more variables, called parameters. In the context of lines, these equations describe how each coordinate of a point on the line changes with the parameter.

For instance, the given line in the problem is:

• \(x = 2 + 5t\),
• \(y = -1 + \frac{1}{3}t\),
• \(z = 9 - 2t\),

where \(t\) is a parameter that varies over all real numbers. As \(t\) changes, different points are described along the line. The starting point of this line when \(t = 0\) is the point \((2, -1, 9)\). As \(t\) changes, the line extends in the direction defined by the coefficients of \(t\).

To find parametric equations for the new line that passes through a different point, we use the same method while substituting the starting point with the given point of the new line.

Direction Vector

The direction vector is crucial for describing the orientation of a line in space. It indicates the line's direction and can be derived from the parametric equations of the line.

In the given problem, the direction vector is

• \(\langle 5, \frac{1}{3}, -2 \rangle\)

These values represent how much the line moves along the x, y, and z axes as the parameter \(t\) increases by 1. This vector ensures that when you create a new line that must be parallel to the existing one, you use the same direction vector.

For the new line that is parallel to the existing line and passes through the point \((4, -11, -7)\), we maintain the direction vector \(\langle 5, \frac{1}{3}, -2 \rangle\), which means the new line will head in the same direction as the original line.

Parallel Lines

Parallel lines are lines in the same plane that never intersect. They have the same direction vector. This indicates they run alongside each other, maintaining a consistent distance throughout their lengths. In terms of parametric equations, parallel lines share the same direction vector, although they likely have different initial points.

In our exercise, we are tasked with finding a line that is parallel to another. To do this, we ensure our new line shares the same direction vector \(\langle 5, \frac{1}{3}, -2 \rangle\) as the existing line. Such parallelism guarantees that the new line will precisely mimic the directionality of the original, effectively making the two lines parallel to each other.

Symmetric Form

The symmetric form of a line is another representation that eliminates the parameter, providing a direct relationship between the coordinates. This form is beneficial as it simplifies the equation to show a line's orientation in space without the explicit dependence on a variable parameter.

From the exercise, the parametric equations \(x = 4 + 5s\), \(y = -11 + \frac{1}{3}s\), and \(z = -7 - 2s\) are solved for \(s\). By equating these separate expressions for \(s\), we derive the symmetric form

• \(\frac{x-4}{5} = \frac{y+11}{1/3} = \frac{z+7}{-2}\).

To make the expression cleaner and easier to understand, we adjust it to

• \(\frac{x-4}{5} = 3(y+11) = \frac{z+7}{-2}\).

The symmetric form helps visualize the relationship and correlation between different coordinates, providing insight into the spatial orientation of the line.