Problem 20 - Essential Calculus Early Transcendentals

Question

\(17-20=\) Determine whether the lines \(L_{1}\) and \(L_{2}\) are parallel, skew, or intersecting. If they intersect, find the point of intersection. $$ \begin{array}{l}{L_{1} : \frac{x}{1}=\frac{y-1}{-1}=\frac{z-2}{3}} \\ {L_{2} : \frac{x-2}{2}=\frac{y-3}{-2}=\frac{z}{7}}\end{array} $$

Step-by-Step Solution

Answer

The lines are skew; they do not intersect and are not parallel.

Step 1: Express Lines in Vector Form

Convert the given symmetric equations of lines into vector form. \[ \text{Line } L_1: \vec{r}_1 = \begin{pmatrix} 0 \ 1 \ 2 \end{pmatrix} + t \begin{pmatrix} 1 \ -1 \ 3 \end{pmatrix} \] \[ \text{Line } L_2: \vec{r}_2 = \begin{pmatrix} 2 \ 3 \ 0 \end{pmatrix} + s \begin{pmatrix} 2 \ -2 \ 7 \end{pmatrix} \] Here, \( t \) and \( s \) are parameters representing any point on their respective lines.

Step 2: Check if Lines are Parallel

Determine if the direction vectors of the lines are scalar multiples of each other, which would indicate parallelism. Compare \( \begin{pmatrix} 1 \ -1 \ 3 \end{pmatrix} \) and \( \begin{pmatrix} 2 \ -2 \ 7 \end{pmatrix} \). Checking the ratios: \[ \frac{1}{2} = \frac{-1}{-2} = \frac{3}{7} \] Since the ratios are not equal, the lines are not parallel.

Step 3: Find if Lines Intersect

Set the parametric equations of the lines equal to find any common point. Solve the system of equations: \[ 0 + t = 2 + 2s \] \[ 1 - t = 3 - 2s \] \[ 2 + 3t = 7s \] From the first equation, \( t = 2 + 2s \). Substituting \( t = 2 + 2s \) into the second, simplify to find \( t \) and \( s \).

Step 4: Solve the System of Equations

Solve \( 1 - (2 + 2s) = 3 - 2s \) and \( 2 + 3(2 + 2s) = 7s \). Simplifying: \[-1 - 2s = 3 - 2s \Rightarrow -1 = 3 \], which is a contradiction. Similarly,\[ 8 + 6s = 7s \Rightarrow s = 8 \] presents inconsistency. Thus, no solution exists, and the lines do not intersect.

Step 5: Conclude Line Relationship

Since the lines are neither parallel nor intersect, they must be skew lines; they do not lie on the same plane and never meet.

Key Concepts

Parametric Equations

Parametric equations are a way of expressing the coordinates of the points on a line as dependent on one or more parameters, usually denoted as "t" or "s". In the context of vector equations, these parameters help describe the entire line and allow every point on the line to be expressed as an extension from a known point with the line's direction vector.

For a line in space, the parametric equations take the form of:

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

Here, \(x_0, y_0, z_0\) represent a point on the line, \(a, b, c\) are the components of the direction vector, and \(t\) is the parameter. For each value of \(t\), you get a corresponding point on the line. The beauty of parametric equations lies in their flexibility to easily represent complex geometric figures. They provide an intuitive way to handle lines, especially when dealing with intersections and parallelism.

Parallel Lines

Two lines are parallel if they have the same direction or, in vector terms, if one direction vector is a scalar multiple of the other. This means the lines do not diverge from one another as they extend in space.

Using the vectors from the exercise:

• Direction vector for \( L_1: \begin{pmatrix} 1 \ -1 \ 3 \end{pmatrix}\)
• Direction vector for \( L_2: \begin{pmatrix} 2 \ -2 \ 7 \end{pmatrix}\)

To check if these vectors are parallel, we compare the ratios of their corresponding components: \( \frac{1}{2}, \frac{-1}{-2}, \frac{3}{7} \).

If all these ratios were equal, the lines would be parallel. However, in our case, the different ratios indicate the lines are not parallel. Thus, there's no constant "stretcher" or "shrinker" that transforms one line into the other along its direction.

Intersection of Lines

For two lines to intersect, they must meet at a single point in the space they occupy. This happens when their parametric equations yield the same solution for the parameter values "t" (for one line) and "s" (for the other).

In the given problem, we attempted to solve them as a system of equations:

• \( 0 + t = 2 + 2s \)
• \( 1 - t = 3 - 2s \)
• \( 2 + 3t = 7s \)

By substituting \( t = 2 + 2s \) into other equations, we aimed to find a common \( t \) and \( s \). The inconsistencies that arose in the solutions, where \(-1 = 3\), reveal no real intersection point.

Since the lines are neither parallel nor intersecting, we conclude they are skew. They exist in different planes and never cross each other's paths.