Problem 61 - Thomas Calculus

Question

Given two lines in space, either they are parallel, they intersect, or they are skew (lie in parallel planes). In Exercises 61 and \(62,\) determine whether the lines, taken two at a time, are parallel, intersect, or are skew. If they intersect, find the point of intersection. Otherwise, find the distance between the two lines. $$L 1 : x=3+2 t y=-1+4 t z=2-t -\infty < t < \infty$$ $$L 2 : x=1+4 s y=1+2 s z=-3+4 s -\infty < s < \infty$$ $$L 3 : x=3+2 r y=2+r z=-2+2 r -\infty < r < \infty$$

Step-by-Step Solution

Answer

All pairs of lines are skew.

Step 1: Identify Line Direction Vectors

The direction vector of a line can be identified from the coefficients of the parameter in its parametric form. For Line 1, the direction vector is \( \mathbf{d}_1 = \langle 2, 4, -1 \rangle \). For Line 2, the direction vector is \( \mathbf{d}_2 = \langle 4, 2, 4 \rangle \). For Line 3, the direction vector is \( \mathbf{d}_3 = \langle 2, 1, 2 \rangle \).

Step 2: Determine Line Parallelism

Two lines are parallel if their direction vectors are scalar multiples of each other. Check \( \mathbf{d}_1 \) and \( \mathbf{d}_2 \) by calculating \( 2 \cdot \langle 4, 2, 4 \rangle = \langle 8, 4, 8 \rangle eq \langle 2, 4, -1 \rangle \). Thus, \( \mathbf{d}_1 \) is not a multiple of \( \mathbf{d}_2 \). Similarly, \( 2 \cdot \langle 2, 1, 2 \rangle = \langle 4, 2, 4 \rangle eq \langle 2, 4, -1 \rangle \) implies \( \mathbf{d}_1 \) is not parallel to \( \mathbf{d}_3 \), and \( \mathbf{d}_2 \) is not parallel to \( \mathbf{d}_3 \). Therefore, none of the lines are parallel.

Step 3: Check for Intersection (Lines 1 and 2)

To determine if Line 1 intersects Line 2, equate the parametric equations: \( x: 3 + 2t = 1 + 4s \), \( y: -1 + 4t = 1 + 2s \), \( z: 2 - t = -3 + 4s \). Solving this system, we find there is no solution that satisfies all three equations simultaneously. Therefore, Lines 1 and 2 are not intersecting.

Step 4: Check for Intersection (Lines 1 and 3)

Set up the system for intersection: \( x: 3 + 2t = 3 + 2r \), \( y: -1 + 4t = 2 + r \), \( z: 2 - t = -2 + 2r \). Solving for \( t \) and \( r \), \( t = -2 \), \( r = 0 \) matches \( z \) equations but not \( y \). This indicates Lines 1 and 3 do not intersect.

Step 5: Check for Intersection (Lines 2 and 3)

Set up the parameter system: \( x: 1 + 4s = 3 + 2r \), \( y: 1 + 2s = 2 + r \), \( z: -3 + 4s = -2 + 2r \). Solving gives no consistent values for \( s \) and \( r \) across all equations, meaning Lines 2 and 3 do not intersect.

Step 6: Determine Lines are Skew

Since none of the pairs of lines are either parallel or intersecting, each pair of lines must be skew. Lines are skew if they do not lie on the same plane and cannot meet at any point, which is confirmed by our calculations.

Key Concepts

Parametric Equations

Parametric equations provide a way to represent a line in three-dimensional space using a parameter. Such a form is helpful in describing points along the line based on a variable, commonly denoted as a parameter like \( t \), \( s \), or \( r \). Each component x, y, and z in the equation varies based on this parameter. For our line \( L_1 \):

• \( x = 3 + 2t \)
• \( y = -1 + 4t \)
• \( z = 2 - t \)

As the parameter \( t \) changes, it identifies different points along the line \( L_1 \). Similar equations apply to lines \( L_2 \) and \( L_3 \), but with parameters \( s \) and \( r \) respectively. The beauty of parametric equations is their ability to easily express lines in space, making it easier to analyze their relationships, such as checking for intersections or parallelism.

Direction Vectors

Direction vectors are derived from the coefficients of the parameter in the parametric equations of the lines. These vectors specify the direction in which the line extends. For example, the direction vectors for the lines are:

• \( L_1 \): \( \mathbf{d}_1 = \langle 2, 4, -1 \rangle \)
• \( L_2 \): \( \mathbf{d}_2 = \langle 4, 2, 4 \rangle \)
• \( L_3 \): \( \mathbf{d}_3 = \langle 2, 1, 2 \rangle \)

Having these vectors allows us to compare their directions. For instance, if two direction vectors are scalar multiples of each other, then the lines they represent are parallel. Direction vectors are crucial in determining whether lines are parallel, skew, or intersecting, and they help in calculations related to projections and intersections.

Line Intersection

To find the intersection of two lines, set their parametric equations equal to each other, point by point. Consider lines \( L_1 \) and \( L_2 \):

• \( 3 + 2t = 1 + 4s \)
• \( -1 + 4t = 1 + 2s \)
• \( 2 - t = -3 + 4s \)

Solving these equations simultaneously would give you any intersecting point by finding values of \( t \) and \( s \) that work across all three equations. However, in this scenario for lines \( L_1 \) and \( L_2 \), no consistent solution exists, which indicates that these lines do not intersect. This same process is used to check for intersections with the other pairings: \( L_1 \) with \( L_3 \), and \( L_2 \) with \( L_3 \). If a solution works only partially or not at all, the lines are not intersecting anywhere in space.

Line Parallelism

Line parallelism involves comparing direction vectors. If the direction vectors of two lines are scalar multiples, the lines are parallel. For example, if we consider lines \( L_1 \) and \( L_2 \), we compute:

• Check if \( \mathbf{d}_1 = \langle 2, 4, -1 \rangle \) is a scalar multiple of \( \mathbf{d}_2 = \langle 4, 2, 4 \rangle \).

This does not hold true as there is no single scalar that satisfies all components simultaneously. We perform similar checks for \( L_1 \) and \( L_3 \), as well as \( L_2 \) and \( L_3 \). None satisfy parallelism conditions, so they are not parallel. Recognizing line parallelism quickly trims down the analysis, confirming whether lines can be skew or intersecting.