Problem 13 - Calculus: Early Transcendentals

Question

Determine whether the lines \(\langle 1,0,2\rangle+t\langle-1,-1,2\rangle\) and \(\langle 4,4,2\rangle+t\langle 2,2,-4\rangle\) are paral lel, intersect, or neither.

Step-by-Step Solution

Answer

The lines neither intersect nor are they parallel.

Step 1: Compare Direction Vectors

Evaluate if the direction vectors of the lines are scalar multiples of each other, indicating parallelism. The direction vector of the first line is \( \langle -1,-1,2 \rangle \) and for the second line \( \langle 2,2,-4 \rangle \). The vectors are not scalar multiples of each other, thus the lines are not parallel.

Step 2: Set Parametric Equations

Set each line's parametric equations using their respective vector equations. Line 1 has \( x = 1-t \), \( y = -t \), \( z = 2+2t \) and Line 2 has \( x = 4+2t \), \( y = 4+2t \), \( z = 2-4t \).

Step 3: Solve for Intersections

Equate the respective parametric equations to find a common solution. Set up equations: \( 1-t = 4+2s \), \( -t = 4+2s \), \( 2+2t = 2-4s \). Solve these equations: From \( 2+2t = 2-4s \), we get \( t = -2s \). Substitute into the first and second equations to check for consistency. \( 1 - (-2s) = 4+2s \) yields \( 1+2s = 4+2s \), giving an impossible scenario (1=4), so there is no intersection.

Key Concepts

Direction Vectors

In line geometry, direction vectors play a crucial role in understanding the orientation and parallelism of lines. Each line in space can be represented by a vector equation that includes a direction vector. For the given problem, we have two lines with direction vectors \( \langle -1, -1, 2 \rangle \) and \( \langle 2, 2, -4 \rangle \).

Direction vectors describe the direction in which the line extends. They are vital in determining whether two lines are parallel. If one direction vector is a scalar multiple of another, the lines are parallel. This means we could multiply the components of one vector by a constant to obtain the corresponding components of the other vector.

In this exercise, the direction vectors are not scalar multiples, as there is no constant that can be uniformly applied to \( \langle -1, -1, 2 \rangle \) to yield \( \langle 2, 2, -4 \rangle \). Therefore, the lines are not parallel.

Scalar Multiples

Scalar multiples involve multiplying a vector by a scalar (a real number). This operation alters the magnitude of the vector but not its direction, unless the scalar is negative, which reverses the direction. In the context of line geometry, scalar multiples help identify whether direction vectors indicate parallel lines.

For our lines, check if direction vector \( \langle 2, 2, -4 \rangle \) is a scalar multiple of \( \langle -1, -1, 2 \rangle \). This involves seeing if we can find a number \( k \) such that:

\[ k \langle -1, -1, 2 \rangle = \langle 2, 2, -4 \rangle \]

Solving these component-by-component:

• \( k(-1) = 2 \)
• \( k(-1) = 2 \)
• \( k(2) = -4 \)

If a consistent \( k \) existed in all equations, the lines would be parallel. But here, we find no such \( k \), confirming the lines aren't parallel.

Parametric Equations

To comprehend how lines behave and relate in space, turning a line's vector equation into parametric form is often useful. Parametric equations break down a line equation into separate expressions for \( x \), \( y \), and \( z \), making the relationships clearer.

For the first line in the exercise, the vector form is given as \( \langle 1,0,2\rangle + t\langle-1,-1,2\rangle \), translating to parametric equations:

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

For the second line \( \langle 4,4,2\rangle + t\langle 2,2,-4\rangle \), we derive:

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

These parametric equations allow us to find out when (or if) the lines meet in 3D space.

Intersection of Lines

When looking to find out if two lines intersect, one must focus on solving their parametric equations for a common point. Here, we take parametric equations from both lines and set them equal to one another, checking if a consistent solution exists.

For the problem, the necessary equations are:

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

By isolating \( t \) and \( s \), we can test whether these conditions hold for specific values. Solving \( 2+2t = 2-4s \) gives \( t = -2s \). Substituting \( t \) in the other equations should lead to consistent answers but here results in an inconsistency like \( 1 = 4 \).

Thus, there's no common solution, meaning these lines do not intersect.