Problem 14 - Calculus: Early Transcendentals

Question

Determine whether the lines \(\langle 1,2,-1\rangle+t\langle 1,2,3\rangle\) and \(\langle 1,0,1\rangle+t\langle 2 / 3,2,4 / 3\rangle\) are parallel, intersect, or neither

Step-by-Step Solution

Answer

The lines are neither parallel nor intersecting.

Step 1: Compare Direction Vectors

The first step is to extract the direction vectors of the lines. The direction vector of the first line is \( \langle 1, 2, 3 \rangle \) and the direction vector of the second line is \( \langle \frac{2}{3}, 2, \frac{4}{3} \rangle \). For the lines to be parallel, these vectors should be scalar multiples of each other.

Step 2: Check for Parallelism

We check if a scalar \( k \) exists such that \( \langle 1, 2, 3 \rangle = k \langle \frac{2}{3}, 2, \frac{4}{3} \rangle \). Comparing components, if they are proportional, then \( \frac{1}{\frac{2}{3}} = \frac{2}{2} = \frac{3}{\frac{4}{3}} \). However, \( \frac{1}{\frac{2}{3}} = \frac{3}{2} \), which does not equal \( 1 \) or \( \frac{9}{4} \), indicating no common scalar \( k \). Thus, the lines are not parallel.

Step 3: Solve for Intersection

To check for intersection, we equate the lines and find if there exists values of \( t \) and \( s \) such that: \( \langle 1 + t, 2 + 2t, -1 + 3t \rangle = \langle 1 + \frac{2}{3}s, 2s, 1 + \frac{4}{3}s \rangle \). Solving these equations: \( 1+t = 1+\frac{2}{3}s \), \( 2+2t = 2s \), and \( -1+3t = 1+\frac{4}{3}s \) gives no consistent solution.

Step 4: Conclusion about Intersecting

Solving the simultaneous equations shows inconsistency, indicating no real intersection point where all components satisfy both lines simultaneously.

Key Concepts

Direction Vectors

In vector calculus, a direction vector defines the direction of a line in space. Each line can be represented using a point on the line and a direction vector. For example, the line \( \langle 1,2,-1\rangle + t\langle 1,2,3\rangle \) has the direction vector \( \langle 1,2,3\rangle \). A direction vector is crucial because it helps determine the relationship between lines, such as being parallel or intersecting.
Understanding direction vectors makes it easier to visualize lines in multi-dimensional environments. When given in component form, the notation \( \langle a, b, c \rangle \) indicates the direction across each respective dimension. This helps in assessing how two vectors might relate to each other.
Direction vectors allow us to conduct further operations, like adding vectors or calculating cross products. Comprehending these operations supports deeper insights into the behavior of lines within vector spaces.

Parallel Lines

Lines are considered parallel if their direction vectors are multiples of each other. Specifically, two lines are parallel if one direction vector is a scalar multiple of another. This means that there exists a scalar \( k \) such that:

• \( \langle a, b, c \rangle = k \langle d, e, f \rangle \)

In the given exercise, checking if the lines are parallel involves comparing their direction vectors: \( \langle 1, 2, 3 \rangle \) and \( \langle \frac{2}{3}, 2, \frac{4}{3} \rangle \).
By examining the ratios \( \frac{1}{\frac{2}{3}}, \frac{2}{2}, \frac{3}{\frac{4}{3}} \), we attempt to find the same scalar \( k \) for all components. The failure to do so signifies that the lines aren't parallel. For lines to be genuinely parallel, the ratios must remain consistent across all components, demonstrating identical directional proportion.

Intersection of Lines

To determine if two lines intersect, solve their respective line equations simultaneously:

• Line 1: \( \langle 1 + t, 2 + 2t, -1 + 3t \rangle \)
• Line 2: \( \langle 1 + \frac{2}{3}s, 2s, 1 + \frac{4}{3}s \rangle \)

Putting this into practice requires equating the components and solving for parameters \( t \) and \( s \). If a solution exists, the lines intersect at the corresponding point.
For this exercise, the equations yield no consistent solution when solved collectively, indicating the lines do not meet at any point. Algebraically, an intersection occurs only if a set of values satisfies all dimensional equations. Failed attempts in finding consistent values mean no positional overlap, and consequentially, no intersection happens for the lines in question.

Scalar Multiples

Scalar multiplication in vector calculus involves scaling a vector by a number, which scales each component of the vector proportionally. Let's illustrate with a vector \( \langle a, b, c \rangle \) and a scalar \( k \):

• When multiplied, the vector becomes \( \langle ka, kb, kc \rangle \)

In the exercise, scalar multiples help compare vector parallelism. Specifically, we checked if \( \langle 1, 2, 3 \rangle \) could be scaled to become \( \langle \frac{2}{3}, 2, \frac{4}{3} \rangle \).
Since no single scalar satisfies all component ratios consistently, the vectors aren't scalar multiples of one another. This approach demonstrates utilizing scalar multiplication to solve vector-related problems, particularly in identifying parallel lines.