Problem 12 - Calculus: Early Transcendentals

Question

Determine whether the lines \(\langle 1,3,-1\rangle+t\langle 1,1,0\rangle\) and \(\langle 0,0,0\rangle+t\langle 1,4,5\rangle\) are parallel, intersect, or neither.

Step-by-Step Solution

Answer

The lines are neither parallel nor intersecting.

Step 1: Identify Direction Vectors

The direction vector of the first line is given by \( \langle 1, 1, 0 \rangle \), and the direction vector of the second line is \( \langle 1, 4, 5 \rangle \). To determine if the lines are parallel, these direction vectors must be scalar multiples of each other.

Step 2: Check for Parallelism

Compare the direction vectors \( \langle 1, 1, 0 \rangle \) and \( \langle 1, 4, 5 \rangle \). For the vectors to be parallel, there must exist a scalar \( k \) such that \( k \langle 1, 1, 0 \rangle = \langle 1, 4, 5 \rangle \). Notice that there is no single \( k \) that satisfies all component equations: \( k \cdot 1 = 1 \), \( k \cdot 1 = 4 \), and \( k \cdot 0 = 5 \). Thus, the lines are not parallel.

Step 3: Set Lines Equal to Find Intersection

To find if the lines intersect, set their parametric equations equal: \( \langle 1 + t, 3 + t, -1 \rangle = \langle s, 4s, 5s \rangle \). This results in three component equations: \( 1 + t = s \), \( 3 + t = 4s \), and \( -1 = 5s \).

Step 4: Solve System of Equations

First, solve \( -1 = 5s \) for \( s \): \( s = -\frac{1}{5} \). Substitute into \( 1 + t = s \) to get \( 1 + t = -\frac{1}{5} \), solving for \( t \) gives \( t = -\frac{6}{5} \). Use this \( t \) in the second equation \( 3 + t = 4s \) to check consistency: \( 3 - \frac{6}{5} = 4\left(-\frac{1}{5}\right) \). Simplify the left side: \( \frac{15}{5} - \frac{6}{5} = \frac{9}{5} \), which does not equal \( -\frac{4}{5} \). Hence, the lines do not intersect.

Key Concepts

Direction Vectors

In vector calculus, direction vectors play a crucial role in understanding the orientation and direction of lines in space. They essentially tell us which direction a line is heading.
A vector is simply an arrow that has a starting point and a direction. In a three-dimensional space, it’s expressed with three components, for example, \( \langle a, b, c \rangle \).

• The direction vector of a line doesn’t indicate the line’s position, only its direction.
• Each of the vectors in the exercise corresponds to a line in space, with the line stretching infinitely in the vector's direction.

Understanding direction vectors helps in determining various line properties, such as parallelism and intersections.

Parametric Equations

Parametric equations provide a way to express the position of any point on a line using a parameter, often denoted as \( t \). This method is quite useful in vector calculus.
A line in vector form is typically expressed as: \( \vec{r}(t) = \vec{r_0} + t\vec{v} \). Here:

• \( \vec{r_0} \) is a point on the line,
• \( \vec{v} \) is the direction vector,
• \( t \) is the parameter, indicating different points along the line.

In the exercise, you have the two parametric equations expressing the different lines.
Understanding these equations helps us analyze how the lines behave as \( t \) changes.

Parallel Lines

Determining whether lines are parallel involves their direction vectors. If two lines are parallel, their direction vectors will be parallel, meaning one vector is a scalar multiple of the other.

• To check for parallelism, compare the components of the direction vectors of the lines.
• If the vector \( \langle a, b, c \rangle \) is a scalar multiple of \( \langle x, y, z \rangle \), then \( a = kx \), \( b = ky \), and \( c = kz \) must be true for some scalar \( k \).

In our exercise, the direction vectors \( \langle 1, 1, 0 \rangle \) and \( \langle 1, 4, 5 \rangle \) were compared, but no such scalar was found, indicating that the lines are not parallel.

Intersection of Lines

Finding the intersection of lines involves checking if there is a common point shared by the lines. In the context of parametric equations, this means finding a set of parameter values (\( t \) and \( s \)) that satisfy both parametric equations simultaneously.
To do this, set the parametric equations of both lines equal and solve the resulting system of equations. For our exercise:
1. Start with the given equations: - First Line: \( \langle 1 + t, 3 + t, -1 \rangle \) - Second Line: \( \langle s, 4s, 5s \rangle \)2. Equate components to form equations. 3. Solve for \( t \) and \( s \).4. Verify the solutions fit all component equations.In this scenario, solving the equations revealed no common solutions, indicating no intersection.