Problem 32 - Thomas Calculus

Question

Line parallel to a vector Show that the vector \(\mathbf{v}=a \mathbf{i}+b \mathbf{j}\) is parallel to the line \(b x-a y=c\) by establishing that the slope of the line segment representing \(\mathbf{v}\) is the same as the slope of the given line.

Step-by-Step Solution

Answer

The vector \(\mathbf{v}\) is parallel to the line since both have the same slope \(\frac{b}{a}\).

Step 1: Identify the Slope of the Line

The given line is defined by the equation \(bx - ay = c\). We can rewrite this in the slope-intercept form \(y = mx + b\) by isolating \(y\). This results in \(ay = bx - c\) and then \(y = \frac{b}{a}x - \frac{c}{a}\). Therefore, the slope \(m\) of the line is \(\frac{b}{a}\).

Step 2: Determine the Slope of the Vector

The vector \(\mathbf{v} = a\mathbf{i} + b\mathbf{j}\) can be thought of as a line segment from the origin \((0, 0)\) to the point \((a, b)\). The slope of this line segment is calculated as the change in \(y\) over the change in \(x\), which is \(\frac{b-0}{a-0} = \frac{b}{a}\).

Step 3: Compare the Slopes

Compare the slope of the line, \(\frac{b}{a}\), with the slope of the vector, \(\frac{b}{a}\). Since both slopes are the same, the vector \(\mathbf{v}\) is parallel to the line.

Key Concepts

Slope of a Line

The slope of a line is a key concept in understanding the relationship between two variables in a linear equation. It's essentially a measure of how steep the line is. Given an equation of the form \(b x - a y = c\), you can find the slope by rearranging the equation into the slope-intercept form \(y = mx + b\), where \(m\) represents the slope.
The initial equation, \(b x - a y = c\), is transformed to isolate \(y\). By rearranging, we get \(ay = bx - c\) and solve further to find \(y = \frac{b}{a}x - \frac{c}{a}\). Here, \(\frac{b}{a}\) is the slope.
In this context, the slope \(\frac{b}{a}\) tells us how much \(y\) changes with a unit change in \(x\). A positive slope suggests an upward trend, while a negative slope indicates a downward slope. This fundamental concept helps to draw parallels between vectors and lines in the coordinate plane.

Vector Representation

Vectors are fundamental tools in mathematics to represent direction and magnitude. A vector in the plane can be described in terms of its components, such as \(\mathbf{v} = a\mathbf{i} + b\mathbf{j}\). Here, \(\mathbf{i}\) and \(\mathbf{j}\) are unit vectors along the \(x\) and \(y\) axes, respectively.

• The component \(a\) determines how far the vector moves in the \(x\) direction.
• Similarly, \(b\) indicates movement in the \(y\) direction.
• The vector itself can be visualized as an arrow pointing from the origin to the point \((a, b)\).

Understanding vector representation allows us to see how vectors form line segments in the coordinate plane.
Since the slope of the vector described from the origin to \((a,b)\) is \(\frac{b}{a}\), it becomes clear that the vector's direction is identical to a line with that same slope. Hence, confirming parallelism between the vector and the line.

Linear Equations

Linear equations are equations that make a straight line when graphed. They show a constant rate of change and can be written in various forms, such as \(bx - ay = c\) or \(y = mx + b\).
With linear equations, the focus is often on determining the slope, intercepts, and finding points of intersection. Solving a linear equation typically involves manipulating the equation to isolate the variable of interest, which often entails rearranging terms on either side of the equation.
Understanding the structure of a linear equation is crucial when analyzing intersections, slopes, and in this case, parallelism.

• In our exercise, the parallelism of the vector and the line indicates that they share the same slope.
• Recognizing that two lines or segments are parallel is a result of them having precisely the same slope.

This knowledge of linear equations and their properties is instrumental in deducing the relationships between vectors and parallel lines.