Problem 65 - College Trigonometry

Question

The goal of this exercise is to use vectors to describe non-vertical lines in the plane. To that end, consider the line \(y=2 x-4 .\) Let \(\vec{v}_{0}=\langle 0,-4\rangle\) and let \(\vec{s}=\langle 1,2\rangle .\) Let \(t\) be any real number. Show that the vector defined by \(\vec{v}=\vec{v}_{0}+t \vec{s}\), when drawn in standard position, has its terminal point on the line \(y=2 x-4\). (Hint: Show that \(\vec{v}_{0}+t \vec{s}=\langle t, 2 t-4\rangle\) for any real number \(t\).) Now consider the non-vertical line \(y=m x+b\). Repeat the previous analysis with \(\vec{v}_{0}=\langle 0, b\rangle\) and let \(\vec{s}=\langle 1, m\rangle .\) Thus any non-vertical line can be thought of as a collection of terminal points of the vector sum of \(\langle 0, b\rangle\) (the position vector of the \(y\) -intercept) and a scalar multiple of the slope vector \(\vec{s}=\langle 1, m\rangle\).

Step-by-Step Solution

Answer

Using given vectors and a scalar parameter, a non-vertical line equation can be expressed as a vector equation.

Step 1: Understanding the Problem

Our task is to express a line, given by the equation \( y = 2x - 4 \), in the form of a vector sum. We are given initial vectors: \( \vec{v}_0 = \langle 0, -4 \rangle \) and \( \vec{s} = \langle 1, 2 \rangle \). Our goal is to show that the vector \( \vec{v} = \vec{v}_{0} + t \vec{s} \) always ends on the line \( y = 2x - 4 \).

Step 2: Setting Up the Vector Equation

The vector equation given is \( \vec{v} = \vec{v}_0 + t \vec{s} \). Substituting the vector values, we have: \( \vec{v} = \langle 0, -4 \rangle + t \langle 1, 2 \rangle \).

Step 3: Solve for the Components of \( \vec{v} \)

Compute \( \vec{v} \) by adding the initial vector \( \vec{v}_0 \) and the scalar multiple of \( \vec{s} \):\[ \vec{v} = \langle 0, -4 \rangle + t \langle 1, 2 \rangle = \langle 0 + t \cdot 1, -4 + t \cdot 2 \rangle = \langle t, 2t - 4 \rangle \].

Step 4: Verifying the Line Equation

Consider the line equation \( y = 2x - 4 \). In our vector \( \langle t, 2t - 4 \rangle \), the first component \( t \) correlates to \( x \) and the second component \( 2t - 4 \) correlates to \( y \). Substituting this into the line equation:\[ y = 2x - 4 \Rightarrow 2t - 4 = 2t - 4 \].This confirms the calculation is consistent with the line equation.

Step 5: Generalizing for the Line \( y = mx + b \)

For a generic line \( y = mx + b \), use the vectors \( \vec{v}_0 = \langle 0, b \rangle \) and \( \vec{s} = \langle 1, m \rangle \). The vector equation becomes:\[ \vec{v} = \vec{v}_0 + t \vec{s} = \langle 0, b \rangle + t \langle 1, m \rangle = \langle t, mt + b \rangle \].Substitute into the line equation:\[ y = mx + b \Rightarrow mt + b = mt + b \], verifying that this form is also correct for any non-vertical line.

Key Concepts

Vectors in 2D Geometry

In the realm of 2D geometry, vectors are essential tools for representing and understanding positions, directions, and magnitudes in the plane. A vector in 2D typically has two components: one representing the horizontal direction (x-axis) and the other representing the vertical direction (y-axis). These components can be written as ordered pairs, such as \( \langle a, b \rangle \), where \( a \) is the x-component and \( b \) is the y-component.

Vectors can be visualized as arrows pointing from one point to another. The starting point of a vector is often called the initial point, and the endpoint is the terminal point. In many math problems, vectors are placed in standard position, meaning their initial point is at the origin \( (0,0) \). This simplification helps us focus on the direction and length of the vector, rather than its position.

Understanding vectors in two-dimensional geometry provides a foundation for analyzing lines, shapes, and motions within the plane. They are fundamental in physics as well, to describe velocities and forces.

Equation of a Line

The equation of a line in a two-dimensional plane is a formula that describes all points that lie on the line. A common form of a line's equation is: \( y = mx + b \). Here, \( m \) represents the slope of the line, which is a measure of its steepness, and \( b \) is the y-intercept, the point where the line crosses the y-axis.

The slope \( m \) is calculated as the change in y divided by the change in x between any two points on the line. It indicates whether and how the line rises or falls as we move from left to right. The y-intercept \( b \) pinpoints exactly where the line meets the vertical axis of the graph.

This form of the equation is particularly convenient when analyzing graphically and algebraically what effects changes in \( m \) or \( b \) have on the line. A positive slope means the line rises to the right, while a negative slope means it falls. Changing \( b \) shifts the line up or down without affecting its slope.

Scalar Multiplication of Vectors

The process of multiplying a vector by a scalar (a real number) is called scalar multiplication. When you multiply a vector by a scalar, you scale or change the vector's magnitude (length) without altering its direction. This is vital in constructing vector equations of lines.

Consider a vector \( \vec{s} = \langle x, y \rangle \) and a scalar \( t \). Scalar multiplication involves multiplying each component of the vector by the scalar:

• \( t \vec{s} = t \langle x, y \rangle = \langle t \cdot x, t \cdot y \rangle \)

This operation is instrumental in creating lines through vectors, where we use scalar multiples to extend or shorten the vector, thereby reaching different points along the direction defined by the vector.

By scalar multiplying a direction vector \( \vec{s} \), we can generate any point on the line defined by that direction, demonstrating the powerful application of this simple operation.

Y-intercept as a Vector

In two-dimensional geometry, the y-intercept is often represented by a position vector. The y-intercept is the point where a line crosses the y-axis, and this point can be expressed as a vector with the coordinates \( \langle 0, b \rangle \), where \( b \) is the y-intercept itself.

This vector, also known as the position vector of the y-intercept, is important because it sets the starting position from which the line extends in the plane. By combining this vector with a direction vector (typically a vector derived from the slope of the line), you form a vector equation that represents every point on the line.

The vector equation \( \vec{v} = \vec{v}_0 + t \vec{s} \) incorporates the y-intercept as the initial vector \( \vec{v}_0 = \langle 0, b \rangle \). Hence, this initial vector anchors the line at the y-axis, and with the addition of a direction vector scaled by any real number \( t \), it describes a complete line in vector form.