Question

Note that two lines \(y=m x+b\) and \(y=n x+c\) are orthogonal provided \(m n=-1\) (the slopes are negative reciprocals of each other). Prove that the condition \(m n=-1\) is equivalent to the orthogonality condition \(\mathbf{u} \cdot \mathbf{v}=0\) where \(\mathbf{u}\) points in the direction of one line and \(\mathbf{v}\) points in the direction of the other line.

Short answer

Question: Prove that the condition for orthogonality between two lines (the slopes of the lines being negative reciprocals) is equivalent to the orthogonality condition between their direction vectors (the dot product being zero). Answer: Given the slopes of the two lines, we defined their direction vectors, found their dot product, and established that the dot product is zero if the product of the slopes is -1, proving the equivalence between the two orthogonality conditions.

Step-by-step solution

Define the slopes and direction vectors

Let the slope of the first line be \(m\), and the slope of the second line be \(n\). Then, the first line can be represented as \(y=mx+b\) and the second line can be represented as \(y=nx+c\). Now we'll define the direction vectors \(\mathbf{u}\) and \(\mathbf{v}\) for the two lines respectively. Since the lines are represented by their slopes, we can assign \(\mathbf{u} = \begin{pmatrix} 1 \\ m \end{pmatrix}\) and \(\mathbf{v} = \begin{pmatrix} 1 \\ n \end{pmatrix}\).

Find the dot product of the direction vectors

Recall that the dot product of two vectors \(\mathbf{u}\) and \(\mathbf{v}\) is given by \(\mathbf{u} \cdot \mathbf{v} = u_1v_1+u_2v_2\). In our case, this translates to \(\mathbf{u} \cdot \mathbf{v} = (1)(1) + (m)(n) = 1 + mn\).

Show equality to the orthogonality condition

According to the givens, the lines are orthogonal if \(mn = -1\). We also know that two vectors are orthogonal if their dot product is 0. To show that the two conditions are equivalent, we must prove that \(1+mn = 0\) if \(mn = -1\). If \(mn = -1\), the dot product becomes \(\mathbf{u} \cdot \mathbf{v} = 1 + mn = 1+(-1) = 0\). This shows that the dot product of the direction vectors of the two lines is indeed 0, which means that the lines are orthogonal. Thus, the condition \(mn = -1\) is equivalent to the orthogonality condition \(\mathbf{u} \cdot \mathbf{v} = 0\).

Key concepts

Dot Product

Understanding the dot product is crucial when dealing with vectors, especially in the context of determining orthogonality. The dot product of two vectors \( \mathbf{u} \) and \( \mathbf{v} \) is calculated as \( \mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 \). This is simply the sum of the products of each corresponding component. When the dot product equals zero, the vectors are orthogonal.
This means they meet at a right angle. In geometric terms, this is like saying the two directions don't overlap; they jut out at a perfect 90 degrees from each other.

• The dot product is easy to compute and is a foundational concept in linear algebra and geometry.
• A zero dot product indicates orthogonality, which is key for identifying perpendicular lines.

Let's take an example: For direction vectors \( \mathbf{u} = \begin{pmatrix} 1 \ m \end{pmatrix} \) and \( \mathbf{v} = \begin{pmatrix} 1 \ n \end{pmatrix} \), the dot product is \( 1 \cdot 1 + m \cdot n = 1 + mn \). If the vectors are orthogonal, \( 1 + mn = 0 \), proving orthogonality.

Direction Vectors

Direction vectors provide a simple way to describe the direction and orientation of a line in 2D space. In the context of algebraic lines given by equations like \( y = mx + b \), the direction vector \( \mathbf{u} \) corresponding to the slope \( m \) can be expressed as \( \begin{pmatrix} 1 \ m \end{pmatrix} \). Similarly, for a line with slope \( n \), the direction vector \( \mathbf{v} \) is \( \begin{pmatrix} 1 \ n \end{pmatrix} \).
These vectors tell us how much the line rises (or falls) for each unit it moves horizontally.

• A direction vector \( \mathbf{u} \) is simply the vector pointing in the direction the line is going, with the slope acting as the vertical change.
• Direction vectors make it easy to manipulate and compare lines, especially when considering interactions like intersection or orthogonality.

For verifying orthogonality, we specifically look at whether the dot product of these vectors is zero. This provides a direct path to understanding their mutual orientation.

Negative Reciprocal Slopes

Negative reciprocal slopes are a unique characteristic of orthogonal lines. If a line has a slope \( m \), then a line perpendicular to it will have a slope \( n \), where \( m \cdot n = -1 \).
This relationship comes from the fact that multiplying a slope by its negative reciprocal effectively rotates the slope by 90 degrees, thus making the lines perpendicular.

• Calculated as \( n = -\frac{1}{m} \), negative reciprocal slopes ensure perpendicularity in a Cartesian plane.
• They provide a quick and reliable check to determine if two lines are orthogonal without the need for further calculations.

In practical terms, if you have two lines with slopes \( m \) and \( n \) such that \( m \cdot n = -1 \), you've confirmed they intersect at a right angle, manifesting as orthogonal lines. The concept of negative reciprocal slopes links cleanly with direction vectors, underscoring that their dot product must be zero, confirming orthogonality.