Problem 26 - Precalculus : Building Concepts and Connections

Question

Determine whether the given pairs of vectors are orthogonal. $$\mathbf{v}=\langle 3,1\rangle, \mathbf{w}=\langle 0,1\rangle$$

Step-by-Step Solution

Answer

No, the vectors \( \mathbf{v}=\langle 3,1\rangle \) and \( \mathbf{w}=\langle 0,1\rangle \) are not orthogonal.

Step 1: Understand Dot Product

Dot product of two vectors \( \mathbf{a}= \langle a1, a2 \rangle \) and \( \mathbf{b}= \langle b1, b2 \rangle \) in the 2-D plane is calculated as \( \mathbf{a} \cdot \mathbf{b} = a1*b1 + a2*b2 \). This formula gives us a single number as a result.

Step 2: Compute Dot Product

Now, using the above formula of dot product for the vectors \( \mathbf{v}=\langle 3,1\rangle \) and \( \mathbf{w}=\langle 0,1\rangle \), we get \( \mathbf{v} \cdot \mathbf{w} = (3*0) + (1*1) = 0 + 1 = 1 \).

Step 3: Conclude if vectors are Orthogonal or not

Since the dot product of vectors \( \mathbf{v} \) and \( \mathbf{w} \) is not zero (it's 1), the vectors are not orthogonal to each other.

Key Concepts

Dot Product

The dot product is a mathematical operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. This operation is also known as the scalar product, hinting at its outcome—a scalar, not a vector. In the context of 2-D plane vectors, you'll typically see it as \( \mathbf{a} \cdot \mathbf{b} = a1 \times b1 + a2 \times b2 \) where \( \mathbf{a} \) and \( \mathbf{b} \) are vectors and \( a1, a2, b1, b2 \) are their respective components.

Understanding the dot product is vital because it conveys how vectors interact with each other. Its value can indicate if two vectors point in the same direction, are orthogonal (perpendicular), or point in opposite directions. A key fact to remember is that if the dot product is zero, it signifies that the two vectors are orthogonal.

2-D Plane Vectors

Vectors in a 2-dimensional plane are defined by two components which correspond to their projection along the x-axis and y-axis. These are usually written in the form of \( \mathbf{v} = \langle x, y \rangle \) and can represent various physical quantities like velocity, force, or displacement. Each component can be thought of as the length of the vector's shadow along the respective axis.

When dealing with 2-D vectors, visualizing them on a graph can often provide a clearer understanding. The behavior of these vectors under various operations, such as addition, scalar multiplication, or dot product, is the foundation of vector algebra in two dimensions.

Vector Orthogonality Concept

The concept of vector orthogonality is directly linked to the idea of perpendicularity in geometry. Two vectors are considered orthogonal if they form a right angle (90 degrees) with each other. This concept is not just a geometric notion but a fundamental principle in vector spaces and various applications in physics and engineering.

One of the most powerful and practical ways to test for orthogonality is by using the dot product. As previously noted, when the dot product of two vectors is zero, they are orthogonal. This property simplifies many problems that would otherwise require more complex geometric or trigonometric methods to solve.

Calculating Dot Product

To calculate the dot product of two vectors in a 2-D space, you multiply their corresponding components and then add the products. The general formula looks like \( \mathbf{a} \cdot \mathbf{b} = a1 \times b1 + a2 \times b2 \). This operation results in a single number.

Going through the actual calculation step-by-step helps crystallize the concept. For instance, given the vectors \( \mathbf{v} = \langle 3, 1 \rangle \) and \( \mathbf{w} = \langle 0, 1 \rangle \), we calculate the dot product as \( \mathbf{v} \cdot \mathbf{w} = (3 \times 0) + (1 \times 1) = 0 + 1 = 1 \). Since the result is not zero, we can conclude that \( \mathbf{v} \) and \( \mathbf{w} \) are not orthogonal. This straightforward method provides a neat, numerical approach to understanding the relationship between vectors.