Problem 18 - Calculus for Biology and Medicine

Question

Find the dot product of \(\mathbf{x}=[2,-3,1]^{\prime}\) and \(\mathbf{y}=[3,1,-2]\).

Step-by-Step Solution

Answer

The dot product is 1.

Step 1: Write Down the Dot Product Formula

The dot product of two vectors \( \mathbf{x} \) and \( \mathbf{y} \) is denoted by \( \mathbf{x} \cdot \mathbf{y} \) and is calculated as follows: \[ \mathbf{x} \cdot \mathbf{y} = x_1y_1 + x_2y_2 + x_3y_3 \] where \( x_1, x_2, x_3 \) are the components of vector \( \mathbf{x} \), and \( y_1, y_2, y_3 \) are the components of vector \( \mathbf{y} \).

Step 2: Identify Components of Vectors

Identify the components of vectors \( \mathbf{x} \) and \( \mathbf{y} \). For vector \( \mathbf{x} = [2, -3, 1]' \), the components are \( x_1 = 2 \), \( x_2 = -3 \), and \( x_3 = 1 \). For vector \( \mathbf{y} = [3, 1, -2] \), the components are \( y_1 = 3 \), \( y_2 = 1 \), and \( y_3 = -2 \).

Step 3: Substitute Components into the Formula

Substitute the identified components into the dot product formula: \[ \mathbf{x} \cdot \mathbf{y} = (2)(3) + (-3)(1) + (1)(-2) \].

Step 4: Calculate Each Term

Calculate each term in the expression from Step 3: 1. \( (2)(3) = 6 \).2. \( (-3)(1) = -3 \).3. \( (1)(-2) = -2 \).

Step 5: Sum the Results

Add up all the calculated terms from Step 4: \[ 6 + (-3) + (-2) = 6 - 3 - 2 \].

Step 6: Simplify the Expression

Simplify the expression to find the dot product: \[ 6 - 3 - 2 = 1 \].

Key Concepts

Vectors

Vectors are fundamental entities in mathematics and physics that represent both a magnitude and a direction. They are often visualized as arrows, where the length of the arrow corresponds to its magnitude, and the direction of the arrow indicates its direction. In a coordinate system, vectors are described using components, each representing the vector's projection along the corresponding axis.

In our example, we have two vectors, \( \mathbf{x} = [2, -3, 1] \) and \( \mathbf{y} = [3, 1, -2] \). These vectors are three-dimensional, meaning they have components along the \( x \), \( y \), and \( z \) axes.

To fully understand vectors, it's important to recognize their properties and how they differ from scalars, which have only magnitude and no direction. Vectors enable you to perform various operations such as addition, subtraction, and particularly the dot product, which we explore in the next sections.

Components

The components of a vector are the building blocks that define its position within a coordinate space. These components tell us how far along each axis the vector extends. Examining a vector's components allows us to conduct operations like addition and the calculation of the dot product.

For instance, in vector \( \mathbf{x} = [2, -3, 1] \):

• Component along x-axis: \( x_1 = 2 \)
• Component along y-axis: \( x_2 = -3 \)
• Component along z-axis: \( x_3 = 1 \)

Similarly, vector \( \mathbf{y} = [3, 1, -2] \) has components:

• Component along x-axis: \( y_1 = 3 \)
• Component along y-axis: \( y_2 = 1 \)
• Component along z-axis: \( y_3 = -2 \)

Each component is crucial because it allows us to apply operations that manipulate the vector mathematically, leading to useful calculations like the dot product.

Mathematical Operations

Mathematical operations involving vectors can greatly vary, with the dot product being one of the most common and useful. The dot product, or scalar product, is an operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number.

For vectors \( \mathbf{x} = [2, -3, 1]' \) and \( \mathbf{y} = [3, 1, -2] \), the dot product is calculated as:

• Multiply the corresponding components: \((2)(3), (-3)(1), (1)(-2)\).
• Sum these products together: \(6 + (-3) + (-2)\).
• The result: \(1\), a scalar.

The dot product is particularly useful in physics for calculating work, as it provides a measure of how much one vector goes in the direction of another. Additionally, it can be used to determine the angle between two vectors or check if two vectors are orthogonal (perpendicular).

Through understanding and performing the dot product, we see how mathematical operations can be used to derive meaningful information from vectors in applied mathematics and physics.