Problem 27 - Thomas Calculus

Question

In Exercises 25–30, express each vector as a product of its length and direction. $$ 5 \mathbf{k} $$

Step-by-Step Solution

Answer

The vector is expressed as \( 5 \cdot \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \).

Step 1: Identify the Vector

The given vector is \( 5 \mathbf{k} \). Here, \( \mathbf{k} \) represents the unit vector in the direction of the positive z-axis. This is already in the form of a magnitude times a direction, but let's break it down further.

Step 2: Determine the Direction

The direction of the vector \( \mathbf{k} \) is straightforward since \( \mathbf{k} \) is a commonly known unit vector along the z-axis: \( \mathbf{k} = \begin{bmatrix} 0 \ 0 \ 1 \end{bmatrix} \). This vector has a length (or magnitude) of 1, as it is a unit vector.

Step 3: Find the Length of the Original Vector

The scalar multiple of the vector \( 5 \mathbf{k} \) indicates the length of the original vector, which is simply the coefficient 5.

Step 4: Express as Length and Direction

The vector \( 5 \mathbf{k} \) is expressed as a product of its length and direction as \( 5 \cdot \begin{bmatrix} 0 \ 0 \ 1 \end{bmatrix} \). This represents a length of 5 along the direction of the unit vector \( \mathbf{k} \).

Key Concepts

Unit Vector

A unit vector is a vector that has a magnitude of 1. It is used to indicate the direction of a vector and is often denoted by a hat on top of the vector symbol or simply identified, like the vector \( \mathbf{k} \) in our exercise.

Unit vectors play a crucial role in vector mathematics because they allow us to separate the direction of a vector from its magnitude. This makes it easier to manipulate and understand the behavior of vectors in different dimensions.

For example, the vector \( \mathbf{k} \) can be expressed as \( \mathbf{k} = \begin{bmatrix} 0 \ 0 \ 1 \end{bmatrix} \), indicating that it points straight up along the z-axis.

Since unit vectors are always of magnitude 1, they simplify calculations involving direction without influencing the vector's length. Thus, they are often used to build other vectors by scaling, which, in this case, involves multiplying the unit vector by the desired magnitude.

Vector Magnitude

The magnitude of a vector is essentially the length or size of that vector. It is determined by calculating the square root of the sum of the squares of its components. For a three-dimensional vector \( \mathbf{v} = \begin{bmatrix} x \ y \ z \end{bmatrix} \), the magnitude \( \| \mathbf{v} \| \) is given by \( \sqrt{x^2 + y^2 + z^2} \).

However, in the case of a unit vector, as we've seen with \( \mathbf{k} \), the magnitude is already 1. Multiplying the unit vector by a scalar modifies its length while maintaining its direction.

In our exercise, the vector \( 5 \mathbf{k} \) has a magnitude of 5. This is because the magnitude is simply the coefficient (5 in this case) multiplying the unit vector \( \mathbf{k} \). Thus, our vector stretches along the z-axis to a length of 5.

Direction of a Vector

Understanding the direction of a vector is crucial in distinguishing it from other vectors and knowing where it points in space. The direction is typically illustrated using unit vectors, which essentially lay the path for the vector without affecting its magnitude.

In the given exercise, the vector \( 5 \mathbf{k} \) is oriented along the z-axis due to the presence of the unit vector \( \mathbf{k} \). This means that any movements or effects represented by this vector will occur along this axis.

Typically, direction in three-dimensional space is defined by unit vectors \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \), which correspond to the x, y, and z axes respectively.

It is important because knowing the direction allows us to apply vector operations like addition, subtraction, and projection effectively. Moreover, it makes visualizing problems in mechanics, physics, and engineering more intuitive, allowing for more accurate analyses and solutions.