Problem 30 - Thomas Calculus

Question

Express each vector as a product of its length and direction. \(\frac{\mathbf{i}}{\sqrt{3}}+\frac{\mathbf{j}}{\sqrt{3}}+\frac{\mathbf{k}}{\sqrt{3}}\)

Step-by-Step Solution

Answer

The vector is already a unit vector: its length is 1, and direction remains the same.

Step 1: Understand Vector Components

The vector given is \( \frac{\mathbf{i}}{\sqrt{3}} + \frac{\mathbf{j}}{\sqrt{3}} + \frac{\mathbf{k}}{\sqrt{3}} \). This means its components are \( \frac{1}{\sqrt{3}} \mathbf{i} \), \( \frac{1}{\sqrt{3}} \mathbf{j} \), and \( \frac{1}{\sqrt{3}} \mathbf{k} \).

Step 2: Calculate the Magnitude of the Vector

The magnitude (or length) of a vector \( \mathbf{v} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k} \) is calculated using the formula \( \sqrt{a^2 + b^2 + c^2} \). For this vector, \( a = \frac{1}{\sqrt{3}}, b = \frac{1}{\sqrt{3}}, c = \frac{1}{\sqrt{3}} \).

Step 3: Simplify the Magnitude

Calculate each square: \( \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3} \). Sum these values: \( \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1 \). Thus, the magnitude is \( \sqrt{1} = 1 \).

Step 4: Express the Vector in Length and Direction Form

A vector can be expressed as the product of its magnitude and a unit vector in its direction. Here, the magnitude is 1, so we multiply the unit vector by 1: \( 1 \cdot \left( \frac{\mathbf{i}}{\sqrt{3}} + \frac{\mathbf{j}}{\sqrt{3}} + \frac{\mathbf{k}}{\sqrt{3}} \right) \). This simplifies to the original vector, showing it is already a unit vector.

Step 5: Verify Unit Vector Nature

A unit vector has a magnitude of 1. As calculated, the magnitude of the given vector is indeed 1, confirming it is already in its length and direction form.

Key Concepts

Vector Magnitude

The magnitude of a vector is essentially its length. Imagine it as the distance from the start of the vector to the endpoint in a coordinate system. To find the magnitude, use the formula \( \sqrt{a^2 + b^2 + c^2} \) for a vector \( \mathbf{v} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k} \). This formula derives from the Pythagorean theorem applied in three dimensions. It considers all components of the vector and finds a cumulative length.

For our example vector \( \frac{\mathbf{i}}{\sqrt{3}} + \frac{\mathbf{j}}{\sqrt{3}} + \frac{\mathbf{k}}{\sqrt{3}} \), each component has the value \( \frac{1}{\sqrt{3}} \). To find the magnitude, you square each component, then add them up:

• \( \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3} \) for the \( \mathbf{i} \) component
• \( \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3} \) for the \( \mathbf{j} \) component
• \( \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3} \) for the \( \mathbf{k} \) component

Adding these gives \( \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1 \). The magnitude is therefore \( \sqrt{1} = 1 \). This straightforward method confirms the vector's length is precisely 1 unit.

Unit Vector

A unit vector is one of the simplest forms of understanding direction in vector mathematics. It has a magnitude of 1, making it a perfect way to describe direction without any scaling factor. Think of it like a pointer that shows the direction but needs to be scaled up to matter in real-world applications.

To convert any vector into a unit vector, you divide its components by its own magnitude. However, in situations where the magnitude is already 1, like in our exercise \( \frac{\mathbf{i}}{\sqrt{3}} + \frac{\mathbf{j}}{\sqrt{3}} + \frac{\mathbf{k}}{\sqrt{3}} \), the vector is already a unit vector. This is an insightful result because it means the direction is described without changing the proportions of its components.

The given vector's magnitude was computed as 1, indicating that it already perfectly points in its direction without needing further division.

Vector Components

Vector components are the building blocks of a vector, represented along each axis of a coordinate system. They show how much of the vector's power is directed along the x-axis, y-axis, and z-axis. Each component is associated with a unit vector in the respective direction \( \mathbf{i}, \mathbf{j}, \mathbf{k} \).

In our example, the vector is \( \frac{\mathbf{i}}{\sqrt{3}} + \frac{\mathbf{j}}{\sqrt{3}} + \frac{\mathbf{k}}{\sqrt{3}} \). Here,

• \( \frac{1}{\sqrt{3}} \mathbf{i} \) represents the x-component
• \( \frac{1}{\sqrt{3}} \mathbf{j} \) represents the y-component
• \( \frac{1}{\sqrt{3}} \mathbf{k} \) represents the z-component

Understanding components is fundamental because each part contributes to the vector's total direction and magnitude.

These sections give you the complete picture of how a vector behaves in space, as every change in a component directly affects the vector's overall direction and action.