Problem 29 - Thomas Calculus

Question

Express each vector as a product of its length and direction. \(\frac{1}{\sqrt{6}} \mathbf{i}-\frac{1}{\sqrt{6}} \mathbf{j}-\frac{1}{\sqrt{6}} \mathbf{k}\)

Step-by-Step Solution

Answer

The vector is \( \frac{1}{\sqrt{2}} \) times the unit vector \( \frac{\mathbf{i} - \mathbf{j} - \mathbf{k}}{\sqrt{3}} \).

Step 1: Identify the Given Vector

We start by identifying the given vector in the problem, which is: \( \frac{1}{\sqrt{6}} \mathbf{i} - \frac{1}{\sqrt{6}} \mathbf{j} - \frac{1}{\sqrt{6}} \mathbf{k} \). The vector components are \( \frac{1}{\sqrt{6}} \) for each of \( \mathbf{i}, \mathbf{j}, \) and \( \mathbf{k} \).

Step 2: Find the Length of the Vector

The length (or magnitude) of a vector \( \mathbf{v} = a \mathbf{i} + b \mathbf{j} + c \mathbf{k} \) is given by \( \| \mathbf{v} \| = \sqrt{a^2 + b^2 + c^2} \). Substituting the components \( a = \frac{1}{\sqrt{6}}, b = -\frac{1}{\sqrt{6}}, c = -\frac{1}{\sqrt{6}} \), we have:\[ \| \mathbf{v} \| = \sqrt{\left(\frac{1}{\sqrt{6}}\right)^2 + \left(-\frac{1}{\sqrt{6}}\right)^2 + \left(-\frac{1}{\sqrt{6}}\right)^2} \]\[ = \sqrt{\frac{1}{6} + \frac{1}{6} + \frac{1}{6}} = \sqrt{\frac{3}{6}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \].

Step 3: Determine the Direction as a Unit Vector

The direction of a vector is represented as a unit vector, which has a length of 1. To find the unit vector in the direction of the given vector, divide each component by the magnitude:\[ \mathbf{u} = \left(\frac{1}{\frac{1}{\sqrt{2}}}\right) \left(\frac{1}{\sqrt{6}} \mathbf{i} - \frac{1}{\sqrt{6}} \mathbf{j} - \frac{1}{\sqrt{6}} \mathbf{k}\right) \].Calculating each component:\[ \mathbf{u} = \sqrt{2} \cdot \left(\frac{1}{\sqrt{6}} \mathbf{i} - \frac{1}{\sqrt{6}} \mathbf{j} - \frac{1}{\sqrt{6}} \mathbf{k}\right) \].

Step 4: Express the Vector as Length and Direction Product

Now, express the original vector as the product of its length and direction:\[ \mathbf{v} = \frac{1}{\sqrt{2}} \cdot \left( \frac{\sqrt{2}}{\sqrt{6}} \mathbf{i} - \frac{\sqrt{2}}{\sqrt{6}} \mathbf{j} - \frac{\sqrt{2}}{\sqrt{6}} \mathbf{k} \right) \]. The expression simplifies to:\[ \mathbf{v} = \frac{1}{\sqrt{2}} \cdot \frac{\mathbf{i} - \mathbf{j} - \mathbf{k}}{\sqrt{3}} \], which represents the vector as a product of its length \( \frac{1}{\sqrt{2}} \) and its direction vector \( \frac{\mathbf{i} - \mathbf{j} - \mathbf{k}}{\sqrt{3}} \).

Key Concepts

Magnitude of a Vector

Understanding the magnitude of a vector is crucial in vector mathematics. It's essentially the vector's length, providing a measure of how long the vector is in space. The formula used to determine the magnitude of a vector \( \mathbf{v} = a \mathbf{i} + b \mathbf{j} + c \mathbf{k} \) is \( \| \mathbf{v} \| = \sqrt{a^2 + b^2 + c^2} \).

This equation comes from the Pythagorean theorem which applies to the multi-dimensional space of vectors. It adds the squares of the components, takes their square root, and yields a positive scalar value. In our example, plugging in \( a = \frac{1}{\sqrt{6}}, b = -\frac{1}{\sqrt{6}}, \) and \( c = -\frac{1}{\sqrt{6}} \), we find:

• Magnitude \( \| \mathbf{v} \| = \sqrt{\frac{1}{6} + \frac{1}{6} + \frac{1}{6}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \).

This result tells us the vector's true length, whether in two-dimensional or three-dimensional space.

Unit Vector

A unit vector has a magnitude of exactly 1. These vectors are crucial for determining direction without adding any additional length. The process to derive a unit vector from an arbitrary vector involves dividing each component of the original vector by the vector's magnitude.

In our exercise, given the magnitude \( \frac{1}{\sqrt{2}} \), the corresponding unit vector \( \mathbf{u} \) is calculated as follows:

• \( \mathbf{u} = \sqrt{2} \left( \frac{1}{\sqrt{6}} \mathbf{i} - \frac{1}{\sqrt{6}} \mathbf{j} - \frac{1}{\sqrt{6}} \mathbf{k} \right) \).

The resulting unit vector indicates directionality with all components maintaining their relative proportions but summed to a magnitude of one.

Vector Components

Vector components represent the projection of a vector along the axes of the coordinate system. They are the building blocks of a vector, giving the complete picture of its influence in each dimension. In three-dimensional space, this typically involves \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \) unit vectors.

For the exercise at hand, our vector components are \( \frac{1}{\sqrt{6}} \) in both positive and negative directions:

• Component along \( \mathbf{i} \): \( \frac{1}{\sqrt{6}} \)
• Component along \( \mathbf{j} \): \( -\frac{1}{\sqrt{6}} \)
• Component along \( \mathbf{k} \): \( -\frac{1}{\sqrt{6}} \)

These values define the vector's spatial placement and its influence along the x, y, and z directions respectively. Understanding each component provides insight into how the vector behaves in space, allowing us to reconstruct or decompose it, as needed for further analysis.