Problem 29 - University Calculus: Early Transcendentals

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 \(\mathbf{d} = \frac{1}{\sqrt{6}}\mathbf{i} - \frac{1}{\sqrt{6}}\mathbf{j} - \frac{1}{\sqrt{6}}\mathbf{k} \).

Step 1: Determine the Vector Length (Magnitude)

To find the magnitude (length) of the vector \(\frac{1}{\sqrt{6}} \mathbf{i}-\frac{1}{\sqrt{6}} \mathbf{j}-\frac{1}{\sqrt{6}} \mathbf{k}\), use the formula for the magnitude of a vector \(\|\mathbf{v}\| = \sqrt{x^2 + y^2 + z^2}\). Given the components are \(\frac{1}{\sqrt{6}}, -\frac{1}{\sqrt{6}}, -\frac{1}{\sqrt{6}}\), we calculate:\[\|\mathbf{v}\| = \sqrt{\left( \frac{1}{\sqrt{6}} \right)^2 + \left( -\frac{1}{\sqrt{6}} \right)^2 + \left( -\frac{1}{\sqrt{6}} \right)^2 } = \sqrt{3 \times \left( \frac{1}{6} \right) } = \sqrt{\frac{3}{6}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}.\]

Step 2: Identify the Direction Vector

The direction vector \(\mathbf{d}\) is a unit vector (vector of length 1) in the same direction as the original vector. For the given vector, the original is already given in a form where each component is divided by \(\sqrt{6}\), which indicates it has a length of 1 when combined. Thus, the original given vector \(\mathbf{d} = \frac{1}{\sqrt{6}}\mathbf{i} - \frac{1}{\sqrt{6}}\mathbf{j} - \frac{1}{\sqrt{6}}\mathbf{k} \) is already a unit vector along its direction.

Step 3: Express the Original Vector as a Product of its Length and Direction

Given the vector's length is \(\frac{1}{\sqrt{2}}\) and its direction vector is \(\mathbf{d} = \frac{1}{\sqrt{6}}\mathbf{i} - \frac{1}{\sqrt{6}}\mathbf{j} - \frac{1}{\sqrt{6}}\mathbf{k} \), the original vector \(\mathbf{v}\) can be expressed as:\[\mathbf{v} = \left( \frac{1}{\sqrt{2}} \right) \left( \frac{1}{\sqrt{6}} \mathbf{i} - \frac{1}{\sqrt{6}} \mathbf{j} - \frac{1}{\sqrt{6}} \mathbf{k} \right)\].

Key Concepts

Vector Magnitude

Understanding the concept of vector magnitude is crucial for grasping vector decomposition. The magnitude of a vector is essentially its length, measuring how far it stretches in space. Imagine a vector as a directed line segment; the magnitude tells you how long that line is. To compute the magnitude, we use the formula for a three-dimensional vector: \[ \|\mathbf{v}\| = \sqrt{x^2 + y^2 + z^2} \]
This means, for a vector with components \(x\), \(y\), and \(z\), you square each component, sum them, and then take the square root of that sum.
In the given problem, our vector is \( \frac{1}{\sqrt{6}} \mathbf{i} - \frac{1}{\sqrt{6}} \mathbf{j} - \frac{1}{\sqrt{6}} \mathbf{k} \). By applying the magnitude formula, we find:

• Find squares: \( \left(\frac{1}{\sqrt{6}}\right)^2 = \frac{1}{6} \)
• Sum the squares: \( 3 \times \frac{1}{6} = \frac{3}{6} = \frac{1}{2} \)
• Take the square root: \( \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \)

Thus, its magnitude is \( \frac{1}{\sqrt{2}} \). This tells us that the vector's length in space is \( \frac{1}{\sqrt{2}} \).

Direction Vector

A direction vector indicates the path or direction in which the vector points. It reveals where the vector is heading in a three-dimensional space without worrying about how far it goes. A direction vector is typically expressed as a unit vector—a vector with a length of 1.
For a given vector, the direction vector is found by dividing each component of the vector by its magnitude. However, in our exercise, the vector \( \frac{1}{\sqrt{6}} \mathbf{i} - \frac{1}{\sqrt{6}} \mathbf{j} - \frac{1}{\sqrt{6}} \mathbf{k} \) is already presenting itself as a unit vector. This is because its components have been divided by \( \sqrt{6} \), normalizing it to length 1.

• This simplification means that the vector itself represents its direction vector.
• Therefore, no additional computation is needed for direction in this specific case.

Understanding the direction vector helps in visualizing how to navigate through space based on the vector's orientation.

Unit Vector

In vector decomposition, transforming a vector into its unit vector version is a pivotal step. A unit vector maintains the direction of the original vector but has a standard magnitude of 1. It's like scaling down the vector to a manageable size while preserving direction.
The process to find a unit vector involves dividing each component of the vector by its magnitude. For instance, given vector \( \mathbf{v} \) with components \((x, y, z)\), the unit vector \( \mathbf{u} \) can be calculated as: \[ \mathbf{u} = \left( \frac{x}{\|\mathbf{v}\|}, \frac{y}{\|\mathbf{v}\|}, \frac{z}{\|\mathbf{v}\|} \right) \]

• In this exercise, the vector \( \frac{1}{\sqrt{6}} \mathbf{i} - \frac{1}{\sqrt{6}} \mathbf{j} - \frac{1}{\sqrt{6}} \mathbf{k} \) is given, and it's already normalized, meaning it's a unit vector as its length measures 1.
• This unit vector is crucial since it indicates the precise direction of the vector irrespective of its length.

Recognizing unit vectors simplifies many mathematical computations, making them integral to understanding vector-related problems and formulas.