Problem 31 - Thomas Calculus

Question

Find the vectors whose lengths and directions are given. Try to do the calculations without writing. $$ \begin{array}{ll}{\frac{\text { Length }}{\text { a. } 2}} & {\text { Direction }} \\ {\text { a. } 2} & {\text { i }} \\ {\text { b. } \sqrt{3}} & {-\mathrm{k}} \\ {\text { c. } \frac{1}{2}} & {\frac{3}{5} \mathrm{j}+\frac{4}{5} \mathbf{k}} \\ {\text { d. } 7} & {\frac{6}{7} \mathbf{i}-\frac{2}{7} \mathbf{j}+\frac{3}{7} \mathbf{k}}\end{array} $$

Step-by-Step Solution

Answer

a. \(2\mathbf{i}\), b. \(-\sqrt{3}\mathbf{k}\), c. \(\frac{3}{10}\mathbf{j} + \frac{2}{5}\mathbf{k}\), d. \(6\mathbf{i} - 2\mathbf{j} + 3\mathbf{k}\)

Step 1: Understanding the Problem

The task is to find vectors based on given lengths (or magnitudes) and directions. Directions are expressed as unit vectors, meaning each direction has a magnitude of 1 and defines the direction of the vector.

Step 1: Compute Vector A

Given: Length = 2 and Direction = \( \mathbf{i} \). Since \( \mathbf{i} \) is already a unit vector in the x-direction, the vector is obtained by multiplying its length with its direction. Thus, Vector \( \mathbf{a} = 2 \mathbf{i} \).

Step 2: Compute Vector B

Given: Length = \( \sqrt{3} \) and Direction = \( -\mathbf{k} \). The direction \( -\mathbf{k} \) is already a unit vector in the negative z-direction. Therefore, Vector \( \mathbf{b} = \sqrt{3} (-\mathbf{k}) = -\sqrt{3} \mathbf{k} \).

Step 3: Compute Vector C

Given: Length = \( \frac{1}{2} \) and Direction = \( \frac{3}{5} \mathbf{j} + \frac{4}{5} \mathbf{k} \). Verify that this is a unit vector: \( \sqrt{(\frac{3}{5})^2 + (\frac{4}{5})^2} = 1 \). Since it is a unit vector, multiply by length: Vector \( \mathbf{c} = \frac{1}{2} (\frac{3}{5} \mathbf{j} + \frac{4}{5} \mathbf{k}) = \frac{3}{10} \mathbf{j} + \frac{2}{5} \mathbf{k} \).

Step 4: Compute Vector D

Given: Length = 7 and Direction = \( \frac{6}{7} \mathbf{i} - \frac{2}{7} \mathbf{j} + \frac{3}{7} \mathbf{k} \). Verify that this is a unit vector: \( \sqrt{(\frac{6}{7})^2 + (-\frac{2}{7})^2 + (\frac{3}{7})^2} = 1 \). Since it is a unit vector, multiply by the length: Vector \( \mathbf{d} = 7(\frac{6}{7} \mathbf{i} - \frac{2}{7} \mathbf{j} + \frac{3}{7} \mathbf{k}) = 6 \mathbf{i} - 2 \mathbf{j} + 3 \mathbf{k} \).

Key Concepts

Vector Magnitude

The magnitude of a vector, often referred to as its length, is a measure of how long or large the vector is. It's calculated using the Pythagorean theorem, which considers the components along each axis. Mathematically, for a vector \( \mathbf{v} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k} \), the magnitude \( \|\mathbf{v}\| \) is given by the formula: \[ \|\mathbf{v}\| = \sqrt{a^2 + b^2 + c^2} \]In the context of our exercise, the vectors are specified by their lengths and directions. The task is to find vectors with given magnitudes, using the unit vectors in the specified direction. Each length given is simply multiplied by a unit vector to produce the final vector with the correct magnitude.

Unit Vectors

A unit vector is a vector that has a magnitude of 1. It is typically used to specify a direction and is essential in constructing vectors of specific lengths. Common unit vectors in a 3D Cartesian plane include:

• \( \mathbf{i} \): the unit vector along the x-axis
• \( \mathbf{j} \): the unit vector along the y-axis
• \( \mathbf{k} \): the unit vector along the z-axis

These unit vectors can be combined to form other unit vectors pointing in different directions. For example, a unit vector \( \mathbf{u} = \frac{3}{5} \mathbf{j} + \frac{4}{5} \mathbf{k} \) has already been normalized, meaning it naturally has a length or magnitude of 1. You can verify this by calculating its magnitude:\[ \sqrt{ \left(\frac{3}{5}\right)^2 + \left(\frac{4}{5}\right)^2 } = 1 \] This kind of normalization ensures that a unit vector can be effortlessly scaled to the desired full length of any vector.

Vector Directions

Vector directions tell us the orientation of the vector in space. They are provided by unit vectors as they inherently describe direction while having a standard magnitude of 1. The direction is crucial for understanding how a vector points relative to each axis.For example:

• \( \mathbf{i} \) defines direction solely along the x-axis.
• \( -\mathbf{k} \) points in the negative z-direction.
• The combination \( \frac{6}{7} \mathbf{i} - \frac{2}{7} \mathbf{j} + \frac{3}{7} \mathbf{k} \) describes a more complex direction within a 3D space.

These unit vectors or combinations like \( \frac{6}{7} \mathbf{i} - \frac{2}{7} \mathbf{j} + \frac{3}{7} \mathbf{k} \) are essential because they form the backbone of constructing fully detailed vectors by attaching lengths to the directions.

Vector Calculations

Vector calculations often involve the use of both magnitudes and directions to develop a new vector from these foundational elements. When you have the length and direction, the actual vector can be formed by simply scaling the unit direction vector by the given magnitude.Let's break down how these calculations work in our examples:- **Compute Vector A**: Given a length of 2 and direction \( \mathbf{i} \), the vector is \( 2 \times \mathbf{i} = 2\mathbf{i} \).
- **Compute Vector B**: With a length of \( \sqrt{3} \) and direction \( -\mathbf{k} \), we have \( \sqrt{3} \times (-\mathbf{k}) = -\sqrt{3}\mathbf{k} \).
- **Compute Vector C**: Using a length of \( \frac{1}{2} \) and direction \( \frac{3}{5}\mathbf{j} + \frac{4}{5}\mathbf{k} \), the resulting vector is \( \frac{1}{2} \times \left( \frac{3}{5}\mathbf{j} + \frac{4}{5}\mathbf{k} \right) = \frac{3}{10}\mathbf{j} + \frac{2}{5}\mathbf{k} \).
- **Compute Vector D**: With a length of 7 and direction \( \frac{6}{7}\mathbf{i} - \frac{2}{7}\mathbf{j} + \frac{3}{7}\mathbf{k} \), the scaled vector computes to \( 7 \times \left( \frac{6}{7}\mathbf{i} - \frac{2}{7}\mathbf{j} + \frac{3}{7}\mathbf{k} \right) = 6\mathbf{i} - 2\mathbf{j} + 3\mathbf{k} \).
This method can be applied universally to any vector derivation task where the direction and magnitude are known.