Problem 31 - University Calculus: Early Transcendentals

Question

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

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: Vector a

To find the vector with length 2 and direction \( \mathbf{i} \), we use the formula \( \text{Vector} = \text{Length} \times \text{Direction} \). Since the direction is \( \mathbf{i} \), the vector is simply \( 2 \mathbf{i} \).

Step 2: Vector b

For the vector with length \( \sqrt{3} \) and direction \( -\mathbf{k} \), use the same formula. Multiply the length by the direction: \( (\sqrt{3})(-\mathbf{k}) = -\sqrt{3}\mathbf{k} \).

Step 3: Vector c

Given the length \( \frac{1}{2} \) and direction \( \frac{3}{5}\mathbf{j} + \frac{4}{5}\mathbf{k} \), scale this direction by the length: \[ \frac{1}{2} \left( \frac{3}{5}\mathbf{j} + \frac{4}{5}\mathbf{k} \right) = \frac{3}{10}\mathbf{j} + \frac{2}{5}\mathbf{k} \].

Step 4: Vector d

The vector has length 7 and direction \( \frac{6}{7}\mathbf{i} - \frac{2}{7}\mathbf{j} + \frac{3}{7}\mathbf{k} \). Multiply the length by this direction to get the vector: \[ 7 \left( \frac{6}{7}\mathbf{i} - \frac{2}{7}\mathbf{j} + \frac{3}{7}\mathbf{k} \right) = 6\mathbf{i} - 2\mathbf{j} + 3\mathbf{k} \].

Key Concepts

Vector Magnitude

In vector calculus, the term vector magnitude refers to the length or size of a vector. It is a crucial concept when it comes to understanding vectors, as it provides a measure of how long or strong the vector is.

To find the magnitude of a vector, we'll use the Pythagorean theorem in a coordinate system. Generally, a vector in three-dimensional space given by components \(a\), \(b\), and \(c\) has a magnitude calculated using the formula: \[|\mathbf{v}| = \sqrt{a^2 + b^2 + c^2}.\]A simpler way to understand it is to imagine a vector as an arrow. The magnitude is simply how long that arrow is. All the vectors given in the step by step solution have specified magnitudes. For example, a vector with length 2 that points in the direction of \(\mathbf{i}\) has a clear magnitude of 2 units.

Whenever we're asked to write a vector with a specified length, we simply multiply the direction unit vector by the given magnitude to achieve this size. This is how the vectors in the exercise are formed.

Vector Direction

Understanding the direction of a vector is key to grasping how vectors work in a multidimensional space. Vector direction is essentially the orientation of the vector in relation to the coordinate axes.

To describe a vector's direction in mathematical terms, we typically use a unit vector. A unit vector is a vector that has a magnitude of 1 and points in a specific direction. By using unit vectors, like \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\), we can easily indicate direction along the x, y, and z axes, respectively.
**How to determine a Unit Vector:**

• Divide each component of the original vector by its magnitude.
• This transforms the original direction into a unit vector.

For instance, when a problem states a direction but doesn't give component values, it often uses pre-calculated unit vector forms. In the exercise, when asked to find vectors with specified directions like \(\mathbf{i}\) or combinations thereof, it's these unit vectors that are being scaled by magnitude. By knowing the direction, we know where our vector is pointing.

Vector Components

The components of a vector describe its impact along each coordinate axis. They break down a vector into its respective influences in the x, y, and z directions. It's like splitting an arrow into parts that show how much it moves horizontally, vertically, and in depth.

Components are written typically in the form of \(a\mathbf{i} + b\mathbf{j} + c\mathbf{k}\), where \(a\), \(b\), and \(c\) are scalars representing the vector's strength in each direction. Breaking vectors into components allows us to analyze and perform calculations like addition, subtraction, and dot product more easily.
**Benefits of Using Components:**

• Helps in accurately visualizing vector direction in multidimensional space.
• Facilitates the calculation of vector operations.

Every vector in the step by step solution is presented in terms of its components. For example, a vector with direction \(\frac{3}{5}\mathbf{j} + \frac{4}{5}\mathbf{k}\) needs these specific components to point in the right direction and have the correct magnitude, ensuring each component complies with the given length. Understanding these components and how vectors are constructed from them is fundamental to vector calculus.