Problem 32 - University Calculus: Early Transcendentals

Question

Find the vectors whose lengths and directions are given. Try to do the calculations without writing. $$\begin{array}{ll} \text { Length } & \text { Direction } \\ \hline \text { a. } 7 & -\mathbf{j} \\ \text { b. } \sqrt{2} & -\frac{3}{5} \mathbf{i}-\frac{4}{5} \mathbf{k} \\ \text { c. } \frac{13}{12} & \frac{3}{13} \mathbf{i}-\frac{4}{13} \mathbf{j}-\frac{12}{13} \mathbf{k} \\ \text { d. } a>0 & \frac{1}{\sqrt{2}} \mathbf{i}+\frac{1}{\sqrt{3}} \mathbf{j}-\frac{1}{\sqrt{6}} \mathbf{k} \end{array}$$

Step-by-Step Solution

Answer

a. \(-7\mathbf{j}\), b. \(-\frac{3\sqrt{2}}{5}\mathbf{i} - \frac{4\sqrt{2}}{5}\mathbf{k}\), c. \(\frac{1}{4}\mathbf{i} - \frac{1}{3}\mathbf{j} - \mathbf{k}\), d. \(\frac{a}{\sqrt{2}}\mathbf{i} + \frac{a}{\sqrt{3}}\mathbf{j} - \frac{a}{\sqrt{6}}\mathbf{k}\)."

Step 1: Understanding the Formula

To find a vector when its length and direction are known, use the formula of multiplying the unit vector (i.e., the directional vector) by the given length. If the direction vector is \( \mathbf{d} \) and the length is \( L \), the desired vector is \( \mathbf{v} = L \cdot \mathbf{d} \).

Step 2: Solving Part (a)

For a vector with length 7 and direction \( -\mathbf{j} \), the vector can be expressed as \( \mathbf{v}_a = 7 \times (-\mathbf{j}) = -7\mathbf{j} \).

Step 3: Solving Part (b)

Given length \( \sqrt{2} \) and direction \( -\frac{3}{5} \mathbf{i} - \frac{4}{5} \mathbf{k} \), the vector is \( \mathbf{v}_b = \sqrt{2} \times \left( -\frac{3}{5} \mathbf{i} - \frac{4}{5} \mathbf{k} \right) = -\frac{3\sqrt{2}}{5} \mathbf{i} - \frac{4\sqrt{2}}{5} \mathbf{k} \).

Step 4: Solving Part (c)

For length \( \frac{13}{12} \) and direction \( \frac{3}{13} \mathbf{i} - \frac{4}{13} \mathbf{j} - \frac{12}{13} \mathbf{k} \), the vector is \( \mathbf{v}_c = \frac{13}{12} \times \left( \frac{3}{13} \mathbf{i} - \frac{4}{13} \mathbf{j} - \frac{12}{13} \mathbf{k} \right) = \frac{3}{12} \mathbf{i} - \frac{4}{12} \mathbf{j} - \frac{12}{12} \mathbf{k} = \frac{1}{4} \mathbf{i} - \frac{1}{3} \mathbf{j} - \mathbf{k} \).

Step 5: Solving Part (d)

Given length \( a > 0 \) and direction \( \frac{1}{\sqrt{2}} \mathbf{i} + \frac{1}{\sqrt{3}} \mathbf{j} - \frac{1}{\sqrt{6}} \mathbf{k} \), the vector is \( \mathbf{v}_d = a \times \left( \frac{1}{\sqrt{2}} \mathbf{i} + \frac{1}{\sqrt{3}} \mathbf{j} - \frac{1}{\sqrt{6}} \mathbf{k} \right) = \frac{a}{\sqrt{2}} \mathbf{i} + \frac{a}{\sqrt{3}} \mathbf{j} - \frac{a}{\sqrt{6}} \mathbf{k} \).

Key Concepts

Unit Vector

A unit vector is a vector that has a magnitude of 1. It is often used to indicate direction without changing the size or length of other vectors.

The general form of a unit vector is obtained by dividing the vector by its magnitude. If a vector \( \mathbf{d} \) has components \( (x, y, z) \), the unit vector \( \mathbf{u} \) can be written as:

• \( \mathbf{u} = \frac{1}{\sqrt{x^2 + y^2 + z^2}} (x, y, z) \)

This calculation effectively standardizes the vector to a length of 1.

Unit vectors are particularly helpful when dealing with directional vectors in problems requiring vector normalization.

Magnitude of a Vector

The magnitude of a vector, often represented as \( \| \mathbf{v} \| \), is the length or size of the vector.

For a vector with components \( (x, y, z) \), the magnitude is calculated using the formula:

• \( \| \mathbf{v} \| = \sqrt{x^2 + y^2 + z^2} \)

Understanding the magnitude of a vector is crucial because it provides information on how extensive the vector is in terms of scale.

Magnitude plays a vital role in many vector computations, including scaling unit vectors and vector addition.

Directional Vector

A directional vector gives the direction of a vector without regard to its magnitude. It tells us in which direction we're pointing, serving as a compass within the vector space.

Often, directional vectors are normalized to become unit vectors, enabling them to maintain accurate directional properties while having a length of 1.

In our exercise, it's the given directions like \(-\mathbf{j}\) in Part (a) or \(-\frac{3}{5} \mathbf{i} - \frac{4}{5} \mathbf{k}\) in Part (b) that act as directional vectors.

Vector Multiplication

Vector multiplication involves scaling a vector by multiplying it with a scalar (a real number).

The multiplication process affects the magnitude of the vector while maintaining its direction. If \( \mathbf{d} \) is a directional vector and \( L \) is the scalar, then the calculated vector \( \mathbf{v} \) is:

• \( \mathbf{v} = L \cdot \mathbf{d} \)

This process is used in multiple scenarios, such as changing the vector's length while preserving the original direction.

In the provided exercise, we used vector multiplication to calculate vectors like \(-7\mathbf{j}\) by multiplying the unit direction vector by the given magnitude. This keeps the direction steady while elongating or shortening the vector.