Problem 32 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

Scale the unit vectors by their respective lengths: vectors are - a: - b: \(-\frac{3\sqrt{2}}{5}\mathbf{i}-\frac{4\sqrt{2}}{5}\mathbf{k}\), - c: \(\frac{1}{4}\mathbf{i}-\frac{1}{3}\mathbf{j}-1\mathbf{k}\), - d: \(\frac{a}{\sqrt{2}}\mathbf{i}+\frac{a}{\sqrt{3}}\mathbf{j}-\frac{a}{\sqrt{6}}\mathbf{k}\).

Step 1: Find the Unit Vector for Direction

To find the vectors with given lengths and directions, we first confirm that the direction vectors are unit vectors. This is necessary because a unit vector retains the direction, and then we can scale it to the required magnitude for each vector.

Step 2: Determine the Given Length of the Vector

Each vector's magnitude is provided. For each given vector problem, identify the provided length: - For part a: length is 7. - For part b: length is \(\sqrt{2}\). - For part c: length is \(\frac{13}{12}\). - For part d: length is \(a\), where \(a>0\).

Step 3: Establish Vector by Scaling Unit Vector

Multiply the unit vector by the given length to obtain the desired vector. - For part a, multiply 7 by the unit vector. - For part b, multiply \(\sqrt{2}\) by \(-\frac{3}{5}\mathbf{i}-\frac{4}{5}\mathbf{k}\) to get \(-\frac{3\sqrt{2}}{5}\mathbf{i}-\frac{4\sqrt{2}}{5}\mathbf{k}\). - For part c, multiply \(\frac{13}{12}\) by the unit vector: \(\frac{3}{13}\mathbf{i}-\frac{4}{13}\mathbf{j}-\frac{12}{13}\mathbf{k}\) to form the vector \(\frac{1}{4}\mathbf{i}-\frac{1}{3}\mathbf{j}-1\mathbf{k}\). - For part d, multiply \(a\) by the unit vector: obtaining \(\frac{a}{\sqrt{2}}\mathbf{i}+\frac{a}{\sqrt{3}}\mathbf{j}-\frac{a}{\sqrt{6}}\mathbf{k}\).

Step 4: Confirm the Calculated Vectors

Verify that each resulting vector has the correct length and maintains the given direction. This ensures our calculations are precise and the vectors align with the specified criteria for each part of the problem.

Key Concepts

unit vectors

Unit vectors are fundamental in the realm of vector mathematics. They serve as a representation of direction and possess a magnitude of exactly 1, irrespective of their orientation. These vectors can be seen as the building blocks for constructing other vectors by scaling. When we discuss a direction vector, like \(\frac{3}{13} \mathbf{i}-\frac{4}{13} \mathbf{j}-\frac{12}{13} \mathbf{k}\), confirming its status as a unit vector ensures it's purely about direction.
Identifying a vector as a unit vector is crucial. It allows us to scale it to any length while maintaining the direction intact.

• Every vector can be transformed into a unit vector by dividing it by its magnitude.
• Unit vectors are denoted by a circumflex or "hat" symbol such as \( \mathbf{\hat{u}} \).
• In Cartesian coordinates, the unit vectors are \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \) along the x, y, and z axes respectively.

To create a vector of a certain magnitude with a given direction, we start with a unit vector. It is then scaled, implying multiplication by the desired vector length, producing a complete vector in the specified direction.

vector length

The length of a vector, also known as its magnitude, gives us an idea of how long the vector is. This quantity informs us about the size or the extent of the vector without considering its actual direction. Calculating vector length is an essential skill when working with vectors.
To find the length of a vector \( \mathbf{v} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k} \), you use the formula:
\[ \| \mathbf{v} \| = \sqrt{a^2 + b^2 + c^2} \]

• This formula derives from the Pythagorean theorem in three-dimensional space.
• The square root of the sum of the squares of its components gives the vector's magnitude.
• For unit vectors, this length is 1.

It's important to correctly calculate length, particularly when verifying the accuracy of a vector after performing operations like scaling. For example, if we need a vector of length \( \frac{13}{12} \) with a given direction, multiplying the unit vector direction components accordingly ensures achieving the desired magnitude.

direction vectors

When dealing with vectors, the direction vector is key in determining the precise path a vector takes. This type of vector has no specific length associated with it, rather it focuses on expressing the orientation aspect of vectors in spaces.
A direction vector, such as \( \frac{1}{\sqrt{2}} \mathbf{i} + \frac{1}{\sqrt{3}} \mathbf{j} - \frac{1}{\sqrt{6}} \mathbf{k} \), can be scaled to meet any magnitude requirement.

• Direction vectors provide valuable information about the trajectory and orientation in space, irrespective of distance.
• They are particularly useful in 3D modeling, physics, and engineering to simulate directional forces.
• To ensure a correct perspective, this orientation must align with the target or need.

Scaling such a vector to achieve a desired length (like \( a \)) while maintaining direction involves multiplying the unit vector by the required magnitude. Maintaining the direction remains central, even as the vector's size is adjusted according to calculations or specific needs in practice.