Problem 22 - University Calculus: Early Transcendentals

Question

Verify that \((\mathbf{u} \times \mathbf{v}) \cdot \mathbf{w}=(\mathbf{v} \times \mathbf{w}) \cdot \mathbf{u}=\) \((\mathbf{w} \times \mathbf{u}) \cdot \mathbf{v}\) and find the volume of the parallelepiped (box) determined by \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{w}\). $$\begin{array}{ccc} \mathbf{u} & \mathbf{v} & \mathbf{w} \\ \hline \mathbf{i}+\mathbf{j}-2 \mathbf{k} & -\mathbf{i}-\mathbf{k} & 2 \mathbf{i}+4 \mathbf{j}-2 \mathbf{k} \end{array}$$

Step-by-Step Solution

Answer

Yes, the expressions are equal and the volume is 8.

Step 1: Express vectors in component form

Firstly, express the vectors \(\mathbf{u}, \mathbf{v}, \text{and } \mathbf{w}\) in component form. We have: \(\mathbf{u} = \begin{pmatrix} 1 \ 1 \ -2 \end{pmatrix}\), \(\mathbf{v} = \begin{pmatrix} -1 \ 0 \ -1 \end{pmatrix}\), and \(\mathbf{w} = \begin{pmatrix} 2 \ 4 \ -2 \end{pmatrix}\).

Step 2: Compute \(\mathbf{u} \times \mathbf{v}\)

To find the cross product \(\mathbf{u} \times \mathbf{v}\), use the determinant:\(\mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 1 & -2 \ -1 & 0 & -1 \end{vmatrix} = \mathbf{i}(1\cdot(-1) - (-2)\cdot0) - \mathbf{j}(1\cdot(-1) - (-2)(-1)) + \mathbf{k}(1\cdot0 - 1\cdot(-1))\ = -\mathbf{i} + 3\mathbf{j} + \mathbf{k} \).Hence, \(\mathbf{u} \times \mathbf{v} = \begin{pmatrix} -1 \ 3 \ 1 \end{pmatrix}\).

Step 3: Compute \(\mathbf{v} \times \mathbf{w}\)

To compute \(\mathbf{v} \times \mathbf{w}\), find the determinant:\(\mathbf{v} \times \mathbf{w} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -1 & 0 & -1 \ 2 & 4 & -2 \end{vmatrix} = \mathbf{i}(0 \cdot (-2) - (-1) \cdot 4) - \mathbf{j}((-1) \cdot (-2) - (-1) \cdot 2) + \mathbf{k}((-1) \cdot 4 - 0 \cdot 2)\ = 4\mathbf{i} + 0\mathbf{j} - 4\mathbf{k} = \begin{pmatrix} 4 \ 0 \ -4 \end{pmatrix}\).

Step 4: Compute \(\mathbf{w} \times \mathbf{u}\)

Find \(\mathbf{w} \times \mathbf{u}\) using the determinant method:\(\mathbf{w} \times \mathbf{u} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2 & 4 & -2 \ 1 & 1 & -2 \end{vmatrix} = \mathbf{i}(4\cdot(-2) - (-2)\cdot1) - \mathbf{j}(2\cdot(-2) - (-2)\cdot1) + \mathbf{k}(2\cdot1 - 4\cdot1)\ = -6\mathbf{i} -2\mathbf{j} - 2\mathbf{k} = \begin{pmatrix} -6 \ -2 \ -2 \end{pmatrix}\).

Step 5: Calculate dot products

Calculate the dot products to check equality:1. \((\mathbf{u} \times \mathbf{v}) \cdot \mathbf{w} = \begin{pmatrix} -1 \ 3 \ 1 \end{pmatrix} \cdot \begin{pmatrix} 2 \ 4 \ -2 \end{pmatrix} = -1\cdot2 + 3\cdot4 + 1\cdot(-2) = 8\)2. \((\mathbf{v} \times \mathbf{w}) \cdot \mathbf{u} = \begin{pmatrix} 4 \ 0 \ -4 \end{pmatrix} \cdot \begin{pmatrix} 1 \ 1 \ -2 \end{pmatrix} = 4\cdot1 + 0\cdot1 - 4\cdot(-2) = 8\)3. \((\mathbf{w} \times \mathbf{u}) \cdot \mathbf{v} = \begin{pmatrix} -6 \ -2 \ -2 \end{pmatrix} \cdot \begin{pmatrix} -1 \ 0 \ -1 \end{pmatrix} = -6\cdot(-1) - 2\cdot0 - 2\cdot(-1) = 8\)

Step 6: Conclusion on equality

All the dot products evaluate to 8. Therefore, \((\mathbf{u} \times \mathbf{v}) \cdot \mathbf{w} = (\mathbf{v} \times \mathbf{w}) \cdot \mathbf{u} = (\mathbf{w} \times \mathbf{u}) \cdot \mathbf{v} = 8\).

Step 7: Volume of the parallelepiped

The volume of the parallelepiped formed by the vectors \(\mathbf{u}, \mathbf{v}, \text{and } \mathbf{w}\) is given by the magnitude of the scalar triple product \(|(\mathbf{u} \times \mathbf{v}) \cdot \mathbf{w}|=|8|=8\).

Key Concepts

Cross Product

The cross product is a way to multiply two vectors in three-dimensional space, resulting in a third vector that is perpendicular to both of the original vectors.
It is especially useful in determining the orientation and area of surfaces like parallelograms.

To calculate the cross product, represented as \( \mathbf{a} \times \mathbf{b} \), you use the determinant of a matrix. This matrix has three rows: the unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) along the x, y, and z axes, and the components of the vectors \( \mathbf{a} \) and \( \mathbf{b} \).
The calculation is as follows:

• Expand the determinant using cofactor expansion to find each component of the resulting vector.
• For example, if \( \mathbf{a} = \begin{pmatrix} a_1 & a_2 & a_3 \end{pmatrix} \) and \( \mathbf{b} = \begin{pmatrix} b_1 & b_2 & b_3 \end{pmatrix} \), then \( \mathbf{a} \times \mathbf{b} = \begin{pmatrix} a_2b_3 - a_3b_2 & a_3b_1 - a_1b_3 & a_1b_2 - a_2b_1 \end{pmatrix} \).

This vector helps in solving problems about forces, rotations, and finding volumes of parallelepipeds.

Dot Product

The dot product is a scalar representation of the multiplication of two vectors.
Unlike the cross product, which results in a vector, the dot product results in a single number (scalar).

It can be calculated by multiplying the corresponding components of two vectors and summing the products.
If \( \mathbf{a} = \begin{pmatrix} a_1 & a_2 & a_3 \end{pmatrix} \) and \( \mathbf{b} = \begin{pmatrix} b_1 & b_2 & b_3 \end{pmatrix} \), then the dot product \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2 b_2 + a_3 b_3 \).
The result gives a measure of how aligned two vectors are, with higher values signifying stronger alignment.

• When the dot product is zero, the vectors are perpendicular to each other.
• It is also used in calculating the angle between vectors.

This concept is pivotal in physics for work calculations and in computer graphics for lighting and shading.

Parallelepiped Volume

The volume of a parallelepiped, which is essentially a 3D box made of parallelograms, can be found using the scalar triple product of vectors.
Given vectors \( \mathbf{a}, \mathbf{b}, \mathbf{c} \) corresponding to the edges of a parallelepiped, the volume \( V \) is computed as the absolute value of one of the scalar triple products:

• \( V = |(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}| \)
• This equals \( |(\mathbf{b} \times \mathbf{c}) \cdot \mathbf{a}| \) or \( |(\mathbf{c} \times \mathbf{a}) \cdot \mathbf{b}| \) due to the properties of cyclic permutation in scalar triple products.

This computation involves first finding the cross product between two vectors to get a vector perpendicular to them, then dot-multiplying this vector with the third vector. The absolute value accounts for direction to ensure volume is non-negative.
This concept is extremely useful in engineering, physics, and computer graphics to determine the capacity and space utilization of constructed structures.

Vector Components

Vectors in mathematics are typically represented by their components, which are numbers indicating magnitude and direction along each axis of a coordinate system.
Breaking a vector down into its components helps in performing mathematical operations like addition, multiplication, and find projections.
For a vector \( \mathbf{a} \) in three-dimensional space, its components are given as \( (a_1, a_2, a_3) \), representing its projection along the x, y, and z axes respectively.

• These components can be used in practical calculations, such as determining the cross or dot product between vectors.
• Each component is essential for understanding the vector in terms of its influence and directionality.
• It simplifies complex spatial problems by breaking them into more manageable parts.

Comprehending how vector components work is crucial in fields such as physics, engineering, computer programming, and anywhere else vectors are applied.