Problem 22 - Thomas Calculus

Question

In Exercises \(19-22,\) 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 \(\mathbf{w} .\) $$ \mathbf{i}+\mathbf{j}-2 \mathbf{k} \quad-\mathbf{i}-\mathbf{k} \quad 2 \mathbf{i}+4 \mathbf{j}-2 \mathbf{k} $$

Step-by-Step Solution

Answer

The volume of the parallelepiped is 8.

Step 1: Understand the Vectors

Given are the vectors \( \mathbf{u} = \mathbf{i} + \mathbf{j} - 2 \mathbf{k} \), \( \mathbf{v} = -\mathbf{i} - \mathbf{k} \), and \( \mathbf{w} = 2\mathbf{i} + 4\mathbf{j} - 2\mathbf{k} \). Identify each vector's components for further calculations.

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

Use the determinant method to find the cross product: \[ \mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 1 & -2 \ -1 & 0 & -1 \end{vmatrix} = ((1)(-1) - (-2)(0))\mathbf{i} - ((1)(-1) - (-2)(-1))\mathbf{j} + ((1)(0) - (1)(-1))\mathbf{k} \]Solve this to get \( \mathbf{u} \times \mathbf{v} = -\mathbf{i} + 3\mathbf{j} + \mathbf{k} \).

Step 3: Calculate \((\mathbf{u} \times \mathbf{v}) \cdot \mathbf{w}\)

Use the dot product with the result from Step 2 and vector \( \mathbf{w} \):\[ (-\mathbf{i} + 3\mathbf{j} + \mathbf{k}) \cdot (2\mathbf{i} + 4\mathbf{j} - 2\mathbf{k}) = -2 + 12 - 2 = 8 \]

Step 4: Calculate \( \mathbf{v} \times \mathbf{w} \)

Find the cross product of \( \mathbf{v} \) and \( \mathbf{w} \):\[ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -1 & 0 & -1 \ 2 & 4 & -2 \end{vmatrix} = ((0)(-2) - (-1)(4))\mathbf{i} - ((-1)(-2) - (-1)(2))\mathbf{j} + ((-1)(4) - (0)(2))\mathbf{k} \]Solve this to get \( \mathbf{v} \times \mathbf{w} = 4\mathbf{i} - 4\mathbf{j} - 4\mathbf{k} \).

Step 5: Calculate \((\mathbf{v} \times \mathbf{w}) \cdot \mathbf{u}\)

Perform the dot product of the vector from Step 4 with \( \mathbf{u} \): \[ (4\mathbf{i} - 4\mathbf{j} - 4\mathbf{k}) \cdot (\mathbf{i} + \mathbf{j} - 2\mathbf{k}) = 4 - 4 + 8 = 8 \]

Step 6: Calculate \( \mathbf{w} \times \mathbf{u} \)

Determine the cross product of \( \mathbf{w} \) and \( \mathbf{u} \):\[ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2 & 4 & -2 \ 1 & 1 & -2 \end{vmatrix} = ((4)(-2) - (-2)(1))\mathbf{i} - ((2)(-2) - (-2)(1))\mathbf{j} + ((2)(1) - (4)(1))\mathbf{k} \]Solve this to get \( \mathbf{w} \times \mathbf{u} = -6\mathbf{i} + 6\mathbf{j} - 2\mathbf{k} \).

Step 7: Calculate \((\mathbf{w} \times \mathbf{u}) \cdot \mathbf{v}\)

Perform the dot product of the vector from Step 6 with \( \mathbf{v} \):\[ (-6\mathbf{i} + 6\mathbf{j} - 2\mathbf{k}) \cdot (-\mathbf{i} - \mathbf{k}) = 6 - 2 = 8 \]

Step 8: Verify "Box Product" Properties and Calculate Volume

Since \( (\mathbf{u} \times \mathbf{v}) \cdot \mathbf{w} = (\mathbf{v} \times \mathbf{w}) \cdot \mathbf{u} = (\mathbf{w} \times \mathbf{u}) \cdot \mathbf{v} = 8 \), the scalar triple product is consistent, confirming the specified property. The volume of the parallelepiped is 8.

Key Concepts

Dot Product

The dot product, often called the "scalar product," focuses on multiplying two vectors to obtain a scalar (a single number). This operation captures how much one vector extends in the direction of another. The dot product of vectors \( \mathbf{a} \) and \( \mathbf{b} \) is expressed as \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \). It involves multiplying corresponding components of each vector together, and then summing those products.

One key property is its relation to the angle \( \theta \) between two vectors:

• If two vectors are perpendicular, their dot product is zero because \( \cos(90^\circ) = 0 \).
• The dot product is positive if the angle between the vectors is acute and negative if it's obtuse.

In simple terms, the dot product tells you how "aligned" two vectors are. When applied in the context of this exercise, the dot product helps verify if the terms obtained from cross products \((\mathbf{u} \times \mathbf{v}) \cdot \mathbf{w} \) have consistent results, maintaining the scalar triple product equality.

Scalar Triple Product

The scalar triple product involves three vectors and is a combination of cross and dot products. For vectors \( \mathbf{a} \), \( \mathbf{b} \), and \( \mathbf{c} \), the scalar triple product is denoted as \(( \mathbf{a} \times \mathbf{b} ) \cdot \mathbf{c} \). This results in a scalar value and is computed by first taking the cross product of two vectors, which produces a vector, and then performing the dot product of the resulting vector with the third vector.

The properties make it vital in verifying identities and properties in vector algebra:

• Cyclic Property: \((\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c} = (\mathbf{b} \times \mathbf{c}) \cdot \mathbf{a} = (\mathbf{c} \times \mathbf{a}) \cdot \mathbf{b} \).
• If the scalar triple product is zero, the vectors are coplanar, meaning they lie in the same plane.

This property was employed in the exercise to confirm the relationship between different products of vectors \( \mathbf{u} \), \( \mathbf{v} \), and \( \mathbf{w} \). The consistency of the scalar triple product across permutations solidifies the reliability of the calculations.

Parallelepiped Volume

A parallelepiped is a six-faced figure (3D analogue of a parallelogram) formed by three vectors. Its volume can be calculated using the scalar triple product, which provides a practical geometric interpretation of this algebraic operation. If vectors \( \mathbf{u} \), \( \mathbf{v} \), and \( \mathbf{w} \) form a parallelepiped, the volume \( V \) is: \[ V = | (\mathbf{u} \times \mathbf{v}) \cdot \mathbf{w} | \]This formula arises because the magnitude of the cross product \( \mathbf{u} \times \mathbf{v} \) gives the area of the parallelogram base, and the dot product with \( \mathbf{w} \) accounts for the height when extended perpendicularly from this base.

The absolute value ensures that the volume is a positive number, even if the scalar triple product itself could be negative due to vector orientation. The exercise illustrates this by revealing the volume of the parallelepiped to be 8, demonstrating an effective application of these vector operations in a spatial context.