Problem 13 - Essential Calculus Early Transcendentals

Question

State whether each expression is meaningful. If not, explain why. If so, state whether it is a vector or a scalar. $$\begin{array}{ll}{\text { (a) } \mathbf{a} \cdot(\mathbf{b} \times \mathbf{c})} & {\text { (b) } \mathbf{a} \times(\mathbf{b} \cdot \mathbf{c})} \\\ {(\mathbf{c}) \mathbf{a} \times(\mathbf{b} \times \mathbf{c})} & {\text { (d) } \mathbf{a} \cdot(\mathbf{b} \cdot \mathbf{c})} \\ { (e) } {(\mathbf{a} \cdot \mathbf{b}) \times(\mathbf{c} \cdot \mathbf{d})} & {\text { (f) }(\mathbf{a} \times \mathbf{b}) \cdot(\mathbf{c} \times \mathbf{d})}\end{array}$$

Step-by-Step Solution

Answer

(a) Scalar. (b) Not meaningful. (c) Vector. (d) Not meaningful. (e) Not meaningful. (f) Scalar.

Step 1: Analyzing Expression (a)

The expression \( \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) \) involves the dot product of a vector \( \mathbf{a} \) and the cross product of vectors \( \mathbf{b} \) and \( \mathbf{c} \). The cross product \( \mathbf{b} \times \mathbf{c} \) results in a vector, and the dot product of two vectors results in a scalar. Therefore, this expression is meaningful and results in a scalar.

Step 2: Analyzing Expression (b)

The expression \( \mathbf{a} \times (\mathbf{b} \cdot \mathbf{c}) \) involves taking the cross product of a vector \( \mathbf{a} \) and the dot product \( \mathbf{b} \cdot \mathbf{c} \). The dot product results in a scalar. However, you cannot take the cross product of a vector and a scalar, making this expression not meaningful.

Step 3: Analyzing Expression (c)

The expression \( \mathbf{a} \times (\mathbf{b} \times \mathbf{c}) \) involves the cross product of \( \mathbf{b} \times \mathbf{c} \) which results in a vector. The subsequent cross product with vector \( \mathbf{a} \) also results in a vector. Hence, this expression is meaningful and results in a vector.

Step 4: Analyzing Expression (d)

The expression \( \mathbf{a} \cdot (\mathbf{b} \cdot \mathbf{c}) \) involves a dot product \( \mathbf{b} \cdot \mathbf{c} \) which results in a scalar. You cannot take the dot product of a vector and a scalar, so this expression is not meaningful.

Step 5: Analyzing Expression (e)

The expression \( (\mathbf{a} \cdot \mathbf{b}) \times (\mathbf{c} \cdot \mathbf{d}) \) involves two dot products, each resulting in a scalar. This results in an attempt to take the cross product of two scalars, which is not possible, making this expression not meaningful.

Step 6: Analyzing Expression (f)

The expression \( (\mathbf{a} \times \mathbf{b}) \cdot (\mathbf{c} \times \mathbf{d}) \) involves two cross products, each resulting in a vector. The subsequent dot product of these vectors results in a scalar. Therefore, this expression is meaningful and results in a scalar.

Key Concepts

Dot Product

The dot product, also known as the scalar product, is a fundamental operation in vector algebra. It involves the multiplication of two vectors to result in a scalar. The mathematical representation of the dot product of two vectors \( \mathbf{a} \) and \( \mathbf{b} \) is expressed as \( \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos(\theta) \), where \( \theta \) is the angle between the two vectors.
- The dot product gives information about the magnitude and direction of vectors relative to each other.
- It is used to determine when two vectors are orthogonal, meaning they are at a right angle to each other. This occurs when the dot product is zero.
- The result is always a scalar, indicating either the scalar component of one vector along the direction of the other or indicating work when dealing with physics problems.
Understanding that the dot product results in a scalar is crucial when analyzing expressions involving vectors, such as confirming the validity of vector expressions like those in the original exercise.

Cross Product

The cross product, or vector product, of two vectors results in a third vector that is perpendicular to both. This operation is unique to three-dimensional space and gives us important information about the spatial relationship between vectors. If \( \mathbf{a} \) and \( \mathbf{b} \) are vectors, the cross product is denoted \( \mathbf{a} \times \mathbf{b} = \mathbf{c} \), where \( \mathbf{c} \) is a vector.
- The magnitude of \( \mathbf{c} \) is given by \( |\mathbf{c}| = |\mathbf{a}| |\mathbf{b}| \sin(\theta) \), where \( \theta \) is the angle between \( \mathbf{a} \) and \( \mathbf{b} \).
- The direction of \( \mathbf{c} \) is determined by the right-hand rule, which means that if your fingers follow the curve of \( \mathbf{a} \) to \( \mathbf{b} \), your thumb points in the direction of \( \mathbf{c} \).
- Unlike the dot product, the cross product is antisymmetric, meaning \( \mathbf{a} \times \mathbf{b} = - (\mathbf{b} \times \mathbf{a}) \).
The cross product's resulting vector nature makes it useful for determining areas of parallelograms and rotations, and is key in vector expressions like those assessed in the original problem, especially when validating vector-to-vector operations.

Scalars and Vectors

In vector algebra, distinguishing between scalars and vectors is crucial for understanding operations like dot and cross products. Scalars are quantities with only magnitude, such as 5 or -3, and are used in multiplication with vectors without any directional attribute.
- Vectors, on the other hand, have both magnitude and direction. Examples include velocity and force, which are represented by arrows.
- Operations involving vectors such as the dot and cross products inherently require understanding their difference. Scalars result from dot products, while vectors result from cross products.
- When combining vectors and scalars within mathematical operations, it's important to remember that you cannot directly cross a scalar with a vector, or take the dot product of two scalars, a mistake that makes some vector operations non-meaningful, as presented in exercises.
By clearly understanding the difference, one can approach problems effectively, especially when deciding the outcome of expressions in vector algebra like those in the given exercise.