Problem 31 - Thomas Calculus

Question

Let \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{w}\) be vectors. Which of the following make sense, and which do not? Give reasons for your answers. $$ \begin{array}{l}{\text { a. }(\mathbf{u} \times \mathbf{v}) \cdot \mathbf{w}} \\\ {\text { b. } \mathbf{u} \times(\mathbf{v} \cdot \mathbf{w})} \\\ {\mathbf{c} \cdot \mathbf{u} \times(\mathbf{v} \times \mathbf{w})} \\ {\text { d. } \mathbf{u} \cdot(\mathbf{v} \cdot \mathbf{w})}\end{array} $$

Step-by-Step Solution

Answer

a makes sense; b, c, and d do not make sense.

Step 1: Analyze (a)

Expression: \((\mathbf{u} \times \mathbf{v}) \cdot \mathbf{w}\) The expression \(\mathbf{u} \times \mathbf{v}\) results in a vector. Then, the expression \((\mathbf{u} \times \mathbf{v}) \cdot \mathbf{w}\) involves the dot product between a vector \((\mathbf{u} \times \mathbf{v})\) and another vector \(\mathbf{w}\), which is valid. Therefore, this expression makes sense.

Step 2: Analyze (b)

Expression: \(\mathbf{u} \times (\mathbf{v} \cdot \mathbf{w})\) The expression \(\mathbf{v} \cdot \mathbf{w}\) results in a scalar. The cross product \(\mathbf{u} \times (\text{scalar})\) is undefined because the cross product requires two vectors. Therefore, this expression does not make sense.

Step 3: Analyze (c)

Expression: \((\mathbf{c} \cdot \mathbf{u}) \times (\mathbf{v} \times \mathbf{w})\) Since there is no vector \(\mathbf{c}\) defined in the problem, assume it is a typo or see if any related vector is mentioned, but without ambiguity, we should treat it invalidly as per the given data. Even if \(\mathbf{c}\) were a valid vector, the expression \(\mathbf{c} \cdot \mathbf{u}\) would result in a scalar, making the expression invalid because a cross product needs two vectors.

Step 4: Analyze (d)

Expression: \(\mathbf{u} \cdot (\mathbf{v} \cdot \mathbf{w})\) The expression \(\mathbf{v} \cdot \mathbf{w}\) results in a scalar. Since a dot product of a vector with a scalar is not defined, i.e., we can't take a dot product between a scalar and a vector, this expression does not make sense.

Key Concepts

Dot Product

The dot product, also known as the scalar product, is a fundamental operation in vector algebra. It calculates the scalar value from two vectors. The primary formula for the dot product of vectors \( \mathbf{u} \) and \( \mathbf{v} \) is given by: \[ \mathbf{u} \cdot \mathbf{v} = \|\mathbf{u}\| \|\mathbf{v}\| \cos \theta \] where \( \|\mathbf{u}\| \) and \( \|\mathbf{v}\| \) are the magnitudes of the vectors, and \( \theta \) is the angle between them. The result of the dot product is always a scalar, which means a single numerical value.
When performing a dot product, it's important to remember that this operation blends two vectors into one number. This can be particularly useful for determining the angle between two vectors or checking if two vectors are perpendicular. If the dot product equals zero, the vectors are orthogonal.

• The dot product is commutative: \( \mathbf{u} \cdot \mathbf{v} = \mathbf{v} \cdot \mathbf{u} \).
• It is also distributive over vector addition: \( \mathbf{u} \cdot (\mathbf{v} + \mathbf{w}) = \mathbf{u} \cdot \mathbf{v} + \mathbf{u} \cdot \mathbf{w} \).
• However, you cannot compute a dot product between a vector and a scalar, as highlighted in step 4 of the solution.

Cross Product

The cross product, or vector product, is another key operation in vector algebra. It is used to find a vector perpendicular to the plane formed by two given vectors. For two vectors \( \mathbf{u} \) and \( \mathbf{v} \), their cross product is defined as:\[ \mathbf{u} \times \mathbf{v} = \|\mathbf{u}\| \|\mathbf{v}\| \sin \theta \cdot \mathbf{n}\] where \( \|\mathbf{u}\| \) and \( \|\mathbf{v}\| \) are magnitudes, \( \theta \) is the angle between them, and \( \mathbf{n} \) is the unit vector perpendicular to the plane containing \( \mathbf{u} \) and \( \mathbf{v} \).
This operation yields another vector, whose direction is given by the right-hand rule. The magnitude of this vector is equal to the area of the parallelogram that \( \mathbf{u} \) and \( \mathbf{v} \) span. Note that the cross product is not defined for scalar quantities. Therefore, expressions like \( \mathbf{u} \times (\mathbf{v} \cdot \mathbf{w}) \) are invalid because \( \mathbf{v} \cdot \mathbf{w} \) produces a scalar.

• The cross product is anti-commutative: \( \mathbf{u} \times \mathbf{v} = - (\mathbf{v} \times \mathbf{u}) \).
• It obeys the distributive law: \( \mathbf{u} \times (\mathbf{v} + \mathbf{w}) = (\mathbf{u} \times \mathbf{v}) + (\mathbf{u} \times \mathbf{w}) \).
• If either vector is zero, the cross product is zero.

Vector Operations

Vector operations encompass various mathematical procedures performed on vectors, including both the dot and cross products. These operations help in manipulating vectors to solve complex problems in geometry, physics, and engineering.
The fundamental vector operations include:

• Vector Addition: Combining two vectors to form a resultant vector. This is done graphically by aligning the tail of one vector to the head of another.
• Scalar Multiplication: Scaling a vector by a real number, affecting its magnitude but not its direction.
• Subtraction: Similar to adding a negative vector, involves finding the resultant from the opposite direction.

Each operation serves a specific application in solving problems related to vector quantities. When evaluating expressions as in the original exercise, understanding what operation results in a vector or scalar is crucial. This allows us to make sense of whether certain operations, like a dot product or cross product, are valid.
Remember, vectors represent both direction and magnitude, which differentiates them from scalars that only possess magnitude. Thus, successful manipulation of vectors often depends on proper application and understanding of these foundational operations.