Problem 27 - Calculus Early Transcendentals: Pearson New International Edition

Question

Which of the following do not make sense? (a) \(\mathbf{u} \cdot(\mathbf{v} \times \mathbf{w})\) (b) \(\mathbf{u}+(\mathbf{v} \times \mathbf{w})\) (c) \((\mathbf{a} \cdot \mathbf{b}) \times \mathbf{c}\) (d) \((\mathbf{a} \times \mathbf{b})+k\) (e) \((\mathbf{a} \cdot \mathbf{b})+k\) (f) \((\mathbf{a}+\mathbf{b}) \times(\mathbf{c}+\mathbf{d})\) (g) \((\mathbf{u} \times \mathbf{v}) \times \mathbf{w}\) (h) \((k \mathbf{u}) \times \mathbf{v}\)

Step-by-Step Solution

Answer

(c) and (d) do not make sense.

Step 1: Understand Dot and Cross Products

In vector algebra, the dot product of two vectors results in a scalar, and the cross product results in a vector. The dot product is performed between two vectors, while the cross product requires two vectors and results in a third vector perpendicular to the original ones. These definitions will help us analyze the given expressions for validity.

Step 2: Analyze Expression (a)

The expression \(\mathbf{u} \cdot(\mathbf{v} \times \mathbf{w})\) is valid. Here, \(\mathbf{v} \times \mathbf{w}\) results in a vector, and \(\mathbf{u} \cdot \) with a vector gives a scalar, which is a valid process.

Step 3: Analyze Expression (b)

In \(\mathbf{u}+(\mathbf{v} \times \mathbf{w})\), \(\mathbf{v} \times \mathbf{w}\) results in a vector, and adding two vectors (\(\mathbf{u}\) and the result of the cross product) is valid. Therefore, this expression makes sense.

Step 4: Analyze Expression (c)

The expression \((\mathbf{a} \cdot \mathbf{b}) \times \mathbf{c}\) does not make sense because \(\mathbf{a} \cdot \mathbf{b}\) is a scalar, and you cannot perform a cross product with a scalar and a vector.

Step 5: Analyze Expression (d)

In \((\mathbf{a} \times \mathbf{b})+k\), \(\mathbf{a} \times \mathbf{b}\) results in a vector, but adding a scalar \(k\) to a vector is not defined in vector algebra, making this expression invalid.

Step 6: Analyze Expression (e)

The expression \((\mathbf{a} \cdot \mathbf{b})+k\) makes sense because both \(\mathbf{a} \cdot \mathbf{b}\) and \(k\) are scalars, and adding two scalars is valid.

Step 7: Analyze Expression (f)

In \((\mathbf{a}+\mathbf{b}) \times(\mathbf{c}+\mathbf{d})\), first calculate vector additions \(\mathbf{a}+\mathbf{b}\) and \(\mathbf{c}+\mathbf{d}\), both of which result in vectors. The cross product can then be calculated between these vectors, making it a valid expression.

Step 8: Analyze Expression (g)

The expression \((\mathbf{u} \times \mathbf{v}) \times \mathbf{w}\) is valid. Here, \(\mathbf{u} \times \mathbf{v}\) gives a vector, and performing a cross product with another vector \(\mathbf{w}\) is legitimate.

Step 9: Analyze Expression (h)

For \((k \mathbf{u}) \times \mathbf{v}\), scalar \(k\) times vector \(\mathbf{u}\) results in a vector. Performing a cross product with another vector \(\mathbf{v}\) is valid.

Key Concepts

Dot Product

In vector algebra,

• the dot product is an operation that takes two vectors and returns a scalar value.
• This operation is a way to measure how much one vector extends in the direction of another.

A mathematical expression for the dot product of two vectors \(\mathbf{a}\) and \(\mathbf{b}\) is given by:\[ \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos(\theta) \]
where \(\theta\) is the angle between the two vectors and \(|\mathbf{a}|\) and \(|\mathbf{b}|\) are the magnitudes of the vectors.
Dot products are useful in determining angles between vectors or in projecting one vector onto another.
In the exercise example,

• Expression (a) \(\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})\) involves a dot product which is valid as it evaluates into a scalar.
• However, expression (c) \((\mathbf{a} \cdot \mathbf{b}) \times \mathbf{c}\) does not make sense because the result of a dot product is a scalar, and a scalar cannot be used in a cross product.

Cross Product

The cross product is another vector operation used in three dimensional space.

• Unlike the dot product, this operation takes two vectors and returns a third vector.
• The resulting vector is perpendicular to the plane formed by the two original vectors.

The mathematical expression for the cross product of vectors \(\mathbf{a}\) and \(\mathbf{b}\) is given as:\[ \mathbf{a} \times \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \sin(\theta) \mathbf{n} \]where \(\mathbf{n}\) is a unit vector perpendicular to both \(\mathbf{a}\) and \(\mathbf{b}\).
This operation is important in physics and engineering for finding torques or rotations.
In the exercise:

• Expression (b) and (f) are valid as the operations are between vectors that result in other vectors.
• Expression (g) is also valid because the operation is between two vectors.

Note that if a cross product expression involves a scalar as a primary operand, such as expression (c), it is invalid.

Vector Operations

In vector algebra, operations can involve adding, subtracting, or otherwise manipulating vectors.

• Addition, for example, combines two vectors to produce another vector.
• This is calculated by adding the corresponding components of each vector.

Mathematically, if \(\mathbf{a} = \langle a_1, a_2, a_3 \rangle\) and \(\mathbf{b} = \langle b_1, b_2, b_3 \rangle\), then:\[ \mathbf{a} + \mathbf{b} = \langle a_1 + b_1, a_2 + b_2, a_3 + b_3 \rangle \]

• These operations are defined using the properties of vectors, ensuring expressions remain consistent within vector spaces.

In the exercise,

• Expression (b) shows valid vector addition after the cross product between \(\mathbf{v}\) and \(\mathbf{w}\) yields another vector to be added to \(\mathbf{u}\).

Scalar Multiplication

Scalar multiplication in vector algebra involves multiplying a vector by a scalar value.

• This operation changes the magnitude of the vector but does not affect its direction, unless the scalar is negative.

If you have a vector \(\mathbf{a} = \langle a_1, a_2, a_3 \rangle\) and a scalar \(k\), then the scalar multiplication is expressed as:\[ k\mathbf{a} = \langle ka_1, ka_2, ka_3 \rangle \]

• This operation is foundational for vector space manipulations.

In the exercise:

• Expression (h) \((k \mathbf{u}) \times \mathbf{v}\) makes sense because scaler multiplication of \(k\) with \(\mathbf{u}\) gives a new vector which can then cross product with another vector \(\mathbf{v}\).
• However, expressions like (d) which attempt to add a scalar to a vector are incorrect as they involve incompatible types.

This showcases the importance of matching vector and scalar types in these operations.