Problem 27 - Advanced Engineering Mathematics

Question

Find the indicated scalar or vector without using \((5),(13)\), or \((15)\). $$ \mathbf{k} \cdot(\mathbf{j} \times \mathbf{k}) $$

Step-by-Step Solution

Answer

The result of the expression is zero.

Step 1: Understand the Expression

The expression to evaluate is \( \mathbf{k} \cdot (\mathbf{j} \times \mathbf{k}) \). This involves a dot product between the vector \( \mathbf{k} \) and the cross product \( \mathbf{j} \times \mathbf{k} \).

Step 2: Recall Cross Product Properties

The cross product \( \mathbf{A} \times \mathbf{B} \) results in a vector perpendicular to both \( \mathbf{A} \) and \( \mathbf{B} \). Specific to unit vectors, \( \mathbf{j} \times \mathbf{k} = \mathbf{i} \).

Step 3: Compute the Dot Product

Substitute the resultant vector from the cross product: \( \mathbf{k} \cdot \mathbf{i} \). Recall that the dot product between two perpendicular vectors, such as \( \mathbf{k} \) and \( \mathbf{i} \), is zero.

Key Concepts

Cross Product

In vector calculus, the cross product is a binary operation on two vectors in three-dimensional space. It is denoted by \( \mathbf{A} \times \mathbf{B} \) and results in a vector that is perpendicular to the plane containing \( \mathbf{A} \) and \( \mathbf{B} \). The magnitude of the cross product can be determined using the formula: \[\| \mathbf{A} \times \mathbf{B} \| = \| \mathbf{A} \| \| \mathbf{B} \| \sin \theta\] where \( \theta \) is the angle between the two vectors. Some important properties include:

• The cross product is not commutative, meaning \( \mathbf{A} \times \mathbf{B} eq \mathbf{B} \times \mathbf{A} \).
• Instead, \( \mathbf{A} \times \mathbf{B} = - (\mathbf{B} \times \mathbf{A}) \).
• The cross product of parallel vectors is zero since \( \sin \theta = 0 \) for \( \theta = 0 \) or \( 180^{\circ} \).

In this exercise, when calculating \( \mathbf{j} \times \mathbf{k} \), we get \( \mathbf{i} \), which is perpendicular to both \( \mathbf{j} \) and \( \mathbf{k} \). This demonstrates how the cross product aids in finding vectors that are orthogonal to a given plane.

Dot Product

The dot product, also known as the scalar product, is an operation that takes two equal-length sequences of numbers and returns a single number. In the context of vectors, the dot product measures the cosine of the angle between them and is given by:\[\mathbf{A} \cdot \mathbf{B} = \| \mathbf{A} \| \| \mathbf{B} \| \cos \theta\]Some essential features of the dot product include:

• The result is a scalar, unlike the vector result of the cross product.
• If two vectors are perpendicular, their dot product is zero since \( \cos 90^{\circ} = 0 \).
• The dot product is commutative, meaning \( \mathbf{A} \cdot \mathbf{B} = \mathbf{B} \cdot \mathbf{A} \).

In this exercise, after determining \( \mathbf{j} \times \mathbf{k} = \mathbf{i} \), we used the result in the dot product \( \mathbf{k} \cdot \mathbf{i} \). Since \( \mathbf{k} \) and \( \mathbf{i} \) are perpendicular, their dot product is zero, providing the answer.

Unit Vectors

Unit vectors are vectors with a magnitude of one and are instrumental in defining directions. In a standard coordinate system, the primary unit vectors are:

• \( \mathbf{i} \): Points in the direction of the x-axis and has coordinates (1, 0, 0).
• \( \mathbf{j} \): Points in the direction of the y-axis with coordinates (0, 1, 0).
• \( \mathbf{k} \): Points in the direction of the z-axis with coordinates (0, 0, 1).

These vectors simplify vector operations as they can be combined to express any vector in three-dimensional space. When dealing with cross and dot products, knowing the relationships between unit vectors is crucial. For example:

• \( \mathbf{i} \times \mathbf{j} = \mathbf{k} \)
• \( \mathbf{j} \times \mathbf{k} = \mathbf{i} \)
• \( \mathbf{k} \times \mathbf{i} = \mathbf{j} \)

The understanding of these vectors helped us solve the exercise efficiently. We used the property \( \mathbf{j} \times \mathbf{k} = \mathbf{i} \) to determine the vector perpendicular to both, and this simplicity is what makes unit vectors a powerful tool in vector calculus.