Problem 20 - Advanced Engineering Mathematics

Question

Find the indicated scalar or vector without using \((5),(13)\), or \((15)\). $$ \mathbf{i} \times(-3 \mathbf{k}) $$

Step-by-Step Solution

Answer

The result is \( 3\mathbf{j} \).

Step 1: Understand the Cross Product

The cross product of two vectors, \( \mathbf{a} \times \mathbf{b} \), results in a vector that is perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \). The cross product of the standard unit vectors \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \) follows specific rules: \( \mathbf{i} \times \mathbf{j} = \mathbf{k} \), \( \mathbf{j} \times \mathbf{k} = \mathbf{i} \), and \( \mathbf{k} \times \mathbf{i} = \mathbf{j} \). Moreover, \( \mathbf{i} \times \mathbf{i} = \mathbf{j} \times \mathbf{j} = \mathbf{k} \times \mathbf{k} = \mathbf{0} \).

Step 2: Simplify the Vectors

Simplify the given expression \( \mathbf{i} \times (-3 \mathbf{k}) \) by taking the scalar \(-3\) outside of the cross product operation. The expression becomes: \(-3(\mathbf{i} \times \mathbf{k})\).

Step 3: Determine the Standard Cross Product

Apply the rule for cross product between standard unit vectors: \( \mathbf{i} \times \mathbf{k} = -\mathbf{j} \). This is because the sequence \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) is cyclic, and a backward operation gives a negative result.

Step 4: Calculate the Final Result

Substitute the result from Step 3 into the simplified expression from Step 2. Calculate: \(-3(-\mathbf{j}) = 3\mathbf{j}\). Therefore, the result of the cross product \( \mathbf{i} \times (-3 \mathbf{k}) \) is \( 3\mathbf{j} \).

Key Concepts

Understanding Cross Product

The cross product is a fundamental operation in vector calculus. It involves two vectors and results in a third vector that is perpendicular to both original vectors. This operation is particularly useful in three-dimensional space where orientation and direction matter.

In mathematics, the cross product of vectors \( \mathbf{a} \) and \( \mathbf{b} \), denoted as \( \mathbf{a} \times \mathbf{b} \), can be calculated using the determinant of a matrix formed by the unit vectors and the components of \( \mathbf{a} \) and \( \mathbf{b} \). However, a simpler way to understand it when using standard unit vectors is through the cyclic rules:

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

Using these rules, the cross product can be computed more intuitively. For any identical vectors like \( \mathbf{i} \times \mathbf{i} \), the result is always zero because a vector cannot be perpendicular to itself within a plane.

The Role of Unit Vectors

Unit vectors are crucial in vector calculus as they provide a basis for describing directions without magnitudes. Standard unit vectors in three-dimensional space are \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \), corresponding to the x, y, and z axes respectively.

The properties of unit vectors make calculations more straightforward, especially when performing operations like the cross product. They allow us to easily define vector directions because each unit vector has a magnitude of one and points in a specific direction.

• They simplify vector arithmetic, making it easier to compute results like the cross product.
• Unit vectors serve as the building blocks for any vector, which can be expressed as a combination of these unit vectors.

In the given exercise, recognizing the standard unit vectors helps to directly apply the cross product rules and get the correct resulting vector.

Understanding Perpendicular Vectors

Perpendicular vectors are vectors that intersect at a 90-degree angle. This concept is fundamental in physics and engineering because it helps determine orientations and planes.

In vector calculus, the result of a cross product, \( \mathbf{a} \times \mathbf{b} \), is a vector that is perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \). Understanding this property is essential because:

• It helps find normals to surfaces, which is crucial in computing surface areas and in graphics for lighting calculations.
• Perpendicular vectors are often used in applications to define axis of rotations and other motions.

In our exercise, the computed vector \( \mathbf{i} \times (-3 \mathbf{k}) \,\), resulting in \( 3\mathbf{j} \), is perpendicular to both \( \mathbf{i} \) and \( \mathbf{k} \). Understanding this makes it clear why the operation is performed and why it results in this particular outcome.