Problem 35 - Advanced Engineering Mathematics

Question

Find the indicated scalar or vector without using \((5),(13)\), or \((15)\). $$ 2 \mathbf{j} \cdot[\mathbf{i} \times(\mathbf{j}-3 \mathbf{k})] $$

Step-by-Step Solution

Answer

The value is 6.

Step 1: Identify the Vectors

Identify the vectors involved in the expression. We have \( \mathbf{i} \) and \( \mathbf{j} \) which are standard unit vectors along the x-axis and y-axis respectively, and \( \mathbf{k} \) is the standard unit vector along the z-axis. The expression inside the brackets is \( \mathbf{i} \times (\mathbf{j} - 3\mathbf{k}) \).

Step 2: Distribute the Cross Product

Recall that the cross product is distributive over addition: \( \mathbf{a} \times (\mathbf{b} + \mathbf{c}) = \mathbf{a} \times \mathbf{b} + \mathbf{a} \times \mathbf{c} \). Apply this property to get \[ \mathbf{i} \times (\mathbf{j} - 3\mathbf{k}) = \mathbf{i} \times \mathbf{j} - 3(\mathbf{i} \times \mathbf{k}). \]

Step 3: Compute the Cross Products

Use the fundamental cross product results: \( \mathbf{i} \times \mathbf{j} = \mathbf{k} \) and \( \mathbf{i} \times \mathbf{k} = -\mathbf{j} \). Substitute these into the expression: \( \mathbf{i} \times \mathbf{j} = \mathbf{k} \) and \( -3(\mathbf{i} \times \mathbf{k}) = 3\mathbf{j} \). So the expression becomes \( \mathbf{k} + 3\mathbf{j} \).

Step 4: Solve the Dot Product

Now consider the dot product: \( 2\mathbf{j} \cdot [\mathbf{k} + 3\mathbf{j}] \). Distribute the dot product: \( 2\mathbf{j} \cdot \mathbf{k} + 2\mathbf{j} \cdot 3\mathbf{j} \). The dot product \( \mathbf{j} \cdot \mathbf{k} = 0 \) and \( \mathbf{j} \cdot \mathbf{j} = 1 \).

Step 5: Simplify the Expression

After simplifying the dot products we have: \[ 2(0) + 2(3)(1) = 6. \] This gives the final scalar value of the expression.

Key Concepts

Cross Product

The cross product is a fundamental operation in vector calculus that results in a vector perpendicular to the plane containing the two input vectors. In three dimensions, it is especially useful for finding perpendicular vectors and calculating areas of parallelograms formed by two vectors. The cross product is denoted as \( \mathbf{a} \times \mathbf{b} \) and is defined only in three-dimensional space. It's important to remember that the order of the vectors matters since the cross product is anti-commutative: \( \mathbf{a} \times \mathbf{b} = - (\mathbf{b} \times \mathbf{a}) \). This operation is governed by the right-hand rule, which helps determine the direction of the resulting vector.
To compute the cross product of two vectors, one typically uses the unit vectors \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \), which correspond to the x, y, and z axes, respectively. Important properties include:

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

These properties are particularly useful for simplifying expressions involving cross products, as shown in the solution where \( \mathbf{i} \times (\mathbf{j} - 3\mathbf{k}) \) simplifies to \( \mathbf{k} + 3\mathbf{j} \). Understanding these properties will greatly enhance your manipulation and understanding of vector quantities.

Dot Product

The dot product, or scalar product, is another crucial operation in vector calculus that results in a scalar value. This operation measures the extent to which two vectors point in the same direction. If the vectors are perpendicular, the dot product is zero. Mathematically, the dot product of two vectors \( \mathbf{a} \) and \( \mathbf{b} \) is defined as:\[ \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos \theta \]where \( \theta \) is the angle between the two vectors.
In Cartesian coordinates, for vectors expressed using unit vectors \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \), the dot product simplifies to multiplying corresponding components and summing them up:\[ \mathbf{a} \cdot \mathbf{b} = a_x b_x + a_y b_y + a_z b_z \]
In the exercise, the dot product \( 2\mathbf{j} \cdot (\mathbf{k} + 3\mathbf{j}) \) was evaluated. Notice how using properties like \( \mathbf{j} \cdot \mathbf{k} = 0 \) and \( \mathbf{j} \cdot \mathbf{j} = 1 \) helps to quickly determine that the scalar product results in a value of 6. This illustrates the practical application of these basic properties in vector operations.

Unit Vectors

Unit vectors are vectors with a magnitude of one and are used to indicate direction. They form the basis of vector space in Cartesian coordinates, where each dimension has its own unit vector: \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \), representing the x-axis, y-axis, and z-axis, respectively. These vectors are crucial for describing directions and performing vector operations.
When working with vectors in standard form, any vector can be expressed as a combination of unit vectors. For example, a vector \( \mathbf{v} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k} \) represents a point in three-dimensional space with components a, b, and c. The unit vectors simplify calculations by breaking down vectors into directional components.
In cross and dot products, unit vectors simplify how vectors interact. For instance:

• Unit vectors maintain orthogonality: \( \mathbf{i} \cdot \mathbf{j} = 0 \), \( \mathbf{i} \times \mathbf{j} = \mathbf{k} \)
• Their magnitudes are useful scalars: \( \mathbf{i} \cdot \mathbf{i} = 1 \)

Having a solid understanding of unit vectors is vital for efficiency in calculating and visualizing vectors in physics and engineering problems.