Problem 3 - University Calculus: Early Transcendentals

Question

Find the length and direction (when defined) of \(\mathbf{u} \times \mathbf{v}\) and \(\mathbf{v} \times \mathbf{u}\). $$\mathbf{u}=2 \mathbf{i}-2 \mathbf{j}+4 \mathbf{k}, \quad \mathbf{v}=-\mathbf{i}+\mathbf{j}-2 \mathbf{k}$$

Step-by-Step Solution

Answer

Both cross products are zero vectors with magnitude 0 and undefined direction.

Step 1: Write the Definition of the Cross Product

The cross product \(\mathbf{u} \times \mathbf{v}\) of two vectors \(\mathbf{u}\) and \(\mathbf{v}\) in 3-dimensional space is defined as a vector that is perpendicular to both \(\mathbf{u}\) and \(\mathbf{v}\). It can be calculated using the determinant:\[\mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2 & -2 & 4 \ -1 & 1 & -2\end{vmatrix}\]

Step 2: Calculate the Cross Product \(\mathbf{u} \times \mathbf{v}\)

To find \(\mathbf{u} \times \mathbf{v}\), solve the determinant:The \(\mathbf{i}\) component is:\[\mathbf{i}((-2)(-2) - 4(1)) = \mathbf{i}(4 - 4) = 0\mathbf{i}\]The \(\mathbf{j}\) component is:\[-\mathbf{j}((2)(-2) - 4(-1)) = -\mathbf{j}(-4 + 4) = 0\mathbf{j}\]The \(\mathbf{k}\) component is:\[\mathbf{k}((2)(1) - (-2)(-1)) = \mathbf{k}(2 -2) = 0\mathbf{k}\]Thus, \(\mathbf{u} \times \mathbf{v} = 0\mathbf{i} + 0\mathbf{j} + 0\mathbf{k} = \mathbf{0}\).

Step 3: Calculate the Cross Product \(\mathbf{v} \times \mathbf{u}\)

By the anticommutative property of the cross product, \(\mathbf{v} \times \mathbf{u} = - (\mathbf{u} \times \mathbf{v})\). Since \(\mathbf{u} \times \mathbf{v} = \mathbf{0}\), it follows that:\(\mathbf{v} \times \mathbf{u} = -\mathbf{0} = \mathbf{0}\).

Step 4: Determine the Length of the Cross Products

The length (magnitude) of a vector \(\mathbf{a}\) is given by \(||\mathbf{a}||\). Since both \(\mathbf{u} \times \mathbf{v}\) and \(\mathbf{v} \times \mathbf{u}\) are zero vectors, their magnitudes are:\[||\mathbf{u} \times \mathbf{v}|| = ||\mathbf{0}|| = 0\]\[||\mathbf{v} \times \mathbf{u}|| = ||\mathbf{0}|| = 0\]

Step 5: Consider the Direction of the Cross Products

The direction of a zero vector is undefined because it has no direction by definition.

Key Concepts

Magnitude of Vectors

The magnitude of a vector is an important concept when working with vectors in mathematics and physics. It describes the length or size of the vector without considering its direction. The magnitude of a vector \( \mathbf{a} = a_1 \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k} \) is calculated using the formula:

• \( || \mathbf{a} || = \sqrt{a_1^2 + a_2^2 + a_3^2} \)

In the problem of finding the length of \( \mathbf{u} \times \mathbf{v} \) and \( \mathbf{v} \times \mathbf{u} \), both cross products resulted in the zero vector \( \mathbf{0} \). The magnitude of a zero vector is always 0 because it has no length:

• \( || \mathbf{0} || = \sqrt{0^2 + 0^2 + 0^2} = 0 \)

Thus, both \( || \mathbf{u} \times \mathbf{v} || \) and \( || \mathbf{v} \times \mathbf{u} || \) are equal to 0.

Direction of Vectors

When analyzing vectors, direction is a key component that, together with magnitude, fully characterizes the vector. However, the direction of the zero vector is undefined. This is because the zero vector has no magnitude or any discernible direction. To better understand why, consider a non-zero vector \( \mathbf{a} \), which points from the origin to a point \( (a_1, a_2, a_3) \) in space. Unlike these vectors, the zero vector points nowhere due to its lack of length.

In our exercise, both cross products, \( \mathbf{u} \times \mathbf{v} \) and \( \mathbf{v} \times \mathbf{u} \), are zero vectors, and thus, they do not have a direction. Zero vectors stand as exceptions in vector mathematics, as they do not comply with the usual directionality that most vectors possess.

Properties of Cross Product

The cross product is a significant operation in vector algebra, especially in three-dimensional space. It involves two vectors, \( \mathbf{u} \) and \( \mathbf{v} \), and produces a third vector that is orthogonal (perpendicular) to both. Here are key properties of the cross product:

• Anticommutative Property: \( \mathbf{u} \times \mathbf{v} = - (\mathbf{v} \times \mathbf{u}) \)
• Distributive Over Addition: \( \mathbf{u} \times (\mathbf{v} + \mathbf{w}) = \mathbf{u} \times \mathbf{v} + \mathbf{u} \times \mathbf{w} \)
• Scalar Multiplication: \( (c \cdot \mathbf{u}) \times \mathbf{v} = c(\mathbf{u} \times \mathbf{v}) \)
• Relation to Area: The magnitude of \( \mathbf{u} \times \mathbf{v} \) corresponds to the area of the parallelogram formed by \( \mathbf{u} \) and \( \mathbf{v} \).

In the given exercise, \( \mathbf{u} \times \mathbf{v} \) and \( \mathbf{v} \times \mathbf{u} \) result in zero vectors. This outcome arises when vectors \( \mathbf{u} \) and \( \mathbf{v} \) are parallel or one of the vectors is zero. Here, the vectors are indeed parallel, resulting in a zero vector by the properties of the cross product.