Problem 9 - University Calculus: Early Transcendentals

Question

Find the length and direction (when defined) of \(\mathbf{u} \times \mathbf{v}\) and \(\mathbf{v} \times \mathbf{u}\). Sketch the coordinate axes and then include the vectors \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{u} \times \mathbf{v}\) as vectors starting at the origin. $$\mathbf{u}=\mathbf{i}, \quad \mathbf{v}=\mathbf{j}$$

Step-by-Step Solution

Answer

\( \mathbf{u} \times \mathbf{v} = \mathbf{k}, \; \mathbf{v} \times \mathbf{u} = -\mathbf{k}, \; \text{Length} = 1 \)

Step 1: Understanding the Cross Product

The cross product, denoted as \( \mathbf{u} \times \mathbf{v} \), is a vector perpendicular to both \( \mathbf{u} \) and \( \mathbf{v} \). For vectors \( \mathbf{u} = \mathbf{i} \) and \( \mathbf{v} = \mathbf{j} \), the cross product is calculated using the determinant of a 3x3 matrix formed by the unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) and the components of \( \mathbf{u} \) and \( \mathbf{v} \).

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

To find \( \mathbf{u} \times \mathbf{v} \), set up the matrix determinant:\[ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 0 & 0 \ 0 & 1 & 0 \end{vmatrix} \]The determinant expands to:\[ \mathbf{u} \times \mathbf{v} = \mathbf{i}(0 \times 0 - 1 \times 0) - \mathbf{j}(1 \times 0 - 0 \times 0) + \mathbf{k}(1 \times 1 - 0 \times 0) = \mathbf{k} \]

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

Use the same method to compute the cross product by switching \( \mathbf{u} \) and \( \mathbf{v} \):\[ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 0 & 1 & 0 \ 1 & 0 & 0 \end{vmatrix} \]The determinant expands to:\[ \mathbf{v} \times \mathbf{u} = \mathbf{i}(0 \times 0 - 0 \times 0) - \mathbf{j}(0 \times 0 - 1 \times 0) + \mathbf{k}(0 \times 1 - 1 \times 0) = -\mathbf{k} \]

Step 4: Determine the Length of the Cross Products

The length of a cross product vector \( \mathbf{a} \times \mathbf{b} \) is given by \( \| \mathbf{a} \| \| \mathbf{b} \| \sin(\theta) \), where \( \theta \) is the angle between the vectors. Here, both vectors are unit vectors and the angle between them is 90 degrees, thus \( \sin(\theta) = 1 \). Therefore, the length is \( 1 \).

Step 5: Sketch the Vectors on Coordinate Axes

Draw the x, y, and z coordinate axes. Place \( \mathbf{u} = \mathbf{i} \) along the x-axis and \( \mathbf{v} = \mathbf{j} \) along the y-axis. The vector \( \mathbf{u} \times \mathbf{v} = \mathbf{k} \) points along the positive z-axis, perpendicular to the plane formed by \( \mathbf{i} \) and \( \mathbf{j} \).

Key Concepts

Determinant

When dealing with the cross product of vectors, understanding determinants is a key concept. The determinant is essentially a mathematical expression that allows us to calculate the cross product easily, especially when it is expressed in terms of a 3x3 matrix.
The matrix for finding the cross product of two vectors employs unit vectors and the components of those vectors. For example, if you have vectors \( \mathbf{u} = \mathbf{i} \) and \( \mathbf{v} = \mathbf{j} \), you can arrange them in the matrix along with the unit vector \( \mathbf{k} \): \[ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 0 & 0 \ 0 & 1 & 0 \end{vmatrix} \] This determinant helps us compute the new vector that will emerge from the cross product.

Unit Vectors

Unit vectors play a critical role in vector mathematics. They are vectors with a magnitude (length) of 1 and are usually used to represent directions. The common unit vectors are \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \). These are used for the x-axis, y-axis, and z-axis, respectively.
When working with the cross product, using unit vectors simplifies calculations. In our specific case, \( \mathbf{u} = \mathbf{i} \) and \( \mathbf{v} = \mathbf{j} \), both represent unit vectors along the x and y axes. They help define the orientation of the vectors in 3D space, allowing us to calculate cross products that are directed along the z-axis when the vectors are perpendicular to each other.

Perpendicular Vector

A fascinating feature of the cross product is that it results in a vector that is perpendicular to the original two vectors. Given two vectors \( \mathbf{u} \) and \( \mathbf{v} \), the vector \( \mathbf{u} \times \mathbf{v} \) will always be perpendicular to both \( \mathbf{u} \) and \( \mathbf{v} \). This is due to the nature of the cross product itself.
In our example, with \( \mathbf{u} = \mathbf{i} \) and \( \mathbf{v} = \mathbf{j} \), the cross product \( \mathbf{u} \times \mathbf{v} = \mathbf{k} \). This \( \mathbf{k} \) vector points along the z-axis, demonstrating perpendicularity to the plane formed by the x and y axes (where \( \mathbf{u} \) and \( \mathbf{v} \) lie). The fact that the result occurs along the positive or negative z-axis highlights the right-hand rule, a handy mnemonic for determining the direction of the cross product vector.

Vector Length

The length of a vector, also known as its magnitude, is important for understanding its size regardless of direction. For a cross product of two vectors \( \mathbf{a} \) and \( \mathbf{b} \), the length (\( \left\| \mathbf{a} \times \mathbf{b} \right\| \)) is computed as follows:

• Take the magnitudes of \( \mathbf{a} \) and \( \mathbf{b} \).
• Multiply these magnitudes by the sine of the angle \( \theta \) between the two vectors.
• Thus, \( \left\| \mathbf{a} \right\| \left\| \mathbf{b} \right\| \sin(\theta) \).

In our exercise, since both are unit vectors and are perpendicular, their angle \( \theta \) is 90 degrees. The sine of 90 degrees is 1, making the length of the resulting vector also 1. This simple calculation helps confirm that even with direction, cross product vectors adhere to specific rules around vector length.