Problem 8 - Thomas Calculus

Question

In Exercises \(1-8,\) find the length and direction (when defined) of \(\mathbf{u} \times \mathbf{v}\) and \(\mathbf{v} \times \mathbf{u} .\) $$ \mathbf{u}=\frac{3}{2} \mathbf{i}-\frac{1}{2} \mathbf{j}+\mathbf{k}, \quad \mathbf{v}=\mathbf{i}+\mathbf{j}+2 \mathbf{k} $$

Step-by-Step Solution

Answer

Magnitudes are \(\sqrt{6.5}\); directions are opposite.

Step 1: Verify Vector Definitions

Ensure the correct components of vectors \( \mathbf{u} \) and \( \mathbf{v} \) are used. \( \mathbf{u} = \frac{3}{2} \mathbf{i} - \frac{1}{2} \mathbf{j} + \mathbf{k} \) and \( \mathbf{v} = \mathbf{i} + \mathbf{j} + 2 \mathbf{k} \).

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

Use the cross product formula for two vectors \( \mathbf{u} = ( \frac{3}{2}, -\frac{1}{2}, 1) \) and \( \mathbf{v} = (1, 1, 2) \). The formula is:\[ \mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ \frac{3}{2} & -\frac{1}{2} & 1 \ 1 & 1 & 2 \end{vmatrix} \]Calculate the determinant to get:\[ \mathbf{u} \times \mathbf{v} = -\frac{5}{2} \mathbf{i} - \frac{1}{2} \mathbf{j} + \frac{5}{2} \mathbf{k} \]

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

Swap the order of \( \mathbf{u} \) and \( \mathbf{v} \). Use the formula:\[ \mathbf{v} \times \mathbf{u} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 1 & 2 \ \frac{3}{2} & -\frac{1}{2} & 1 \end{vmatrix} \]Calculate the determinant to get:\[ \mathbf{v} \times \mathbf{u} = \frac{5}{2} \mathbf{i} + \frac{1}{2} \mathbf{j} - \frac{5}{2} \mathbf{k} \]

Step 4: Determine Magnitude of \(\mathbf{u} \times \mathbf{v}\)

The magnitude of a vector \( \mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k} \) is given by \( \sqrt{a_1^2 + a_2^2 + a_3^2} \). Hence, the magnitude of \( \mathbf{u} \times \mathbf{v} \) is:\[ \sqrt{\left(-\frac{5}{2}\right)^2 + \left(-\frac{1}{2}\right)^2 + \left(\frac{5}{2}\right)^2} = \sqrt{6.5} \approx 2.55 \]

Step 5: Determine Magnitude of \(\mathbf{v} \times \mathbf{u}\)

Using the same formula for magnitude, the magnitude of \( \mathbf{v} \times \mathbf{u} \) is also:\[ \sqrt{\left(\frac{5}{2}\right)^2 + \left(\frac{1}{2}\right)^2 + \left(-\frac{5}{2}\right)^2} = \sqrt{6.5} \approx 2.55 \]

Step 6: Verify Direction Relationship

The direction of \(\mathbf{u} \times \mathbf{v}\) and \(\mathbf{v} \times \mathbf{u}\) are opposite since:\[ \mathbf{u} \times \mathbf{v} = - (\mathbf{v} \times \mathbf{u}) \]

Key Concepts

Determinant of a Matrix

In the context of vector calculus, the determinant of a matrix is a crucial concept used to calculate the cross product of two vectors. When you arrange the components of vectors \(\mathbf{u}\) and \(\mathbf{v}\) into a 3x3 matrix, the first row is

• the unit vectors \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\), representing the three-dimensional space,
• the second row contains the components of the first vector, and
• the third row contains the components of the second vector.

The cross product is then found by calculating the determinant of this matrix.

For example, the determinant of the matrix:\[\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\frac{3}{2} & -\frac{1}{2} & 1 \1 & 1 & 2 \end{vmatrix}\] yields the vector \(-\frac{5}{2} \mathbf{i} - \frac{1}{2} \mathbf{j} + \frac{5}{2} \mathbf{k}\).

This operation involves finding the area or volume of forms made by vectors, giving a new vector perpendicular to the original two. It simplifies the job of computing the cross product by providing a method to "calculate" a sense of orientation in 3D space.

Direction of Vectors

Vectors are known for their ability to indicate direction in space. With the cross product, we specifically look at the direction of the resulting vector, which is something easily determined due to the nature of the cross product operation.

It's important to note that the direction of the vector \(\mathbf{u} \times \mathbf{v}\) is perpendicular to both original vectors \(\mathbf{u}\) and \(\mathbf{v}\). This orthogonal direction follows the right-hand rule:

• if you point your thumb in the direction of \(\mathbf{u}\) and your index finger in the direction of \(\mathbf{v}\),
• then your middle finger points in the direction of the cross product \(\mathbf{u} \times \mathbf{v}\).

Similarly, for \(\mathbf{v} \times \mathbf{u}\), the resulting vector will have the opposite direction compared to \(\mathbf{u} \times \mathbf{v}\). This is evident from the solution as \(\mathbf{v} \times \mathbf{u} = -\left(\mathbf{u} \times \mathbf{v}\right)\).

This property emphasizes how direction is a fundamental aspect of vectors, ensuring proper orientation and assignments in vector spaces.

Magnitude of a Vector

Understanding the magnitude of a vector is essential when working with the cross product. It provides insight into the size or length of the vector.

The magnitude of a vector \(\mathbf{a} = a_1 \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k}\) is calculated using the formula:\[\sqrt{a_1^2 + a_2^2 + a_3^2}\]This computes the "length" of the vector in three-dimensional space, treating the components like the hypotenuse in Pythagoras' theorem applied in a three-dimensional space.

For instance, in the solution, the magnitude of both \(\mathbf{u} \times \mathbf{v}\) and \(\mathbf{v} \times \mathbf{u}\) is \(\sqrt{6.5}\), or approximately 2.55. This shows us how the same magnitude reflects the fact that despite the direction being opposite, the "length" or "size" of the vector remains constant.

• Magnitude measures how far from the origin the endpoint of the vector lies.
• It describes the vector's "strength," irrespective of its direction.