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

Length: \(2\sqrt{3}\); Directions: \(\mathbf{u} \times \mathbf{v}\) is \(-\mathbf{i} - \mathbf{j} + \mathbf{k}\), \(\mathbf{v} \times \mathbf{u}\) is \(\mathbf{i} + \mathbf{j} - \mathbf{k}\).

Step 1: Write the Vectors

Given the vectors, \( \mathbf{u} = \frac{3}{2}\mathbf{i} - \frac{1}{2}\mathbf{j} + \mathbf{k} \) and \( \mathbf{v} = \mathbf{i} + \mathbf{j} + 2\mathbf{k} \), start by writing them down clearly to avoid any confusion in later steps.

Step 2: Set Up the Cross Product Formula

The cross product of two vectors \( \mathbf{u} \) and \( \mathbf{v} \) is given by the determinant:\[\mathbf{u} \times \mathbf{v} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\frac{3}{2} & -\frac{1}{2} & 1 \1 & 1 & 2\end{vmatrix}\]

Step 3: Calculate the Determinant

Expand the determinant to calculate \( \mathbf{u} \times \mathbf{v} \):\[\begin{align*}\mathbf{u} \times \mathbf{v} &= \mathbf{i} \left((-\frac{1}{2})(2) - (1)(1)\right) - \mathbf{j} \left(\frac{3}{2}(2) - (1)(1)\right) + \mathbf{k} \left(\frac{3}{2}(1) + \frac{1}{2}(1)\right) \&= \mathbf{i}(-1 - 1) - \mathbf{j}(3 - 1) + \mathbf{k}(\frac{3}{2} + \frac{1}{2}) \&= \mathbf{i}(-2) - \mathbf{j}(2) + \mathbf{k}(2) \&= -2\mathbf{i} - 2\mathbf{j} + 2\mathbf{k}\end{align*}\]

Step 4: Calculate the Length of \( \mathbf{u} \times \mathbf{v} \)

The length of \( \mathbf{u} \times \mathbf{v} \) is given by:\[ ||\mathbf{u} \times \mathbf{v}|| = \sqrt{(-2)^2 + (-2)^2 + 2^2} = \sqrt{4 + 4 + 4} = \sqrt{12} = 2\sqrt{3}\]

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

The direction of \( \mathbf{u} \times \mathbf{v} \), when defined, is given by the unit vector:\[ \mathbf{n} = \frac{-2\mathbf{i} - 2\mathbf{j} + 2\mathbf{k}}{2\sqrt{3}} = \frac{-\mathbf{i} - \mathbf{j} + \mathbf{k}}{\sqrt{3}} \]

Step 6: Set Up \( \mathbf{v} \times \mathbf{u} \) Using Anticommutativity

Use the property \( \mathbf{v} \times \mathbf{u} = - (\mathbf{u} \times \mathbf{v}) \). Therefore, \( \mathbf{v} \times \mathbf{u} = 2\mathbf{i} + 2\mathbf{j} - 2\mathbf{k} \).

Step 7: Confirm the Length of \( \mathbf{v} \times \mathbf{u} \)

Since the cross product remains the same in magnitude, the length of \( \mathbf{v} \times \mathbf{u} \) is also \( 2\sqrt{3} \).

Step 8: Determine the Direction of \( \mathbf{v} \times \mathbf{u} \)

The direction is also similar, but opposite to \( \mathbf{u} \times \mathbf{v} \):\[ \mathbf{n} = \frac{2\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}}{2\sqrt{3}} = \frac{\mathbf{i} + \mathbf{j} - \mathbf{k}}{\sqrt{3}} \]

Key Concepts

Vector Calculus

In vector calculus, understanding how vectors interact is essential, and the cross product is fundamental to this. A cross product takes two vectors in three-dimensional space and returns a new vector that is orthogonal (or perpendicular) to both original vectors. This new vector's properties help in applications like determining the area of a parallelogram formed by two vectors or finding torque in physics.

When dealing with vectors such as \( \mathbf{u} = \frac{3}{2}\mathbf{i} - \frac{1}{2}\mathbf{j} + \mathbf{k} \) and \( \mathbf{v} = \mathbf{i} + \mathbf{j} + 2\mathbf{k} \), the cross product \( \mathbf{u} \times \mathbf{v} \) involves calculation that brings insights into their geometric and spatial relationship.

Here are some key points about the cross product:

• The order of multiplication matters, meaning \( \mathbf{u} \times \mathbf{v} eq \mathbf{v} \times \mathbf{u} \). In fact, \( \mathbf{v} \times \mathbf{u} = - (\mathbf{u} \times \mathbf{v}) \).
• The resulting vector is perpendicular to both \( \mathbf{u} \) and \( \mathbf{v} \), providing a handy tool for finding normals to planes.
• This operation is especially relevant in physics and engineering, as it describes rotational forces and magnetic field directions.

Understanding the cross product extends our capability to navigate multidimensional interactions and apply them in practical scenarios.

Determinants

The calculation of a cross product hinges on the use of determinants. In the case of the vectors \( \mathbf{u} = \frac{3}{2}\mathbf{i} - \frac{1}{2}\mathbf{j} + \mathbf{k} \) and \( \mathbf{v} = \mathbf{i} + \mathbf{j} + 2\mathbf{k} \), we use the determinant of a 3x3 matrix to find the cross product:

\[ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ \frac{3}{2} & -\frac{1}{2} & 1 \ 1 & 1 & 2 \end{vmatrix} \]
The process of expanding this determinant involves selecting a row or column and combining products of elements, alternately adding and subtracting these product terms.

Here's a brief outline of the steps involved in solving a 3x3 determinant:

• Select a row or column for expansion, usually the first row with unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \).
• Multiply each element by the determinant of the 2x2 matrix that remains when the row and column of that element are removed.
• Combine these products, alternating between addition and subtraction based on their positions.

Understanding determinants helps to bridge the gap between numerical computation and geometric interpretation, thereby enhancing problem-solving capabilities in vector operations.

Magnitude and Direction

The magnitude and direction of vectors resulting from a cross product give us a wealth of information about their spatial arrangement. When we compute \( \mathbf{u} \times \mathbf{v} \), we get an outcome of \( -2\mathbf{i} - 2\mathbf{j} + 2\mathbf{k} \).

The magnitude, or length, of this vector can be determined using the formula:
\[ ||\mathbf{u} \times \mathbf{v}|| = \sqrt{(-2)^2 + (-2)^2 + 2^2} = 2\sqrt{3} \]
This represents the area of the parallelogram spanned by \( \mathbf{u} \) and \( \mathbf{v} \).

The direction is offered by a unit vector, which is the normalized form of the cross product:

• In the case of \( \mathbf{u} \times \mathbf{v} \), the directional vector is \( \frac{-\mathbf{i} - \mathbf{j} + \mathbf{k}}{\sqrt{3}} \).
• This unit vector denotes the axis along which the result is oriented, ensuring that it remains consistently perpendicular in space.

By grasping both magnitude and direction, one can not only quantify but also clearly visualize the relationship and interaction between the original vectors.