Problem 10 - Thomas Calculus

Question

In Exercises \(9-14,\) 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. $$ \mathbf{u}=\mathbf{i}-\mathbf{k}, \quad \mathbf{v}=\mathbf{j} $$

Step-by-Step Solution

Answer

Plot \( \mathbf{u} = (1,0,-1) \), \( \mathbf{v} = (0,1,0) \), and \( \mathbf{u} \times \mathbf{v} = (0,0,1) \) starting at the origin.

Step 1: Identify the Given Vectors

We are given two vectors: \( \mathbf{u} = \mathbf{i} - \mathbf{k} \) and \( \mathbf{v} = \mathbf{j} \). Let's convert these into the coordinate form. \( \mathbf{u} = (1, 0, -1) \) and \( \mathbf{v} = (0, 1, 0) \).

Step 2: Set Up the Coordinate Axes

Draw a 3D coordinate system with axes labeled as \( x \), \( y \), and \( z \) representing \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \) respectively. Ensure that the positive directions are clearly indicated.

Step 3: Plot the Vector \( \mathbf{u} \)

Starting from the origin \((0,0,0)\), draw the vector \( \mathbf{u} = (1, 0, -1) \). Move 1 unit along the positive \( x \)-axis and 1 unit along the negative \( z \)-axis. This vector lies in the \( xz \)-plane.

Step 4: Plot the Vector \( \mathbf{v} \)

Starting again from the origin, draw the vector \( \mathbf{v} = (0, 1, 0) \). This leads to moving 1 unit along the positive \( y \)-axis. This vector is along the \( y \)-axis.

Step 5: Calculate \( \mathbf{u} \times \mathbf{v} \)

Use the cross product formula to find \( \mathbf{u} \times \mathbf{v} \):\[ \mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 0 & -1 \ 0 & 1 & 0 \end{vmatrix} = (1)(0) - (0)(-1)\mathbf{i} - ((1)(0) - (0)(-1))\mathbf{j} + ((1)(1) - (0)(0))\mathbf{k} \] \[ = 1\mathbf{k} = \mathbf{k} \] so, \( \mathbf{u} \times \mathbf{v} = (0, 0, 1) \).

Step 6: Plot the Vector \( \mathbf{u} \times \mathbf{v} \)

Draw the result of the cross product \( \mathbf{u} \times \mathbf{v} \) starting from the origin. This vector is \( (0, 0, 1) \), which means 1 unit in the positive \( z \)-direction. It lies on the \( z \)-axis.

Step 7: Final Step: Confirm the Perpendicularity

Ensure that the plotted vectors \( \mathbf{u} \) and \( \mathbf{v} \) are perpendicular to \( \mathbf{u} \times \mathbf{v} \).\( \mathbf{u} \), \( \mathbf{v} \), and \( \mathbf{u} \times \mathbf{v} \) should all be mutually perpendicular vectors.

Key Concepts

3D Coordinate System

Understanding the 3D coordinate system is essential when working with vectors in three-dimensional space. Imagine it as a framework consisting of three axes: the x-axis, y-axis, and z-axis. Each axis is perpendicular to the other two, allowing for a complete representation of any point or vector in 3D space.
The x-axis usually points horizontally, the y-axis points either horizontally or vertically depending on the drawing convention, and the z-axis is perpendicular to both. It seems complicated, but think of it as just adding depth to the usual 2D graph.

• To locate a point in this system, use a combination of three numbers representing movements along the x, y, and z axes.
• This allows for accurate depiction and manipulation of vectors, objects, and their interactions within a three-dimensional realm.

Positioning is often visualized with positive and negative directions along each axis, adding to the understanding of vector direction and interactions. This system forms the foundation for discussing vector representation and cross products.

Vector Representation

Vectors in a 3D coordinate system are represented using coordinates that will indicate their direction and magnitude from the origin point. For instance, the vector \( \mathbf{u} = \mathbf{i} - \mathbf{k} \) can be expressed with coordinates as \( \mathbf{u} = (1, 0, -1) \), meaning it moves 1 unit along the positive x-axis and 1 unit along the negative z-axis.
In the exercise, the vector \( \mathbf{v} = \mathbf{j} \) is represented as \( \mathbf{v} = (0, 1, 0) \), moving 1 unit along the y-axis.
Vector representation is key to understanding their interactions, such as the cross product which shows how vectors relate spatially.

• Each vector component reflects how far and in which direction you move along each axis.
• Understanding these movements helps you visualize the vector in 3D space.

These representations serve as the stepping stone to delve into more complex operations like vector addition, subtraction, and finding vector cross products.

Vector Perpendicularity

In 3D, vectors can be perpendicular, meaning they intersect at a right angle (90 degrees). This notion of perpendicularity is critical when evaluating the cross product. The cross product of two vectors results in a vector that is perpendicular to the plane formed by the original vectors. In our example, \( \mathbf{u} = (1, 0, -1) \) and \( \mathbf{v} = (0, 1, 0) \) lead to \( \mathbf{u} \times \mathbf{v} = (0, 0, 1) \).
Let's break it down:

• \( \mathbf{u} \) is in the xz-plane.
• \( \mathbf{v} \) is along the y-axis.
• \( \mathbf{u} \times \mathbf{v} \) is along the z-axis, confirming it is perpendicular to both \( \mathbf{u} \) and \( \mathbf{v} \).

The resultant perpendicular vector provides fascinating insights into how forces or movement occur in physics and engineering, ensuring equilibrium or stability in systems. Understanding vector perpendicularity also deepens appreciation for how space is structured and the interactions of forces within it.