Problem 11 - 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. $$ \mathbf{u}=\mathbf{i}-\mathbf{k}, \quad \mathbf{v}=\mathbf{j}+\mathbf{k} $$

Step-by-Step Solution

Answer

The vectors are \(\mathbf{u} = \langle 1, 0, -1 \rangle\), \(\mathbf{v} = \langle 0, 1, 1 \rangle\), and \(\mathbf{u} \times \mathbf{v} = \langle 1, -1, 1 \rangle\).

Step 1: Identify the Vectors and Their Components

Begin by identifying the given vectors and their respective components.Vector \(\mathbf{u}\) is \(\langle 1, 0, -1 \rangle\), where the components correspond to the coefficients of \(\mathbf{i}, \mathbf{j},\) and \(\mathbf{k}\) respectively.Vector \(\mathbf{v}\) is \(\langle 0, 1, 1 \rangle\).

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

To find the cross product \(\mathbf{u} \times \mathbf{v}\), set up the determinant:\[\mathbf{u} \times \mathbf{v} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \1 & 0 & -1 \0 & 1 & 1\end{vmatrix}\]Calculate the determinant:\[\mathbf{u} \times \mathbf{v} = (0-(-1))\mathbf{i} - (1-0)\mathbf{j} + (1\cdot1 - 0\cdot(-1))\mathbf{k}\]This simplifies to:\[\mathbf{u} \times \mathbf{v} = \mathbf{i} - \mathbf{j} + \mathbf{k}\]

Step 3: Draw the Coordinate Axes

Draw a three-dimensional coordinate system with axes labeled \(x\), \(y\), and \(z\). Ensure that each axis is properly oriented and scaled.

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

From the origin \((0, 0, 0)\), draw the vector \(\mathbf{u} = \langle 1, 0, -1 \rangle\). This vector moves one unit in the \(x\)-direction and one unit back in the \(z\)-direction.

Step 5: Plot the Vector \(\mathbf{v}\)

Starting at the origin \((0, 0, 0)\), draw the vector \(\mathbf{v} = \langle 0, 1, 1 \rangle\). This vector moves one unit in the \(y\)-direction and one unit up in the \(z\)-direction.

Step 6: Plot the Cross Product Vector \(\mathbf{u} imes \mathbf{v}\)

Finally, starting from the origin \((0, 0, 0)\), draw the cross product vector \(\mathbf{u} \times \mathbf{v} = \langle 1, -1, 1 \rangle\). This vector moves one unit in the \(x\)-direction, one unit back in the \(y\)-direction, and one unit up in the \(z\)-direction.

Key Concepts

Coordinate Axes

The coordinate axes are a crucial part of vector mathematics. They help us understand where points lie in a three-dimensional space. Typically, these axes are labeled as \(x\), \(y\), and \(z\), representing three perpendicular directions.
When we plot vectors like \(\mathbf{u}\) and \(\mathbf{v}\), we use these axes to locate their positions in the space. For instance, in the exercise, vector \(\mathbf{u}\) moves along the \(x\) and \(-z\) directions. Similarly, vector \(\mathbf{v}\) corresponds to the \(y\) and \(z\) axes.
By visualizing these vectors on the coordinate axes, we can better comprehend their direction and magnitude.

• The \(x\)-axis is usually horizontal.
• The \(y\)-axis is typically vertical.
• The \(z\)-axis comes out towards you, perpendicular to the other two axes.
  

In understanding coordinate axes, we gain the foundational knowledge needed to perform operations like cross products, which rely on this structured three-dimensional space.

Vector Components

Vector components are the projections of a vector along the coordinate axes. These components give us a clear understanding of how much a vector extends in each direction. In our exercise, the vector \(\mathbf{u}\) is expressed as \(\langle 1, 0, -1 \rangle\), meaning it has:

• 1 unit in the \(x\)-direction
• 0 units in the \(y\)-direction
• -1 unit in the \(z\)-direction
  

Similarly, vector \(\mathbf{v}\) is \(\langle 0, 1, 1 \rangle\), with projections:

• 0 units in the \(x\)-direction
• 1 unit in the \(y\)-direction
• 1 unit in the \(z\)-direction
  

Understanding these components is essential for calculating operations like the cross product, as the magnitude of each direction directly influences the result.
It helps us visualize how a vector moves through 3D space.

Determinant

The determinant is an important concept when dealing with matrices and vectors. When calculating the cross product of two vectors, the determinant provides a systematic way to find a vector that is perpendicular to both.
To compute the cross product \(\mathbf{u} \times \mathbf{v}\), we organize their components in a 3x3 matrix form, with the unit vectors \(\mathbf{i}, \mathbf{j}, \mathbf{k}\) forming the first row. The vector components of \(\mathbf{u}\) and \(\mathbf{v}\) follow in subsequent rows:
\[\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 0 & -1 \ 0 & 1 & 1\end{vmatrix}\]
This matrix is solved by expanding the determinant along the first row, yielding each component of the resulting vector: \(\mathbf{i} - \mathbf{j} + \mathbf{k}\).
This systematic approach not only aids in finding the cross product but also enhances our understanding of vector direction and orientation. It's a foundational skill in linear algebra and vector calculus.