Problem 24 - Precalculus Mathematics for Calculus

Question

Find the area of the parallelogram determined by the given vectors. $$\mathbf{u}=\mathbf{i}-\mathbf{j}+\mathbf{k}, \quad \mathbf{v}=\mathbf{i}+\mathbf{j}-\mathbf{k}$$

Step-by-Step Solution

Answer

The area of the parallelogram is \(2\sqrt{2}\).

Step 1: Understand Vectors and Parallelogram Area

The given exercise involves finding the area of a parallelogram defined by two vectors \( \mathbf{u} \) and \( \mathbf{v} \). Recall that the area of the parallelogram formed by two vectors is equal to the magnitude of their cross product. So, our task is to calculate \( | \mathbf{u} \times \mathbf{v} | \).

Step 2: Set Up the Cross Product

For vectors \( \mathbf{u} = \langle 1, -1, 1 \rangle \) and \( \mathbf{v} = \langle 1, 1, -1 \rangle \), the cross product \( \mathbf{u} \times \mathbf{v} \) can be calculated using the determinant. Set up the determinant with the unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \):\[\mathbf{u} \times \mathbf{v} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \1 & -1 & 1 \1 & 1 & -1\end{vmatrix}\]

Step 3: Calculate the Determinant

Compute the determinant of the matrix:\[\mathbf{u} \times \mathbf{v} = \mathbf{i}((-1)(-1) - (1)(1)) - \mathbf{j}((1)(-1) - (1)(1)) + \mathbf{k}((1)(1) - (-1)(1))\]Simplify each expression:- \( \mathbf{i}(1 - 1) = \mathbf{i}(0) = 0 \)- \( -\mathbf{j}(-1 - 1) = \mathbf{j}(2) \)- \( \mathbf{k}(1 + 1) = \mathbf{k}(2) \)Thus, \( \mathbf{u} \times \mathbf{v} = 0\mathbf{i} + 2\mathbf{j} + 2\mathbf{k} \).

Step 4: Calculate the Magnitude

The magnitude of the cross product vector \( \mathbf{u} \times \mathbf{v} = \langle 0, 2, 2 \rangle \) is:\[|\mathbf{u} \times \mathbf{v}| = \sqrt{0^2 + 2^2 + 2^2} = \sqrt{0 + 4 + 4} = \sqrt{8}\]\[|\mathbf{u} \times \mathbf{v}| = 2\sqrt{2}\]

Step 5: Conclusion

Thus, the area of the parallelogram determined by vectors \( \mathbf{u} \) and \( \mathbf{v} \) is \( 2\sqrt{2} \).

Key Concepts

Cross Product

In vector geometry, the cross product is a way to find a vector that is perpendicular to two given vectors in 3-dimensional space. It is denoted by \( \mathbf{u} \times \mathbf{v} \). Unlike the dot product, which produces a scalar, the cross product results in a vector.

When calculating the cross product, it involves a determinant made up of the unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) and the components of the given vectors. For example, given vectors \( \mathbf{u} = \langle 1, -1, 1 \rangle \) and \( \mathbf{v} = \langle 1, 1, -1 \rangle \), the setup is:

• The first row comprises the unit vectors: \( \mathbf{i}, \mathbf{j}, \mathbf{k} \)
• The second row has components of \( \mathbf{u} \): 1, -1, 1
• The third row includes components of \( \mathbf{v} \): 1, 1, -1

This arrangement leads us to a determinant calculation, which provides us with a new vector as the result of the cross product.

Determinant

To compute the cross product of two vectors, we solve a determinant. This determinant is a 3x3 matrix involving unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \), and components of the vectors.

• The first part of the matrix: top row has \( \mathbf{i}, \mathbf{j}, \mathbf{k} \)
• Middle row: vector \( \mathbf{u} \)'s components: 1, -1, 1
• Bottom row: vector \( \mathbf{v} \)'s components: 1, 1, -1

To solve the determinant, break it down into smaller 2x2 determinants and look at each component:

• \( \mathbf{i} \) component: solve \(((-1)(-1) - (1)(1)) \)
• \( \mathbf{j} \) component: solve \(((1)(-1) - (1)(1)) \)
• \( \mathbf{k} \) component: solve \(((1)(1) - (-1)(1)) \)

The results are then combined to form the new vector. For our exercise, the cross product results in a vector \( 0\mathbf{i} + 2\mathbf{j} + 2\mathbf{k} \).

Magnitude of a Vector

The magnitude of a vector is its length and is found using a formula that involves the square root of the sum of the squares of its components. This length (or magnitude) is crucial, especially when we're interested in scalar quantities like lengths and areas in vector geometry.

After computing the cross product vector \( \mathbf{u} \times \mathbf{v} = \langle 0, 2, 2 \rangle \), find the magnitude using the formula:

• Square each component:
• \( 0^2 = 0 \)
• \( 2^2 = 4 \)
• \( 2^2 = 4 \)
• Sum them: \( 0 + 4 + 4 = 8 \)
• Take the square root: \( \sqrt{8} = 2\sqrt{2} \)

This results in the magnitude, \( 2\sqrt{2} \), which represents the length of the vector resulting from the cross product.

Parallelogram Area

The area of a parallelogram formed by two vectors is given by the magnitude of their cross product. This is a crucial application of both cross product and vector magnitude. Understanding this helps solve problems where vectors define shapes.

Given vectors \( \mathbf{u} \) and \( \mathbf{v} \), where:

• The cross product is \( \mathbf{u} \times \mathbf{v} = \langle 0, 2, 2 \rangle \)
• The magnitude is found as \( 2\sqrt{2} \)

Thus, the area of the parallelogram they define is \( 2\sqrt{2} \), demonstrating a neat geometric interpretation of the cross product. This concept reinforces the understanding of vectors and their applications in geometry, making vector geometry a versatile tool for various mathematical computations and spatial interpretations.