Problem 21 - Precalculus Mathematics for Calculus

Question

Find the area of the parallelogram determined by the given vectors. $$\mathbf{u}=\langle 3,2,1\rangle, \quad \mathbf{v}=\langle 1,2,3\rangle$$

Step-by-Step Solution

Answer

The area of the parallelogram is \( 4\sqrt{6} \).

Step 1: Understand the Formula

To find the area of a parallelogram formed by two vectors \( \mathbf{u} \) and \( \mathbf{v} \), we use the cross product. The area is given by the magnitude of the cross product of the two vectors, \( \| \mathbf{u} \times \mathbf{v} \| \).

Step 2: Compute the Cross Product

Calculate the cross product \( \mathbf{u} \times \mathbf{v} \) using the formula:\[ \mathbf{u} \times \mathbf{v} = \left| \begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 3 & 2 & 1 \ 1 & 2 & 3 \end{array} \right| \]This determinant calculation involves:\[ \mathbf{u} \times \mathbf{v} = \mathbf{i}(2 \cdot 3 - 1 \cdot 2) - \mathbf{j}(3 \cdot 3 - 1 \cdot 1) + \mathbf{k}(3 \cdot 2 - 2 \cdot 1) \]Solve the determinants: \( = 4\mathbf{i} - 8\mathbf{j} + 4\mathbf{k} \).

Step 3: Find the Magnitude of the Cross Product

The magnitude of the cross product \( \mathbf{u} \times \mathbf{v} \) represents the area of the parallelogram. Calculate:\[ \| \mathbf{u} \times \mathbf{v} \| = \sqrt{4^2 + (-8)^2 + 4^2} \]This simplifies to \( \sqrt{16 + 64 + 16} = \sqrt{96} \).

Step 4: Simplify the Magnitude

Simplify the magnitude \( \sqrt{96} \) further to find the final area in the simplest form. \[ \sqrt{96} = \sqrt{16 \times 6} = 4 \sqrt{6} \]

Step 5: Write the Final Answer

The area of the parallelogram formed by the vectors \( \mathbf{u} = \langle 3, 2, 1 \rangle \) and \( \mathbf{v} = \langle 1, 2, 3 \rangle \) is \( 4\sqrt{6} \).

Key Concepts

Cross Product

The cross product, also known as the vector product, is a mathematical operation that is used to find a vector perpendicular to two given vectors in three-dimensional space. This is key in vector calculus, as it allows us to calculate important geometric quantities such as the area of parallelograms.

• Given two vectors \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \), their cross product \( \mathbf{a} \times \mathbf{b} \) is expressed as:

\[\mathbf{a} \times \mathbf{b} = \left| \begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{array} \right|\]Here, \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \) are unit vectors in the x, y, and z directions. This determinant calculates each component of the resulting vector. The cross product cannot be performed in two dimensions, as it inherently requires a third dimension to take place.

Parallelogram Area

The area of a parallelogram can be effortlessly computed in three-dimensional space using the cross product of two vectors that define the parallelogram. The formula used is:

• The area \( A \) is simply the magnitude of the cross product of the vectors:

\[ A = \| \mathbf{u} \times \mathbf{v} \| \]This works because the magnitude of the cross product vector gives the size of the area spanned by the original two vectors.
Let's break it down:

• A larger cross product magnitude indicates a larger spanning area.
• The vectors \( \mathbf{u} \) and \( \mathbf{v} \) are seen as two sides meeting to form the parallelogram.
• This is analogous to a base-height calculation, but instead with vectors.

This method is efficient because it circumvents the need to directly measure angles or additional geometry.

Determinant

The concept of a determinant arises when calculating the cross product using its determinant form. It's a compact way to encapsulate the necessary operations among the components of vectors in three-dimensional space.

• Determinants can be seen as a scale factor, providing area or volume under specific conditions.
• In the cross product calculation, the determinant helps evaluate the components of the resulting vector.

The use of determinants in cross product calculations is fundamental because they handle the arithmetic involved in finding the perpendicular vector efficiently. Understanding this builds a strong foundation for tackling more complex topics in vector calculus beyond just cross products, as determinants are vital in other matrix-related computations as well.