Problem 21 - Algebra and Trigonometry

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 Problem

To find the area of a parallelogram determined by two vectors \( \mathbf{u} \) and \( \mathbf{v} \), you need to find the magnitude of the cross product of these vectors.

Step 2: Compute the Cross Product

Calculate the cross product \( \mathbf{u} \times \mathbf{v} \). Use the formula for the cross product of two vectors in three dimensions:\[ \mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 3 & 2 & 1 \ 1 & 2 & 3 \end{vmatrix} \] This determinant expands to: \[ \mathbf{u} \times \mathbf{v} = \mathbf{i}(2 \cdot 3 - 1 \cdot 2) - \mathbf{j}(3 \cdot 3 - 1 \cdot 1) + \mathbf{k}(3 \cdot 2 - 2 \cdot 1) \] = \( 4\mathbf{i} - 8\mathbf{j} + 4\mathbf{k} \).

Step 3: Calculate the Magnitude of the Cross Product

Find the magnitude of the cross product vector \( 4\mathbf{i} - 8\mathbf{j} + 4\mathbf{k} \). The magnitude is given by: \[ |\mathbf{u} \times \mathbf{v}| = \sqrt{(4)^2 + (-8)^2 + (4)^2} \] = \( \sqrt{16 + 64 + 16} = \sqrt{96} = 4\sqrt{6} \).

Step 4: Conclusion

The magnitude of the cross product \( 4\sqrt{6} \) represents the area of the parallelogram formed by vectors \( \mathbf{u} \) and \( \mathbf{v} \).

Key Concepts

Cross Product

The cross product is a fascinating operation from vector calculus, crucial for several applications in physics and engineering. It takes two vectors and returns a third vector that is perpendicular to both. This makes the cross product particularly useful when determining geometrical properties like area or volume.

To compute the cross product of two vectors \( \mathbf{u} = \langle u_1, u_2, u_3 \rangle \) and \( \mathbf{v} = \langle v_1, v_2, v_3 \rangle \), we can express the operation through the determinant of a 3x3 matrix:

• \( \mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ u_1 & u_2 & u_3 \ v_1 & v_2 & v_3 \end{vmatrix} \)

This determinant expands computationally to:

• \( \mathbf{i}(u_2v_3 - u_3v_2) - \mathbf{j}(u_1v_3 - u_3v_1) + \mathbf{k}(u_1v_2 - u_2v_1) \)

In our exercise, the cross product yields \( \mathbf{u} \times \mathbf{v} = 4\mathbf{i} - 8\mathbf{j} + 4\mathbf{k} \). This vector is orthogonal to both \( \mathbf{u} \) and \( \mathbf{v} \), and encodes essential spatial information for further calculations.

Magnitude

Understanding the magnitude of a vector is essential as it represents the 'size' or 'length' of the vector, without regard to direction. For a given vector \( \mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k} \), its magnitude is calculated using the Euclidean norm:

• \( |\mathbf{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2} \)

This formula ensures that regardless of the direction the vector points towards, its magnitude will always be a positive scalar.

In the provided exercise, after calculating the cross product resulting in \( 4\mathbf{i} - 8\mathbf{j} + 4\mathbf{k} \), we use this very concept to determine its magnitude:

• \( |\mathbf{u} \times \mathbf{v}| = \sqrt{4^2 + (-8)^2 + 4^2} = \sqrt{16 + 64 + 16} = \sqrt{96} \)

This computes to \( 4\sqrt{6} \), which is the magnitude of the cross product vector, conveying crucial information about the size of related geometric figures.

Parallelogram Area

When two vectors define a parallelogram, the cross product's magnitude gives us this area. The idea relies on the parallelogram having sides formed by the vectors and its orientation naturally derived from the cross product.

• To find the area, calculate \( |\mathbf{u} \times \mathbf{v}| \).

For our vectors \( \mathbf{u} = \langle 3, 2, 1 \rangle \) and \( \mathbf{v} = \langle 1, 2, 3 \rangle \), the cross product is \( 4\mathbf{i} - 8\mathbf{j} + 4\mathbf{k} \), and its magnitude is \( 4\sqrt{6} \).

This magnitude, \( 4\sqrt{6} \), directly tells us that the area of the parallelogram defined by these vectors is precisely \( 4\sqrt{6} \) square units. This formula comes from the geometric interpretation of the cross product as the area of a parallelogram spanned by two vectors. Thus, mastering these concepts equips you with the ability to evaluate complex geometrical forms easily.