Problem 22 - Algebra and Trigonometry

Question

Find the area of the parallelogram determined by the given vectors. $$ \mathbf{u}=\langle 0,-3,2\rangle, \quad \mathbf{v}=\langle 5,-6,0\rangle $$

Step-by-Step Solution

Answer

The area is \( \sqrt{469} \).

Step 1: Understand the Problem

The problem is asking for the area of a parallelogram defined by two vectors, \( \mathbf{u} = \langle 0, -3, 2 \rangle \) and \( \mathbf{v} = \langle 5, -6, 0 \rangle \). This is given by the magnitude of the cross product of these two vectors.

Step 2: Calculate the Cross Product

The cross product \( \mathbf{u} \times \mathbf{v} \) is calculated using the determinant:\[\mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 0 & -3 & 2 \ 5 & -6 & 0 \end{vmatrix}\]This gives:\[\mathbf{u} \times \mathbf{v} = (12\mathbf{i} + 10\mathbf{j} + 15\mathbf{k})\]

Step 3: Calculate the Magnitude of the Cross Product

The magnitude of \( \mathbf{u} \times \mathbf{v} \) is calculated by:\[\| \mathbf{u} \times \mathbf{v} \| = \sqrt{12^2 + 10^2 + 15^2}\]Simplifying:\[= \sqrt{144 + 100 + 225} = \sqrt{469}\]

Step 4: Final Answer

The area of the parallelogram formed by vectors \( \mathbf{u} \) and \( \mathbf{v} \) is \( \sqrt{469} \).

Key Concepts

Cross Product

The cross product is a mathematical operation that takes two vectors and returns a new vector that is perpendicular to the plane formed by the original vectors. To find the cross product of two vectors in three-dimensional space, you organize the vectors into a determinant form. This uses unit vectors \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \).
To compute the cross product \( \mathbf{u} \times \mathbf{v} \) for the vectors \( \mathbf{u} = \langle 0, -3, 2 \rangle \) and \( \mathbf{v} = \langle 5, -6, 0 \rangle \), you arrange them in a 3x3 matrix like this:
\[\mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 0 & -3 & 2 \ 5 & -6 & 0 \end{vmatrix}\]

• The entries of the first row are the unit vectors \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \).
• The entries of the second and third rows are the components of vectors \( \mathbf{u} \) and \( \mathbf{v} \), respectively.

To solve the determinant, you apply the cofactor expansion method, resulting in the vector:
\[\mathbf{u} \times \mathbf{v} = (12\mathbf{i} + 10\mathbf{j} + 15\mathbf{k})\]

Vector Magnitude

The magnitude of a vector gives us a measure of its length or size. For a cross product vector, this magnitude can be interpreted as the strength of the perpendicularity between the two original vectors. Calculating the magnitude involves taking the square root of the sum of the squares of the components of the vector.
For our vector \( \mathbf{u} \times \mathbf{v} = 12\mathbf{i} + 10\mathbf{j} + 15\mathbf{k} \), its magnitude is calculated as follows:

• Square each of the components: \( 12^2 \), \( 10^2 \), \( 15^2 \).
• Add these squares together: \( 144 + 100 + 225 \).
• Take the square root of the sum: \( \sqrt{469} \).

The magnitude, \( \| \mathbf{u} \times \mathbf{v} \| = \sqrt{469} \), reflects how long the resulting vector from the cross product is.

Parallelogram Area

The area of a parallelogram defined by two vectors \( \mathbf{u} \) and \( \mathbf{v} \) can be found using the cross product of these vectors. The magnitude of the cross product gives the exact area.
Why does this work?

• A parallelogram's area is equivalent to the base times the height.
• In a vector representation, one vector acts as the base, and the cross product magnitude gives the height when you project onto a perpendicular axis.

For our specific case of vectors \( \mathbf{u} = \langle 0, -3, 2 \rangle \) and \( \mathbf{v} = \langle 5, -6, 0 \rangle \), we determined the cross product \( \mathbf{u} \times \mathbf{v} \) and found its magnitude:
\[\sqrt{469}\]This result is the area of the parallelogram spanned by \( \mathbf{u} \) and \( \mathbf{v} \). It shows that even with vector calculations, we can discover geometric properties like area through algebraic operations.