Problem 24 - Algebra and Trigonometry

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 the Cross Product

To find the area of a parallelogram formed by two vectors, we need to take the cross product of the two vectors \( \mathbf{u} \) and \( \mathbf{v} \). The magnitude of this cross product gives the area. Recall that for 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 \( \mathbf{a} \times \mathbf{b} = \langle a_2 b_3 - a_3 b_2, a_3 b_1 - a_1 b_3, a_1 b_2 - a_2 b_1 \rangle \). Apply this formula in the next step.

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

Substitute \( \mathbf{u} = \langle 1, -1, 1 \rangle \) and \( \mathbf{v} = \langle 1, 1, -1 \rangle \) into the cross product formula. Compute each component:\( \mathbf{u} \times \mathbf{v} = \langle (-1)(-1) - (1)(1), (1)(1) - (1)(-1), (1)(1) - (-1)(1) \rangle \).Simplify to obtain \( \mathbf{u} \times \mathbf{v} = \langle 1 - 1, 1 + 1, 1 + 1 \rangle = \langle 0, 2, 2 \rangle \).

Step 3: Calculate the Magnitude of the Cross Product

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

Step 4: State the Area of the Parallelogram

The area of the parallelogram is equal to the magnitude of the cross product vector. Therefore, the area is \( 2\sqrt{2} \).

Key Concepts

Parallelogram Area

The area of a parallelogram in three-dimensional space can be elegantly calculated using vectors. If you have two vectors that define the sides of the parallelogram, the area is the magnitude of their cross product. This is because the cross product of two vectors gives a third vector, orthogonal to the plane containing the first two. The length of this new vector represents the area of the parallelogram. To break it down:

• Take two vectors, say \( \mathbf{u} \) and \( \mathbf{v} \).
• Compute their cross product \( \mathbf{u} \times \mathbf{v} \).
• The magnitude of \( \mathbf{u} \times \mathbf{v} \) gives the area.

In our example, vectors \( \mathbf{u} \) and \( \mathbf{v} \) were used, and after computing the cross product, we found the resultant vector’s magnitude to be \( 2\sqrt{2} \). Thus, the area of the parallelogram is \( 2\sqrt{2} \).

Magnitude of Vectors

Understanding vector magnitude is key when dealing with cross products, especially in calculations like finding the area of a parallelogram. The magnitude of a vector \( \langle a, b, c \rangle \) is calculated using the formula:\[|\mathbf{v}| = \sqrt{a^2 + b^2 + c^2} \]This formula helps find the length of the vector, which is crucial in understanding its 'size' or 'strength'.In context, when you compute the cross product of two vectors, what you obtain is another vector. To find out how much 'area' this vector covers in space, you calculate the magnitude. For our problem, the cross product vector was \( \langle 0, 2, 2 \rangle \), and its magnitude was \( 2\sqrt{2} \). This involves squaring each component, summing them, and finding the square root of the total:

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

This value directly represents the area in our parallelogram problem.

Three-Dimensional Vectors

In three-dimensional geometry, vectors are not only about direction but also about defining planes and spaces. A vector in this realm is expressed with three components, typically \( \langle a, b, c \rangle \), corresponding to the \( x \), \( y \), and \( z \) axes respectively.These vectors can describe positions and movements in the 3D world, and operations on them, like the cross product, reveal deeper insights.

• The cross product is a tool that translates vectors into a new vector perpendicular to the original pair, suitable for applications like finding areas.
• Cross products are unique to three-dimensional vectors, enjoying properties such as being non-commutative—meaning \( \mathbf{u} \times \mathbf{v} eq \mathbf{v} \times \mathbf{u} \).

Working with 3D vectors offers a rich area for exploration, like in problems involving volumes and surfaces, where calculating areas such as that of a parallelogram hinges on understanding and manipulating these fundamental components. In this specific exercise, vectors \( \mathbf{u} \) and \( \mathbf{v} \) were given in three dimensions, leading us to leverage the cross product for identifying the area of the structured parallelogram.