Problem 28 - Essential Calculus Early Transcendentals

Question

Find the area of the parallelogram with vertices \(K(1,2,3),\) \(L(1,3,6), M(3,8,6),\) and \(N(3,7,3) .\)

Step-by-Step Solution

Answer

The area of the parallelogram is \( \sqrt{265} \).

Step 1: Understand the Concept

The area of a parallelogram can be found using the vector cross product. If two vectors define a plane, the area is the magnitude of their cross product.

Step 2: Select Vectors from Vertices

Choose two vectors from the given vertices to represent the sides of the parallelogram. For instance, take \( \vec{KL} \) and \( \vec{KM} \) as the vectors. The coordinates give us: \( \vec{KL} = L - K = (0, 1, 3) \) and \( \vec{KM} = M - K = (2, 6, 3) \).

Step 3: Compute the Cross Product

Calculate the cross product of vectors \( \vec{KL} = (0, 1, 3) \) and \( \vec{KM} = (2, 6, 3) \). The cross product formula gives us \( \vec{KL} \times \vec{KM} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 0 & 1 & 3 \ 2 & 6 & 3 \end{vmatrix} = \hat{i}(1 \cdot 3 - 3 \cdot 6) - \hat{j}(0 \cdot 3 - 3 \cdot 2) + \hat{k}(0 \cdot 6 - 1 \cdot 2) \).

Step 4: Simplify the Cross Product

Continue simplifying the determinant: \( \vec{KL} \times \vec{KM} = \hat{i}(3 - 18) - \hat{j}(0 - 6) + \hat{k}(0 - 2) = \hat{i}(-15) + \hat{j}(6) - \hat{k}(2) = (-15, 6, -2) \).

Step 5: Find the Magnitude of the Cross Product

The area of the parallelogram is the magnitude of the cross product vector. Compute the magnitude: \( \| (-15, 6, -2) \| = \sqrt{(-15)^2 + 6^2 + (-2)^2} = \sqrt{225 + 36 + 4} = \sqrt{265} \).

Key Concepts

Vector Cross Product

The vector cross product is a critical operation in vector algebra that produces a new vector perpendicular to the plane formed by the initial two vectors. In a three-dimensional space, if you have vectors \( \vec{a} \) and \( \vec{b} \), the cross product, denoted as \( \vec{a} \times \vec{b} \), results in a vector \( \vec{c} \).

• The direction of \( \vec{c} \) is perpendicular to both \( \vec{a} \) and \( \vec{b} \).
• The magnitude of \( \vec{c} \) is equal to the area of the parallelogram spanned by \( \vec{a} \) and \( \vec{b} \).

The cross product can be computed using the determinant of a matrix formed by unit vectors \( \hat{i}, \hat{j}, \hat{k} \), and the components of the vectors. For example, for vectors \( \vec{a} = (a_1, a_2, a_3) \) and \( \vec{b} = (b_1, b_2, b_3) \), the cross product is given by:\[\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix}\]This results in a new vector \( (a_2b_3 - a_3b_2)\hat{i} - (a_1b_3 - a_3b_1)\hat{j} + (a_1b_2 - a_2b_1)\hat{k} \). Calculating this helps find the area of the parallelogram.

Magnitude of Vector

The magnitude of a vector, often considered its "length," is determined using the Pythagorean theorem in three-dimensional space. It's especially useful for calculating areas and distances in geometry.

• The magnitude \( \| \vec{v} \| \) of a vector \( \vec{v} = (v_1, v_2, v_3) \) can be calculated using the formula: \( \| \vec{v} \| = \sqrt{v_1^2 + v_2^2 + v_3^2} \).
• To find the area of a parallelogram using vectors, find the magnitude of the cross product vector.

For instance, if a cross product results in the vector \( (-15, 6, -2) \), its magnitude is: \[\| (-15, 6, -2) \| = \sqrt{(-15)^2 + 6^2 + (-2)^2} = \sqrt{265}\]This gives the area of the parallelogram, making this method a practical use of vector magnitude.

Coordinate Geometry

Coordinate geometry, or analytical geometry, allows us to geometrically represent algebraic equations using a coordinate system. This approach is fundamental in linking geometrical shapes and their numerical properties.

• Coordinates of points, such as vertices of a shape, help define vectors between those points. For example, the vector from point \( K(1, 2, 3) \) to point \( L(1, 3, 6) \) is found by subtracting the coordinates to obtain \( (0, 1, 3) \).
• By using coordinates, vectors defining sides of polygons like parallelograms can be determined and manipulated to calculate areas, angles, and other properties.

Coordinate geometry makes it easy to apply methods like the cross product. By choosing appropriate vectors based on the coordinates of vertices, we efficiently solve complex problems in 3D space. Its straightforward and structured approach helps identify relationships between the shapes and formulas you are working with.