Problem 48 - University Calculus: Early Transcendentals

Question

Find the volume of a parallelepiped with one of its eight vertices at \(A(0,0,0)\) and three adjacent vertices at \(B(1,2,0), C(0,-3,2)\) and \(D(3,-4,5)\).

Step-by-Step Solution

Answer

The volume of the parallelepiped is 5 cubic units.

Step 1: Identify Vectors

Identify three vectors starting from point \(A\) to \(B\), \(C\), and \(D\). The vectors are \(\mathbf{AB} = B - A = \langle 1, 2, 0 \rangle\), \(\mathbf{AC} = C - A = \langle 0, -3, 2 \rangle\), and \(\mathbf{AD} = D - A = \langle 3, -4, 5 \rangle\).

Step 2: Calculate the Cross Product

Calculate the cross product of vectors \(\mathbf{AC}\) and \(\mathbf{AD}\): \[ \mathbf{AC} \times \mathbf{AD} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 0 & -3 & 2 \ 3 & -4 & 5 \end{vmatrix} = \mathbf{i}((-3) \cdot 5 - 2 \cdot (-4)) - \mathbf{j}(0 \cdot 5 - 2 \cdot 3) + \mathbf{k}(0 \cdot (-4) - (-3) \cdot 3) \] \[ = \mathbf{i}(-15 + 8) - \mathbf{j}(0 - 6) + \mathbf{k}(0 + 9) \]\[ = \langle -7, 6, 9 \rangle \]

Step 3: Calculate the Scalar Triple Product

Calculate the scalar triple product which is the dot product of \(\mathbf{AB}\) with \(\mathbf{AC} \times \mathbf{AD}\): \[ \mathbf{AB} \cdot (\mathbf{AC} \times \mathbf{AD}) = \langle 1, 2, 0 \rangle \cdot \langle -7, 6, 9 \rangle \] \[ = (1)(-7) + (2)(6) + (0)(9) \] \[ = -7 + 12 + 0 \] \[ = 5 \]

Step 4: Find the Volume

The volume of the parallelepiped is the absolute value of the scalar triple product. Therefore, the volume is \[ |5| = 5 \].

Key Concepts

Parallelepiped Volume

A parallelepiped is a three-dimensional geometric figure with six parallelogram faces. Think of it as a skewed box with opposite faces parallel and equal in area. The volume of a parallelepiped is the three-dimensional space it occupies.
To find its volume, you use vectors because it is efficiently determined by three vectors that start from one vertex. In this exercise, the region extends from vertex \((0,0,0)\) to each of the points \((1,2,0)\), \((0,-3,2)\), and \((3,-4,5)\). The vectors reached from these adjacent vertices are crucial.
By utilizing these vectors, one can calculate the volume using the scalar triple product. This approach is powerful because it compactly incorporates vector cross and dot product operations to deduce the area embedded within a seeming complexity.

• The steps involve identifying three edge vectors originating from a common vertex.
• Then determining the cross product of any two vectors.
• Finally computing the dot product of the third vector with the obtained cross product.

Cross Product

The cross product is a fundamental operation in vector calculus employed to get a vector that is perpendicular to two given vectors in three-dimensional space. Given two vectors, \(\mathbf{a}\) and \(\mathbf{b}\), their cross product \(\mathbf{a} \times \mathbf{b}\) results in a new vector. In this problem, you use the cross product as part of the process to find the volume.
In this exercise, let us focus on the two vectors \(\mathbf{AC} = \langle 0, -3, 2 \rangle\) and \(\mathbf{AD} = \langle 3, -4, 5 \rangle\). The cross product formula involves setting up a determinant:

1. Place unit vectors \(\mathbf{i}, \mathbf{j}, \mathbf{k}\) along the top row of a 3x3 matrix.
2. Fill in the components of \(\mathbf{AC}\) in the second row and of \(\mathbf{AD}\) in the final row.
3. Compute the determinant to find the resulting vector.

This perpendicular vector is instrumental in calculating the parallelepiped volume since it forms part of the scalar triple product operation.

Scalar Triple Product

The scalar triple product is a concise computation to determine the volume of a parallelepiped and is expressed as \((\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}))\). Here, the result is a scalar, a plain number, representing the volume.
For the volume of the given parallelepiped, \(\mathbf{AB}\) represents the vector used for the dot product, \(\mathbf{AC} \times \mathbf{AD}\) encapsulates the cross product:

• First, execute the cross product as earlier reviewed to derive a vector.
• Next, conduct the dot product by matching corresponding components from \(\mathbf{AB}\) with the cross product vector.

This operation yields a scalar that might be positive or negative due to geometry orientation, emphasizing the use of absolute value. Therefore, always take the absolute value to reflect the positive volume measurement of space occupied without direction and negative signs involved. In this exercise, the result was simply \(|5|\).