Problem 27 - Thomas Calculus

Question

The tetrahedron in the first octant bounded by the coordinate planes and the plane passing through \((1,0,0),(0,2,0),\) and \((0,0,3) .\) (GRAPH NOT COPY)

Step-by-Step Solution

Answer

The volume of the tetrahedron is 1 cubic unit.

Step 1: Understand the Problem

We need to calculate the volume of the tetrahedron located in the first octant of a 3D coordinate system. The tetrahedron is bounded by the coordinate planes (the planes represented by the x-axis, y-axis, and z-axis) and another plane defined by three given points: \[(1,0,0), (0,2,0), \text{ and } (0,0,3)\].

Step 2: Determine the Equation of the Plane

To find the equation of the plane passing through the three points \((1,0,0), (0,2,0), \text{ and } (0,0,3)\), we use the general form of a plane equation: \[ax + by + cz = d\].First, find the normal vector to the plane using the cross product of vectors formed by the three points:Vector \(PQ\): \((0-1, 2-0, 0-0) = (-1, 2, 0)\)Vector \(PR\): \((0-1, 0-0, 3-0) = (-1, 0, 3)\) Calculate \(\text{cross product} = (6, 3, 2)\).The plane equation becomes \(6x + 3y + 2z = d\).Use point \((1,0,0)\) to find \(d: \)\(6\times1 + 3\times0 + 2\times0 = 6\Rightarrow d= 6\).Thus, the equation of the plane is \[6x + 3y + 2z = 6\].

Step 3: Find the Volume of the Tetrahedron

The formula for the volume of a tetrahedron is \[ V = \frac{1}{6} \times \text{Base Area} \times \text{Height} \].Use the determinant method for four points \((0,0,0), (1,0,0), (0,2,0), (0,0,3)\)Volume \(V = \frac{1}{6}\left|\begin{array}{cccc}1 & 0 & 0 & 0 \1 & 1 & 0 & 0 \1 & 0 & 2 & 0 \1 & 0 & 0 & 3 \\end{array}\right|\)The determinant simplifies to 6.Thus, the volume \(V = \frac{1}{6} \times 6 = 1\).

Step 4: Interpret the Result

We've calculated the volume of the tetrahedron as 1 cubic unit. The tetrahedron is defined by the coordinate planes in the first octant and the plane passing through the three defined points. The use of determinants confirms that this simple shape has a volume that can easily be calculated once the bounding plane is known.

Key Concepts

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, is a branch of geometry where we use algebraic equations to describe geometric figures and their relationships. It is particularly useful for finding the equations of lines and planes, distances, midpoints, and angles between lines in various geometries, including three-dimensional space.

In 3D space, each point is described by three coordinates, (x, y, z), representing its position along the x, y, and z axes respectively. For instance, the points (1,0,0), (0,2,0), and (0,0,3) in the exercise are such coordinates. These coordinates not only specify a point's location but also become essential in determining the shapes, like the tetrahedron.

Understanding how these points relate to each other geometrically is crucial. You can determine the lengths, angles, and even the volume seen in a tetrahedron using these calculations. This makes coordinate geometry an invaluable tool in both theoretical and practical applications, especially for solving real-world problems involving space and dimensions.

Determinants in Calculus

Determinants play a significant role in calculus, especially in the contexts of linear algebra and multivariable calculus. A determinant is a calculated value that can help determine properties of matrices and systems of equations.

In this exercise, we use the determinant to find the volume of a tetrahedron. The points defining the tetrahedron can be represented in a matrix, for which the determinant gives a scalar value representing the volume when scaled appropriately. The formula used for the volume of a tetrahedron involves a one-sixth factor and the determinant of a matrix that includes these points:

• Matrix involves the coordinates of the points.
• The determinant is calculated as part of finding the volume.
• This method highlights the symmetry and geometric interpretation of determinants.

Understanding and calculating determinants require practice, but they provide deep insights into the geometric properties, making them a powerful tool in multivariable calculus and geometry.

Surface Equation in 3D Space

The equation of a surface in 3D space typically requires understanding how points relate to a given plane. In coordinate geometry, a plane can be defined by the equation \(ax + by + cz = d\). This is derived using the coordinates of points lying on the plane and vectors calculated between these points.

In the example problem, the plane passing through points (1,0,0), (0,2,0), and (0,0,3) results in the equation \(6x + 3y + 2z = 6\). Here's how it works:

• First, vectors between given points are calculated.
• The normal vector to the plane is found using the cross product of these vectors.
• The plane equation is assembled using this normal vector and checked using any of the original points.

This gives a clear mathematical description of the plane. Understanding how to derive and utilize these equations is essential in fields like physics and engineering, where the dimensions and shapes of various objects need to be precisely calculated and understood.