Problem 14 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems \(11-14\), find the equation of the plane through the given points. \((a, 0,0),(0, b, 0)\), and \((0,0, c),(\) None of \(a, b\), and \(c\) is zero. \()\)

Step-by-Step Solution

Answer

The equation of the plane is \( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \).

Step 1: Identify the Given Points

We are given three points through which the plane passes: - Point 1: \((a, 0, 0)\)- Point 2: \((0, b, 0)\)- Point 3: \((0, 0, c)\).

Step 2: Find Two Vectors on the Plane

Use the given points to find two vectors that lie on the plane. Let \( \vec{v_1} \) be the vector from Point 1 to Point 2: \[ \vec{v_1} = (0, b, 0) - (a, 0, 0) = (-a, b, 0) \]Let \( \vec{v_2} \) be the vector from Point 1 to Point 3:\[ \vec{v_2} = (0, 0, c) - (a, 0, 0) = (-a, 0, c) \]

Step 3: Find the Normal Vector to the Plane

The normal vector to the plane, \( \vec{n} \), can be found using the cross product of \( \vec{v_1} \) and \( \vec{v_2} \).Calculate the cross product:\[ \vec{n} = \vec{v_1} \times \vec{v_2} = | \begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \-a & b & 0 \-a & 0 & c \\end{array} | \]Expanding the determinant gives:\[ \vec{n} = (bc)\hat{i} - (ac)\hat{j} + (ab)\hat{k} \]

Step 4: Equation of the Plane

The general equation of a plane with normal vector \( \vec{n} = (A, B, C) \) is\[ Ax + By + Cz = D \]Substitute \( A = bc \), \( B = -ac \), and \( C = ab \).Use point \((a, 0, 0)\) to find \(D\):\[ bc(a) + (-ac)(0) + ab(0) = D \]\[ D = abc \]The equation of the plane is:\[ bcx - acy + abz = abc \]

Step 5: Simplify the Equation

Simplifying the equation, we divide through by \(abc\), assuming \(abc eq 0\):\[ \frac{bcx - acy + abz}{abc} = 1 \]The simplified plane equation is:\[ \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \]

Key Concepts

Cross Product

The cross product is a powerful tool in vector mathematics, often used to find a perpendicular vector to two given vectors. This is especially useful in three-dimensional spaces, like when determining the equation of a plane. To compute the cross product of two vectors, you need to use the determinant of a matrix formed by these vectors. For vectors \(\vec{v_1} = (-a, b, 0)\) and \(\vec{v_2} = (-a, 0, c)\), their cross product is calculated using the following method:

• Create a 3x3 matrix with unit vectors \((\hat{i}, \hat{j}, \hat{k})\) in the first row, components of \(\vec{v_1}\) in the second row, and components of \(\vec{v_2}\) in the third row.
• Calculate the determinant of this matrix.

This results in a vector \(\vec{n} = (bc)\hat{i} - (ac)\hat{j} + (ab)\hat{k}\), which is perpendicular to both vectors. The cross product provides a way to establish a normal vector for the plane.

Vectors

Vectors are fundamental elements in the study of geometry and physics, extending beyond simple arrows with a direction and magnitude. In three-dimensional space, they represent entities such as positions, displacements, or velocities.

• A vector has both a direction and a length (magnitude).
• Vectors can be represented as coordinates \((x, y, z)\) in three-dimensional space.

In the plane equation problem, vectors were used to represent the direction between given points. By determining the vectors \(\vec{v_1}\) and \(\vec{v_2}\) that lie on the plane, we set the stage to find the normal vector. These vectors \((-a, b, 0)\) and \((-a, 0, c)\) each define linear paths between specific points, providing essential input for the cross product used later to find the normal vector.

Normal Vector

The normal vector is crucial when defining a plane in three-dimensional space. It is perpendicular to the plane and helps determine its orientation. This perpendicular vector is often found using the cross product of two other vectors that lie on the plane.

• The normal vector is expressed as \(\vec{n} = (A, B, C)\).
• It provides the necessary coefficients for the plane equation \(Ax + By + Cz = D\).

In the given solution, the normal vector \(\vec{n} = (bc, -ac, ab)\) was found by taking the cross product of two vectors on the plane. This vector not only gives insight into the orientation of the plane but also plays a key role in forming the plane's general equation.

Determinants

Determinants are mathematical objects that can provide valuable information about matrices, especially in the context of transformations and orientation in space. Determinants are used to calculate the cross product when finding a normal vector for a plane.

• A determinant is calculated from a square matrix.
• The sign of a determinant helps to understand the orientation of the vector cross product.

For our plane problem, we arranged components of two vectors and unit vectors into a 3x3 matrix. The determinant of that matrix gives the components \((bc, -ac, ab)\) of the normal vector. Determinants are thus vital in evolving vector calculations into effective plane equations, and understanding them is key to grasping how those transformations reflect on spatial geometry.