Find the equation of the plane through (3,2,1) and (3,1,-5) and is perpendi čular to the plane 6x+ 7y+ 2z-103D0

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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### Problem Statement:
**Find the equation of the plane through \((3,2,1)\) and \((3,1,-5)\) and is perpendicular to the plane \(6x + 7y + 2z - 10 = 0\)**.

### Detailed Explanation:

To determine the equation of the desired plane, consider the following steps:

1. **Understand the Geometry**:
   - We are given two points \((3, 2, 1)\) and \((3, 1, -5)\) that lie on the plane.
   - The plane is perpendicular to another plane whose equation is \(6x + 7y + 2z - 10 = 0\).

2. **Normal Vectors**:
   - The normal vector to the given plane \(6x + 7y + 2z - 10 = 0\) is \( \mathbf{n_1} = (6, 7, 2) \).
   - The normal vector to the plane we're seeking (let's call it \(\mathbf{n_2}\)) will be perpendicular to \(\mathbf{n_1}\).
   
3. **Vector Direction**:
   - Compute the direction vector \(\mathbf{d}\) between points \((3, 2, 1)\) and \((3, 1, -5)\):
     \[
     \mathbf{d} = (3 - 3, 1 - 2, -5 - 1) = (0, -1, -6)
     \]

4. **Formulation**:
   - The desired plane will have a normal vector \( \mathbf{n_2} \), which will be orthogonal to both \(\mathbf{n_1}\) and \(\mathbf{d}\). This can be obtained using the cross product:
     \[
     \mathbf{n_2} = \mathbf{n_1} \times \mathbf{d}
     \]

5. **Cross Product Calculation**:
   - Calculate the cross product: 
     \[
     \mathbf{n_2} = (6, 7, 2) \times (0, -1, -6)
     \]
     \[
     \mathbf{n_2} = \left| \begin{matrix}
     \mathbf{i} & \mathbf{
Transcribed Image Text:### Problem Statement: **Find the equation of the plane through \((3,2,1)\) and \((3,1,-5)\) and is perpendicular to the plane \(6x + 7y + 2z - 10 = 0\)**. ### Detailed Explanation: To determine the equation of the desired plane, consider the following steps: 1. **Understand the Geometry**: - We are given two points \((3, 2, 1)\) and \((3, 1, -5)\) that lie on the plane. - The plane is perpendicular to another plane whose equation is \(6x + 7y + 2z - 10 = 0\). 2. **Normal Vectors**: - The normal vector to the given plane \(6x + 7y + 2z - 10 = 0\) is \( \mathbf{n_1} = (6, 7, 2) \). - The normal vector to the plane we're seeking (let's call it \(\mathbf{n_2}\)) will be perpendicular to \(\mathbf{n_1}\). 3. **Vector Direction**: - Compute the direction vector \(\mathbf{d}\) between points \((3, 2, 1)\) and \((3, 1, -5)\): \[ \mathbf{d} = (3 - 3, 1 - 2, -5 - 1) = (0, -1, -6) \] 4. **Formulation**: - The desired plane will have a normal vector \( \mathbf{n_2} \), which will be orthogonal to both \(\mathbf{n_1}\) and \(\mathbf{d}\). This can be obtained using the cross product: \[ \mathbf{n_2} = \mathbf{n_1} \times \mathbf{d} \] 5. **Cross Product Calculation**: - Calculate the cross product: \[ \mathbf{n_2} = (6, 7, 2) \times (0, -1, -6) \] \[ \mathbf{n_2} = \left| \begin{matrix} \mathbf{i} & \mathbf{
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