Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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![### Determining the Equation of a Plane
To find an equation of the plane passing through three given points, we use the points provided and the formula for the plane.
#### Given Points:
- Point A: \( (9, 5, -5) \)
- Point B: \( (9, -5, 5) \)
- Point C: \( -9, -5, -5) \)
### Steps to Find the Equation of a Plane:
1. **Identify the Vectors:**
- **Vector AB:** Connects Point A to Point B.
- **Vector AC:** Connects Point A to Point C.
2. **Find the Cross Product of these Vectors:**
- The cross product of the vectors will give a normal vector \( \mathbf{n} \) to the plane.
3. **Use the Normal Vector and a Point to Find the Plane Equation:**
- The equation of the plane in general form is:
\[ ax + by + cz = d \]
- Where \( (a, b, c) \) are the components of the normal vector \( \mathbf{n} \).
4. **Substitute the Coordinates of a Point:**
- To find \( d \), substitute the coordinates of one of the given points into the plane equation.
### Sample Calculation (Illustrative):
1. **Vectors Calculation:**
- \( \overrightarrow{AB} = B - A = (0, -10, 10) \)
- \( \overrightarrow{AC} = C - A = (-18, -10, 0) \)
2. **Cross Product \( \mathbf{n} = \overrightarrow{AB} \times \overrightarrow{AC} \):**
- \( \mathbf{n} = (100, -180, 180) \)
3. **Plane Equation:**
- Using the normal vector:
\[ 100x - 180y + 180z = d \]
- Substitute \( (9, 5, -5) \):
\[ 100(9) - 180(5) + 180(-5) = d \]
\[ 900 - 9000 - 900 = d \]
\[ d = 0 \]
So, the equation of the plane is:
\[ 10x - 18](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4bd1af25-6093-4425-b8ce-aaaaeb1acf7d%2Fd137e697-3638-47c6-af12-84d8f745e02b%2F554vfz_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Determining the Equation of a Plane
To find an equation of the plane passing through three given points, we use the points provided and the formula for the plane.
#### Given Points:
- Point A: \( (9, 5, -5) \)
- Point B: \( (9, -5, 5) \)
- Point C: \( -9, -5, -5) \)
### Steps to Find the Equation of a Plane:
1. **Identify the Vectors:**
- **Vector AB:** Connects Point A to Point B.
- **Vector AC:** Connects Point A to Point C.
2. **Find the Cross Product of these Vectors:**
- The cross product of the vectors will give a normal vector \( \mathbf{n} \) to the plane.
3. **Use the Normal Vector and a Point to Find the Plane Equation:**
- The equation of the plane in general form is:
\[ ax + by + cz = d \]
- Where \( (a, b, c) \) are the components of the normal vector \( \mathbf{n} \).
4. **Substitute the Coordinates of a Point:**
- To find \( d \), substitute the coordinates of one of the given points into the plane equation.
### Sample Calculation (Illustrative):
1. **Vectors Calculation:**
- \( \overrightarrow{AB} = B - A = (0, -10, 10) \)
- \( \overrightarrow{AC} = C - A = (-18, -10, 0) \)
2. **Cross Product \( \mathbf{n} = \overrightarrow{AB} \times \overrightarrow{AC} \):**
- \( \mathbf{n} = (100, -180, 180) \)
3. **Plane Equation:**
- Using the normal vector:
\[ 100x - 180y + 180z = d \]
- Substitute \( (9, 5, -5) \):
\[ 100(9) - 180(5) + 180(-5) = d \]
\[ 900 - 9000 - 900 = d \]
\[ d = 0 \]
So, the equation of the plane is:
\[ 10x - 18
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