5. Determine the angle between each of the following pairs of planes. = 4 and (2, 4, −5) · (x − 2, y − 1, z + 9) = 0. (a) 3x +2y — 7z (b) 4x+2z+y = 1 and −2z + x − 7y = 3.
5. Determine the angle between each of the following pairs of planes. = 4 and (2, 4, −5) · (x − 2, y − 1, z + 9) = 0. (a) 3x +2y — 7z (b) 4x+2z+y = 1 and −2z + x − 7y = 3.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![### Problem Statement
5. **Determine the angle between each of the following pairs of planes.**
#### (a)
- Plane 1: \( 3x + 2y - 7z = 4 \)
- Plane 2: \( \langle 2, 4, -5 \rangle \cdot \langle x - 2, y - 1, z + 9 \rangle = 0 \)
#### (b)
- Plane 1: \( 4x + 2z + y = 1 \)
- Plane 2: \( -2z + x - 7y = 3 \)
### Explanation
In this problem, we are given pairs of plane equations. Each part requires us to find the angle between the given planes.
**Method to find the angle between two planes:**
The general equation of a plane in 3D space can be written as:
\[ Ax + By + Cz + D = 0 \]
Here, \([A, B, C]\) represents the normal vector to the plane.
The angle \( \theta \) between two planes can be determined using the dot product of their normal vectors. If \( \mathbf{n_1} \) and \( \mathbf{n_2} \) are normal vectors to the two planes, then the cosine of the angle \( \theta \) between the planes is given by:
\[ \cos(\theta) = \frac{\mathbf{n_1} \cdot \mathbf{n_2}}{\|\mathbf{n_1}\| \|\mathbf{n_2}\|} \]
Where:
- \( \mathbf{n_1} \cdot \mathbf{n_2} \) is the dot product of the normal vectors.
- \( \|\mathbf{n_1}\| = \sqrt{A_1^2 + B_1^2 + C_1^2} \) is the magnitude of the first normal vector.
- \( \|\mathbf{n_2}\| = \sqrt{A_2^2 + B_2^2 + C_2^2} \) is the magnitude of the second normal vector.
### Detailed Explanation of Each Part
#### (a)
For the given planes:
- First plane: \( 3x + 2y - 7z = 4 \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F717f4d87-5a32-4cdf-855f-90bb556622e3%2Fa78a9508-53fc-4003-850c-47f7daf751b3%2Fmkclvr_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement
5. **Determine the angle between each of the following pairs of planes.**
#### (a)
- Plane 1: \( 3x + 2y - 7z = 4 \)
- Plane 2: \( \langle 2, 4, -5 \rangle \cdot \langle x - 2, y - 1, z + 9 \rangle = 0 \)
#### (b)
- Plane 1: \( 4x + 2z + y = 1 \)
- Plane 2: \( -2z + x - 7y = 3 \)
### Explanation
In this problem, we are given pairs of plane equations. Each part requires us to find the angle between the given planes.
**Method to find the angle between two planes:**
The general equation of a plane in 3D space can be written as:
\[ Ax + By + Cz + D = 0 \]
Here, \([A, B, C]\) represents the normal vector to the plane.
The angle \( \theta \) between two planes can be determined using the dot product of their normal vectors. If \( \mathbf{n_1} \) and \( \mathbf{n_2} \) are normal vectors to the two planes, then the cosine of the angle \( \theta \) between the planes is given by:
\[ \cos(\theta) = \frac{\mathbf{n_1} \cdot \mathbf{n_2}}{\|\mathbf{n_1}\| \|\mathbf{n_2}\|} \]
Where:
- \( \mathbf{n_1} \cdot \mathbf{n_2} \) is the dot product of the normal vectors.
- \( \|\mathbf{n_1}\| = \sqrt{A_1^2 + B_1^2 + C_1^2} \) is the magnitude of the first normal vector.
- \( \|\mathbf{n_2}\| = \sqrt{A_2^2 + B_2^2 + C_2^2} \) is the magnitude of the second normal vector.
### Detailed Explanation of Each Part
#### (a)
For the given planes:
- First plane: \( 3x + 2y - 7z = 4 \
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