Find the angle of intersection of the plane 46-99-2z=1 with the plane 2x-3y +27= 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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Question
Can anyone please help me to solve this problem please? I am stuck !
![**Problem Statement:**
Find the angle of intersection of the plane given by the equation:
\[ 6x - 4y - 2z = 1 \]
with the plane:
\[ 2x - 3y + 2z = 3 \]
**Explanation:**
To find the angle of intersection between two planes, we will use the concept of the dot product of their normal vectors. The normal vector to a plane \( ax + by + cz = d \) is \( \langle a, b, c \rangle \).
1. **Identify Normal Vectors:**
- For the plane \( 6x - 4y - 2z = 1 \), the normal vector \( \mathbf{n_1} \) is \( \langle 6, -4, -2 \rangle \).
- For the plane \( 2x - 3y + 2z = 3 \), the normal vector \( \mathbf{n_2} \) is \( \langle 2, -3, 2 \rangle \).
2. **Use the Dot Product:**
The dot product \( \mathbf{n_1} \cdot \mathbf{n_2} \) is calculated as follows:
\[
\mathbf{n_1} \cdot \mathbf{n_2} = (6)(2) + (-4)(-3) + (-2)(2)
\]
3. **Calculate Magnitudes:**
- Magnitude of \( \mathbf{n_1} \):
\[
\|\mathbf{n_1}\| = \sqrt{6^2 + (-4)^2 + (-2)^2}
\]
- Magnitude of \( \mathbf{n_2} \):
\[
\|\mathbf{n_2}\| = \sqrt{2^2 + (-3)^2 + 2^2}
\]
4. **Find the Angle \( \theta \):**
Use the formula for the angle between two vectors:
\[
\cos \theta = \frac{\mathbf{n_1} \cdot \mathbf{n_2}}{\|\mathbf{n_1}\| \|\mathbf{n_2}\|}
\]
Then solve for \( \theta \) using](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F99d15f92-0bff-4b4d-a47e-2ac33d144271%2F105ced22-5b47-401b-a95b-16b12bb5f891%2Fcqx54q_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find the angle of intersection of the plane given by the equation:
\[ 6x - 4y - 2z = 1 \]
with the plane:
\[ 2x - 3y + 2z = 3 \]
**Explanation:**
To find the angle of intersection between two planes, we will use the concept of the dot product of their normal vectors. The normal vector to a plane \( ax + by + cz = d \) is \( \langle a, b, c \rangle \).
1. **Identify Normal Vectors:**
- For the plane \( 6x - 4y - 2z = 1 \), the normal vector \( \mathbf{n_1} \) is \( \langle 6, -4, -2 \rangle \).
- For the plane \( 2x - 3y + 2z = 3 \), the normal vector \( \mathbf{n_2} \) is \( \langle 2, -3, 2 \rangle \).
2. **Use the Dot Product:**
The dot product \( \mathbf{n_1} \cdot \mathbf{n_2} \) is calculated as follows:
\[
\mathbf{n_1} \cdot \mathbf{n_2} = (6)(2) + (-4)(-3) + (-2)(2)
\]
3. **Calculate Magnitudes:**
- Magnitude of \( \mathbf{n_1} \):
\[
\|\mathbf{n_1}\| = \sqrt{6^2 + (-4)^2 + (-2)^2}
\]
- Magnitude of \( \mathbf{n_2} \):
\[
\|\mathbf{n_2}\| = \sqrt{2^2 + (-3)^2 + 2^2}
\]
4. **Find the Angle \( \theta \):**
Use the formula for the angle between two vectors:
\[
\cos \theta = \frac{\mathbf{n_1} \cdot \mathbf{n_2}}{\|\mathbf{n_1}\| \|\mathbf{n_2}\|}
\]
Then solve for \( \theta \) using
![**Problem:**
Find the angle of intersection of the plane
\[ 4x - 5y - 2z = 1 \]
with the plane
\[ 2x - 3y + 7z = 5 \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F99d15f92-0bff-4b4d-a47e-2ac33d144271%2F105ced22-5b47-401b-a95b-16b12bb5f891%2Fwipai9e_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem:**
Find the angle of intersection of the plane
\[ 4x - 5y - 2z = 1 \]
with the plane
\[ 2x - 3y + 7z = 5 \]
Expert Solution

Step 1
The given problem is to find the angle of intersection of the given planes.
Given planes 4x-4y-2z=1 and 2x-3y+2z=3.
Step by step
Solved in 2 steps with 1 images

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