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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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
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

\[ 2x - 3y + 7z = 5 \]
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.

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