Find the distance from the point Q = (4, -3,5) to the plane - 2x-3y + 2z=6 □
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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"Find the distance from the point Q = (4, -3, 5) to the plane -2x - 3y + 2z = 6."
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![### Distance from a Point to a Plane
**Problem:**
Find the distance from the point \( Q = (4, -3, 5) \) to the plane given by the equation \(-2x - 3y + 2z = 6\).
**Explanation:**
To find the distance \( d \) from a point \( (x_1, y_1, z_1) \) to a plane given by the equation \( ax + by + cz = d_0 \), use the formula:
\[
d = \frac{|ax_1 + by_1 + cz_1 - d_0|}{\sqrt{a^2 + b^2 + c^2}}
\]
**Steps:**
1. **Identify the components:**
- Point \( Q(x_1, y_1, z_1) = (4, -3, 5) \)
- Plane equation coefficients: \( a = -2 \), \( b = -3 \), \( c = 2 \), and \( d_0 = 6 \).
2. **Substitute into the formula:**
- Compute \( ax_1 + by_1 + cz_1: \)
\[
-2(4) - 3(-3) + 2(5) = -8 + 9 + 10 = 11
\]
- Substitute into the distance formula:
\[
d = \frac{|11 - 6|}{\sqrt{(-2)^2 + (-3)^2 + 2^2}}
\]
3. **Calculate:**
\[
d = \frac{|5|}{\sqrt{4 + 9 + 4}} = \frac{5}{\sqrt{17}}
\]
**Final Answer:**
The distance from point \( Q = (4, -3, 5) \) to the plane \(-2x - 3y + 2z = 6\) is \( \frac{5}{\sqrt{17}} \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F99d15f92-0bff-4b4d-a47e-2ac33d144271%2Fd139edbd-a48b-4eb6-89a1-57c4350ed249%2Fpmfcx9d_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Distance from a Point to a Plane
**Problem:**
Find the distance from the point \( Q = (4, -3, 5) \) to the plane given by the equation \(-2x - 3y + 2z = 6\).
**Explanation:**
To find the distance \( d \) from a point \( (x_1, y_1, z_1) \) to a plane given by the equation \( ax + by + cz = d_0 \), use the formula:
\[
d = \frac{|ax_1 + by_1 + cz_1 - d_0|}{\sqrt{a^2 + b^2 + c^2}}
\]
**Steps:**
1. **Identify the components:**
- Point \( Q(x_1, y_1, z_1) = (4, -3, 5) \)
- Plane equation coefficients: \( a = -2 \), \( b = -3 \), \( c = 2 \), and \( d_0 = 6 \).
2. **Substitute into the formula:**
- Compute \( ax_1 + by_1 + cz_1: \)
\[
-2(4) - 3(-3) + 2(5) = -8 + 9 + 10 = 11
\]
- Substitute into the distance formula:
\[
d = \frac{|11 - 6|}{\sqrt{(-2)^2 + (-3)^2 + 2^2}}
\]
3. **Calculate:**
\[
d = \frac{|5|}{\sqrt{4 + 9 + 4}} = \frac{5}{\sqrt{17}}
\]
**Final Answer:**
The distance from point \( Q = (4, -3, 5) \) to the plane \(-2x - 3y + 2z = 6\) is \( \frac{5}{\sqrt{17}} \).
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