2. Let A(2, 4,2), B(3,7,-2), and C(1,3,3). (a) Which point is closest to the y-axis? (b) Write the equation of the plane passing through A and orthogonal to the y-axis. (c) Are the points A, B, and C collinear?
2. Let A(2, 4,2), B(3,7,-2), and C(1,3,3). (a) Which point is closest to the y-axis? (b) Write the equation of the plane passing through A and orthogonal to the y-axis. (c) Are the points A, B, and C collinear?
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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![### Example Exercise from Geometry Class:
**Problem Statement:**
Given the points \( A(2, 4, 2) \), \( B(3, 7, -2) \), and \( C(1, 3, 3) \):
(a) Determine which point is closest to the y-axis.
(b) Write the equation of the plane passing through point A and orthogonal to the y-axis.
(c) Assess whether points A, B, and C are collinear.
1. **Finding the Closest Point to the y-axis:**
To determine proximity to the y-axis, only the x and z coordinates need to be considered because the y-axis occurs where \(x = 0 \) and \( z = 0 \). So the Euclidean distances for points to the y-axis will be calculated like this:
\[
\text{Distance} = \sqrt{x^2 + z^2}
\]
Compute the distance for each point:
- \( A(2, 4, 2) \)
\[
\text{Distance for A} = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}
\]
- \( B(3, 7, -2) \)
\[
\text{Distance for B} = \sqrt{3^2 + (-2)^2} = \sqrt{13}
\]
- \( C(1, 3, 3) \)
\[
\text{Distance for C} = \sqrt{1^2 + 3^2} = \sqrt{10}
\]
Comparing the distances, the closest point to the y-axis is the one with the smallest distance.
2. **Equation of the Plane:**
The plane that passes through point \( A(2, 4, 2) \) and is orthogonal to the y-axis implies that the normal vector to the plane is parallel to the y-axis, which is defined by the direction vector \( (0, 1, 0) \):
Therefore, the general form of the equation of this plane will not involve y, resulting in:
\[
Ax + By + Cz = D
\]
Where \( B = 0 \):](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdb24b717-8ee5-4ca9-8061-74ddb7e91c1a%2Fc8299902-6e61-4dbc-9bc6-f9530207e1a8%2Flnp47u_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Example Exercise from Geometry Class:
**Problem Statement:**
Given the points \( A(2, 4, 2) \), \( B(3, 7, -2) \), and \( C(1, 3, 3) \):
(a) Determine which point is closest to the y-axis.
(b) Write the equation of the plane passing through point A and orthogonal to the y-axis.
(c) Assess whether points A, B, and C are collinear.
1. **Finding the Closest Point to the y-axis:**
To determine proximity to the y-axis, only the x and z coordinates need to be considered because the y-axis occurs where \(x = 0 \) and \( z = 0 \). So the Euclidean distances for points to the y-axis will be calculated like this:
\[
\text{Distance} = \sqrt{x^2 + z^2}
\]
Compute the distance for each point:
- \( A(2, 4, 2) \)
\[
\text{Distance for A} = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}
\]
- \( B(3, 7, -2) \)
\[
\text{Distance for B} = \sqrt{3^2 + (-2)^2} = \sqrt{13}
\]
- \( C(1, 3, 3) \)
\[
\text{Distance for C} = \sqrt{1^2 + 3^2} = \sqrt{10}
\]
Comparing the distances, the closest point to the y-axis is the one with the smallest distance.
2. **Equation of the Plane:**
The plane that passes through point \( A(2, 4, 2) \) and is orthogonal to the y-axis implies that the normal vector to the plane is parallel to the y-axis, which is defined by the direction vector \( (0, 1, 0) \):
Therefore, the general form of the equation of this plane will not involve y, resulting in:
\[
Ax + By + Cz = D
\]
Where \( B = 0 \):
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