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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:**
Find the equation of a plane through the point \((-6, 4, 3)\) which is orthogonal to the line.
**Parametric Equations of the Line:**
- \(x = 5t - 1\)
- \(y = 3t + 7\)
- \(z = -6t + 3\)
**Explanation:**
To find the equation of a plane that is orthogonal (perpendicular) to a given line and passes through a specified point, we need the direction vector of the line. The direction vector can be derived from the parametric equations:
\[
\vec{d} = \langle 5, 3, -6 \rangle
\]
This direction vector will act as the normal vector for the plane. The general equation of a plane in space is given by:
\[
a(x - x_0) + b(y - y_0) + c(z - z_0) = 0
\]
Where \((a, b, c)\) is the normal vector and \((x_0, y_0, z_0)\) is a point on the plane, in this case, \((-6, 4, 3)\).
**Diagram Explanation:**
The small diagram appears to be a rectangle or just a simple rectangular shape, which may indicate a placeholder or a separate note, but it is not relevant to the main problem presented.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F99d15f92-0bff-4b4d-a47e-2ac33d144271%2F4aa10fc2-3ada-47f2-afc1-70d1eb3e8bda%2Figipxsg_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find the equation of a plane through the point \((-6, 4, 3)\) which is orthogonal to the line.
**Parametric Equations of the Line:**
- \(x = 5t - 1\)
- \(y = 3t + 7\)
- \(z = -6t + 3\)
**Explanation:**
To find the equation of a plane that is orthogonal (perpendicular) to a given line and passes through a specified point, we need the direction vector of the line. The direction vector can be derived from the parametric equations:
\[
\vec{d} = \langle 5, 3, -6 \rangle
\]
This direction vector will act as the normal vector for the plane. The general equation of a plane in space is given by:
\[
a(x - x_0) + b(y - y_0) + c(z - z_0) = 0
\]
Where \((a, b, c)\) is the normal vector and \((x_0, y_0, z_0)\) is a point on the plane, in this case, \((-6, 4, 3)\).
**Diagram Explanation:**
The small diagram appears to be a rectangle or just a simple rectangular shape, which may indicate a placeholder or a separate note, but it is not relevant to the main problem presented.
![**Problem Statement:**
Find the equation of a plane through the point (-6, 4, -3) which is orthogonal to the line:
\[
x = 5t - 1, \quad y = 3t + 7, \quad z = -6t + 3
\]
**Diagram:**
A rectangle is drawn, possibly as a placeholder or representation of the plane in discussion.
**Explanation:**
To determine the equation of the specified plane, it is important to note:
- A plane orthogonal to a line implies that the normal vector of the plane is parallel to the direction vector of the line.
- The direction vector of the given line can be derived from its parametric equations as \((5, 3, -6)\).
- Using the point \((-6, 4, -3)\) through which the plane passes, and adopting the direction vector as the normal vector, the plane's equation can be constructed accordingly.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F99d15f92-0bff-4b4d-a47e-2ac33d144271%2F4aa10fc2-3ada-47f2-afc1-70d1eb3e8bda%2Fqlglf6f_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find the equation of a plane through the point (-6, 4, -3) which is orthogonal to the line:
\[
x = 5t - 1, \quad y = 3t + 7, \quad z = -6t + 3
\]
**Diagram:**
A rectangle is drawn, possibly as a placeholder or representation of the plane in discussion.
**Explanation:**
To determine the equation of the specified plane, it is important to note:
- A plane orthogonal to a line implies that the normal vector of the plane is parallel to the direction vector of the line.
- The direction vector of the given line can be derived from its parametric equations as \((5, 3, -6)\).
- Using the point \((-6, 4, -3)\) through which the plane passes, and adopting the direction vector as the normal vector, the plane's equation can be constructed accordingly.
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