Complete the parametric equations for the line where the planes - 8x +12y - 5z = -5 and -9x + 11y - 5z = -5 intersect. x(t) y(t) = z(t) = - - 5t: Question Help: Video
Complete the parametric equations for the line where the planes - 8x +12y - 5z = -5 and -9x + 11y - 5z = -5 intersect. x(t) y(t) = z(t) = - - 5t: Question Help: Video
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![**Topic: Parametric Equations for Line of Intersection Between Planes**
In this section, we will learn how to complete the parametric equations for the line where two planes intersect. The planes given are:
\[ -8x + 12y - 5z = -5 \]
\[ -9x + 11y - 5z = -5 \]
To find the parametric equations of the line of intersection of these planes, we can proceed as follows.
### Given:
1. The first plane: \(-8x + 12y - 5z = -5\)
2. The second plane: \(-9x + 11y - 5z = -5\)
These equations can be rewritten as the following system of linear equations:
\[ -8x + 12y - 5z = -5 \]
\[ -9x + 11y - 5z = -5 \]
### Parametric Equations Form:
The solution to this system can be expressed parametrically as:
\[ x(t) = -5t \]
\[ y(t) = \] \[ \text{(to be filled using appropriate algebraic steps)}\]
\[ z(t) = \] \[ \text{(to be filled using appropriate algebraic steps)}\]
### Steps to Find \(y(t)\) and \(z(t)\):
1. **Express one variable in terms of another using one of the equations.**
2. **Substitute back into the other equation to find the relationships between variables.**
3. **Assign a parameter (usually \( t \)) and express each variable as a function of this parameter.**
### Completing the Equations:
From the equation provided:
\[ x(t) = -5t \]
To find \( y(t) \) and \( z(t) \), we substitute \( t \) and solve the resultant system. This step involves algebraic manipulations which are typically performed to find the values of \( y(t) \) and \( z(t) \).
**Example Solution Steps (for illustration):**
1. **Select a common parameter (typically \( t \) ).**
2. **Solve for \( y \) and \( z \) after substituting \( x = -5t \).**
Thus, the parametric equations are written as:
\[ x(t) = -5t \]
\[ y(t) = \text{expression](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F406ee3bf-a3e1-420c-814a-008c3fe57146%2F450e8ff6-145f-43a9-bf29-ba8f2963c9e7%2F1czb2rc_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Topic: Parametric Equations for Line of Intersection Between Planes**
In this section, we will learn how to complete the parametric equations for the line where two planes intersect. The planes given are:
\[ -8x + 12y - 5z = -5 \]
\[ -9x + 11y - 5z = -5 \]
To find the parametric equations of the line of intersection of these planes, we can proceed as follows.
### Given:
1. The first plane: \(-8x + 12y - 5z = -5\)
2. The second plane: \(-9x + 11y - 5z = -5\)
These equations can be rewritten as the following system of linear equations:
\[ -8x + 12y - 5z = -5 \]
\[ -9x + 11y - 5z = -5 \]
### Parametric Equations Form:
The solution to this system can be expressed parametrically as:
\[ x(t) = -5t \]
\[ y(t) = \] \[ \text{(to be filled using appropriate algebraic steps)}\]
\[ z(t) = \] \[ \text{(to be filled using appropriate algebraic steps)}\]
### Steps to Find \(y(t)\) and \(z(t)\):
1. **Express one variable in terms of another using one of the equations.**
2. **Substitute back into the other equation to find the relationships between variables.**
3. **Assign a parameter (usually \( t \)) and express each variable as a function of this parameter.**
### Completing the Equations:
From the equation provided:
\[ x(t) = -5t \]
To find \( y(t) \) and \( z(t) \), we substitute \( t \) and solve the resultant system. This step involves algebraic manipulations which are typically performed to find the values of \( y(t) \) and \( z(t) \).
**Example Solution Steps (for illustration):**
1. **Select a common parameter (typically \( t \) ).**
2. **Solve for \( y \) and \( z \) after substituting \( x = -5t \).**
Thus, the parametric equations are written as:
\[ x(t) = -5t \]
\[ y(t) = \text{expression
![### Problem Statement
**Goal:** Find a plane containing the point \((-3, 4, 3)\) and the line given by the parametric equations:
\[
\begin{cases}
x(t) = 5 + t \\
y(t) = -1 - t \\
z(t) = 7 - 6t
\end{cases}
\]
---
**Solution:**
To solve this problem, you'll need to:
1. Identify the direction vector of the given line.
2. Identify another vector lying on the plane using the given point.
3. Use the cross product to find the normal vector to the plane.
4. Use the normal vector and the given point to write the equation of the plane.
### Steps:
1. **Direction Vector of the Line:**
- Given parametric equations:
\[
x(t) = 5 + t, \quad y(t) = -1 - t, \quad z(t) = 7 - 6t
\]
- The direction vector \(\vec{d}\) can be extracted from these equations:
\[
\vec{d} = (1, -1, -6)
\]
2. **Vector from the Given Point to a Point on the Line:**
- A point on the line can be found by setting \( t = 0 \):
\[
(5, -1, 7)
\]
- The vector \(\vec{v}\) from the given point \((-3, 4, 3)\) to this point on the line:
\[
\vec{v} = (5 - (-3), -1 - 4, 7 - 3) = (8, -5, 4)
\]
3. **Normal Vector to the Plane:**
- The normal vector \(\vec{n}\) is found by taking the cross product of \(\vec{d}\) and \(\vec{v}\):
\[
\vec{n} = \vec{d} \times \vec{v}
\]
- Calculation of the cross product:
\[
\vec{d} \times \vec{v} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
1 &](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F406ee3bf-a3e1-420c-814a-008c3fe57146%2F450e8ff6-145f-43a9-bf29-ba8f2963c9e7%2Fmjcm93w_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement
**Goal:** Find a plane containing the point \((-3, 4, 3)\) and the line given by the parametric equations:
\[
\begin{cases}
x(t) = 5 + t \\
y(t) = -1 - t \\
z(t) = 7 - 6t
\end{cases}
\]
---
**Solution:**
To solve this problem, you'll need to:
1. Identify the direction vector of the given line.
2. Identify another vector lying on the plane using the given point.
3. Use the cross product to find the normal vector to the plane.
4. Use the normal vector and the given point to write the equation of the plane.
### Steps:
1. **Direction Vector of the Line:**
- Given parametric equations:
\[
x(t) = 5 + t, \quad y(t) = -1 - t, \quad z(t) = 7 - 6t
\]
- The direction vector \(\vec{d}\) can be extracted from these equations:
\[
\vec{d} = (1, -1, -6)
\]
2. **Vector from the Given Point to a Point on the Line:**
- A point on the line can be found by setting \( t = 0 \):
\[
(5, -1, 7)
\]
- The vector \(\vec{v}\) from the given point \((-3, 4, 3)\) to this point on the line:
\[
\vec{v} = (5 - (-3), -1 - 4, 7 - 3) = (8, -5, 4)
\]
3. **Normal Vector to the Plane:**
- The normal vector \(\vec{n}\) is found by taking the cross product of \(\vec{d}\) and \(\vec{v}\):
\[
\vec{n} = \vec{d} \times \vec{v}
\]
- Calculation of the cross product:
\[
\vec{d} \times \vec{v} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
1 &
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