a) Find a vector equation of the line through the point (1,5,8) that is perpendicular to the plane x-y + 4z = 8 b) At which point does the line intersect the xy-plane?
a) Find a vector equation of the line through the point (1,5,8) that is perpendicular to the plane x-y + 4z = 8 b) At which point does the line intersect the xy-plane?
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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![### Vector Equation and Intersection with the xy-Plane
#### Problem:
Consider a point in space and a plane defined by the following elements:
a) **Vector Equation**
Find a vector equation of the line through the point \((1,5,8)\) that is perpendicular to the plane \(x - y + 4z = 8\).
b) **Line and Plane Intersection**
Determine the point at which this line intersects the \(xy\)-plane.
### Solution:
**a) Finding the Vector Equation:**
Given a point \(P = (1,5,8)\) and a plane defined by the equation \(x - y + 4z = 8\), we need to find the vector equation of the line that goes through the point and is perpendicular to the plane.
The normal vector (perpendicular) to the plane \(x - y + 4z = 8\) can be directly derived from the plane's equation coefficients. The normal vector \(\mathbf{n}\) is:
\[
\mathbf{n} = \begin{pmatrix} 1 \\ -1 \\ 4 \end{pmatrix}
\]
A line perpendicular to the plane and passing through point \(P\) can be described by the following parametric equations:
\[
\mathbf{r(t)} = \mathbf{r_0} + t \mathbf{n}
\]
Where:
- \(\mathbf{r(t)}\) is the position vector of any point on the line,
- \(\mathbf{r_0} = \begin{pmatrix} 1 \\ 5 \\ 8 \end{pmatrix}\) is the position vector of the given point \(P\),
- \(t\) is a scalar parameter,
- \(\mathbf{n} = \begin{pmatrix} 1 \\ -1 \\ 4 \end{pmatrix}\) is the direction vector (normal to the plane).
So the vector equation of the line is:
\[
\mathbf{r(t)} = \begin{pmatrix} 1 \\ 5 \\ 8 \end{pmatrix} + t \begin{pmatrix} 1 \\ -1 \\ 4 \end{pmatrix}
\]
Expanded, this gives us the parametric form:
\[
\begin{cases}
x = 1 + t \\
y = 5 - t](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff06024b0-5ab7-436b-ba03-1a05fb23d390%2F43c951eb-4f33-4240-a0dd-7386f957a526%2Fylz1kfj_processed.png&w=3840&q=75)
Transcribed Image Text:### Vector Equation and Intersection with the xy-Plane
#### Problem:
Consider a point in space and a plane defined by the following elements:
a) **Vector Equation**
Find a vector equation of the line through the point \((1,5,8)\) that is perpendicular to the plane \(x - y + 4z = 8\).
b) **Line and Plane Intersection**
Determine the point at which this line intersects the \(xy\)-plane.
### Solution:
**a) Finding the Vector Equation:**
Given a point \(P = (1,5,8)\) and a plane defined by the equation \(x - y + 4z = 8\), we need to find the vector equation of the line that goes through the point and is perpendicular to the plane.
The normal vector (perpendicular) to the plane \(x - y + 4z = 8\) can be directly derived from the plane's equation coefficients. The normal vector \(\mathbf{n}\) is:
\[
\mathbf{n} = \begin{pmatrix} 1 \\ -1 \\ 4 \end{pmatrix}
\]
A line perpendicular to the plane and passing through point \(P\) can be described by the following parametric equations:
\[
\mathbf{r(t)} = \mathbf{r_0} + t \mathbf{n}
\]
Where:
- \(\mathbf{r(t)}\) is the position vector of any point on the line,
- \(\mathbf{r_0} = \begin{pmatrix} 1 \\ 5 \\ 8 \end{pmatrix}\) is the position vector of the given point \(P\),
- \(t\) is a scalar parameter,
- \(\mathbf{n} = \begin{pmatrix} 1 \\ -1 \\ 4 \end{pmatrix}\) is the direction vector (normal to the plane).
So the vector equation of the line is:
\[
\mathbf{r(t)} = \begin{pmatrix} 1 \\ 5 \\ 8 \end{pmatrix} + t \begin{pmatrix} 1 \\ -1 \\ 4 \end{pmatrix}
\]
Expanded, this gives us the parametric form:
\[
\begin{cases}
x = 1 + t \\
y = 5 - t
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