### 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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Answered: a) Find a vector equation of the line…

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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Question

### 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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