1) Find the vector and parametric equation of the line in R³ through the point P = (1, 2,-4) 5 parallel to the vector A=-1 H 3 2) Find the normal and general forms of the equation of the plane that contains the point 1 P = (6, 0,1) and has normal vector 2 -1.
1) Find the vector and parametric equation of the line in R³ through the point P = (1, 2,-4) 5 parallel to the vector A=-1 H 3 2) Find the normal and general forms of the equation of the plane that contains the point 1 P = (6, 0,1) and has normal vector 2 -1.
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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![### Vector and Parametric Equations in 3D Space
#### Problem 1
Find the vector and parametric equation of the line in \( \mathbb{R}^3 \) through the point \( P = (1, 2, -4) \) parallel to the vector
\[
\mathbf{A} = \begin{bmatrix} 5 \\ -1 \\ 3 \end{bmatrix}
\]
**Solution:**
The vector equation of a line can be expressed as:
\[ \mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{d} \]
where \( \mathbf{r}_0 \) is a point on the line and \( \mathbf{d} \) is a direction vector. Here, \( \mathbf{r}_0 = \begin{bmatrix} 1 \\ 2 \\ -4 \end{bmatrix} \) and \( \mathbf{d} = \begin{bmatrix} 5 \\ -1 \\ 3 \end{bmatrix} \).
The parametric equations of the line can thus be written as:
- \( x = 1 + 5t \)
- \( y = 2 - t \)
- \( z = -4 + 3t \)
#### Problem 2
Find the normal and general forms of the equation of the plane that contains the point \( P = (6, 0, 1) \) and has normal vector
\[
\begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix}
\]
**Solution:**
The general format for the equation of a plane is:
\[ ax + by + cz = d \]
Given the normal vector \( \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix} \), the equation becomes:
\[ 1(x - 6) + 2(y - 0) - 1(z - 1) = 0 \]
Simplifying, we get:
\[ x + 2y - z = 5 \]
This is the general form of the plane’s equation.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9cb769e4-ba9f-4168-862a-bb68b5e0d9ff%2F1d9e5579-d420-4a95-a788-e890e26f3e48%2Fzjx11hw_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Vector and Parametric Equations in 3D Space
#### Problem 1
Find the vector and parametric equation of the line in \( \mathbb{R}^3 \) through the point \( P = (1, 2, -4) \) parallel to the vector
\[
\mathbf{A} = \begin{bmatrix} 5 \\ -1 \\ 3 \end{bmatrix}
\]
**Solution:**
The vector equation of a line can be expressed as:
\[ \mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{d} \]
where \( \mathbf{r}_0 \) is a point on the line and \( \mathbf{d} \) is a direction vector. Here, \( \mathbf{r}_0 = \begin{bmatrix} 1 \\ 2 \\ -4 \end{bmatrix} \) and \( \mathbf{d} = \begin{bmatrix} 5 \\ -1 \\ 3 \end{bmatrix} \).
The parametric equations of the line can thus be written as:
- \( x = 1 + 5t \)
- \( y = 2 - t \)
- \( z = -4 + 3t \)
#### Problem 2
Find the normal and general forms of the equation of the plane that contains the point \( P = (6, 0, 1) \) and has normal vector
\[
\begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix}
\]
**Solution:**
The general format for the equation of a plane is:
\[ ax + by + cz = d \]
Given the normal vector \( \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix} \), the equation becomes:
\[ 1(x - 6) + 2(y - 0) - 1(z - 1) = 0 \]
Simplifying, we get:
\[ x + 2y - z = 5 \]
This is the general form of the plane’s equation.
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