Chapter 10: Problem 9
\(6-10=\) Find parametric equations and symmetric equations for the line. The line through \((1,-1,1)\) and parallel to the line \(x+2=\frac{1}{2} y=z-3\)
Short Answer
Expert verified
Parametric: \(x=1+t, y=-1+2t, z=1+t\); Symmetric: \(x-1=\frac{y+1}{2}=z-1\).
Step by step solution
01
Determine Direction Vector
The given symmetric equations \( x+2 = \frac{1}{2} y = z-3 \) imply specific relationships for the direction vector of the line. Rewriting them in terms of parameter \( t \), we find the direction vector. From \( x+2 = t \), \( \frac{1}{2}y = t \), and \( z-3 = t \), the directional vector is \( \langle 1, 2, 1 \rangle \).
02
Create Parametric Equations
Using the point \((1,-1,1)\) and the direction vector \(\langle 1, 2, 1 \rangle\), we write the parametric equations for the line. Let the parameter be \( t \). Then, \(x = 1 + 1t, y = -1 + 2t, z = 1 + 1t\). Thus, the parametric equations are: \(x = 1 + t\), \(y = -1 + 2t\), and \(z = 1 + t\).
03
Form Symmetric Equations
To form symmetric equations from the parametric equations \(x = 1 + t\), \(y = -1 + 2t\), and \(z = 1 + t\), solve for \(t\) in terms of each variable: \(t = x - 1\), \(t = \frac{y + 1}{2}\), and \(t = z - 1\). Equate these expressions for \(t\) to get the symmetric equation: \(x - 1 = \frac{y + 1}{2} = z - 1\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding the Direction Vector
In the world of parametric equations, the direction vector is a vital concept. It tells us the direction in which a line moves through 3D space. This vector is made up of values that represent the changes in x, y, and z coordinates. It is often denoted as \( \langle a, b, c \rangle \), where each component shows how much each coordinate should change with respect to time or another parameter like \( t \).
To find the direction vector when given symmetric equations such as \( x+2 = \frac{1}{2} y = z-3 \), we need to express each variable in terms of a common parameter \( t \). This yields mutual relationships, allowing us to see the pattern of movements in these coordinates.
Finding these expressions results in a direction vector for the line, in our example, this process leads to the direction vector \( \langle 1, 2, 1 \rangle \). This means as the line progresses in space, the x-coordinate increases by 1 for every unit of \( t \), the y-coordinate changes by 2, and the z-coordinate by 1.
To find the direction vector when given symmetric equations such as \( x+2 = \frac{1}{2} y = z-3 \), we need to express each variable in terms of a common parameter \( t \). This yields mutual relationships, allowing us to see the pattern of movements in these coordinates.
Finding these expressions results in a direction vector for the line, in our example, this process leads to the direction vector \( \langle 1, 2, 1 \rangle \). This means as the line progresses in space, the x-coordinate increases by 1 for every unit of \( t \), the y-coordinate changes by 2, and the z-coordinate by 1.
The Beauty of Symmetric Equations
Symmetric equations present a unique format expressing a line in space without directly involving a parameter such as \( t \). These equations show the relationship between coordinates more implicitly.
For instance, once we derived the parametric equations from the direction vector and point information, such as \( x = 1 + t \), \( y = -1 + 2t \), and \( z = 1 + t \), we can formulate symmetric equations. These equations eliminate the parameter by equating the terms for \( t \) obtained from each parametric equation.
For instance, once we derived the parametric equations from the direction vector and point information, such as \( x = 1 + t \), \( y = -1 + 2t \), and \( z = 1 + t \), we can formulate symmetric equations. These equations eliminate the parameter by equating the terms for \( t \) obtained from each parametric equation.
- From the x equation: \( t = x - 1 \).
- From the y equation: \( t = \frac{y + 1}{2} \).
- From the z equation: \( t = z - 1 \).
Lines in 3D Space
The exploration of lines in 3D space can be both fascinating and challenging. Unlike a 2D plane where lines are simply described by slope and y-intercept, lines in 3D require more complex descriptions.
In three dimensions, lines must account for changes in three distinct directions: x, y, and z. This complexity is why parametric and symmetric equations are used. Such equations capture how each coordinate on the line changes concerning the parameter \( t \).
In our specific exercise, we sought to identify the line passing through the point \( (1, -1, 1) \) and parallel to another one described by the symmetric equations \( x+2 = \frac{1}{2} y = z-3 \).
To build an understanding:
In three dimensions, lines must account for changes in three distinct directions: x, y, and z. This complexity is why parametric and symmetric equations are used. Such equations capture how each coordinate on the line changes concerning the parameter \( t \).
In our specific exercise, we sought to identify the line passing through the point \( (1, -1, 1) \) and parallel to another one described by the symmetric equations \( x+2 = \frac{1}{2} y = z-3 \).
To build an understanding:
- A direction vector provides the directional path in 3D space.
- Parametric equations use this path to describe the points on the line dynamically.
- Symmetric equations present a static, yet insightful perspective focusing on coordinate relationships.