Chapter 11: Problem 57
Find parametrizations for the lines in which the planes. $$x+y+z=1, \quad x+y=2$$
Short Answer
Expert verified
Parametric equation: \( x = 0, y = 2 - t, z = 1 + t \).
Step by step solution
01
Understand Plane Equations
The first plane is described by the equation \(x + y + z = 1\) and the second by \(x + y = 2\). These are equations of planes in 3D space. We need to find the line of intersection between these two planes.
02
Find a Point of Intersection
To locate a point in the intersection of the planes, set \(z = 0\) to simplify the system. Substituting into the two plane equations gives \(x + y = 1\) and \(x + y = 2\). Clearly, these equations are inconsistent for \(z = 0\). Try \(z = 1\): Substitute \(z = 1\) into the first equation to get \(x + y = 0\), and substitute \(x + y = 2\) into the second still holds for any \(z\). Letting \(x = 0\), \(y = 2\) and \(z = 1\) gives a point (0, 2, 1).
03
Determine Direction Vector for the Line
Substitute \(z = t\) for parameterization. Use the first plane equation \(x + y + z = 1\) to find other constraints, \(x + y = 1 - t\). From the second \(x + y = 2\), solve \(x + y - t = 2 - z\). Both planes have solutions that differ by a common factor, eliminating one, leaving us to choose \(x = 0\). then find \(y = 2 - t\), giving a direction vector (0, -1, 1).
04
Form the Parametric Equation for the Line
Combine the point (0, 2, 1) found in Step 2 and the direction vector \( (0, -1, 1) \) to form the parametric line equation: \( \mathbf{r}(t) = (0, 2, 1) + t(0, -1, 1) \). This results in the parametric equations: \( x = 0 \), \( y = 2 - t \), \( z = 1 + t \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Line of Intersection
When two planes intersect in three-dimensional space, they typically form a line. Visualize two sheets of paper crossing each other; their common edge represents the line of intersection. To find this line, we solve a system of plane equations simultaneously. The line exists where both plane equations hold true simultaneously.
The line can be represented in a parametric form, which involves determining a specific point on this line and a direction vector that shows its orientation. The key steps for finding this line include:
The line can be represented in a parametric form, which involves determining a specific point on this line and a direction vector that shows its orientation. The key steps for finding this line include:
- Identifying points satisfying both plane equations, which help us find a specific location on the line.
- Determining the direction vector, which provides the line's path or orientation.
Plane Equations
Plane equations describe flat surfaces in 3D space and have a general form of \( ax + by + cz = d \). Each plane equation consists of coefficients that multiply the variables \( x, y, \text{and} z \), and a constant \( d \). These coefficients determine the plane's orientation and position in space.
In exercises involving intersection, you'll usually see two plane equations, similar to the originals given: \( x + y + z = 1 \) and \( x + y = 2 \). Here's how to understand them:
In exercises involving intersection, you'll usually see two plane equations, similar to the originals given: \( x + y + z = 1 \) and \( x + y = 2 \). Here's how to understand them:
- The first equation implies a three-dimensional surface, where any point \( (x, y, z) \) satifying it is on the plane.
- The second equation can be visualized as a tilted plane where every combination of \( (x, y, z) \) that makes it true lies on this plane.
Direction Vector
A direction vector is crucial in defining the line of intersection between two planes. It indicates the line's orientation and direction in three-dimensional space. Think of it as an arrow showing the line's path.
To determine the direction vector in such problems, we typically parameterize one of the variables, like \( z = t \), and solve the plane equations with this parameter. Once you find how the other variables relate to \( t \), you'll notice a pattern or a direction vector.
In the exercise, the direction vector \( (0, -1, 1) \) was found by observing how \( x \), \( y \), and \( z \) change as \( t \) varies. This vector suggests that moving one unit positively in the \( z \)-direction accompanies a one-unit negative change in the \( y \)-direction, while \( x \) remains constant.
This part of the exercise highlights the beauty of geometry: a set of numbers beautifully describes how lines behave and extend in space.
To determine the direction vector in such problems, we typically parameterize one of the variables, like \( z = t \), and solve the plane equations with this parameter. Once you find how the other variables relate to \( t \), you'll notice a pattern or a direction vector.
In the exercise, the direction vector \( (0, -1, 1) \) was found by observing how \( x \), \( y \), and \( z \) change as \( t \) varies. This vector suggests that moving one unit positively in the \( z \)-direction accompanies a one-unit negative change in the \( y \)-direction, while \( x \) remains constant.
This part of the exercise highlights the beauty of geometry: a set of numbers beautifully describes how lines behave and extend in space.