Problem 57 - Thomas Calculus

Question

Find parametrizations for the lines in which the planes in Exercises 57–60 intersect. $$ x+y+z=1, \quad x+y=2 $$

Step-by-Step Solution

Answer

The parametrization of the line is \( \mathbf{r}(t) = \langle t, 2-t, -1 \rangle \).

Step 1: Understand the Intersection of Planes

When two planes intersect, their intersection is either a line or the planes are parallel and do not intersect at all. For the given planes, x+y+z=1 and x+y=2, we will find where they intersect.

Step 2: Express Intersection as Equations

We have two planes:\[ \begin{aligned} x+y+z &= 1 \ x+y &= 2 \end{aligned} \] Subtract the second equation from the first:\ \( x+y+z - (x+y) = 1 - 2 \) which simplifies to \( z = -1 \).

Step 3: Solve for Parameters

Use the equation \( x+y=2 \): Let \( x = t \) (a parameter), then \( y = 2 - t \). Since \( z = -1 \) does not depend on \( t \), we have all components for the line parametrized in terms of \( t \).

Step 4: Write Parametrization of the Line

The line of intersection can be given by the vector \( \langle x, y, z \rangle \), where substituting our previous findings we get \( x = t \), \( y = 2 - t \), and \( z = -1 \). Therefore, the parametrization of the line is: \[ \mathbf{r}(t) = \langle t, 2-t, -1 \rangle \]

Key Concepts

Intersection of Planes

The intersection of two planes occurs when they meet along a line or are parallel without intersecting. For instance, consider the two planes defined by the equations \(x+y+z=1\) and \(x+y=2\). In this case, these equations describe two planes in three-dimensional space.
Subtracting the second equation from the first gives us: \(x+y+z - (x+y) = 1 - 2\), which simplifies to \(z = -1\).
This computation indicates that the planes intersect along a line where every point has a \(z\)-coordinate of \(-1\). Understanding this relationship is significant since it allows us to see that the intersection, rather than being a point, is indeed a line.
When addressing intersection problems with planes, consider:

• If the resultant is a non-zero constant, they run parallel without intersecting.
• If they intersect, they do so along a line determined by manipulating the original equations.

Line Parametrization

Line parametrization is about expressing a line using a parameter, often denoted as \(t\). A parametrized line gives us equations for each coordinate of a point on that line in terms of \(t\).
Given the planes \(x+y+z=1\) and \(x+y=2\), we determined that \(z = -1\) remains constant. We can express \(x\) and \(y\) in terms of a parameter: let \(x = t\), which makes it easier to solve for \(y\).
Using the equation \(x+y=2\), substitute \(x = t\) to get \(y = 2 - t\). Thus, each point on the line of intersection is given by \((t, 2-t, -1)\).
When calculating line parametrizations, remember:

• Choose a parameter for one variable.
• Solve other variables with the given equations.
• Ensure the final expression accounts for all involved coordinates.

Vector Equations

Vector equations are a compact way to describe lines, especially in multidimensional spaces. A vector equation of a line shows each coordinate as a function of a parameter, typically in vector notation.
For our intersecting planes, the parametrization \(x = t\), \(y = 2 - t\), and \(z = -1\) can be written as the vector equation \(\mathbf{r}(t) = \langle t, 2-t, -1 \rangle\).
This notation allows for an elegant representation of all possible points on the line using the parameter \(t\).
When using vector equations, keep in mind:

• It identifies the direction of the line through incremental changes in \(t\).
• It provides a clear geometric interpretation of line characteristics.
• Understanding vector equations can simplify working with lines, especially when extending to higher dimensions.