Problem 56 - Thomas Calculus

Question

In Exercises 53–56, find the point in which the line meets the plane. $$ x=-1+3 t, \quad y=-2, \quad z=5 t ; \quad 2 x-3 z=7 $$

Step-by-Step Solution

Answer

The line meets the plane at the point (-4, -2, -5).

Step 1: Parameterize the Point on the Line

The line is given by the parametric equations: \[ x = -1 + 3t, \quad y = -2, \quad z = 5t \]Any point on the line is represented by these equations in terms of parameter \( t \).

Step 2: Substitute in Plane Equation

The plane is described by the equation:\[ 2x - 3z = 7 \]Substitute the expressions for \( x \) and \( z \) from the parameterized line equations into this plane equation:\[ 2(-1 + 3t) - 3(5t) = 7 \]

Step 3: Simplify and Solve for t

Simplify the expression to solve for the parameter \( t \):\[2(-1) + 6t - 15t = 7 \-2 + 6t - 15t = 7 \-2 - 9t = 7 \-9t = 9 \t = -1\]

Step 4: Find the Intersection Point

Substitute \( t = -1 \) back into the parametric equations of the line to find the specific point:\[x = -1 + 3(-1) = -4, \quad y = -2, \quad z = 5(-1) = -5\]Thus, the coordinates of the point of intersection are \((-4, -2, -5)\).

Key Concepts

Parametric Equations

Parametric equations describe a set of equations that express the coordinates of points on a line or curve as functions of a parameter, usually denoted by a letter like \( t \). In three-dimensional space, a line can be described by three parametric equations, one for each coordinate: \( x \), \( y \), and \( z \).
In our example, the following parametric equations are given for the line:

• \( x = -1 + 3t \)
• \( y = -2 \)
• \( z = 5t \)

Each value of \( t \) produces a specific point on the line. The parameter \( t \) usually ranges over all real numbers, allowing us to see how the line extends infinitely in both directions. Parametric equations are particularly useful in describing lines when orientation and position in space are important.

Plane Equation

A plane in three-dimensional space is described using a plane equation. A typical form for this equation is \( Ax + By + Cz = D \), where \( A \), \( B \), \( C \), and \( D \) are constants, and \( x \), \( y \), \( z \) are the variables representing points on the plane.
In our problem, the plane is given by:

• \( 2x - 3z = 7 \)

This equation tells us that for any point \((x, y, z)\) on the plane, the relationship between the \( x \) and \( z \) coordinates must satisfy this equation. Unlike lines, which usually require three equations in 3D space, identifying a plane needs only one equation since it is a two-dimensional surface.

Parameterization

Parameterization in mathematics involves expressing a geometric object using one or more parameters to describe it. When parameterizing a line, we use a parameter \( t \) to define every point on the line via the parametric equations provided.
The process involves:

• Identifying the direction vector, essentially showing the direction of the line.
• Pointing out a specific initial point through which the line passes.
• Setting equations that capture changes in \( x \), \( y \), and \( z \) as \( t \) changes.

In our exercise, using parameter \( t \):

• \( x = -1 + 3t \)
• \( y = -2 \)
• \( z = 5t \)

This form showcases the line’s parameterization, transforming geometric problems into a simple algebraic form that is easier to work with, especially when intersecting with surfaces like planes.

Solving Equations

Solving equations involves finding the values of variables that make an equation true. In this exercise, the goal is to find the intersection of a line and a plane, which requires solving a system of equations established by their relationships.
Here, we substitute the parametric equations of the line into the plane equation to connect the parameter \( t \) with the specified plane:

• Substitute: \( 2(-1 + 3t) - 3(5t) = 7 \)
• This simplifies to: \( -2 + 6t - 15t = 7 \)

Following simplification, solve for \( t \):

• \( -2 - 9t = 7 \)
• \( -9t = 9 \)
• \( t = -1 \)

This \( t \) value specifies the point at which the line intersects the plane. Substituting \( t = -1 \) back into the line's parametric equations gives you the exact intersection point, making this approach straightforward through clear, step-by-step algebraic manipulation.