Problem 53 - Thomas Calculus

Question

In Exercises \(53-56,\) find the point in which the line meets the plane. $$ x=1-t, \quad y=3 t, \quad z=1+t ; \quad 2 x-y+3 z=6 $$

Step-by-Step Solution

Answer

The intersection point is \( \left(\frac{3}{2}, -\frac{3}{2}, \frac{1}{2}\right) \).

Step 1: Parametrize the Line

The line is given by the parametric equations \( x = 1 - t \), \( y = 3t \), and \( z = 1 + t \). In these equations, \( t \) is the parameter that determines the position on the line.

Step 2: Set Up the Plane Equation

The equation of the plane is given as \( 2x - y + 3z = 6 \). To find where the line intersects the plane, substitute the parametric equations of the line into this plane equation.

Step 3: Substitute and Solve for t

Substitute \( x = 1 - t \), \( y = 3t \), and \( z = 1 + t \) into the plane's equation: \[ 2(1 - t) - 3t + 3(1 + t) = 6 \]Simplifying the left side gives:\[ 2 - 2t - 3t + 3 + 3t = 6 \]Combine like terms:\[ 5 - 2t = 6 \]Solving for \( t \) gives:\[ -2t = 1 \Rightarrow t = -\frac{1}{2} \].

Step 4: Find Intersection Point

Substitute \( t = -\frac{1}{2} \) back into the parametric equations:\[ x = 1 - \left(-\frac{1}{2}\right) = 1 + \frac{1}{2} = \frac{3}{2} \]\[ y = 3\left(-\frac{1}{2}\right) = -\frac{3}{2} \]\[ z = 1 + \left(-\frac{1}{2}\right) = \frac{1}{2} \]The intersection point is \( \left(\frac{3}{2}, -\frac{3}{2}, \frac{1}{2}\right) \).

Key Concepts

Parametric Equations

Parametric equations are a powerful tool to describe geometric objects using parameters. In the context of lines in 3D space, they provide a way to express the coordinates of any point on the line using a variable, commonly denoted as \( t \). This variable allows you to trace the full length of the line by changing its value.
The parametric equations given in the exercise are:

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

Each part of the equation consists of a constant term and a multiple of \( t \). As \( t \) changes, these equations describe every point on the line. For example, when \( t = 0 \), the point becomes \((1, 0, 1)\). By increasing \( t \) to 1, the point shifts to \((0, 3, 2)\). This method of using parametric equations streamlines working within dimensional spaces by temporarily ignoring fixed axes like \( x \), \( y \), and \( z \), focusing instead on the direction and length along the line.

Plane Equation

Plane equations characterize flat surfaces in 3D space. They are often expressed in the general form \( ax + by + cz = d \). This linear equation relates the coordinates of any point lying on the plane, adhering to a consistent ratio between \( x \), \( y \), and \( z \).
In the exercise, the plane equation is given as \( 2x - y + 3z = 6 \). Each coefficient \((2, -1, 3)\) represents the plane's orientation, which dictates how the surface extends in three dimensions. The coefficient on the right-hand side, \( 6 \), shifts this plane parallel to its original orientation in space.
To find where a line intersects this plane, substitute the parametric equations of the line into the plane's equation. This substitution involves replacing \( x \), \( y \), and \( z \) with their parametric equivalents, simplifying the plane equation into an expression solely in terms of \( t \). This transformation reveals whether the line crosses the plane and, if so, identifies the specific \( t \) value at the intersection point.

Solving Linear Equations

Solving linear equations involves finding values for variables that satisfy a given equation. It's a critical skill for algebraic problems. In the context of intersecting a line and a plane, the process shows where the line meets the plane.
Using the exercise's context, we substitute expressions from the parametric equations into the plane equation, resulting in the equation \[2(1 - t) - 3t + 3(1 + t) = 6\]This substitution gives a single equation in one variable. The goal is to solve this linear equation for \( t \), simplifying step by step:

• Distribute: \( 2 - 2t - 3t + 3 + 3t = 6 \)
• Combine like terms: \( 5 - 2t = 6 \)
• Isolate \( -2t = 1 \)
• Solve for \( t \): \( t = -\frac{1}{2} \)

After finding \( t \), substitute back into the parametric equations to find the coordinates at this point. This example shows how solving linear equations can uncover key intersections in geometric problems, providing practical solutions to spatial queries.