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 point of intersection is (1.5, -1.5, 0.5).

Step 1: Parameterize the line equations

The line is given in parametric form: \[ x = 1 - t, \quad y = 3t, \quad z = 1 + t \]We can use these equations to represent any point on the line in terms of parameter \( t \).

Step 2: Substitute line equations into the plane equation

Substitute the parametric equations of the line into the plane equation, \( 2x - y + 3z = 6 \).Start by substituting \( x = 1 - t \), \( y = 3t \), and \( z = 1 + t \):\[ 2(1-t) - 3t + 3(1+t) = 6 \]

Step 3: Simplify the equation

Simplify the substituted plane equation:\[ 2 - 2t - 3t + 3 + 3t = 6 \]The equation simplifies to: \[ 5 - 2t = 6 \]

Step 4: Solve for the parameter t

Solve the simplified equation for \( t \):\[ 5 - 2t = 6 \] \Subtract 5 from both sides: \[ -2t = 1 \] \Divide by -2: \[ t = -\frac{1}{2} \]

Step 5: Find the intersection point

Substitute \( t = -\frac{1}{2} \) back into the parametric equations to find the point of intersection.\[ x = 1 - (-\frac{1}{2}) = 1.5 \] \\[ y = 3(-\frac{1}{2}) = -1.5 \] \\[ z = 1 + (-\frac{1}{2}) = 0.5 \] \Therefore, the point where the line intersects the plane is \( (1.5, -1.5, 0.5) \).

Key Concepts

Parametric Equations

Parametric equations are used to describe a line in three-dimensional space by expressing the coordinates \( x, y, \) and \( z \) in terms of a single variable, commonly denoted as \( t \). This parameter, \( t \), acts as a variable that traces the position of any point along the line.
For example, given the equations:

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

these parametric equations define a line by indicating how the coordinates of points on the line change with respect to \( t \). As \( t \) varies, different points are described along the path of the line.
Thus, parametric equations are a powerful tool in geometry as they allow us to easily manipulate and use lines within various contexts, such as intersection problems with planes.

Plane Equation

A plane equation describes a flat, two-dimensional surface that extends infinitely in three-dimensional space. The standard form of a plane equation is expressed as \( Ax + By + Cz = D \), where \( A, B, \) and \( C \) are constants that define the orientation of the plane, and \( D \) is a constant that shifts the plane along the perpendicular axis.
In the exercise, the plane equation given is \( 2x - y + 3z = 6 \). Here:

• \( A = 2 \)
• \( B = -1 \)
• \( C = 3 \)
• \( D = 6 \)

These coefficients can be interpreted as the components of a normal vector to the plane \( (2, -1, 3) \), which indicates the direction perpendicularly away from the plane's surface.
Understanding the plane equation is crucial as it helps us determine how the plane intersects with other geometric entities like lines, which is essential for solving intersection problems.

Solving Equations

Solving equations involves finding the values of the unknown variables that satisfy the given equations. In the context of the intersection of a line and a plane, we substitute the parametric line equations into the plane equation to find the point where they meet.
Firstly, we substitute the values from the parametric equations \( x = 1 - t, y = 3t, z = 1 + t \) into the plane equation \( 2x - y + 3z = 6 \). This results in:

• \( 2(1 - t) - 3t + 3(1+t) = 6 \)

By simplifying this equation, we isolate \( t \) through arithmetic operations:

• Simplify to \( 5 - 2t = 6 \)
• Rearrange to solve for \( t \):
• Subtract 5 from both sides: \( -2t = 1 \)
• Divide by -2: \( t = -\frac{1}{2} \)

Finally, substituting \( t = -\frac{1}{2} \) back into the original parametric equations provides the coordinates of the intersection point \((1.5, -1.5, 0.5)\). This method is a straightforward application of algebra to determine where a line intersects a plane.