Problem 54 - Thomas Calculus

Question

In Exercises \(53-56,\) find the point in which the line meets the plane. $$ x=2, \quad y=3+2 t, \quad z=-2-2 t ; \quad 6 x+3 y-4 z=-12 $$

Step-by-Step Solution

Answer

The line intersects the plane at the point \((2, -\frac{20}{7}, \frac{27}{7})\).

Step 1: Understand the Problem

We need to find the intersection point of a line and a plane. The line is given in parametric form: \(x = 2\), \(y = 3 + 2t\), \(z = -2 - 2t\). The plane is given by the equation \(6x + 3y - 4z = -12\).

Step 2: Substitute the Line Equations into the Plane Equation

To find the intersection, substitute the parametric line equations into the plane equation. For the given plane equation \(6x + 3y - 4z = -12\), substitute: \(x = 2\), \(y = 3 + 2t\), and \(z = -2 - 2t\).

Step 3: Simplify and Solve for t

Substitute the values into the plane equation:\[6(2) + 3(3 + 2t) - 4(-2 - 2t) = -12\]This simplifies to:\[12 + 9 + 6t + 8 + 8t = -12\]Combine like terms to get:\[29 + 14t = -12\]Solve for \(t\):\[14t = -12 - 29 \14t = -41 \t = -\frac{41}{14}\]

Step 4: Find the Intersection Point

Substitute \(t = -\frac{41}{14}\) back into the parametric equations to find the point coordinates:\(x = 2\)\(y = 3 + 2(-\frac{41}{14}) = 3 - \frac{82}{14} = 3 - \frac{41}{7} = \frac{21}{7} - \frac{41}{7} = -\frac{20}{7}\)\(z = -2 - 2(-\frac{41}{14}) = -2 + \frac{82}{14} = -2 + \frac{41}{7} = -\frac{14}{7} + \frac{41}{7} = \frac{27}{7}\)

Step 5: Write the Final Answer

The point of intersection where the line meets the plane is:\(\begin{pmatrix}2, -\frac{20}{7}, \frac{27}{7}\end{pmatrix}\)

Key Concepts

Parametric Equations

Parametric equations are a way to express geometric figures, like lines and curves, using parameters. In the context of lines, parametric equations describe every point on the line as a function of one or two parameters, often called \( t \). This parameter can represent different values, often ranging over the real numbers, which generate coordinates for points along the line.
For the line described in this exercise, the parametric equations are:\[ x = 2, \quad y = 3 + 2t, \quad z = -2 - 2t \]
Each of these equations tells us how any point on the line corresponds to a specific value of \( t \).

• The \( x \)-coordinate is constant, \( x = 2 \), meaning every point on this line has the same \( x \) value.
• For the \( y \)-coordinate, as \( t \) increases, \( y \) increases since \( y = 3 + 2t \).
• For the \( z \)-coordinate, as \( t \) increases, \( z \) decreases because \( z = -2 - 2t \).

This means the line moves in space along the \( y \) and \( z \) directions as \( t \) changes.

Plane Equation

A plane equation defines a flat surface in three-dimensional space using a linear equation involving \( x \), \( y \), and \( z \). The general form of a plane equation is \( ax + by + cz = d \), where \( a \), \( b \), \( c \), and \( d \) are constants. These constants determine the plane's orientation and position in space.
For the problem at hand, the given plane equation is:
\[ 6x + 3y - 4z = -12 \]
This equation means that any point \((x, y, z)\) on this plane will satisfy the equation. Here:

• The coefficient \( 6 \) is associated with \( x \), affecting how changes in \( x \) impact the plane position.
• The \( 3 \) coefficient is for \( y \), indicating how \( y \) variations alter the equation's balance.
• \( -4 \) is the coefficient for \( z \), suggesting the change opposite to \( y \) and \( x \).

The negative \(-12\) on the right-hand side shifts the plane within the coordinate system.

Solving Linear Equations

Solving linear equations is crucial for determining specific variable values that satisfy given mathematical conditions. In the context of the line meeting the plane, we substitute the line’s parametric equations into the plane equation to find the intersection point. The intersection occurs where the values of \( t \) satisfy both the plane and line equations simultaneously.
Starting with our substituted equation:
\[ 6(2) + 3(3 + 2t) - 4(-2 - 2t) = -12 \]
You simplify the equation by distributing and combining like terms:
\[ 12 + 9 + 6t + 8 + 8t = -12 \]
This reduces to:
\[ 29 + 14t = -12 \]
To find \( t \), solve the equation:
\[ 14t = -12 - 29 \ 14t = -41 \ t = -\frac{41}{14} \]
The value \( t = -\frac{41}{14} \) is then used to find the \((x, y, z)\) coordinates of the intersection point by substituting it back into each parametric equation. This process demonstrates how linear equation solving helps us find critical values that define intersections in geometry.