Problem 58 - University Calculus: Early Transcendentals

Question

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 meets the plane at the point \((2, -2.857, 3.857)\).

Step 1: Express the Line in Parametric Form

We are given the parametric equations for the line as: \( x = 2 \), \( y = 3 + 2t \), and \( z = -2 - 2t \). This means any point \((x, y, z)\) on the line can be represented using the parameter \( t \).

Step 2: Substitute into the Plane Equation

The plane equation is given by \( 6x + 3y - 4z = -12 \). Substitute \( x = 2 \), \( y = 3 + 2t \), and \( z = -2 - 2t \) into the plane equation:\[ 6(2) + 3(3 + 2t) - 4(-2 - 2t) = -12 \]

Step 3: Simplify

Simplify the equation:\[ 12 + 9 + 6t + 8 + 8t = -12 \]Combine like terms:\[ 29 + 14t = -12 \]

Step 4: Solve for the Parameter t

Solve \( 29 + 14t = -12 \) for \( t \):\[ 14t = -12 - 29 \]\[ 14t = -41 \]\[ t = -\frac{41}{14} \]

Step 5: Find the Intersection Point

Substitute \( t = -\frac{41}{14} \) back into the parametric equations to find the intersection point:For \( y \):\[ y = 3 + 2\left(-\frac{41}{14}\right) = 3 - \frac{82}{14} = 3 - 5.857 \approx -2.857 \]For \( z \):\[ z = -2 - 2\left(-\frac{41}{14}\right) = -2 + \frac{82}{14} = -2 + 5.857 \approx 3.857 \]The intersection point is \((2, -2.857, 3.857)\).

Step 6: Verify the Point Lies on the Plane

To verify, substitute \( x = 2 \), \( y = -2.857 \), and \( z = 3.857 \) back into the plane equation:\[ 6(2) + 3(-2.857) - 4(3.857) \approx -12 \]Since this holds true, the calculation is verified.

Key Concepts

Parametric Equations

Parametric equations are a way to express the coordinates of the points that form a geometric object, such as a line, in terms of a parameter. Imagine this like a slider that, when moved, traces out the entire line. In our example, the line is represented by:

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

These equations indicate that as we change the value of \(t\), we move along the line, obtaining different points \((x, y, z)\) that belong to it.
For the line given, \(x\) is constant at \(2\), while \(y\) and \(z\) change based on the parameter \(t\). This helps us track exactly which part of the line we're interested in, especially important when intersecting with other geometric objects like planes.

Plane Equation

A plane equation represents a flat, two-dimensional surface that extends infinitely in three-dimensional space. The equation given is:

• \(6x + 3y - 4z = -12\)

Each plane is defined by a linear equation in the form \(ax + by + cz = d\), where \(a\), \(b\), \(c\), and \(d\) are constants. The coefficients \(a\), \(b\), and \(c\) determine the orientation of the plane, while \(d\) adjusts its position along the normal vector.
In this specific equation, substituting the parametric equations into the plane equation helps us find out whether and where a given line intersects the plane. By solving for \(t\), we can determine the exact point on the line that lies within the plane.

Solve for Parameter

Solving for the parameter \(t\) involves substituting the parametric equations into the plane equation and simplifying. Let's break this down.
Start by substituting \(x = 2\), \(y = 3 + 2t\), and \(z = -2 - 2t\) into the plane equation \(6x + 3y - 4z = -12\). This substitution helps us transition from three variables to a single variable problem where we only deal with \(t\).
Here is how it simplifies:

• First, plug in the values:\[6(2) + 3(3 + 2t) - 4(-2 - 2t) = -12\]
• Simplify to get:\[12 + 9 + 6t + 8 + 8t = -12\]
• Combine like terms and solve for \(t\):\[29 + 14t = -12\]\[14t = -12 - 29\]\[14t = -41\]\[t = -\frac{41}{14}\]

This value of \(t\) specifies the point on the line that also resides on the plane. By inserting this \(t\) back into the parametric equations, you find the coordinates of the point of intersection.