Problem 69 - Advanced Engineering Mathematics

Question

In Problems, find the point of intersection of the given plane and line. $$ 2 x-3 y+2 z=-7 ; x=1+2 t, y=2-t, z=-3 t $$

Step-by-Step Solution

Answer

The point of intersection is (-5, 5, 9).

Step 1: Understand the Problem

We are given a plane with the equation \(2x - 3y + 2z = -7\) and a line with parametric equations \(x = 1 + 2t, y = 2 - t, z = -3t\). Our goal is to find the point \((x, y, z)\) where the line intersects the plane.

Step 2: Substitute Parametric Equations into Plane Equation

Substitute the parametric equations of the line into the equation of the plane: \[2(1 + 2t) - 3(2 - t) + 2(-3t) = -7\].

Step 3: Simplify the Equation

Simplify the left-hand side of the equation: \[2 + 4t - 6 + 3t - 6t = -7\]. Combine like terms to get: \[4t + 3t - 6t + 2 - 6 = -7\].

Step 4: Further Simplification

Further simplify the equation by combining like terms on the left: \[t - 4 = -7\].

Step 5: Solve for t

Solve the equation for \(t\): Add 4 to both sides to get \[t = -3\].

Step 6: Find the Intersection Point

With \(t = -3\), substitute back into the parametric equations of the line: \[x = 1 + 2(-3) = -5\], \[y = 2 - (-3) = 5\], \[z = -3(-3) = 9\].

Step 7: Conclusion

The point of intersection of the line and the plane is \((-5, 5, 9)\).

Key Concepts

Parametric Equations

Parametric equations are a powerful tool in mathematics, particularly when dealing with lines in three-dimensional space. These equations express the coordinates of the points on the line as functions of a parameter, typically denoted as \( t \).
In the given example, the parametric equations for the line are:

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

These equations tell us that the x-coordinate changes by increments of 2 as \( t \) increases. At the same time, the y-coordinate decreases by 1, and the z-coordinate is scaled by -3. This representation is handy because it provides a complete description of all the points on a line, merely by varying the parameter \( t \). This becomes particularly useful when we are tasked with finding where such a line might intersect with a plane.

Plane Equation

The equation of a plane is a fundamental concept in geometry, representing a flat, two-dimensional surface that extends infinitely in three-dimensional space. In this exercise, we have a plane described by the equation: \[ 2x - 3y + 2z = -7 \].
This equation is in a standard linear form \( Ax + By + Cz = D \), where \( A = 2 \), \( B = -3 \), \( C = 2 \), and \( D = -7 \). These coefficients define the orientation and position of the plane in space.

• \(A,B,C\) represent the components of a vector normal to the plane, which is crucial because it indicates the direction perpendicular to the plane.
• The constant \(D\) provides specific positioning of the plane, offsetting it by the value \(-7\) from the origin based on the plane's orientation vector.

In problems involving the intersection of lines and planes, we often utilize these concepts of the plane equation to determine the relationship between the line and the plane.

Algebraic Manipulation

Algebraic manipulation is the process of rearranging and simplifying equations to solve for unknown variables. In our exercise, this involves substituting the parametric equations into the plane equation.
The given substitution is: \[ 2(1 + 2t) - 3(2 - t) + 2(-3t) = -7 \] In this step, we distribute the multiplication and simplify:

• \(2 + 4t - 6 + 3t - 6t = -7\)
• Combine like terms: \(4t + 3t - 6t + 2 - 6 = -7\)
• Simplify to: \(t - 4 = -7\)

These techniques are often used to isolate one variable and are essential for finding the specific value of \( t \) that will give us the point of intersection. Simplifying expressions can make a seemingly complex problem much more manageable.

Intersection Point Calculation

Finding the intersection point is a key step in understanding how a line meets a plane. Once we have simplified our equation to isolate the parameter \( t \), we found that \( t = -3 \).
We can then calculate the exact coordinates for the intersection point. By substituting \( t = -3 \) back into the parametric equations of the line:

• For x: \( x = 1 + 2(-3) = -5 \)
• For y: \( y = 2 - (-3) = 5 \)
• For z: \( z = -3(-3) = 9 \)

Thus, the point of intersection of the line and the plane is \((-5, 5, 9)\). Calculating the intersection point helps visualize the precise location where the line 'pierces' the plane. This intersection point is crucial in applications like computer graphics, engineering, and physics.