Problem 72 - Advanced Engineering Mathematics

Question

In Problems, find the point of intersection of the given plane and line. $$ x-3 y+2 z=0 ; x=4+t, y=2+t, z=1+5 t $$

Step-by-Step Solution

Answer

The point of intersection is (4, 2, 1).

Step 1: Understanding the Plane Equation

The equation of the plane given is \( x - 3y + 2z = 0 \). This equation represents a flat surface in three-dimensional space.

Step 2: Understanding the Line Equation

The parametric equations of the line are given as \( x = 4 + t \), \( y = 2 + t \), and \( z = 1 + 5t \). Here, \( t \) is the parameter that describes points on the line.

Step 3: Substitute Line Equations into Plane Equation

To find the intersection, substitute the expressions for \( x, y, \) and \( z \) from the line into the plane equation: \( (4+t) - 3(2+t) + 2(1+5t) = 0 \).

Step 4: Simplify Substitution Equation

Simplifying the equation: \( 4 + t - 6 - 3t + 2 + 10t = 0 \), we combine like terms to get \( 8t = 0 \).

Step 5: Solve for the Parameter t

From \( 8t = 0 \), solving for \( t \) gives \( t = 0 \).

Step 6: Calculate Intersection Point

Use \( t = 0 \) in the parametric equations to find the point of intersection. Substitute \( t = 0 \) into \( x = 4 + t \), \( y = 2 + t \), and \( z = 1 + 5t \) to get \( x = 4 \), \( y = 2 \), \( z = 1 \).

Key Concepts

Plane Equation

The equation of a plane in 3D space can be represented as \( ax + by + cz = d \). In this formula, \( a, b, \) and \( c \) are the coefficients that define the orientation of the plane, whereas \( x, y, \) and \( z \) are the coordinates of any point on the plane.
A plane is essentially a flat, two-dimensional surface that extends infinitely in three-dimensional space.
To understand a plane equation fully, let's consider the example from the exercise, which is \( x - 3y + 2z = 0 \). Here:

• \( a = 1 \)
• \( b = -3 \)
• \( c = 2 \)
• \( d = 0 \)

This specific equation implies that the plane passes through the origin (0,0,0) because the constant term \( d \) is zero. With the coefficients, we understand the plane's tilt in space, determining its orientation.

Parametric Line Equation

A parametric line equation describes a line through expressions that relate coordinates to a parameter, commonly \( t \).
This kind of equation makes it easy to find different points on a line by simply varying the parameter \( t \).
For example, a line in our problem is described by the parametric equations: \( x = 4 + t \), \( y = 2 + t \), and \( z = 1 + 5t \).
Here's what these equations mean:

• \( x = 4 + t \) indicates the line moves along the x-axis starting at point 4, advancing by 1 unit per each increment of \( t \).
• \( y = 2 + t \) tells us it starts at point 2 on the y-axis and moves similarly.
• \( z = 1 + 5t \) informs that it starts at point 1 on the z-axis and moves more steeply, as indicated by the multiplier 5 on \( t \).

Parametric equations are potent tools for depicting trajectories and motions within space.

Intersection Point Calculation

To find the intersection point of a plane and a line, a key technique is substituting the line's parametric equations into the plane equation.
This substitution enables us to calculate the specific parameter value where the line penetrates the plane.
In our problem, after substituting the line equations \( x = 4 + t \), \( y = 2 + t \), \( z = 1 + 5t \) into the plane equation \( x - 3y + 2z = 0 \), we arrive at the equation \( (4+t) - 3(2+t) + 2(1+5t) = 0 \).
Simplifying it gives \( 8t = 0 \). This tells us that the solution is \( t = 0 \).
To discover the exact coordinates of the intersection point, substitute \( t = 0 \) back into the parametric equations of the line:

• For \( x \), \( x = 4 + 0 = 4 \)
• For \( y \), \( y = 2 + 0 = 2 \)
• For \( z \), \( z = 1 + 5 \times 0 = 1 \)

Thus, the point of intersection is \( (4, 2, 1) \). Through this systematic substitution, we locate how the line intersects the plane in 3D space.