In this article, we will explore the process of finding the equation of a plane that passes through a specific point and is perpendicular to two given planes. This involves understanding the role of normal vectors, the cross product, and how to write the equation of a plane using these concepts. Follow the detailed steps and calculations to derive the final equation.

Introduction to the Problem

We need to determine the equation of a plane that passes through the point (1, -2, 1) and is perpendicular to the planes given by the equations:

The goal is to understand the geometric and algebraic methods to solve such a problem.

Step 1: Finding the Normal Vectors of the Given Planes

A plane in 3D space can be represented by the general equation:

Ax By Cz D

The normal vector of the plane is given by the coefficients of x, y, and z. Therefore, for the first plane:

2X - 2Y Z 0

The normal vector is:

n1 [2, -2, 1]

Similarly, for the second plane:

X - Y 2Z 4

The normal vector is:

n2 [1, -1, 2]

Step 2: Finding the Normal Vector of the New Plane

The new plane we are interested in must be perpendicular to both given planes. This means that the normal vector of the new plane n is perpendicular to both n1 and n2. We can find n using the cross product:

n n1 × n2

The cross product can be computed as:

n1 × n2

| i j k|
|2 -2 1 |
|1 -1 2 |

This results in:

n1 × n2 i ( -2 * 2 - 1 * -1 ) - j ( 2 * 2 - 1 * 1 ) k ( 2 * -1 - 1 * -2 )

Solving the 2x2 determinants:

1. | -2 * 2 - 1 * -1 | -4 - 1 -3

2. | 2 * 2 - 1 * 1 | 4 - 1 3

3. | 2 * -1 - 1 * -2 | -2 2 0

So, the normal vector n is:

n [ -3, -3, 0 ]

Step 3: Writing the Equation of the New Plane

The general equation of a plane with normal vector [a, b, c] passing through the point (x0, y0, z0) is given by:

Ax - x0 By - y0 Cz - z0 0

Here, the normal vector [a, b, c] is [ -3, -3, 0 ] and the point (x0, y0, z0) is (1, -2, 1).

The equation becomes:

-3x - 1 - 3y - 2 0z - 1 0

Expanding this, we get:

-3x 3 - 3y - 6 0

-3x - 3y - 3 0

x y 1 0

The final equation of the plane is:

x y 1 0

Geometric and Algebraic Insight

Understanding the problem geometrically, we can represent the normal vectors as:

For the first plane: P 2i - 2j k

For the second plane: Q i - j 2k

A vector R perpendicular to both P and Q can be found using the cross product:

Using the determinant formula:

|i j k|
|2 -2 1|
|1 -1 2|

This results in:

R -3i - 3j 0k [-3, -3, 0]

The point S is (1, -2, 1) and T is (x, y, z).

The vector ST is:

ST [x - 1, y 2, z - 1]

Since ST is perpendicular to R, the dot product is zero:

[-3, -3, 0] ? [x - 1, y 2, z - 1] 0

Solving the equation:

-3(x - 1) - 3(y 2) 0(z - 1) 0

Simplifying:

-3x 3 - 3y - 6 0

-3x - 3y - 3 0

x y 1 0

Hence, the equation of the plane is:

boxed{x y 1 0}

This method provides a detailed step-by-step solution to a complex geometric problem, demonstrating the interplay between normal vectors, cross products, and plane equations.