Finding the Equation of a Plane Through the Intersection of Two Planes and a Point

The problem asks for the equation of a plane that satisfies two conditions:

• It passes through the line formed by the intersection of two given planes.
• It passes through a specific point.

Understanding the Concept: Family of Planes

The equation of any plane passing through the line of intersection of two given planes \(P_1: A_1x + B_1y + C_1z + D_1 = 0\) and \(P_2: A_2x + B_2y + C_2z + D_2 = 0\) is given by the equation:

\(P_1 + \lambda P_2 = 0\)

or

\((A_1x + B_1y + C_1z + D_1) + \lambda (A_2x + B_2y + C_2z + D_2) = 0\)

where \(\lambda\) is a constant parameter. This represents a 'family' of planes that all contain the line of intersection of \(P_1\) and \(P_2\).

Applying the Concept to the Given Planes

The given planes are:

• Plane 1 (\(P_1\)): \(2x + y + 2z = 9\), which can be written as \(2x + y + 2z - 9 = 0\).
• Plane 2 (\(P_2\)): \(4x - 5y - 4z = 1\), which can be written as \(4x - 5y - 4z - 1 = 0\).

Using the formula for the family of planes, the equation of a plane passing through their intersection is:

\((2x + y + 2z - 9) + \lambda (4x - 5y - 4z - 1) = 0\) (Equation 1)

Using the Given Point to Find \(\lambda\)

We are given that the required plane also passes through the point \((3, 2, 1)\). Since this point lies on the plane represented by Equation 1, substituting the coordinates \(x=3\), \(y=2\), and \(z=1\) into Equation 1 must satisfy the equation.

Substitute \(x=3, y=2, z=1\) into Equation 1:

\((2(3) + 2 + 2(1) - 9) + \lambda (4(3) - 5(2) - 4(1) - 1) = 0\)

Now, let's simplify the terms inside the parentheses:

• First parenthesis: \(2(3) + 2 + 2(1) - 9 = 6 + 2 + 2 - 9 = 10 - 9 = 1\)
• Second parenthesis: \(4(3) - 5(2) - 4(1) - 1 = 12 - 10 - 4 - 1 = 2 - 4 - 1 = -2 - 1 = -3\)

Substitute these simplified values back into the equation:

\(1 + \lambda (-3) = 0\)

\(1 - 3\lambda = 0\)

\(3\lambda = 1\)

\(\lambda = \frac{1}{3}\)

Substituting \(\lambda\) Back into the Plane Equation

Now that we have the value of \(\lambda\), substitute \(\lambda = \frac{1}{3}\) back into Equation 1:

\((2x + y + 2z - 9) + \frac{1}{3} (4x - 5y - 4z - 1) = 0\)

Simplifying the Equation of the Plane

To get rid of the fraction, multiply the entire equation by 3:

\(3(2x + y + 2z - 9) + 1(4x - 5y - 4z - 1) = 0 \times 3\)

\((6x + 3y + 6z - 27) + (4x - 5y - 4z - 1) = 0\)

Combine like terms (terms with x, y, z, and constants):

• x terms: \(6x + 4x = 10x\)
• y terms: \(3y - 5y = -2y\)
• z terms: \(6z - 4z = 2z\)
• Constant terms: \(-27 - 1 = -28\)

So the equation becomes:

\(10x - 2y + 2z - 28 = 0\)

Rearranging the equation to match the options:

\(10x - 2y + 2z = 28\)

Comparing with Options

Let's compare our derived equation \(10x - 2y + 2z = 28\) with the given options:

• Option 1: \(10x - 2y + 2z = 28\)
• Option 2: \(10x + 2y + 2z = 28\)
• Option 3: \(10x + 2y - 2z = 28\)
• Option 4: \(10x - 2y - 2z = 24\)

Our derived equation matches Option 1.

Conclusion

The equation of the plane passing through the intersection of the given planes \(2x + y + 2z = 9\), \(4x - 5y - 4z = 1\) and the point \((3, 2, 1)\) is \(10x - 2y + 2z = 28\).

Revision Table: Key Concepts for Plane Equations

Additional Information: Planes in 3D Geometry

Planes are fundamental objects in three-dimensional geometry. They are flat, two-dimensional surfaces that extend infinitely. Understanding their equations and properties is crucial for solving problems involving lines and surfaces in space.

• Normal Vector: The coefficients A, B, C in the general equation \(Ax + By + Cz + D = 0\) represent the components of a vector \(\vec{n} = A\hat{i} + B\hat{j} + C\hat{k}\) which is normal (perpendicular) to the plane.
• Intercept Form: If a plane intersects the axes at \((a, 0, 0)\), \((0, b, 0)\), and \((0, 0, c)\) (where a, b, c are non-zero), its equation can be written as \(\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1\).
• Parallel and Perpendicular Planes:
  • Two planes \(A_1x + B_1y + C_1z + D_1 = 0\) and \(A_2x + B_2y + C_2z + D_2 = 0\) are parallel if their normal vectors are parallel, i.e., \(\frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2}\).
  • They are perpendicular if their normal vectors are orthogonal, i.e., \(A_1A_2 + B_1B_2 + C_1C_2 = 0\).
• Angle Between Two Planes: The angle between two planes is defined as the angle between their normal vectors. If \(\theta\) is the angle between the planes, then \(\cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}\).

The method used in this problem, \(P_1 + \lambda P_2 = 0\), is a powerful technique for finding the equation of a plane under certain conditions, especially when the plane is constrained to pass through the line of intersection of two other planes.