Question

The equation of the plane passing through the point \( (1, 1, 1) \) and perpendicular to the planes \( 2x + y - 2z = 5 \) and \( 3x - 6y - 2z = 7 \) is:

• A. \( 14x + 2y - 15z = 1 \)
• B. \( -14x + 2y + 15z = 3 \)
• C. \( 14x - 2y + 15z = 27 \)
• D. \( 14x + 2y + 15z = 31 \)

Hint:

The normal vector to a plane is perpendicular to every vector lying within the plane. The cross product of two normal vectors gives a vector perpendicular to both.

Answer 1

The Correct Option is D

Step 1: Problem Definition: Determine the equation of a plane that passes through the point \( (1, 1, 1) \) and is orthogonal to the planes \( 2x + y - 2z = 5 \) and \( 3x - 6y - 2z = 7 \).

Step 2: Identify Normal Vectors: The normal vectors for the provided planes are: - \( \langle 2, 1, -2 \rangle \) for \( 2x + y - 2z = 5 \). - \( \langle 3, -6, -2 \rangle \) for \( 3x - 6y - 2z = 7 \).

Step 3: Compute Cross Product: The normal vector for the desired plane is the cross product of the identified normal vectors: \[ \langle 2, 1, -2 \rangle \times \langle 3, -6, -2 \rangle = \langle 14, 2, 15 \rangle. \]

Step 4: Formulate Plane Equation: The equation of the plane is derived using the point-normal form: \[ 14(x - 1) + 2(y - 1) + 15(z - 1) = 0 \]. Simplified, this yields: \[ 14x + 2y + 15z = 31 \].

Step 5: Final Result: The equation of the plane is \( 14x + 2y + 15z = 31 \).