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: Understand the problem. We are asked to find the equation of a plane passing through the point \( (1, 1, 1) \) and perpendicular to the planes \( 2x + y - 2z = 5 \) and \( 3x - 6y - 2z = 7 \). 

Step 2: Find the normal vectors. The normal vectors of the given planes are: - \( \langle 2, 1, -2 \rangle \) for the plane \( 2x + y - 2z = 5 \) - \( \langle 3, -6, -2 \rangle \) for the plane \( 3x - 6y - 2z = 7 \) 

Step 3: Find the cross product of the normal vectors. The normal vector to the required plane is the cross product of the normal vectors of the given planes: \[ \langle 2, 1, -2 \rangle \times \langle 3, -6, -2 \rangle = \langle 14, 2, 15 \rangle. \] 

Step 4: Equation of the plane. The equation of the plane is: \[ 14(x - 1) + 2(y - 1) + 15(z - 1) = 0 \] Simplifying, we get: \[ 14x + 2y + 15z = 31. \] 

Step 5: Conclusion. Thus, the equation of the plane is \( 14x + 2y + 15z = 31 \).