<p>Distance between two parallel planes \(2x + y + 2z = 8\) and \(4x + 2y + 4z + 5 = 0\) is</p>
Step-by-Step Solution
Key Concept: Convert both planes to the same normal vector form, then use the formula: distance = |d₁ - d₂|/√(a² + b² + c²) for parallel planes ax + by + cz = d₁ and ax + by + cz = d₂.
Step 1: Write both plane equations with identical normal vectors. Plane 1: 2x + y + 2z = 8 Plane 2: 4x + 2y + 4z + 5 = 0 → Divide by 2: 2x + y + 2z = -5/2 Step 2: Verify planes are parallel by checking normal vectors are proportional. Both planes have normal vector (2, 1, 2) ✓ Step 3: Apply distance formula for parallel planes: d = |d_1 - d_2|/√(a^2 + b^2 + c^2) d = |8 - (-5/2)|/√(2^2 + 1^2 + 2^2) d = |8 + 5/2|/√(4 + 1 + 4) d = |16/2 + 5/2|/√9 d = |21/2|/3 d = 21/(2 × 3) = 21/6 = 7/2 ∴ Answer: C (7/2 or 3.5)
Correct Answer: C