<p><strong>83.</strong> We have three planes<br>\(P_1: 2x - y + 2z + 3 = 0\)<br>\(P_2: 2x - y + 2z + \lambda/2 = 0\)<br>\(P_3: 2x - y + 2z + \mu = 0\)<br>Given that the distance between \(P_1\) and \(P_2\) is \(\frac{1}{3}\) and the distance between \(P_1\) and \(P_3\) is \(\frac{2}{3}\). Find \(\lambda_{\max} + \mu_{\max}\).</p>
Step-by-Step Solution
Key Concept: For parallel planes with equations ax + by + cz + d₁ = 0 and ax + by + cz + d₂ = 0, the distance between them is |d₁ - d₂|/√(a² + b² + c²). Use this formula to find λ and μ from given distances, then identify maximum values considering both possible sign combinations.
Step 1: Identify that all three planes are parallel (same normal vector coefficients 2, -1, 2). Step 2: Apply distance formula between parallel planes. For P_1 and P_2: Distance = |3 - λ/2|/√(4 + 1 + 4) = |3 - λ/2|/3 = 1/3 Therefore: |3 - λ/2| = 1 This gives: 3 - λ/2 = ±1 So: λ/2 = 2 or λ/2 = 4 Thus: λ = 4 or λ = 8 Step 3: Apply distance formula between P_1 and P_3: Distance = |3 - μ|/3 = 2/3 Therefore: |3 - μ| = 2 This gives: 3 - μ = ±2 So: μ = 1 or μ = 5 Step 4: Find maximum values: λ_max = 8 (larger of {4, 8}) μ_max = 5 (larger of {1, 5}) ∴ λ_max + μ_max = 8 + 5 = 13
Correct Answer: 13