3D Geometry
Distance between parallel planes
Grade 12
Question:
<p>The distance between the parallel planes \(2x + y + 2z = 8\) and \(2x + y + 2z = -\frac{5}{2}\) is:</p>
<p>\(\frac{5}{2}\) units</p>
<p>\(\frac{7}{2}\) units</p>
<p>\(\frac{21}{2}\) units</p>
<p>\(7\) units</p>
Step-by-Step Solution
Key Concept: For parallel planes ax + by + cz = d₁ and ax + by + cz = d₂, the distance formula is |d₁ - d₂|/√(a² + b² + c²), which directly gives the perpendicular distance between them.
Step 1: Verify planes are parallel. Both planes have the form 2x + y + 2z = d, so they are parallel with normal vector n = (2, 1, 2). Step 2: Apply the distance formula for parallel planes: d = |d_1 - d_2|/√(a^2 + b^2 + c^2) Step 3: Calculate the numerator: |8 - (-5/2)| = |8 + 5/2| = |16/2 + 5/2| = |21/2| = 21/2 Step 4: Calculate the denominator: √(2^2 + 1^2 + 2^2) = √(4 + 1 + 4) = √9 = 3 Step 5: Compute the distance: d = (21/2)/3 = 21/(2·3) = 21/6 = 7/2 ∴ Answer: B (distance = 7/2)
Correct Answer: B