<p>Two parallel planes are given by \(x + y + z = 1\) and \(x + y + z = \dfrac{9}{2}\). A third plane that intersects them is given by \(2x - 5y + z = -5\), resulting in two parallel lines of intersection. If the distance \(d\) between these two parallel lines can be expressed as \(d = \sqrt{\dfrac{a}{b}}\), where \(a\) and \(b\) are co-prime positive integers, then find the value of \([d]\).</p><p>[Note: Where \([k]\) denotes greatest integer function less than or equal to \(k\).]</p>
Step-by-Step Solution
Key Concept: The distance between two parallel lines formed by intersecting a plane with two parallel planes equals the perpendicular distance from a point on one line to the other line, which can be found using the formula: d = |distance between parallel planes| × sin(θ), where θ is the angle between the intersecting plane and the parallel planes.
Step 1: Identify the normal vectors. The two parallel planes have normal n_1 = (1, 1, 1) and the intersecting plane has normal n_2 = (2, -5, 1). Step 2: Calculate the perpendicular distance between the two parallel planes: distance = |9/2 - 1|/√3 = (7/2)/√3 = 7√3/6. Step 3: Find cos(θ) where θ is angle between planes: n_1 · n_2 = 2 - 5 + 1 = -2. | n_1 | = √3, | n_2 | = √30. So cos(θ) = |-2|/(√3·√30) = 2/√90 = 2/(3√10) = 2√10/30 = √10/15. Step 4: The distance between the two parallel lines is: d = (perpendicular distance between planes) / |cos(θ)| = (7√3/6) / (√10/15) = (7√3/6) × (15/√10) = 105√3/(6√10) = 105√(3/10)/6 = 35√30/20 = 7√30/4. Step 5: Express as d = √(a/b): d = 7√30/4 = √(49·30/16) = √(1470/16) = √(735/8). Since gcd(735, 8) = 1, we have a = 735, b = 8. Step 6: Calculate d = √(735/8) ≈ √91.875 ≈ 9.586... ∴ Answer: [d] = 9
Correct Answer: 9