The line l₁ passes through the points A (2, 3, −5) and B (8, 7, −13). The line l₂ passes through the points C (-2, 1, 8) and D (3, −1, 4). Find the shortest distance between the lines l₁ and l₂. The plane Π₁ passes through the points A, B and D. The plane Π₂ passes though the points A, C and D. Find the acute angle between Π₁ and Π₂, giving your answer in degrees.
- The line l₁ passes through the points A (2, 3, −5) and B (8, 7, −13). The line l₂ passes through the points C (-2, 1, 8) and D (3, −1, 4). Find the shortest distance between the lines l₁ and l₂.[5]
- The plane Π₁ passes through the points A, B and D. The plane Π₂ passes though the points A, C and D. Find the acute angle between Π₁ and Π₂, giving your answer in degrees.[6]
Mark scheme
Marking points from the official mark scheme (May/June 2014). The mark or mark type is shown in brackets.
- [B1] AB = (6, 4, -8)^T, CD = (5, -2, -4)^T
- [M1A1] Common perpendicular = AB × CD = (-32, -16, -32)^T (or scalar multiple, e.g., (2, 1, 2)^T)
- [M1A1] AD = (1, -4, 9)^T. Shortest distance = | AD . (AB × CD) / |AB × CD| | = |(1, -4, 9) . (2, 1, 2)| / |(2, 1, 2)| = |2 - 4 + 18| / √(4+1+4) = 16 / 3 (or 5.33)
- [M1A1] Normal to Π₁ is AB × AD. AB = (6, 4, -8)^T, AD = (1, -4, 9)^T. Normal = (4, -62, -28)^T (or scalar multiple, e.g., (2, -31, -14)^T)
- [A1] (2, -31, -14)^T
- [M1A1] Normal to Π₂ is AC × AD. AC = (-4, -2, 13)^T, AD = (1, -4, 9)^T. Normal = (34, 49, 18)^T (or scalar multiple)
- [A1] (34, 49, 18)^T
- [M1A1] cos θ = | n₁ . n₂ / (|n₁| |n₂|) | = | (2)(34) + (-31)(49) + (-14)(18) | / (√(2² + (-31)² + (-14)²) * √(34² + 49² + 18²))
- [A1] ⇒ θ = 36.7° (CAO)
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