The plane Π has equation r = 2i + 3j – k + λ(i – 2j + 2k) + μ(3i + j – 2k). The line l, which does not lie in Π, has equation r = 3i + 6j + 12k + t(8i + 5j – 8k).
- Show that l is parallel to Π.[4]
- Find the position vector of the point at which the line with equation r = 5i – 4j + 7k + s(2i – j + k) meets Π.[4]
- Find the perpendicular distance from the point with position vector 9i + 11j + 2k to l.[4]
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2012). The mark or mark type is shown in brackets.
- [M1A1] Finds vector normal to Π: n = (i – 2j + 2k) × (3i + j – 2k) = 2i + 8j + 7k
- [A1] Dot product of this with general point on l: (8i + 5j – 8k) ⋅ (2i + 8j + 7k) = 16 + 40 - 56 = 0
- [A1] Deduce result: Independent of t ⇒ parallel, or ⇒ parallel.
- [B1] Cartesian equation of Π: 2x + 8y + 7z = 21
- [M1A1] Substitutes general point of line 2 into Cartesian equation: 2(5+2s) + 8(-4-s) + 7(7+s) = 21 ⇒ s = -2
- [A1] Finds p.v. of intersection: i – 2j + 5k
- [B1] Finds distance from point to known point on l: Take (9,11,2) as A, (3,6,12) as B and let C be foot of perpendicular from A to l. AB = √(6² + 5² + 10²) = √161
- [M1] Finds distance along l from known point to foot of perpendicular from given point to l: BC = |(6i + 5j –10k)⋅(8i + 5j – 8k)| / √(8² + 5² + 8²) = 153 / √153 = √153
- [A1] AC = √(AB² – BC²) = √(161 – 153) = √8 or 2√2 (= 2.83)
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