OR The lines l₁ and l₂ have equations r = 8i + 2j + 3k + λ(i – 2j) and r = 5i + 3j – 14k + μ(2j – 3k) respectively. The point P on l₁ and the point Q on l₂ are such that PQ is perpendicular to both l₁ and l₂. Find the position vector of the point P and the position vector of the point Q. The points with position vectors 8i + 2j + 3k and 5i + 3j – 14k are denoted by A and B respectively. Find

Mark scheme

Marking points from the official mark scheme (May/June 2015). The mark or mark type is shown in brackets.

1. [M1A1] p = [[8+λ],[2-2λ],[3]], q = [[5],[3+2μ],[-14-3μ]]
2. [M1A1] QP = [[3+λ],[-1-2λ-2μ],[17+3μ]]
3. [M1A1] QP ⋅ [[1],[-2],[0]] = 0 ⇒ 5λ + 4μ = -5
4. [A1] QP ⋅ [[0],[2],[-3]] = 0 ⇒ -4λ - 13μ = 53
5. [M1A1] Solving : λ = 3, μ = -5
6. [A1] Whence: p = [[11],[-4],[3]], q = [[5],[-7],[1]]
7. [M1A1] Alternative for marks 3,4,5,6 and 7 in part (i): i j k / 1 -2 0 / 0 2 -3 = [[6],[3],[2]]
8. [A1] 6k – λ = 3 / 3k + 2λ + 2 = -1 / 2k – 3μ = 17
9. [M1A1] Solving; k = 1, λ = 3 and μ = -5 (If k missing, or assumed to be 1, deduct 1 mark.)
10. [B1] (i) AP = [[3],[-3],[0]], AQ = [[-3],[-9],[-2]]
11. [M1A1] AP × AQ = [[6],[12],[-45]] (CAO)
12. [A1] Area = 1/2 √2205 (=23.5)
13. [M1A1] (ii) AB = [[-3],[1],[-17]]
14. [M1A1] Volume = 1/3 × 1/2 √2205 ⋅ |[[6],[12],[-45]] ⋅ [[-3],[1],[-17]]| / √2205 = 735/6 (= 122.5)