Answer only one of the following two alternatives. OR The position vectors of the points A, B, C, D are 2i + 4j – 3k, –2i + 5j – 4k, i + 4j + k, i + 5j + mk, respectively, where m is an integer. It is given that the shortest distance between the line through A and B and the line through C and D is 3. Show that the only possible value of m is 2. Find the shortest distance of D from the line through A and C. Show that the acute angle between the planes ACD and BCD is cos⁻¹ (1/√3).

Mark scheme

Marking points from the official mark scheme (May/June 2012). The mark or mark type is shown in brackets.

1. [M1A1] Obtains direction of common perpendicular. n = i j k | 4 -1 1 | 0 1 m-1 | = -mi + 4(1–m)j + 4k
2. [M1A1] Uses result for shortest distance between lines. CA = [1,0,4] Distance = |(CA ⋅ n)| / |n| = 3
3. [A1] Solves equation. 19m² – 40m + 4 = 0
4. [M1] (19m-2)(m-2) = 0
5. [A1] m = 2, since m is an integer. (AG)
6. [B1] Finds relevant vectors. CA = A-C = <1,0,-4> and AD = D-A = <-1,1,m+3>
7. [M1] Use of cross-product. |CA x AD| / |CA|
8. [A1] Obtains shortest distance. √18 / √17 = √(18/17) (= 1.03)
9. [B1] Finds 2ⁿᵈ vector in BCD (CD may already have been found.) BC = 3i - j +5k
10. [M1] Finds normal vector to BCD. (Normal to ACD already found.) n_BCD = BC x CD = <-6,-3,3> (~2i+j-k)
11. [M1] Finds angle between planes = angle between normal vectors. cos θ = | (4i-j+k) ⋅ (-6i-3j+3k) | / (|4i-j+k| | -6i-3j+3k|)
12. [A1] = 18 / (√18 √54) = 1/√3 (AG)