The lines l and m have equations l: r = ai + 3j + bk + λ(ci – 2j + 4k), m: r = i +2j + 3k + μ(2i – 3j + k). Relative to the origin O, the position vector of the point P is 4i + 7j – 2k.

Mark scheme

Marking points from the official mark scheme (May/June 2023). The mark or mark type is shown in brackets.

1. [M1] Perform scalar product of direction vectors and set result equal to zero; 2c+6+4=0.
2. [M1] Use P to find the value of λ; 3-2λ = 7 ⇒ λ = −2 [a + λ c = 4, b + 4 λ = − 2]. Equation for line l may contain – instead of + leading to λ = 2 all marks available.
3. [A1] Obtain c = -5 or b = 6
4. [A1] a = -6, b = 6 and c = -5 all correct; SC1: Use P to find the value of λ M1 Substitute λ = −2 into point P, so a − 2c = 4, and put μ = - 1 and λ = - 1 into l so a - c = - 1, then solve to obtain a=-6, b=6 and c = -5. All 3 values correct A1. Max 2/4.
5. [B1] Find PQ (or QP) for a general point Q on m; = ± ((1+ 2 μ)i + (2-3μ)j + (3+ μ)k) – (4i + 7j – 2k) = (-3+2μ)i + (-5-3μ)j + (5+ μ)k; Could be their a, b, c and λ values provided M1 M1 gained in (a). Allow expression in answer column.
6. [M1*] Equate the scalar product of PQ (or QP) and a direction vector for m to zero and obtain an equation in μ; (2(-3+2μ)-3(-5-3μ)+(5+μ)) = 0. Allow PQ = OQ + OP sign problem.
7. [A1] Solve and obtain μ=-1; PQ² = (−3+2μ)² + (−5−3μ)² + (5 + μ)². [= 14( μ + 1)² + 45]. Min when μ = - 1 or by differentiation.
8. [A1] Obtain OQ = −i+5j+2k or PQ= −5i – 2j + 4k; Must be labelled correctly; The working may be in (a) provided at least this result is used in (b).
9. [DM1] Carry out a method to find the position vector of R; e.g. Use OR = OP + (5/2)PQ or OR = OQ + (3/2)PQ or OR = (5/2)OQ - (3/2)OP or 2QR = 2(OR-OQ) = 3 PQ where OR = (x, y, z). PQ used in all these approaches, may be incorrect, must be in the correct direction, i.e. not using QP for PQ. Alternative method for DM1: OR = (4, 7, - 2) + t (− 5, − 2, 4) QR = OR – OQ Solve |QR|² = (9/4)|PQ|² or |QR| = (3/2)|PQ| t = 2.5
10. [A1] Obtain (17/2)i+2j+8k from correct working; Accept coordinates. Don't accept (17/2)i + (4/2)j + (16/2)k. SC2 Equate lines, attempt to find μ = −1 or λ = -1 M1* OQ = −i+5j+2k A1. Attempt to find OQ using other parameter value DM1. OQ=−i+5j+2k therefore intersect A1. Then use main scheme for the final DM1 A1. First DM1 A1 are available if they show the 3 coordinates are consistent for the 2 parameter values instead of attempting to find OQ using the other parameter value and then showing intersection