In the diagram, OABCD is a solid figure in which OA = OB = 4 units and OD = 3 units. The edge OD is vertical, DC is parallel to OB and DC = 1 unit. The base, OAB, is horizontal and angle AOB = 90°. Unit vectors i, j and k are parallel to OA, OB and OD respectively. The midpoint of AB is M and the point N on BC is such that CN = 2NB. [Figure 11.1]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2022). The mark or mark type is shown in brackets.

1. [B1] (a) State OM = 2i + 2j or equivalent
2. [B1] (a) Obtain MD = – 2i – 2j + 3k
3. [M1] (a) Use a correct method to find ON
4. [A1] (a) Obtain answer 3j + k
5. [M1] (b) Use the correct process for evaluating the scalar product of MD and ON
6. [M1] (b) Using the correct process for the moduli, divide the scalar product by the product of the moduli and reach the inverse cosine of the result
7. [A1] (b) Obtain final answer 103.3°
8. [M1] (c) Taking a general point P of ON to have position vector λ(3j+k), form an equation in λ by either equating the scalar product of ON and MP to zero, or applying Pythagoras to triangle OMP, or equating the derivative of |MP| to zero
9. [A1] (c) Solve and obtain λ = 3/5
10. [M1] (c) Substitute for λ and calculate MP
11. [A1] (c) Obtain √(22/5) (AG)