The diagram shows a three-dimensional shape OABCDEFG. The base OABC and the upper surface DEFG are identical horizontal rectangles. The parallelograms OAED and CBFG both lie in vertical planes. Points P and Q are the mid-points of OD and GF respectively. Unit vectors i and j are parallel to OA and OC respectively and the unit vector k is vertically upwards. The position vectors of A, C and D are given by OA = 6i, OC = 8j and OD = 2i + 10k. [Figure 7]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2019). The mark or mark type is shown in brackets.

1. [B2,1,0] (PB) = 5i + 8j - 5k B2 all correct, B1 for two correct components.
2. [B2,1,0] (PQ) = 4i + 8j + 5k B2 all correct, B1 for two correct components.
3. Accept column vectors.
4. SC B1 for each vector if all components multiplied by -1.
5. [M1] (Length of PB =) √(5² + 8² + 5²) = (√114 ≈ 10.7) (Length of PQ =) √(4² + 8² + 5²) = (√105 ≈ 10.2) Evaluation of both lengths. Other valid complete comparisons can be accepted.
6. [A1] P is nearer to Q. WWW
7. [M1] (PB.PQ) = 20 + 64 - 25 Use of x1x2 + y1y2 + z1z2 on their PB and PQ
8. [M1] (Their √114)(their √105) cos BPQ = (their 59) All elements present and in correct places.
9. [A1] BPQ = 57.4(°) or 1.00 (rad) AWRT Calculating the obtuse angle and then subtracting gets A0.