The diagram shows a three-dimensional shape. The base OAB is a horizontal triangle in which angle AOB is 90°. The side OBCD is a rectangle and the side OAD lies in a vertical plane. Unit vectors i and j are parallel to OA and OB respectively and the unit vector k is vertical. The position vectors of A, B and D are given by OA = 8i, OB = 5j and OD = 2i + 4k. [Figure 6]

Mark scheme

Marking points from the official mark scheme (May/June 2018). The mark or mark type is shown in brackets.

1. [B1] (i) DA = OA - OD = 8i - (2i + 4k) = 6i - 4k.
2. [B1] (i) To find CA, first find OC. Since OBCD is a rectangle, OC = OB + OD = 5j + (2i + 4k) = 2i + 5j + 4k. Then CA = OA - OC = 8i - (2i + 5j + 4k) = 6i - 5j - 4k.
3. [*M1] (ii) Method marks awarded only for their vectors ± CA & ± DA. Full marks can be obtained using AC & AD. Using DA = 6i - 4k and CA = 6i - 5j - 4k, calculate the scalar product DA . CA = (6)(6) + (0)(-5) + (-4)(-4) = 36 + 0 + 16 = 52.
4. [*M1] (ii) Uses modulus twice: |DA| = √(6² + (-4)²) = √(36 + 16) = √52. |CA| = √(6² + (-5)² + (-4)²) = √(36 + 25 + 16) = √77.
5. [*M1] (ii) All linked correctly using the scalar product formula: DA . CA = |DA||CA| cos(CAD). So, 52 = √52 √77 cos(CAD) oe.
6. [A1] (ii) cos(CAD) = 52 / (√52 √77) ≈ 0.82178. Angle CAD = arccos(0.82178) ≈ 34.7° or 0.606 radians (awrt). Answer must come from +ve cosine ratio.