With respect to the origin O, the points A and B have position vectors given by OA = 6i + 2j and OB = 2i + 2j + 3k. The midpoint of OA is M. The point N lying on AB, between A and B, is such that AN = 2NB.

Mark scheme

Marking points from the official mark scheme (May/June 2020). The mark or mark type is shown in brackets.

1. [B1] State that the position vector of M is 3i + j
2. [M1] Use a correct method to find the position vector of N
3. [A1] Obtain answer (10/3)i + (2/3)j + 2k
4. [M1] Use a correct method to form an equation for MN
5. [A1] Obtain correct answer in any form, e.g. r = 3i + j + λ((1/3)i + (1/3)j + 2k)
6. [B1] State or imply r = μ(2i + 2j + 3k) as equation for OB
7. [M1] Equate sufficient components of MN and OB and solve for λ or for μ
8. [A1] Obtain λ = 3 or μ = 2 and position vector 4i + 4j + 6k for P
9. [M1] Carry out correct process for evaluating the scalar product of direction vectors for OP and MP, or equivalent
10. [M1] Using the correct process for the moduli, divide the scalar product by the product of the moduli and evaluate the inverse cosine of the result
11. [A1] Obtain answer 21.6°