Relative to the origin O, the position vectors of the points A, B and C are OA = 4i−2j, OB = 2i+8j+4k, and OC = -2i+6k. The midpoint of AB is M, as shown in the diagram. [Figure 11.1]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2025). The mark or mark type is shown in brackets.

1. [1] (a) MB = −i+5j+2k. Do not ISW. Accept as a column vector.
2. [1] (a) MC=-5i-3j+4k. Do not ISW. Accept as a column vector. SC: B0B1 for column vectors with i, j, k (or coordinates) (allow one slip).
3. [1] (b) Carry out the correct process for evaluating the scalar product of MB and MC. Or BM and CM. Need some evidence of an attempt to evaluate the scalar product.
4. [1] (b) Using the correct process for the moduli, divide the scalar product by the product of the moduli for their two vectors.
5. [1] (b) Obtain -1 / (5√15). Or exact equivalent. A0 if they never have a statement "cosθ=....". ISW if they have correct "cosθ=...." and then state the angle.
6. [1] (b) Alternative Method: Carry out the correct process for finding the lengths of the sides of triangle BMC. MB = √30, MC = √50, BC = √84.
7. [1] (b) Alternative Method: Correct use of the cosine rule as far as cosθ=.... cos M = (|their MB|² + |their MC|² - |BC|²) / (2|their MB|×|their MC|). cos M = (30+50-84) / (2√30√50).
8. [1] (b) Alternative Method: Obtain -1 / √375. Or exact equivalent.
9. [1] (c) Working with triangle MBC: Correct method to calculate the exact value of sin(∠CMB) (e.g., using sin²θ + cos²θ = 1).
10. [1] (c) sin(∠CMB) = √(374/375). Or equivalent, e.g. √(5400/5625). A0 XP if they had changed the sign in their working for the cosine in (b) and for the final answer.
11. [1] (c) Correct method to calculate the area of triangle ABC using MB and MC (e.g., 2 x area of triangle CMB).
12. [1] (c) Obtain 2√374. Or exact equivalent, e.g. 2√2√11√17.