The diagram shows the triangle OAB, where OA = a and OB = b. The point P lies on OA such that OP = (3/4)OA. The point Q lies on AB such that AQ = (1/3)AB. The straight line through P and Q meets the straight line through O and B at the point R. It is given that OR = λb and PR = µPQ, where λ and µ are constants.

Mark scheme

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

1. [M1] OQ = OA + AQ = a + (1/3)(b-a) or OQ = OB + BQ = b + (2/3)(a-b) (for attempt to use OQ=a+k(b-a) or OQ=b+k(a−b), where k can be positive or negative. Allow unsimplified)
2. [A1] OQ = (2/3)a + (1/3)b (Allow unsimplified)
3. [B1] OP = (3/4)a (Allow anywhere)
4. [M1] PQ = PO + OQ = -(3/4)a + (2/3)a + (1/3)b (for PQ)
5. [A1] PQ = -(1/12)a + (1/3)b (Allow unsimplified)
6. [B1] OR = (3/4)a + µ(-(1/12)a + (1/3)b) oe (Must be simplified)
7. [B1] λ = (1/3)µ (For equating b vectors)
8. [B1] (3/4) - (1/12)µ = 0 oe (For equating a vectors to zero)
9. [B1] µ=9, λ=3