O-LevelAdditional MathematicsVectors in two dimensions4037/21 · May/June 2015 · Q78 Marks

The four points O, A, B and C are such that OA = 5a, OB = 15b, OC = 24b-3a.

(a)Show that B lies on the line AC.[3]
(b)(i)Relative to an origin O, the position vector of the point P is i-4j and the position vector of the point Q is 3i + 7j. Find |PQ|.[2]
(b)(ii)the unit vector in the direction PQ,[1]
(b)(iii)the position vector of M, the mid-point of PQ.[2]

Mark scheme

Marking points from the official mark scheme (May/June 2015). The mark or mark type is shown in brackets.

[B1] AB = 15b – 5a = 5(3b – a) or BC = 24b-3a – 15b = 3(3b – a) or AC = 24b-3a – 5a = 8(3b – a) Any correct simplified vector
[B1] Any second simplified vector
[B1dep] Comment: e.g. the vectors are scalar multiples of each other AND they have a common point (A, B or C as appropriate)
[B1] 2i + 11j soi
[B1ft] √2² + 11² = √125 or 5√5 or 11.2 (3 s.f.) or better) ft their 2i +11j (not OP or OQ)
[B1ft] (1/√125)(2i+11j) isw ft their answers from (i)
[M1] (P+Q)/2 or (i-4j) + (2i+11j)/2 or (3i+7j) - (2i+11j)/2
[A1] 2i + 1.5j

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About this question

Topic

Vectors in two dimensions, O-Level Additional Mathematics (syllabus 4037). Structured question, 8 marks.

Source

O-Level Additional Mathematics, May/June 2015 Paper 2 Variant 1, question 7.

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