Kuala Lumpur Trial 2025 Add Maths Paper 2, Question 3Vectors

Solved on video in English · Form 4, Vectors · 5:37 video

The question

(a)(i)

Diagram 2 shows two triangles APB and EPB. C is a point on AB such that AC = 2CB and Q is the midpoint of PB. It is given that vector PA = p and vector PB = q.

(a) Express in terms of p and/or q: (i) vector QA.

(a)(ii)

(a) Express in terms of p and/or q: (ii) vector PC.

(b)(i)

C is a point on AB such that AC = 2CB and Q is the midpoint of PB, with vector PA = p and vector PB = q. It is given that vector PE = h·vector PC and vector BE = k·vector QA, where h and k are constants.

(b) Express vector PE: (i) in terms of h, p and/or q.

(b)(ii)

(b) Express vector PE: (ii) in terms of k, p and/or q. Hence, find the value of h and of k.

Kuala Lumpur Trial 2025 Add Maths Paper 2, Question 3: the question as drawn in Eduly's video5:37

Part (a)(i)

Step 1

Use the triangle law: vector QA = vector QP + vector PA.

Why this step

We only know vectors from P (PA = p, PB = q), so we travel from Q to A by going through P.

2 more steps, each with the reason for it, in the full solution

Final answer

QA = p − ½q

Where students lose marks

Watch out: QP points opposite to PQ, so QP = −½q, not +½q. Mixing directions in QB + BA also loses marks.

Part (a)(ii)

4 more steps, each with the reason for it, in the full solution

Final answer

PC = (1/3)p + (2/3)q

Where students lose marks

Watch out: AC:CB = 2:1 means AC = (2/3)AB, not ½AB or 2AB. Also AB = −p + q.

Part (b)(i)

3 more steps, each with the reason for it, in the full solution

Final answer

PE = (1/3)hp + (2/3)hq

Where students lose marks

Watch out: h multiplies both terms, so PE = (1/3)hp + (2/3)hq. Use PC from part (a)(ii), not a guess.

Part (b)(ii)

7 more steps, each with the reason for it, in the full solution

Final answer

PE = kp + (1 − ½k)q; h = 6/5, k = 2/5

Where students lose marks

Watch out: go P → B → E, mind the sign in (1 − ½k)q, and equate both forms of PE.

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