Selangor Trial 2025 Add Maths Paper 2, Question 6Vectors

Solved on video in English · Form 4, Vectors · 6:55 video

The question

(a)(i)

Diagram 3 shows a triangle BOA. The straight lines OX and BY intersect at point P. It is given that OA = 2a, OB = 3b, 5BX = 3BA and OY = (5/7)OA. Given that OP = m·OX and BP = n·BY. (a)(i) Express OP in terms of m, a and/or b.

(a)(ii)

(a)(ii) Express OP in terms of n, a and/or b.

(b)

(b) Hence, find the value of m and of n.

Selangor Trial 2025 Add Maths Paper 2, Question 6: the question as drawn in Eduly's video

6:55

Part (a)(i)

Step 1

Find OX. First BA = OA − OB = 2a − 3b, and 5BX = 3BA gives BX = (3/5)BA = (3/5)(2a − 3b) = (6/5)a − (9/5)b.

Why this step

To go from B to A, take the end minus the start: BA = OA − OB. Writing OB − OA flips the direction and loses marks.

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

Final answer

OP = (6/5)m·a + (6/5)m·b

Where students lose marks

Watch out: BA = OA − OB, not OB − OA, and multiply the whole vector OX by m, not just one term.

Part (a)(ii)

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

Final answer

OP = (10/7)n·a + (3 − 3n)b

Where students lose marks

Watch out: OY is 5/7 of OA = 2a, not 5/7 of a, and remember to add OB = 3b to get OP.

Part (b)

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

Final answer

m = 25/31, n = 21/31

Where students lose marks

Watch out: compare the coefficients of a and b separately to get two equations, and double-check your fraction arithmetic.

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Selangor Trial 2025 Add Maths Paper 2, Question 6Vectors

The question

Part (a)(i)

Part (a)(ii)

Part (b)

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