Melaka Trial 2025 Add Maths Paper 1, Question 13Vectors

Solved on video in English · Form 4, Vectors · 6:16 video

The question

(a)

Diagram 5 shows a triangle PQR, where point S is the midpoint of line PR. It is given that vector QP = 4p and vector QR = 5r. Find vector QS.

(b)(i)

Given that point D(−2, 1) and point E(−7, 13) are two points on the Cartesian plane, find:

(b)(i) vector DE.

(b)(ii)

Given that point D(−2, 1) and point E(−7, 13) are two points on the Cartesian plane, find:

(b)(ii) the unit vector of DE, in terms of i and j.

Step 1

Find PR using the triangle law: PR = PQ + QR = −QP + QR = −4p + 5r.

Why this step

We're given QP but need PQ, which points the opposite way. Reversing a vector flips its sign, so PQ = −QP. Mixing them up is the classic slip.

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

Final answer

QS = 2p + 5r/2

Where students lose marks

Watch out: PQ = −QP, and RS points from R to S (the same way as RP). Check every direction before adding vectors.

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

Final answer

DE = −5i + 12j (or (−5, 12))

Where students lose marks

Watch out: DE = OE − OD (end minus start). Subtracting the wrong way reverses the signs, and −(−2) becomes +2.

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

Final answer

−(5/13)i + (12/13)j

Where students lose marks

Watch out: square each component before adding, then take the root. Keep the negative sign on the i-component of the unit vector.

Watch every step of this question solved on video (6:16), with the reason behind each step and a box that checks your own answer.

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