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→ → 16
12. PS = 2QM (b) 4fx = k x
→
→
= 2(QP + PM) 4
3
2
1
= 2 –2i – 3j + i + j f = k
2
= –i – 4j 12(k – 1) y = 18(1 – f )y
[3] k
→ → 1 4 2
13. 3rp + 5q = 4s(OP + PA) 2(k – 1) = 3k 1 – k
3 →
2
1
3rp + 5q = 4s p + PQ k = 10
5
= 1 3 → → f = 2
3
4
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2 3rp + 5q = 4s p + (PO + OQ) 5 [4]
5
= √3 3
3
4
3rp + 5q = 4s p + (–p + 5q) (c) AD = √OA + OD
2
2
5
= √(7) – 3 AD = √(4x) + (18y)
2
2
2
4
3
3rp + 5q = 4s p – p + 3q AD = √16x + 324y
5
2
2
2
2
2
= 1 8 + 3 3rp + 5q = 4s 1 2 5 p + 3q AD = √16 + 324(1.5)
2
2
1 8 AD = 27.295
= 2 2 3rp + 5q = sp + 12sq 2
5
= 5q = 12sq CD = (27.295)
5
s = 5 = 10.92
12
8 [3]
3rp = sp → → →
5 15. (a) (i) SP = SQ + QP
1 2
3r = 8 5 = –8y + 4x
5 12 [1]
→
→
→
r = 2 (ii) RP = RQ + QP
9
[7] 1 → →
= SQ + QP
→ → → 4
14. (a) (i) OC = OE + EC = 1 (–8y) + 4x
2 → 1 → 4
= OD + EB = –2y + 4x
3 k [2]
1 →
→
= 2 (18y) + (EO + OB) → →
3 k (b) PT = kPR
1
= 12y + (–12y + 16x) = 2ky – 4kx
k → → →
PT = PS + ST
= 12(k – 1) y + 16 x = 8y – 4x + h(2x – 3y)
k k = (8 – 3h)y + (2h – 4)x
[3]
→ → → 8 – 3h = 2k
(ii) OC = OA + AC 2h – 4 = –4k
1 → → → 2 – h = 2k
= OB + AD – CD
4 h = 3
1 → → → 1
= OB + AD – fAD k = –
4 2 [4]
1 → →
= OB + (1 – f )AD m –6
4 (c) 2 = –4
1 → → → m = 3
= OB + (1 – f )(AO + OD)
4 [1]
= 1 (16x) + (1 – f )(–4x + 18y)
4 16. Velocity vector
2
= 4fx + 18(1 – f )y = 1 −7.5 − 3.3 i + –2 – 1.4 j km/h
[3] 1.5 1.5
= (−7.2i − 2.267j) km/h
[2]
Cambridge IGCSE
TM
200 Ace Your Additional Mathematics
Answers Add Math.indd 200 14/03/2022 12:29 PM