PROGRAM 2Q PPDHL SPM 2011
ADDITIONAL MATHEMATICS
PAPER 2
TOPIC:
VECTORS
1. 2003 / SECTION A / Q6
A
5
(a) OA = OB − AB Use triangle law 7
= 2 − 5 √M1
3 7
B
2
−3
= −4
3
A (−3, −4) √ A1
O
(b) |OA| = √(−3)2 + (−4)2 = 5 √M1
1 −3
Unit vector OA = − −4
5
√ A1
(c) OA = λCD ( when OA and CD are parallel )
−3 k
−4 =λ 5
−3 = kλ √M1 and −4 = 5λ
k = −3/(−4/5) λ = −4/5
= 15/4 √ A1
2. 2005 / SECTION A / Q6
D Use triangle law
(a) (i) BD AB AD
= −20 x + 4 AE
32 y
= −20 x + 4(8 y )
= −20 x + 32 y √ A1 B
A
20 x
D
(ii) EC ED DC
√M1 = ¾ (32 y ) + (25 x − 24 y )
= 25 x √ A1 24y
E C
(b) BD = λFD (if B, F and D are collinear)
−20x + 32y = λFD D
−20x + 32y = λ(−15x + 24y) √M1
= −15λx + 24λy F
Compare:
B
20 = 15λ or 32 = 24λ
λ = 4/3 √M1 λ = 4/3 FD = −EF + ED
= 3/5(−15x) + ¾(32y)
BD = 4/3 FD √A1 = −15x + 24y
Thus, B, F and D are collinear
(c) BD = −20x + 32y
|BD| = √ (20|x|)2 + (32|y|)2
= √ [20(2)]2 + [32(3)]2 √M1
= 104 √ A1
3. 2006 / SECTION A / Q5
5 Use triangle law
(a) AC AB BC √K1 − (6x)
6 B 5x
C
2y 5x √A1 2y
A
(b)(i) 2EF mAB either one
m 2 2
EF ( 2y) my √M1 AE AB ( 6x) 4x
2 3 3
F
AF AE EF
4x my my
√A1
A 4x E
(b)(ii) AC = λAF (if A, F and C are collinear)
5x + 2y = λ(4x + my) √M1 C
5x + 2y = 4λx + mλy F
Compare:
5 = 4λ √M1 A
either one λ = 5/4 or AF = kAC
4x + my = k(5x + 2y)
and mλ = 2 4x + my = 5kx + 2ky
Compare:
m = 8/5 √A1
5k = 4
k = 4/5
m = 2k
m = 8/5
4. 2008 / SECTION A / Q6
Use triangle law D
(a)(i) DB = −AD + AB 3y
= −3y + x √A1 A B
x
(ii) AR = AB – BR
R
= x – ⅓(x – 3y) √M1 ⅓(DB)
= ⅔x + y √A1 A x
B
D
or AR = AD + DR 3y ⅔(DB)
= 3y + ⅔(x – 3y) R
= ⅔x + y A
(b) DC = kx – y AC = AD + DC
AR = hAC = 3y + kx – y
= h(2y + kx) = 2y + kx
⅔x + y = 2hy + hkx √M1 D
kx – y
Compare: 3y
C
2h = 1 √M1 A
h = ½ √A1
and hk = ⅔
k = 4/3 √A1
5. 2004 / SECTION B / Q8
A
(a)(i) AP = −OA + OP 2y
= −2y + 6x √A1
O P
6x
(ii) OQ = OA + AQ OB = 3OP = 18x
AB = −OA + OB
= 2y + ¼ AB √M1 = −2y + 18x
= 2y + ¼(−2y + 18x) A
¼ AB √K1
= 2y − ½ y + 18/4 x 2y Q
= 3/2 y + 9/2 x √A1 O
(b)(i) AR = hAP = h(−2y + 6x)
= 6hx − 2hy √K1
A
RQ = kOQ = k(3/2 y + 9/2 x)
Q
= 9/2 kx + 3/2 ky √K1
(ii) AQ = AR + RQ Use triangle law R
9/2 x − ½ y = 6hx − 2hy + 9/2 kx + 3/2 ky
9/2 x − ½ y = (6h + 9/2 k)x + (3/2 k − 2h)y
Compare: √M1
9/2 k + 6h = 9/2 .....(1) k = ⅓ √A1
+ 3/2 k – 2h = −½ .....(2) 3/2(⅓) − 2h = −½
(2)x3 9/2 k – 6h = −3/2 .....(3) − 2h = −1
(1)+(3) 9k = 3 √M1 h = ½ √A1
6. 2009 / SECTION A / Q5
C
(a)(i) BC = − AB + AC 4y
= − 3x + 4y √A1
A B
3x
(ii) AQ = AC – QC C
⅓(BC)
= 4y – ⅓(− 3x + 4y) √M1 4y
Q
= 4y + x – 4/3 y
AQ = x + 8/3 y √A1 A
Q
Or AQ = AB + BQ
⅔(BC)
= 3x + ⅔(−3x + 4y)
= x + 8/ 3 y A 3x B
(b) AP = hAQ Use triangle law
= h(x + 8/3 y) √K1 RB = −AR + AB
AP = AR + kRB = −3y + 3x
= 3y + k(−3y + 3x) √K1
h(x + 8/3 y) = 3y + k(−3y + 3x)
hx + 8/3 hy = (3 – 3k)y + 3kx √M1
Compare:
h = 3k and 8/3 h = 3 – 3k
k = 9/33 √A1 8/3 h = 3 – h
h = 9/11 √A1