CHAPTER 4: VECTOR PAPER 1
1. Diagram below shows two vectors, OP and QO
Q(-8,4) P(5,3) O Express
x (a) OP in the form , y (b) QO in the form x i + y j [2 marks] p = 2a + 3b q = 4a – b r = ha + ( h – k ) b, where h and k are constants 2. Use the above information to find the values of h and k when r = 3p – 2q. [3 marks]
3. Diagram below shows a parallelogram ABCD with BED as a straight line. D
C
E A
B
Given that AB = 6p , AD = 4q and DE = 2EB, express, in terms of p and q (a) BD (b) EC [4 marks]
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prepared by: cik shila 11/4/2012
4. Given that O(0,0), A(-3,4) and B(2, 16), find in terms of the unit vectors, i and j, (a) AB (b) the unit vector in the direction of AB [4 marks]
5. Given that A(-2, 6), B(4, 2) and C(m, p), find the value of m and of p such that
AB + 2 BC = 10i – 12j. [4 marks] 6. Diagram below shows vector OA drawn on a Cartesian plane. y 6 A 4 2
0
2
4
6
8
10
12
x
x y (b) Find the unit vector in the direction of OA (a) Express OA in the form
[3 marks]
7. Diagram below shows a parallelogram, OPQR, drawn on a Cartesian plane. y Q
R
P
O
x
It is given that OP = 6i + 4j and PQ = - 4i + 5j. Find PR . [3 marks]
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prepared by: cik shila 11/4/2012
8. Diagram below shows two vectors, OA and AB .
y A(4,3)
O
-5
x
B
Express (a) (b)
x in the form y AB in the form xi + yj
OA
[2 marks] 9. The points P, Q and R are collinear. It is given that PQ = 4a – 2b and
QR 3a (1 k )b , where k is a constant. Find (a) the value of k (b) the unit vector in the direction of PQ [4 marks] 10. Given that a 6i 7 j and b pi 2 j, find the possible value (or values) of p for following cases:a) a and b are parallel b) a b [5 marks]
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PAPER 2
1.
5 2 k Give that AB , OB and CD , find 7 3 5 (a) the coordinates of A, (b) the unit vector in the direction of OA , (c) the value of k, if CD is parallel to AB
2.
[2 marks] [2 marks] [2 marks]
Diagram below shows triangle OAB. The straight line AP intersects the straight line OQ at R. It is given that OP = 1/3 OB, AQ = ¼ AB, OP 6 x and OA 2y. A Q R
O
P
B
(a) Express in terms of x and/or y: (i) AP (ii) OQ (b)
[4 marks]
(i) Given that AR h AP, state AR in terms of h, x and y. (ii) Given that RQ k OQ, state RQ in terms of k, x and y. [2 marks]
(c) Using AR and RQ from (b), find the value of h and of k. 3.
[4 marks] In diagram below, ABCD is a quadrilateral. AED and EFC are straight lines. D
E
C F
A
B 4
prepared by: cik shila 11/4/2012
It is given that AB 20x, AE 8y, DC = 25x – 24y, AE = ¼ AD 3 and EF = EC. 5 (a) Express in terms of x and/or y: (i) BD (ii) EC
4.
[3 marks]
(b) Show that the points B, F and D are collinear.
[3 marks]
(c) If | x | = 2 and | y | = 3, find | BD |.
[2 marks]
Diagram below shows a trapezium ABCD. B
C F •
A
• E
It is given that AB =2y, AD = 6x, AE =
D
2 5 AD and BC = AD 3 6
(a) Express AC in terms of x and y
[2 marks]
(b) Point F lies inside the trapezium ABCD such that 2 EF = m AB , and m is a constant. (i) Express AF in terms of m , x and y (j) Hence, if the points A, F and C are collinear, find the value of m. [5 marks]
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prepared by: cik shila 11/4/2012
ANSWERS (PAPER 1) 1. a) b) 2.
3 a) b)
5 3
1
8i – 4j
1
r = - 2a + 11b r = ha + (h – k)b
1
h = -2 (h – k) = 11 k = −13
1
BD = −6p + 4q DB = − BD = 6p −4q
1
EB =
1
1
DB 3
4 2p q 3
1
EC EB BC 8 2p q 3
4. a)
b)
6. a) b)
1 1 1
1 (5i 12 j ) 13
AB 2BC (6i 4 j ) 2((m 4)i 2( p 2) j ) = (-2+2m)i + (-8+2p)j m=6 p = -2 12 OA 5 1 u (12i 5 j ) 2 2 12 5
7.
1
AB (2 (3))i (16 4) j = 5i + 12j 1 u (5i 12 j ) 5 2 12 2
5.
1
1 (12i 5 j ) 13
1 1 1 1 1 1
1
PO OP
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6i 4 j
1 1
OR PQ 4i 5 j
8. a) b)
9. a)
b)
PR PO OR 10i j
1
4 OA 3
1
AB 4i 8 j
1
QR m PQ 3 4 m 1 k 2 3 = m(4) 3 m 4 1+ k = m(-2) 5 k 2 PQ u 2 4 (2) 2
10 a)
b)
1
1
1
1 (4a 2b) 20
1
a kb 6i 7 j k ( pi 2 j ) 7 k 2 12 p 7 a b 62 7 2 p 9
1 1 1 1
p 2 22
1
7
prepared by: cik shila 11/4/2012
ANSWERS (PAPER 2)
3 AO 4 3 OA 4 A = (-3,-4)
1 (a)
(b)
u
1
1
OA OA
u
(c)
32 4 2
1 OA
1
1 3i 4 j 5
CD m AB k 5 m 5 7 5 m 7 25 k 7
2 (a) (i) (ii)
(b) (i)
(c)
1
1
1
AP 6 x 2 y
1
AB 18x 2 y 9 1 AQ x y 2 2 9 3 OQ x y 2 2
1
AR 6hx 2hy 9 3 RQ kx ky 2 2
1
1 1
1
AR RQ AQ 9 3 9 1 6h k x 2h k y x y 2 2 2 2 1 k 3 1 h 2
8
1 1 1
prepared by: cik shila 11/4/2012
3 (a) (i) (ii)
1
BD 20 x 32 y 3 ED AD 4 = 24y
1 1
EC 25x
(b)
FC 10 x BC BD DC 5x 8 y
1
BF BC CF 5x 8 y
1
BD 20 x 32 y 4(5x 8 y) 4( BF ) (c)
BD
1
20 x 32 y 2
2
(20 2) 2 32 3 = 104
1
2
4 (a)
(b) (i)
1
5 AD 6 = 5x
1
AC AB BC = 5x + 2y
1
BC
2EF m AB EF my 2 AE AD 3 = 4x
1
1
AF AE EF = 4x + my (ii)
1
AC 5x 2 y 4 4 AC 5 x 2 y 5 5 4 8 AC 4 x y 5 5 Assume A, F, C collinear, 4 AC AF 5 = 4x + my 8 m 5
1
1
9
prepared by: cik shila 11/4/2012
10
prepared by: cik shila 11/4/2012