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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]

1

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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prepared by: cik shila 11/4/2012


 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

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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

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prepared by: cik shila 11/4/2012


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prepared by: cik shila 11/4/2012


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