vectors sl marks

8/7/2019 vectors sl marks

1/47

1. (a) OCODCD = (A1)(C1)

(b)

CDOA

2

1=

=)(

2

1OCOD

(A1)

(C1)

(c) OAODAD =

=)(

2

1OCODOD

(A1)

=

OCOD

2

1

2

1+

(A1)(C2)

Note:Deduct [1 mark](once only) if appropriate vectornotation is omitted.

[4]

2. (a)jiu 2+= jiv 53 +=

jivu 1252 +=+(A1)

(C1)

(b)22

1252 +=+ vu

= 13 (A1)

Vector)125(

13

26jiw

+=(A1)

=ji 2410 +

(A1) (C3)[4]

3. (a)OA

= 6 A is on the circle (A1)OB

= 6 B is on the circle. (A1)

=

11

5OC

= 1125 +

= 6 Cis on the circle. (A1) 3

1


2/47

(b) OAOCAC =

=

0

6

11

5

(M1)

=

111

(A1) 2

(c)ACAO

ACAOCAO

=cos(M1)

= 1116

11

1.

0

6

+

= 1266

(A1)

=6

3

32

1=

(A1)

OR 1262

6)12(6cos222

+=CAO

(M1)(A1)

= 12

1

as before (A1)

ORusing the triangle formed by AC and its horizontal andvertical components:

12=AC(A1)

12

1cos =CAO(M1)(A1)

3

Note: The answer is 0.289 to 3 sf

2


3/47

(d) A number of possible methods here

OBOCBC =

=

0

6

11

5

(A1)

=

11

11

(A1)

BC = 132

ABC =12132

2

1

(A1)

= 116 (A1)

OR ABChas baseAB = 12 (A1)

and height = 11 (A1)

area =1112

2

1

(A1)

= 116 (A1)

ORGiven 6

3cos =CAB

6

331212

2

1

6

33sin == ABCCAB(A1)(A1)(A1)

= 116 (A1) 4[12]

4. (a)

=

5

10OB

(A1)

(C1)

= 6

3AC

(A1) (C1)

(b) ACOB = (10 (3)) + (5 6) = 0 (M1)Angle = 90 (A1) (C2)

[4]

3


4/47

5. u + v = 4i+ 3j (A1)

Then a(4i+ 3j) =8i+ (b 2)j4a = 83a = b 2 (A1)

Whence a = 2 (A1) (C2)

b = 8 (A1) (C2)[4]

6. Required vector will be parallel to

4

1

1

3

(M1)

=

54

(A1)

Hence required equation is r=

+

54

41 t

(A1)(A1) (C4)

Note: Accept alternative answers, eg

+

5

4

1

3s

.[4]

7. (a)

24

18

= 30 km h1 (A1)

22 )16(3616

36+=

= 39.4(A1) 2

(b) (i) After hour, position vectors are

12

9

and

818

(A1)(A1)

(ii) At 6.30 am, vector joining their positions is

=

20

9

8

18

12

9

(or

20

9

) (M1)

20

9

(M1)

= 481 (= 21.9 km to 3 sf) (A1)5

4


5/47

(c) The Toyundai must continue until its position vector is

k

18

(M1)

Clearly k= 24, ie position vector

24

18

. (A1)To reach this position, it must travel for 1 hour in total. (A1)

Hence the crew starts work at 7.00 am (A1)

4

(d) Southern (Chryssault) crew lays 800 5 = 4000 m (A1)

Northern (Toyundai) crew lays 800 4.5 = 3600 m (A1)

Total by 11.30 am = 7.6 km

Their starting points were 24 (8) = 32 km apart (A1)

Hence they are now 32 7.6 = 24.4 km apart (A1)

4

(e) Position vector of Northern crew at 11.30 am is

=

4.20

18

6.324

18

(M1)(A1)

Distance to base camp =

4.20

18

(A1)

= 27.2 km

Time to cover this distance = 30

2.27

60 (A1)= 54.4 minutes

= 54 minutes (to the nearest minute) (A1)

5[20]

8. Vector equation of a line r= a + t (M1)

a =

0

0

, t=

3

2

(M1)(M1)r= (2i+ 3j) (A1) (C4)[4]

5


6/47

9. (a)

A

B

C

4

3

2

1

0

1

2

3

1 2 3 4 5 x

y

(A3)(C3)

Note: Award (A1) for B at (5, 1); (A1) for BC perpendicular toAB; (A1) for AC parallel to the y-axis.

(b)

=

25.3

2OC

(A1)

(C1)

Note: Accept correct readings from diagram (allow 0.1).[4]

10. (a) (i) r1 =

+

5

12

12

16t

t= 0 r1 =

12

16

(M1)

| r1| =)1216( 22 +

= 20 (A1)

(ii) Velocity vector =

512

speed = ))5(12(22 +

(M1)

= 13 (A1) 4

6


7/47

(b)

+

=

5

12

12

16t

y

x

+

=

5

12.

12

5

12

6.

12

5.

12

5t

y

x

(M1)5x + 12y = 80 + 144 (A1)

5x + 12y = 224 (A1)(AG)

OR

5

12

12

16

= yx

(M1)

5x 80 = 144 12y (A1)

5x + 12y = 224 (A1)(AG)

OR

x = 16 + 12t,y = 12 5tt= 512 y

(M1)

x = 16 + 12

5

12 y

(A1)

5x = 80 + 144 12y5x + 12y = 224 (A1)(AG)

3

(c) v1 =

512

v2 =

6

5.2

(M1)

v1.v2 =

6

5.2.

5

12

(M1)

= 30 30

v1.v2 = 0 (A1) = 90 (A1) 4

(d) (i)

=

5

12.

5

23

5

12.

y

x

(M1)

12x 5y = 23 12 + 25 = 301 (A1)

OR

6

5

5.2

23 +=

yx

6x 138 = 2.5y + 12.5 (M1)12x 276 = 5y + 25

12x 5y = 301 (A1)

7


8/47

(ii)

==+

==+

361260144

11206025

301512

224125

yx

yx

yx

yx

(M1)

169x = 4732x = 28,y = (12 28 301) 5 = 7(28, 7) (A1)(A1) 5

Note: Accept any correct method for solving simultaneousequations.

(e) 16 + 12t= 23 + 2.5t9.5t= 7 (M1)12 5t= 5 + 6t 17 = 11t (M1)

11

17

5.9

7

(A1)

planes cannot be at the same place at the same time (R1)

OR

r1 =

+

=

5

12

12

16

7

28

7

28t

(M1)

==

55

1212

t

t

t= 1 (A1)

When t= 1 r2 =

=

+

7

28

1

5.25

6

5.2

5

23

(A1)(R1)

OR

r2 =

+

=

6

5.2

5

23

7

28

7

28t

(M1)

t= 2 (A1) 4[20]

8


9/47

11.

8

6.

2

1

= 6 16 = 10 (A1)

10086

8

6,521

2

1 2222 =+=

=+=

= 10 (A1)

=

8

6

2

1

8

6.

2

1

cos

10 = 5 10 cos cos = 51

510

10=

= arccos 51

(M1)

117 (A1)[4]

12.

+

1

4.

3

2

y

x

(M1) (M1)

Notes: Award (M1) for using scalar product.

Award (M1) for

+

1

4

y

x

.

2(x 4) + 3(y + 1) = 0 (A1)

2x 8 + 3y + 3 = 0

2x + 3y = 5 (A1)

OR

Gradient of a line parallel to the vector

3

2

is 2

3

(M1)

Gradient of a line perpendicular to this line is 3

2

(M1)

So the equation isy + 1 = 3

2

(x 4) (A1)3y + 3 = 2x + 82x + 3y = 5 (A1)

[4]

13. (a) At 13:00, t= 1 (M1)

=

+

=

20

6

8

61

28

0

y

x

(A1)

2

9


10/47

(b) (i) Velocity vector: 01 ==

tty

x

y

x

(M1)

=

=

8

6

28

0

20

6

(km h1

) (A1)

(ii) Speed =))8(6( 22 +

; (M1)

= 10; 10 km h1

(A1)

4

(c) EITHER

==

ty

tx

828

6

(M1)

Note: Award (M1) for both equations.

y = 28 8

6x

(M1)(A1)

Note: Award (M1) for elimination, award (A1) for equation inx, y.

4x + 3y = 84 (a1)4

OR

= 86

.28

0

8

6.y

x

(M1)

=

6

8.

28

0

6

8.

y

x

(M1)(A1)

4x + 3y = 84 (A1)4

10


11/47

(d) They collide if

4

18

lies on path; (R1)

EITHER(18, 4) lies on 4x + 3y = 84

4 18 + 3 4 = 84 72 + 12 = 84; OK; (M1)x = 18 (M1)

18 = 6tt= 3, collide at 15:00 (A1)4

OR

+

=

8

6

28

0

4

18t

for some t,

==

t

t

8284

618

and(A1)

==

248

3

t

t

and(A1)

==

3

3

t

t

and

They collide at 15:00 (A1)

4

(e)

+

=

12

5)1(

4

18t

y

x

(M1)

=

++12124

5518

t

t

(M1)

2

=

+

12

5

8

13t

(AG)

(f) At t= 3, (M1)

=

+

+=

28

28

1238

5313

y

x

(A1)

=

24

10

4

18

28

28

(A1)

)676()2410( 22 =+= 26

26 km apart (A1) 4[20]

11


12/47

14. cos =ba

a.b

(M1)

=5020

144 +

(A1)

= 1010

10

= 10

1

(= 0.3162) (A1)

= 72 (to the nearest degree) (A1) (C4)

Note: Award (C2) for a radian answer between 1.2 and 1.25.[4]

15. (a) At t= 2,

=

+

2

4.3

1

7.02

0

2

(M1)

Distance from (0, 0) =22 24.3 + = 3.94 m (A1)

2

(b)

22 17.01

7.0+=

(M1)

= 1.22 m s1 (A1)2

(c) x = 2 + 0.7 tandy = t (M1)x 0.7y = 2 (A1) 2

(d) y = 0.6x + 2 andx 0.7y = 2 (M1)

x = 5.86 andy = 5.52

==

29

160and

29

170or yx

(A1)(A1)

3

12


13/47

(e) The time of the collision may be found by solving

+

=

1

7.0

0

2

52.5

86.5

tfort (M1)

t= 5.52 s (A1)[ie collision occurred 5.52 seconds after the vehicles set out].Distance dtravelled by the motorcycle is given by

d=

22 )52.3()86.5(2

0

52.5

86.5+=

(M1)

= 73.46

= 6.84 m (A1)

Speed of the motorcycle = 52.5

84.6=

t

d

= 1.24 m s1

(A1) 5

[14]

16. Direction vector =

3

1

5

6

(M1)

=

2

5

(A1)

+

=

2

5

3

1t

y

x

(A2)

OR

+

=

2

5

5

6t

y

x

(A2) (C4)[4]

17. (a)

+

5

1

3

2 x

x

x

= 0 (M1)(M1)

2x(x + 1) + (x 3)(5) = 0 (A1)2x2 + 7x 15 = 0

(C3)

13


14/47

(b) METHOD 1

2x2

+ 7x 15 = (2x 3)(x + 5) = 0

x = 23

orx = 5 (A1)(C1)

METHOD 2

x =)2(2

)15)(2(477 2

x = 23

orx = 5 (A1)(C1)

[4]

18. (a) (i)

=

70

240OA

OA =22 70240 + = 250 (A1)

unit vector =

=

28.0

96.0

70

240

250

1

(M1)(AG)

(ii)

=

=

84

288

28.0

96.0300v

(M1)(A1)

(iii) t= 6

5

288

240=

hr (= 50 min) (A1)

5

(b)

=

=

180

240

70250

240480AB

(A1)

AB =22 180240 + = 300

cos =)300)(250(

)180)(70()240)(240(

ABOA

ABOA +=

(M1)

= 0.936 (A1)

= 20.6 (A1) 4

(c) (i)

=

=

168

99

70238

240339AX

(A1)

(ii)

180

240

4

3

= 720 + 720 = 0 (M1)(A1)

n AB (AG)

14


15/47

(iii) Projection ofAX in the direction ofn is

XY =5

672297

4

3

168

99

5

1 +=

= 75 (M1)(A1)(A1)

6

(d) AX =22 16899 + = 195 (A1)

AY =22 75195 = 180 km (M1)(A1)

3[18]

19. x = l 2t (A1)

y = 2 + 3t (A1)

3

2

2

1 yx=

(M1)

3x + 2y = 7 (A1)(A1)(A1) (C6)[6]

15


16/47

20.

T

S

U

V

y

x

(a) ST = ts (M1)

=

22

77

=

9

9

(A1)

VU = ST (M1)

u v =

9

9

v = u

9

9

=

=

6

4

9

9

15

5

(A1)

V(4, 6) (A1)5

(b) Equation of (UV): direction is =

1

1or

9

9k

(A1)

r=+ 9

9

15

5

or+ 1

1

15

5

(A1)

OR

r=

+

9

9

6

4

or

+

1

1

6

4

(A1)

2

16


17/47

(c)

11

1

is on the line because it gives the same value of , for both thexandy coordinates. (R1)

For example, 1 = 5 + 9 = 94

11 = 15 + 9 = 9

4

(A1) 2

(d) (i)

=

11

1

17EW

a

(M1)

=

6

1a

(A1)

EW = 2 13 ( ) 3612 +a

= 2 13 (or (a 1)2

+ 36 = 52) (M1)

a2

2a + 1 +36 = 52

a2

2a 15 = 0 (A1)

a = 5 or a = 3 (A1)(AG)

(ii) For a = 3

EW =

6

4

ET = t e =

4

6

(A1)(A1)

cos TEW =ETEW

ETEW

(M1)

= 5252

2424

(A1)

= 13

12

Therefore, TE

W = 157 (3 sf) (A1)10

[19]

17


18/47

21. Angle between lines = angle between direction vectors. (M1)

Direction vectors are

3

4

and

1

1

. (A1)

34

.

1

1

=3

4

11

cos (M1)

4(1) + 3(1) =( ) ( )

++ 2222 1134

cos (A1)

cos = 25

1

= 0.1414 (A1)

= 81.9 (3 sf), (1.43 radians) (A1) (C6)

Note: If candidates find the angle between the vectors

1

4

and

4

2

, award marks as below:

Angle required is between

1

4

and

4

2

(M0)(A0)

1

4

.

4

2

=

1

4

4

2

cos (M1)

4(2) + (1) 4 =( ) ( )

2222 4214 + + cos (A1)

2017

4

= cos = 0.2169 (A1)

= 77.5 (3sf), (1.35 radians) (A1) (C4)[6]

22. (i) a=22 512 + = 13 (A1)

(ii) b=22

86 + = 10 (A1)

=> unit vector in direction ofb = 10

1

(6i+ 8j) (A1)

= 0.6i + 0.8j

18


19/47

(iii) a.b = a b cos (M1)

=> cos =

( ) ( )( )1013

85612 +

(A1)

= 65

56

130

112

= (A1)6

[6]

23. METHOD 1

At point of intersection:

5 + 3 = 2 + 4t (M1)l 2 = 2 + t (M1)

Attempting to solve the linear system (M1)

= l (ort= 1) (A1)

=

3

2OP

(A1)(A1) (C6)

METHOD 2

(changing to Cartesian coordinates)

2x + 3y = 13,x 4y = 10 (M1)(A1)(A1)Attempt to solve the system (M1)

=

3

2OP

(A1)(A1) (C6)

Note: Award(C5)forthepointP(2, 3).

24. (a) c d= 3 5 + 4 (12) (M1)= 33 (A1) (C2)

[2]

25. (a) PQOR== qp

=

3

7

1

10

(A1)(A1)

=

2

3

(A1)

3

19


20/47

(b) cosPQPO

PQPOQPO

=

(A1)

( ) ( ) 22 37PO +== 58 ,

( )22 23PQ +== 13 (A1)(A1)

PQPO = 21 + 6 = 15 (A1)

cos 754

15

1358

15QPO ==

(AG)

4

(c) (i) Since QPO + RQP = 180 (R1)

cos RQP = cos QPO

=

754

15

(AG)

(ii) sin RQP =

2

754

151

(M1)

= 754

529

(A1)

= 754

23

(AG)

OR

cos = 754

15

1 57 5 4

Px

(M1)

thereforex2

= 754 225 = 529 x = 23 (A1)

sin = 75423

(AG)

Note: Award(A1)(A0)forthefollowingsolution.

cos = 754

15

= 56.89sin = 0.8376

754

23

= 0.8376 sin = 75423

(iii) Area of OPQR = 2 (area of triangle PQR) (M1)

20


21/47

= 2 RQPsinQRPQ

2

1

(A1)

= 2 754

235813

2

1

(A1)

= 23 sq units. (A1)

ORArea of OPQR = 2 (area of triangle OPQ) (M1)

= 2

( )103172

1

(A1)(A1)

= 23 sq units. (A1)

7

Notes: Othervalidmethodscanbeused.Awardfinal(A1)fortheintegeranswer.

[14]

26. B, orr=

+

2

6

4

4t

(C3)

D, orr=

+

1

3

5

7t

(C3)

Note: AwardC4forB,Dandoneincorrect,C3foronecorrectandnothingelse,C1foronecorrectandone

incorrect,C0foranythingelse.[6]

27. (a)

40

30

25

60

= 60 (30) + 25 40 (M1)

= 800 (A1) (C2)

(b) cos =( ) 2222 40302560

800

++

(M1)(A1)

Note: Trigsolutions:AwardM1forattempttouseacorrectstrategy,A1forcorrectvalues.

cos = 0.246... (A1)= 104.25... (or 255.75...) (A1)

(C4)

She turns through 104 (or 256)

Note: Acceptanswersinradiansie1.82or4.46.[6]

21


22/47

28. (a)

=

7

1OB

=

9

8OC

(A1)(A1)

2

(b)OBOCBCAD ==

(M1)

=

=

2

9

7

1

9

8

(A1)

=

+

=

+

=+=

4

11

5

3

9

8or

4

11

2

9

2

2ADOAOD

d= 11

4

11accept

(A1)

3

(c)

=

=

3

12

7

1

4

11BD

(A1)

1

(d) (i) l:

+

+

=

1

4

7

1or

3

12

7

1tt

y

x

(A2)

(ii) At B, t= 0 by observation (A1)

OR

+

=

3

12

7

1

7

1t

t= 0 (A1) 3

(e)

+

=

3

12

7

1

5

7t

7 + 1 = 12t= 8

t= 32

(A1)

Note:Theequation

+

=

1

4

7

1t

y

x

leadstot=2.

when t= 3

2

,y = 7 +

3

2

(3) (M1)

= 7 2 = 5 (A1)

ie P on line (AG)

22


23/47

OR

5 7 = 3t= 2

t= 32

(A1)

when t= 3

2

,x = 1 + 3

2

12 (M1)

= 1 + 8 = 7 (A1)

ie P on line (AG) 3

(f)

=

=

4

1

9

8

5

7CP

(A1)

3

12

4

1

= 12 +12 = 0 (M1)(A1)Scalar product of non-zero vectors = 0 are perpendicular (R1)(AG)

OR

Geometric approach

CP: m = 4 (A1)

BD: m1 = 4

1

(A1)

mm1 = 4

4

1

= 1 (A1)

Product of gradients is 1 lines (vectors) are perpendicular (R1)(AG)4[16]

29. Direction vectors are a = i 3jand b = ij. (A2)

a b = (1 + 3) (A1) a = 10 , b = 2 (A1)

cos =

=

210

4

ba

ba

(M1)

cos = 20

4

(A1) (C6)[6]

30. (a) (i)

==2

2

1

3OAOBAB

(M1)

=1

5

(A1) (N2)

2

23


24/47

(ii)125AB +=

(M1)

= 26 (= 5.10 to 3 sf) (A1) (N2)2

Note:An answer of 5.1 is subject toAP.

(b) OAODAD =

=

2

2

23

d

=

25

2d

(A1)(A1) 2

(c) (i) EITHER

ADAB90DAB = = 0 or mention of scalar (dot) product. (M1)

25

2

1

5 d

= 0

5d+ 10 + 25 = 0 (A1)

d= 7 (AG)

OR

=

=

2

25ADofGradient

51ABofGradient

d (A1)

51

2

25

d = 1 (A1)

d= 7 (AG)

(ii)

=

23

7OD

(correct answer only) (A1)

3

24


25/47

(d) BCAD = (M1)

=

25

5BC

(A1)

BCOBOC += (M1)

+

=25

5

1

3OC

=

24

2

(A1) (N3)

4

Note: Many other methods, including scale drawing, areacceptable.

(e)650255BCorAD 22 =+=

(A1)

Area = 65026 =( 5.099 25.5)

= 130 (A1) 2[15]

31. (a) (i) BC OC OB

=

6 2= i j (A1)(A1) (N2)

(ii) OD OA BC

= +

2 0 ( 2 )= + = i j i (A1)(A1) (N2) 4

25


26/47

(b) BD OD OB

= 3 3= +i j

(A1)

AC OC OA

= 9 7= i j (A1)

Let be the angle between BD

and AC

+

+=

jiji

jiji

79)33(

)79()33(cos

(M1)

numerator = + 27 21 (= 6) (A1)

denominator

( )18 130 2340= =(A1)

therefore,

6cos

2340 =

82.9 =o

(1.45 rad) (A1) (N3)

6

(c)3 (2 7 )= + +r i j t i j

( )(1 2 ) ( 3 7 )t= + + +i t j

(A1)

(N1) 1

(d) EITHER

)72(3)4(24 jijijiji ++=+++ ts (may be implied) (M1)

4 1 2

2 4 3 7

s t

s t

+ = + + = + (A1)

7 and/or 11t s= = (A1)

Position vector of P is15 46+i j

(A1)

(N2)

OR

7 2 13 or equivalentx y = (A1)

4 14 or equivalentx y = (A1)

15 , 46x y= =(A1)

Position vector of P is 15 46+i j

(A1)

(N2) 4[15]

26


27/47

32. (a) OG = 5i+ 5j 5k A22

(b) BD = 5i+ 5k A2

2

(c) EB = 5i+ 5j 5k A22

Note: Award A0(A2)(A2) if the 5 is consistently omitted.[6]

33. (a) Finding correct vectors, AB =

3

4

AC =

1

3

A1A1

Substituting correctly in the scalar product

ACAB = 4(3) + 3(1) A1= 9 AG 3

(b) | AB | = 5 | AC | = 10 (A1)(A1)

Attempting to use scalar product formula cos BAC = 105

9

M1

= 0.569 (3 s.f) AG

3[6]

34. (a) Attempting to find unit vector (eb) in the direction ofb (M1)

Correct values =

++ 04

3

043

1222

A1

=

0

8.0

6.0

A1

Finding direction vector forb, vb = 18 eb (M1)

b =

0

4.14

8.10

A1

Using vector representation b = b0 + tvb (M1)

27


28/47

=

+

0

4.14

8.10

5

0

0

t

AG 6

(b) (i) t= 0

(49, 32, 0) A1

1

(ii) Finding magnitude of velocity vector (M1)

Substituting correctly vh =222

6)24()48( ++A1

= 54(km h1

) A1 3

(c) (i) At R,

=

t

t

t

t

t

6

2432

4849

5

4.14

8.10

A1

t= 65

(= 0.833) (hours) A1

2

(ii) For substituting t= 65

into expression forb orh M1

(9,12,5) A2 3[15]

35. METHOD 1

Using a b = ab cos (may be implied) (M1)cos

1

2

4

3

1

2

4

3

=

(A1)

Correct value of scalar product

( ) ( ) 214231

2

4

3=+=

(A1)

Correct magnitudes( )

3 225 5 , 5

4 1 = = = (A1)(A1)

2cos

125

=

(A1) (C6)

28


29/47

METHOD 2

325

4

= (A1)

25

1

= (A1)

534

3

=

(A1)

Using cosine rule (M1)

34 25 5 25 5 cos= + (A1)

2

cos 125

= (A1) (C6)[6]

36. (a) (i)

200 600AB

400 200

= (A1)

800

600

=

(A1) (N2)

(ii)2 2AB 800 600 1000

= + = (must be seen) (M1)

unit vector

8001

6001000

=

(A1)

0.8

0.6

=

(AG) (N0) 4Note:A reverse method is not acceptable in show thatquestions.

(b) (i)

0.8250

0.6

=

v

(M1)

200

150

=

(AG) (N0)Note:A correct alternative method is using the given vectorequation with t = 4.

29


30/47

(ii) at 13:00, t= 1

600 2001

200 150

x

y

= + (M1)

400

50 = (A1) (N1)

(iii) AB 1000

=

Time

10004 (hours)

250= =

(M1)(A1)

over town B at 16:00 (4 pm, 4:00 pm)

(Do not accept 16 or 4:00 or 4) (A1)

(N3) 6

(c) Note: There are a variety of approaches. The table shows some of them,with the mark allocation. Use discretion, following this allocation as closelyas possible.

Time for A to B to C

= 9 hours

Distance from A to B to

C

= 2250 km

Fuel used from A to B

= 1800 4 7200 = litres(A1)

Light goes on after

16000 litres

Light goes on after

16000 litres

Fuel remaining

= 9800 litres (A1)

Time for 16 000

litres

)889.8(9

88

1800

16000

==

=

Time remaining is

= )111.0(9

1

= hour

Distance on 16000 litres

2501800

16000 =

)22.2222(9

22222 ==

km

Hours before light

8800

1800

( )8

4 4.8899

= =

Time remaining is

( )1 0.1119

= =hour

(A1)

(A1)

(A1)

Distance

1250

9=

= 27.8 km

Distance to C

= 2250 2222.22

= 27.8 km

Distance

1250

9=

= 27.8 km (A2) (N4) 7

[17]

37. (a) 916 + = 25 = 5 (M1)(A1)(C2)

30


31/47

(b)

=

+

7

6

3

42

1

2

(so B is (6, 7) ) (M1)(A1)

(C2)

(c) r=

+

3

4

1

2t

(not unique) (A2)

(C2)

Note: Award (A1) if r= is omitted, ienot

an equation.[6]

38. (a)

DE =

511

412

=

6

8

(M1)(A1)

(N2)

(b)

DE =22 68 + 3664+=

(M1)

= 10 (A1) (N2)

31


32/47

(c) Vector geometry approach

Using DG = 10 (M1)

(x4)2 + (y 5)2 = 100 (A1)

Using (DG) perpendicular to (DE) (M1)

Leading to

DG =

8

6

,

DG =

8

6

(A1)(A1)

Using

DG =

+ OGDO

(O is the origin) (M1)

G (2, 13), G (10, 3) (accept position vectors) (A1)(A1)

Algebraic approach

gradient of DE = 8

6

(A1)

gradient of DG = 6

8

(A1)

equation of line DG isy 5 =4)(

3

4 x

(A1)

Using DG = 10 (M1)

(x4)2 + (y5)2 = 100 (A1)

Solving simultaneous equation (M1)

G (2, 13), G (10, 3) (accept position vectors) (A1)(A1)Note: Award full marks for anappropriatelylabelled diagram (eg showing that DG =10 ,displacements of 6 and 8), or an accurate

diagram leading to the correct answers.[12]

39. (a) p = 2

+

3

52

12

0

(A1)

=

6

10

(accept any other vector notation, including (10, 6) ) (A1)

(N2)

32


33/47

(b) METHOD 1

(i) equating components (M1)

0 + 5p = 14 + q , 12 3p = 0 + 3q (A1)

p = 3, q =1 (A1)(A1) (N1)(N1)(ii) The coordinates of P are (15, 3) (acceptx = 15,y = 3 ) (A1)

(A1) (N1)(N1)

METHOD 2

(i) Setting up Cartesian equations (M1)

x = 5p x = 14 +q

y =12 3p y = 3q

giving 3x + 5y = 60 3xy = 42 (A1)

Solving simultaneously givesx = 15,y = 3

Substituting to findp and q

,3

1

0

14

3

15,

3

5

12

0

3

15

+

=

+

=

qp

p = 3 q = 1 (A1)(A1)

(N1)(N1)

(ii) From above, P is (15, 3) (acceptx = 15,y = 3 seen above)(A1)(A1) (N1)(N1)

[8]

40. (a)

PQ =

35

A1A1

N2

(b) Using r= a + tb

+

=

3

5

6

1t

y

x

A2A1A1 N4[6]

33


34/47

41. (a) (i) Evidence of subtracting all three components in the correct order M1

eg( ) ( )kjikji +++==

322154OAOBAB

= 2i8j+ 20k AGN0

(ii)

AB = ( ) ( )6.2111721364682082222 ====++

(A1)

u =

( )kji 2082468

1+

A1 N2

++= etc.,925.0370.00925.0,

468

20

468

8

468

2kjikji

(iii) If the scalar product is zero, the vectors are perpendicular. R1

Note: Award R1 for stating the relationshipbetween

the scalar product andperpendicularity, seen

anywhere in the solution.

Finding an appropriate scalar product

OAABorOAu

M1

eg

1468

203

468

82

468

2OA

+

+

=

u

+=

468

20244

( ) 1203822OAAB ++=

0OAABor0OA ==

uA1 N0

34


35/47

(b) (i) EITHER

++

2

211,

2

53,

2

42S

(M1)(A1)

Therefore,OS = 3ij+ 11k (accept (3, 1, 11)) A1 N3

OR

+= AB

2

1OAOS

(M1)

= (2i+ 3j+ k) + 2

1

(2i+ 8j+ 20k) (A1)

OS = 3ij+ 11k A1 N3

(ii) L1 : r= (3ij+ 11k) + t(2i+ 3j+ 1k) A1N1

(c) Using direction vectors (eg2i+ 3j+ 1kand 2i+ 5j3k) (M1)Valid explanation of whyL1 is not parallel toL2 R1

N2

eg. Direction vectors are not scalar multiples of each other.Angle between the direction vectors is not zero or 180.

Finding the angle

d1 d2 d1 d2 .Note: Award R0 for direction vectors arenot equal.

(d) Setting up any two of the three equations (M1)

For each correct equationA1A1

eg3 + 2t= 5 2s, 1 + 3t= 10 + 5s, 11 + t= 10 3s

Attempt to solve these equations (M1)

Finding one correct parameter (s = 1, t= 2) (A1)

P has position vector 7i+ 5j+ 13k A2N4

Notes: Award (M1)A2 if the same parameter is used

for both lines in the initial correctequations.

Award no further marks.[19]

35


36/47

42. (a) (i)

= OAOBAB (A1)

=

5

3

7

5

2

17

=

10

5

10

A1 N2

(ii)222 10510AB ++=

(M1)

= 15 A1 N2

(b) Evidence of correct calculation of scalar product (may be in (i), (ii)

or (iii)) A1

(i)0AEAB =

((6)(2) + 6(4) + 3(4)) A1N1

(ii)0ADAB =

((10)(6) + 5(6) + 10(3)) A1N1

(iii)

0AEAB =

((10)(2) + 5(4) + 10(4)) A1N1

(iv) 90

2or

A1 N1

(c) Volume =

AB

AD

AE (A1)

= 15 9 6

= 810 (cubic units) A1 N2

36


37/47

(d) Setting up a valid equation involving H. There are many possibilities.

eg

=

++=+=

10

5

10

12

4

9

,EHAEOAOH,GHOGOH

z

y

x

(M1)

Using equal vectors (M1)

eg

== ADEH,ABGH

=

+

+

=

=

=

2

1

1

3

6

6

4

4

2

5

3

7

OH,

2

1

1

10

5

10

12

4

9

OH

coordinates of H are (1, 1, 2) A1N3

(e)

=

3

3

18

HB

A1

Attempting to use formula cos

=

HBAG

HBAGP

(M1)

=

=

++++

++342342

108

33181772

31737182

222222

A1

= 0.31578... (A1)

= 6.71P(= 1.25 radians) A1

N3[19]

43. (a)

==

1

2

3

3

5

1

OAOBAB

(M1)

=

2

3

2

AB

A2 N3

37


38/47

(b) Using r= a + tb

+

=

+

=

2

3

2

3

5

1

or

2

3

2

1

2

3

t

z

y

x

t

z

y

x

A1A1A1N3

[6]

44. (a)

=

=

x

x

33OR,

3

1AB

A1A1

N2

(b)( )xx 333ORAB

=

A1

( )09100ORAB ==

x M1

R is

10

3,

10

9

A1A1 N2[6]

45. (a) uv = 8 + 3 +p (A1)

For equating scalar product equal to zero (M1)

8 + 3 +p = 0

p = 11 A1 N3

(b)u

=( ) 74.3,14132 222 =++

(M1)

1414 =qA1

q = ( )74.314 = A1 N2[6]

38


39/47

46. Note: In this question, accept any correct vector notation, including row

vectors eg (1, 2, 3).

(a) (i)PQ

= OQ OP (M1)

= i2j+ 3k A1 N2

(ii) r=

OP +s PQ (M1)

= 5i+ 11j8k+s(i2j+ 3k) A1

= (5 +s) i+ (11 2s)j+ (8 + 3s) k AGN0

(b) If (2,y1,z1) lies onL1 then 5 +s = 2 (M1)

s = 7 A1

y1 = 3,z1 = 13 A1A1 N3

(c) Evidence of correct approach (M1)

eg(5 +s)i+ (11 2s)j+ (8 + 3s) k= 2i+ 9j+13k+ t(i+2j+ 3k)

At least two correct equations A1A1

eg5 +s = 2 + t, 11 2s = 9 + 2t, 8 + 3s = 13 + 3t

Attempting to solve their equations (M1)

One correct parameter (s = 4, t= 3) A1

OT = i+ 3j+ 4k A2 N4

(d) Direction vector forL1 is d1 = i2j+ 3k (A1)Note: Award A1FT for their vector from (a)(i).

Direction vector forL2 is d2 = i2j+ 3k (A1)

d1d2 = 6,1d = ,14 2

d= ,14 (A1)(A1)(A1)

=== 73

14

6

1414

6cos

A1

= 64.6 (= 1.13radians) A1N4

Note: Award marks as per the markschemeif their (correct) direction vectors

give

d1d2 = 6, leading to = 115

(= 2.01 radians).[22]

39


40/47

47. (a) speed =222 1043 ++

(M1)

= 125 = 55 , 11.2, (metres per minute) A1N2

(b) Let the velocity vector be

c

b

a

Finding a velocity vector A2

eg

39

16

3

=

23

10

5

+ 2

c

b

a

,

39

16

3

23

10

5

Dividing by 2 to give

8

3

4

A1

z

y

x

=

23

10

5

+ t

8

3

4

AG

N0

(c) (i) At Q,

7

2

3

+ t

10

4

3

=

23

10

5

+ t

8

3

4

(M1)

Setting up one correct equation A1

eg3 + 3t= 5 + 4t, 2 + 4t= 10 + 3t, 7 + 10t= 23 + 8t

t= 8 (A1)

Correct answer A1

egafter 8 minutes, 13:08 N3

(ii) Substituting fort (M1)

z

y

x

=

7

2

3

+ 8

10

4

3

, or

z

y

x

=

23

10

5

+ 8

8

3

4

x = 27,y = 34,z= 87 or (27, 34, 87), or

87

34

27

A1

N2

40


41/47

(d) For choosing both direction vectors d1 =

10

4

3

and d2 =

8

3

4

(A1)

d1d2 = 104, 1d

= 125 , 2d

= 89 (A1)(A1)(A1)

cos =89125

104

= 0.98601... A1

= 0.167 (radians)(accept = 9.59) A1 N3[17]

48. (a) (i) Evidence of approach

eg

JQ

=

+=

OQJOJQ,

0

0

6

10

7

0

M1

=

10

7

6

JQ

AG N0

(ii)

=

10

7

6

MK

A1 N1

41


42/47

(b) (i) r=

0

0

6

+ t

10

7

6

orr=

10

7

0

+ t

10

7

6

A2

N2

Note: Award A1 if r= is missing.

(ii) Evidence of choosing correct vectors

10

7

6

,

10

7

6

(A1)(A1)

Evidence of calculating magnitudes (A1)(A1)

eg( ) ( ) 222222 10761851076 ++=++

185=

10

7

6

10

7

6

= 36 49 + 100 (= 15) (accept 15) (A1)

For evidence of substitution into the correct formula M1

egcos =

== 0811.0

185

15

185185

15

185185

15accept

= 1.49 (radians), 85.3 A1 N4

(c) METHOD 1

Geometric approach (M1)

Valid reasoning A2

egdiagonals bisect each other,

+= MK

2

1OMOD

Calculation of mid point (A1)

eg

+++

2

100,

2

70,

2

06

=

5

5.3

3

OD

(accept (3,3.5,5)) A1

N3

42


43/47

METHOD 2

Correct approach (M1)

eg

00

6

+ t

107

6

=

107

0

+s

107

6

Two correct equations A1

eg6 6t= 6s,7t= 7 7s, 10t= 10s

Attempt to solve (M1)

One correct parameter

s = 0.5 t= 0.5 A1

=

55.3

3

OD

(accept (3, 3.5, 5)) A1

N3

METHOD 3

Correct approach (M1)

eg

10

7

0

+ t

10

7

6

=

0

7

0

+s

10

7

6

Two correct equations A1

eg6t= 6s, 7 + 7t= 7 7s, 10 + 10t= 10s

Attempt to solve (M1)

One correct parameter

s = 0.5 t= 0.5 A1

=

5

5.3

3

OD

(accept (3, 3.5, 5)) A1

N3[16]

43


44/47

49. (a) (i) evidence of combining vectors (M1)

eg

AB =

OB

OA (or

AD =

AO +

OD in part (ii))

AB =

2

4

2

A1 N2

(ii)

AD =

2

5

2

k

A1 N1

(b) evidence of using perpendicularity scalar product = 0 (M1)

0

2

52

2

42

.. =

kge

4 4(k5) + 4 = 0 A1

4k+ 28 = 0 (accept any correct equation clearly leading to k= 7) A1

k = 7 AG

N0

(c)

AD =

222

(A1)

BC =

11

1

A1

evidence of correct approach (M1)

eg

=

+

+=

1

1

1

2

1

3

,

1

1

1

2

1

3

,BCOBOC

z

y

x

OC

=

1

2

4

A1 N3

44


45/47

(d) METHOD 1

choosing appropriate vectors,

BC,BA

(A1)

finding the scalar productM1

eg2(1) + 4(1) + 2(1), 2(1) + (4)(1) + (2)(1)cos CBA = 0 A1 N1

METHOD 2

BC parallel to

AD

(may show this on a diagram with points labelled) R1

BC

AB(may show this on a diagram with points labelled) R1

CBA = 90

cos CBA = 0 A1 N1[13]

50. pw=pi+ 2pj3pk(seen anywhere) (A1)

attempt to find v +pw (M1)

eg3i+ 4j+ k+ p(i+ 2j3k)

collecting terms (3 +p)i+ (4 + 2p)j+ (1 3p) k A1

attempt to find the dot product (M1)eg1(3 +p) + 2(4 + 2p) 3(1 3p)

setting their dot product equal to 0 (M1)

eg1(3 +p) + 2(4 + 2p) 3(1 3p) = 0

simplifying A1

eg3 +p + 8 + 4p3 + 9p = 0, 14p + 8 = 0

P= 0.571

14

8

A1

N3[7]

45


46/47

51. (a) (i) evidence of approach M1

eg

AO +

OB =

AB, B A

AB =

1

6

4

AG N0

(ii) for choosing correct vectors, (

AO with

AB , or

OA with

BA ) (A1)(A1)

Note: Using

AO with

BA will lead to

0.799. If they then say B A O= 0.799, this is a correct solution.

calculating

AO

AB ,

AB,AO

(A1)(A1)(A1)

egd1d2 = (1)(4) + (2)(6) + (3)(1) (= 19)

( ) ( ) ( ) ,14321d 2221 =++=

( ) ( ) ( )53164d 2222 =++=

evidence of using the formula to find the angle M1

egcos =

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ,164321

136241

222222 ++++

++

...69751.0,5314

19

OAB = 0.799 radians (accept 45.8) A1N3

(b) two correct answers A1A1

eg(1, 2, 3), (3, 4, 2), (7, 10, 1), (11, 16, 0)N2

46


47/47

(c) (i) r=

+

2

4

3

3

2

1

t

A2

N2

(ii) C onL2, so

+

=

2

4

3

3

2

1

5

tk

k

(M1)

evidence of equating components (A1)

eg1 3t= k, 2 + 4t= k, 5 = 3 + 2t

one correct value t= 1, k= 2 (seen anywhere) (A1)

coordinates of C are (2, 2, 5) A1 N3

(d) for setting up one (or more) correct equation using

+

=

1

2

1

0

8

3

5

2

2

p

(M1)

eg3 +p = 2, 8 2p = 2, p = 5

p = 5 A1 N2[18]

52. evidence of equating vectors (M1)

egL1 =L2

for any two correct equations A1A1

eg2 +s = 3 t, 5 + 2s = 3 + 3t, 3 + 3s = 8 4t

attempting to solve the equations (M1)

finding one correct parameter (s = 1, t= 2) A1

the coordinates of T are (1, 3, 0) A1

N3[6]

vectors sl marks

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