SOLUTIONS - Vectors Review 2012

8/3/2019 SOLUTIONS - Vectors Review 2012

1/15

1

1. Required vector will be parallel to (M1)

= (A1)

Hence required equation is r= (A1)(A1) (C4)

Note: Accept alternative answers, eg . [4]

2. (a) = 30 km h1 (A1)

= 39.4 (A1) 2

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

and (A1)(A1)

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

(or ) (M1)

(M1)

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

(c) The Toyundai must continue until its position vector is (M1)

Clearly k= 24, ie position vector . (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

(M1)(A1)

Distance to base camp = (A1)

= 27.2 km

Time to cover this distance = 60 (A1)

= 54.4 minutes = 54 minutes (to the nearest minute)(A1) 5[20]

4

1

1

3

5

4

5

4

4

1t

5

4

1

3s

24

18

22 )16(3616

36!

129

8

18

!

20

9

8

18

12

9

20

9

20

9

481

k

18

24

18

!

4.20

18

6.324

18

4.20

18

30

2.27


2/15

2

3. Vector equation of a line r= a + Pt (M1)

a = , t= (M1)(M1)

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

4. (a) (i) r1 =

t= 0 r1 = (M1)

`r1` = = 20 (A1)

(ii) Velocity vector =

speed = (M1)

= 13 (A1) 4

(b)

(M1)

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

OR

(M1)

5x 80 = 144 12y (A1)

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

OR

x = 16 + 12t,y = 12 5tt= (M1)

x = 16 + 12 (A1)

5x = 80 + 144 12 y

5x + 12y = 224 (A1)(AG) 3

(c) v1 = v2 = (M1)

v1.v2 = (M1)

= 30 30

v1.v2 = 0 (A1)

U = 90 (A1) 4

(d) (i) (M1)

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

OR

0

0

3

2

-

-

5

12

12

16t

-

12

16

)1216( 22

-

5

12

))5(12( 22

-

-

!

-

5

12

12

16

ty

x

-

-

-

-

!

-

-

5

12.

12

5

12

6.

12

5.

12

5t

y

x

5

12

12

16

!

yx

5

12 y

5

12 y

-

5

12

-

6

5.2

-

-

6

5.2

.5

12

-

-

!

-

-

5

12.

5

23

5

12.

y

x


3/15

3

6x 138 = 2.5y + 12.5 (M1)

12x 276 = 5y + 25

12x 5y = 301 (A1)

(ii) (M1)

169x = 4732

x = 28, y = (12 28 301) 5 = 7

(28, 7) (A1)(A1) 5

Note: Accept any correct method for solving simultaneous

equations.

(e) 16 + 12t= 23 + 2.5t 9.5t= 7 (M1)

12 5t= 5 + 6t 17 = 11t (M1)

(A1)

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

OR

r1 = (M1)

t= 1 (A1)

When t= 1 r2 = (A1)(R1)

OR

r2 = (M1)

t= 2 (A1) 4[20]

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

(A1) 2

(b) (i) Velocity vector: (M1)

= (km h1

) (A1)

(ii) Speed = ; (M1)

= 10; 10 km h1 (A1) 4

(c) EITHER (M1)

Note: Award (M1) for both equations.

y = 28 8 (M1)(A1)

6

5

5.2

23 !

yx

!

!

!

!

361260144

11206025

301512

224125

yx

yx

yx

yx

11

17

5.9

7{

-

-

!

-

-

5

12

12

16

7

28

7

28t

!

!

55

1212

t

t

-

{

-

!

-

-

7

28

1

5.25

6

5.2

5

23

-

-

!

-

-

6

5.2

5

23

7

28

7

28t

!

v

!

20

6

8

61

28

0

y

x

01 !!

tty

x

y

x

!

8

6

28

0

20

6

))8(6( 22

!

!

ty

tx

828

6

6

x


4/15

4

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

4x + 3y = 84 (a1) 4

OR

(M1)

(M1)(A1)

4x + 3y = 84 (A1) 4

(d) They collide if 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 for some t,

(A1)

(A1)

They collide at 15:00 (A1) 4

(e) (M1)

= (M1) 2

= (AG)

(f) At t= 3, (M1)

(A1)

(A1)

= 26

26 km apart (A1) 4[20]

7. (a) At t= 2, (M1)

BB

!

8

6.

28

0

8

6.

y

x

!

6

8.

28

0

6

8.

y

x

4

18

!

8

6

28

0

4

18t

!

!

t

t

8284

618

and

!

!

248

3

t

t

and

!

!

3

3

t

t

and

!

12

5)1(

4

18t

y

x

12124

5518

t

t

12

5

8

13t

!

v

v!

28

28

1238

5313

y

x

!

24

10

4

18

28

28

)676()2410( 22 !

!

2

4.3

1

7.02

0

2


5/15

5

Distance from (0, 0) = = 3.94 m (A1) 2

(b) (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 and x 0.7y = 2 (M1)

x = 5.86 and y = 5.52 (A1)(A1) 3

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

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= (M1)

=

= 6.84 m (A1)

Speed of the motorcycle =

= 1.24 m s1

(A1) 5[14]

8. Direction vector = (M1)

= (A1)

(A2)

OR

(A2)(C4) [4]

9. (a) (i) OA = = 250 (A1)

unit vector = (M1)(AG)

(ii) (M1)(A1)

(iii) t= hr (= 50 min) (A1) 5

22 24.3

2217.0

1

7.0!

!!

29

160and

29

170or yx

!

1

7.0

0

2

52.5

86.5

22 )52.3()86.5(2

0

52.5

86.5

!

73.46

52.5

84.6!

t

d

3

1

5

6

2

5

!

2

5

3

1t

y

x

!

2

5

5

6t

y

x

!

70

240OA 22 70240

!

28.096.0

70240

2501

!

!

84

288

28.0

96.0300v

6

5

288

240!


6/15

6

(b) (A1)

AB = = 300

cos = (M1)

= 0.936 (A1)

= 20.6 (A1) 4

(c) (i) (A1)

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

nB (AG)

(iii) Projection of in the direction ofn is

XY = = 75 (M1)(A1)(A1) 6

(d) AX = = 195 (A1)

AY = = 180 km (M1)(A1) 3

[18]

10. x = l 2t (A1)

y = 2 + 3t (A1)

(M1)

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

11.

(a) = ts (M1)

=

= (A1)

= (M1)

u v =

!

!

180

240

70250

240480AB

22 180240

)300)(250(

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

ABOA

ABOA !

v

y

!

!

168

99

70238

240339AX

y

180

240

4

3

AB

AX

5

672297

4

3

168

99

5

1 !

y

22 16899

22 75195

3

2

2

1 yx!

T

S

U

V

y

x

ST

2

27

7

9

9

VU ST

9

9


7/15

7

v =u

= (A1)

V(4, 6) (A1) 5

(b) Equation of (UV): direction is = (A1)

r= or (A1)

OR

r= or (A1) 2

(c) is on the line because it gives the same value ofP, for both thex

and ycoordinates. (R1)

For example, 1 = 5 + 9P P =

11 = 15 + 9P P = (A1) 2

(d) (i) (M1)

= (A1)

= 2 = 2 (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) Fora = 3

= = te = (A1)(A1)

cos = (M1)

= (A1)

=

Therefore, = 157 (3 sf) (A1) 10[19]

9

9

!

6

4

9

9

15

5

11or

99 k

9

9

15

5P

1

1

15

5P

9

9

6

4P

1

1

6

4P

11

1

9

4

9

4

!

11

1

17EW

a

6

1a

EW 13 361 2 a 13

EW

6

4ET

4

6

TEW

ETEW

ETEW

5252

2424

13

12

TEW


8/15

8

13. 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)

(A1)(A1) (C6)

METHOD 2

(changing to Cartesian coordinates)

2x + 3y = 13, x 4y = 10 (M1)(A1)(A1)

Attempt to solve the system (M1)

(A1)(A1) (C6)

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

14. B, orr= (C3)

D, orr= (C3)

Note: AwardC4forB,Dandoneincorrect,

C3foronecorrectandnothingelse,C1foronecorrectandoneincorrect,C0foranythingelse.

[6]

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

ab = (1 + 3) (A1)

a = , b = (A1)

cos = (M1)

cos = (A1) (C6)

[6]

17. (a) (i)

(A1)(A1) (N2)

(ii) (A1)(A1) (N2) 4

(b)

(A1)

(A1)

Let be the angle between and

!

3

2OP

!

3

2OP

2

6

4

4t

1

3

5

7t

10 2

!

210

4

ba

ba

20

4

BC OC OBp p p

!

6 2! i j

OD OA BCp p p

! 2 0 ( 2 )! ! i j i

BD OD OBp p p

! 3 3! i j

AC OC OAp p p

! 9 7! i j

U BDp

ACp


9/15

9

(M1)

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

denominator (A1)

therefore,

(1.45 rad) (A1) (N3) 6

(c) (A1) (N1) 1

(d) EITHER

(may be implied) (M1)

(A1)

(A1)

Position vector of P is (A1) (N2)

OR

(A1)

(A1)

(A1)

Position vector of P is (A1) (N2) 4

[15]

19.

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

Correct values = A1

= A1

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

b = A1

Using vector representation b = b0 + tvb (M1)

= AG 6

(b) (i) t= 0 (49, 32, 0) A1 1

y!

jiji

jiji

79)33(

)79()33(cos

18 130 2340! !

6cos2340

U !

82.9U ! $

3 (2 7 )! r i j t i j (1 2 ) ( 3 7 )t! i t j

)72(3)4(24 jijijiji ! ts

4 1 2

2 4 3 7

s t

s t

!

!

7 and/or 11t s! !

15 46i j

7 2 13 or equivalentx y !

4 14 or equivalentx y !

15 , 46x y! !

15 46i j

0

4

3

043

1222

0

8.0

6.0

0

4.14

8.10

0

4.14

8.10

5

0

0

t


10/15

10

(ii) Finding magnitude of velocity vector (M1)

Substituting correctly vh = A1

= 54(km h1

) A1 3

(c) (i) At R, A1

t= (= 0.833) (hours) A1 2

(ii) For substituting t= into expression forb orh M1

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

20. METHOD 1

Using ab = ab cosU (may be implied) (M1)

(A1)

Correct value of scalar product (A1)

Correct magnitudes (A1)(A1)

(A1) (C6)

METHOD 2

(A1)

(A1)

(A1)

Using cosine rule (M1)

(A1)

(A1) (C6)

[6]

21. (a) (i) (A1)

222 6)24()48(

!

t

t

t

t

t

6

2432

4849

5

4.14

8.10

65

65

Ucos1

2

4

3

1

2

4

3

!

y

214231

2

4

3!vv!

y

3 2

25 5 , 54 1

! ! !

2cos

125U

!

325

4

!

25

1

!

534

3

!

34 25 5 25 5cosU!

2cos

125U !

200 600AB

400 200

p !


11/15

11

(A1) (N2)

(ii) (must be seen) (M1)

unit vector (A1)

(AG) (N0) 4

(b) (i) (M1)

(AG) (N0)

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

(ii) at 13:00, t= 1

(M1)

(A1) (N1)

(iii)

Time (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 closely as 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= litres

(A1)

Light goes on after16000 litres

Light goes on after16000 litres

Fuel remaining= 9800 litres (A1)

Time for 16 000

litres

Time remaining is

= hour

Distance on 16000 litres

km

Hours before light

Time remaining is

hour

(A1)(A

1)

(A1)

800

600

!

2 2AB 800 600 1000p

! !

8001

6001000

!

0.8

0.6

!

0.8250

0.6

!

v

200

150

!

600 2001

200 150

x

y

!

400

50

!

AB 1000p

!

10004 (hours)

250! !

1800 4 7200v !

)889.8(9

88

1800

16000

!!

!

)111.0(9

1!

2501800

16000v!

)22.2222(9

22222 !!

8800

1800

8

4 4.8899

! !

1 0.1119

! !


12/15

12

Distance

= 27.8 km

Distance to C= 2250 2222.22

= 27.8 km

Distance

= 27.8 km (A2) (N4) 7

[17]

22. (a) = = 5 (M1)(A1) (C2)

(b) (so B is (6, 7) ) (M1)(A1) (C2)

(c) r= (not unique) (A2) (C2)

Note: Award (A1) if r= is omitted, ie not

an equation.[6]

24. (a) = A1A1 N2

(b) Using r= a + tb

A2A1A1 N4

[6]

25. (a) (M1)

A2 N3

(b) Using r= a + tb

A1A1A1 N3

[6]

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

For equating scalar product equal to zero (M1)

8 + 3 +p = 0

p = 11 A1 N3

(b) = (M1)

A1

q = A1 N2

[6]

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

1250

9! v

1250

9! v

916 25

!

76

342

12

3

4

1

2t

p

PQ

3

5

!

3

5

6

1t

y

x

!!ppp

1

2

3

3

5

1

OAOBAB

!p

2

3

2

AB

!

!

2

3

2

3

5

1

or

2

3

2

1

2

3

t

z

y

x

t

z

y

x

u 74.3,14132 222 !

1414 !q

74.314 !


13/15

13

eg = (or = + in part (ii))

= A1 N2

(ii) = A1 N1

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

4 4(k 5) + 4 = 0 A1

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

k = 7 AG N0

(c) = (A1)

= A1

evidence of correct approach (M1)

eg

= A1 N3

(d) METHOD 1

choosing appropriate vectors, (A1)

finding the scalar product M1

eg2(1) + 4(1) + 2(1), 2(1) + (4)(1) + (2)(1)

cos = 0 A1 N1

METHOD 2

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

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

= 90r

p

ABp

OBp

OAp

ADp

AOp

OD

p

AB

2

4

2

p

AD

2

5

2

k

0

2

5

2

2

4

2

.. !

y

kge

p

AD

2

2

2

p

BC

1

1

1

!

!

ppp

1

1

1

2

1

3

,

1

1

1

2

1

3

,BCOBOC

z

y

x

p

OC

1

2

4

pp

BC,BA

CBA

p

BCp

AD

p

BCp

AB

CBA


14/15

14

cos = 0 A1 N1[13]

30. pw =pi+ 2pj 3pk(seen anywhere) (A1)

attempt to find v +pw (M1)

eg3i+ 4j+ k+ p(i+ 2j 3k)

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 + 4p 3 + 9p = 0, 14p + 8 = 0

P= 0.571 A1 N3

[7]

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

eg + = B A

= AG N0

(ii) for choosing correct vectors, ( with , or with

) (A1)(A1)

Note: Using with will lead to

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

calculating , (A1)(A1)(A1)

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

evidence of using the formula to find the angle M1

egcosU =

= 0.799 radians (accept 45.8r) A1 N3

(b) two correct answers A1A1

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

CBA

14

8

p

AOp

OBp

AB,

p

AB

1

6

4

p

AOp

ABp

OAp

BA

p

AOp

BA

A

p

AOp

ABpp

AB,AO

,14321d 2221 !! 53164d 2222 !!

,

164321

136241222222

...69751.0,5314

19

OAB


15/15

15

(c) (i) r= A2 N2

(ii) C on L2, so (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

(M1)

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

p = 5 A1 N2[18]

32. 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]

2

4

3

3

2

1

t

!

2

4

3

3

2

1

5

tk

k

!

1

2

1

0

8

3

5

2

2

p

Documents

SOLUTIONS - Vectors Review 2012