1.) (a) correct equation in the form r = a + tb A2 N2 2

2

3

1

2

1

1

tr

(b) (i) attempt to substitute t = 2 into the equation (M1)

e.g.

2

3

1

2

2

1

1

,

4

6

2

OP

2

5

33 A1 N2

(ii) correct substitution into formula for magnitude A1

e.g. 222222 253,)2(53

38OP A1 N14

[6]

2.) (a) (i)

4

0

3

BA A1 N1

(ii) evidence of combining vectors (M1)

e.g.

4

0

3

6

2

2

,ACBA,ACBCAB

2

2

1

BC A1 N23

(b) (i) METHOD 1

finding BC,BA,BCBA (A1)(A1)(A1)

e.g. 3BC,43BA,24013BCBA 22

substituting into formula for cos M1

e.g. 35

5,

4033

2401322

cos

3

1

15

5CBA

A1 N3

METHOD 2

finding BC,BA,AC (A1)(A1)(A1)

e.g. 3BC,43AB,622AC 22222

substituting into cosine rule M1

e.g. 30

44925,

352

4435222

cos

3

1

30

10CBA

A1 N3

(ii) evidence of using Pythagoras (M1)

e.g. right-angled triangle with values, sin2 x + cos2 x =1

sin

3

22

3

8CBA

A1 N27

(c) (i) attempt to find an expression for CD (M1)

e.g. 2222

222 54CD,5)4( pp

correct equation A1

e.g. 5054,505)4( 222222 pp

92 p A1

3p AG N0

(ii) evidence of scalar product (M1)

e.g. BCCD,

2

2

1

3

5

4

correct substitution

e.g. –4 × 1+ 5 × 2 + 3 × –2, –4 + 10 – 6 A1

0BCCD A1

CD is perpendicular to BC AG N06[16]

3.) (a) evidence of appropriate approach (M1)

e.g.

14

34ABAC ,

31BC A1 N22

(b) METHOD 1

31AD (A1)

correct approach A1

e.g.

13

31ABAD ,

2

2BD AG N02

METHOD 2

Recognizing BACD (A1)

correct approach A1

e.g.

13

31CDBC ,

2

2BD AG N02

(c) METHOD 1

evidence of scalar product (M1)

e.g.

4

422ACBD ,

correct substitution A1

e.g. (–2)(4) + (2)(4), –8 + 8

0ACBD A1

therefore vectors BD and AC are perpendicular AG N03

METHOD 2

attempt to find angle between two vectors (M1)

e.g.ab

ba

correct substitution A1

e.g.

0cos,328

4242

°90 A1

therefore vectors BD and AC are perpendicular AG N0[7]

4.) (a) appropriate approach (M1)

e.g. AB, OBAD

1

1

1

AB A1 N22

(b) any correct equation in the form r = a + tb A2 N22

where b is a scalar multiple of

1

1

1

e.g. kjikjirrr

t

t

t

t

t 522,

5

2

2

,

1

1

1

4

1

1

(c) choosing correct direction vectors

1

1

1

,

3

1

2

(A1)(A1)

finding scalar product and magnitudes (A1)(A1)(A1)

scalar product =1× 2 + –1×1+1×3 (= 4)

magnitudes 74.3914,.73.1111 222

substitution into

vu

vu.

vu

v.u

vu

vu. sinnotbut,accept M1

e.g. 42

4cos,

312111

311121cos

222222

θ = 0.906 (51.9˚) A1 N57

(d) METHOD 1

1

1

1

4

1

1

from tr

appropriate approach (M1)

e.g. p = r, st

3

1

2

7

4

2

1

1

1

4

1

1

,

two correct equations A1A1

e.g. 1 + t = 2 + 2s, –1 – t = 4 + s, 4 + t = 7 + 3s

attempt to solve (M1)

one correct parameter A1

e.g. t = –3, s = –2

C is (–2, 2, 1) A1 N36

METHOD 2

1

1

1

5

2

2

from tr 1

appropriate approach (M1)

e.g. p = r ,

3

1

2

7

4

2

1

1

1

5

2

2

st

two correct equations A1A1

e.g. 2 + t = 2 + 2s, –2 – t = 4 + s, 5 + t = 7 + 3s

attempt to solve (M1)

one correct parameter A1

e.g. t = –4, s = –2

C is (–2, 2, 1) A1 N36[17]

5.) (a) any correct equation in the form r = a + tb (accept any parameter) A2 N2

e.g. r =

8

1

2

25

5

8

t

Note: Award A1 for a + tb, A1 for L = a + tb, A0 for r = b + ta.

(b) recognizing scalar product must be zero (seen anywhere) R1e.g. a • b = 0

evidence of choosing direction vectors

k

2

7

,

8

1

2

(A1)(A1)

correct calculation of scalar product (A1)e.g. 2(–7) + 1(–2) – 8k

simplification that clearly leads to solution A1e.g. –16 – 8k, –16 – 8k =0

k = –2 AGN0

(c) evidence of equating vectors (M1)

e.g. L1 = L3,

2

2

7

3

0

5

8

1

2

25

3

3

qp

any two correct equations A1A1e.g. –3 + 2p = 5 – 7q, –1 + p = –2q, –25 – 8p = 3 –2q

attempting to solve equations (M1)finding one correct parameter (p = –3, q = 2) A1

the coordinates of A are (–9, –4, –1) A1N3

(d) (i) evidence of appropriate approach (M1)

e.g.

1

4

9

25

5

8

AB,OBABOA

26

1

1

AB A1 N2

(ii) finding

2

2

7

AC A1

evidence of finding magnitude (M1)

e.g.222 227AC

57AC A1N3

[18]

6.) (a) evidence of appropriate approach (M1)

e.g.

3

2

6

2

3

2

,ACBABC

1

1

8

BC A1N2

(b) attempt to find the length of AB (M1)

)7499436(3)2(6AB 222 (A1)

unit vector is

7

37

27

6

3

2

6

7

1A1N2

(c) recognizing that the dot product or cos θ being 0 implies perpendicular (M1)

correct substitution in a scalar product formula A1

e.g. (6) × (–2) + (–2) × (–3) + (3) × (2), cos θ =177

6612

correct calculation A1e.g. ACAB = 0, cos θ = 0

therefore, they are perpendicular AGN0[8]

7.) (a) (i) (3, –4, 0) A1 N1

(ii) choosing velocity vector

1

3

2

(M1)

finding magnitude of velocity vector (A1)

e.g. 194,13)2( 222

speed = 3.74 ( 14 ) A1N2

(b) (i) substituting p = 7 (M1)B = (–11, 17, 7) A1 N2

(ii) METHOD 1

appropriate method to find BAorAB (M1)

e.g. OBAO , A – B

7

21

14

BAor

7

21

14

AB (A1)

distance = 26.2 )147( A1N3

METHOD 2

evidence of applying distance is speed × time (M2)e.g. 3.74 × 7distance = 26.2 )147( A1N3

METHOD 3

attempt to find AB2, AB (M1)

e.g. (3 – (–11))2 + (–4 – 17)2 + (0 – 7)2, 222 )70()174())11(3(

AB2 = 686, AB = 686 (A1)distance AB = 26.2 ( 147 ) A1N3

(c) correct direction vectors

a

2

1–and

1

3

2

(A1)(A1)

a

a

a

2

1

1

3

2

,52

12

= a + 8 (A1)(A1)

substituting M1

e.g. cos 40° =514

82

a

a

a = 3.21, a = –0.990 A1A1N3[16]

8.) (a) (i) correct approach A1

e.g.

0

1

2

5,OAOC

2

4AC AG N0

(ii) appropriate approach (M1)

e.g. D – B,

5

1

1

4, move 3 to the right and 6 down

6

3BD A1N2

(iii) finding the scalar product A1e.g. 4(3) + 2(–6), 12 – 12

valid reasoning R1e.g. 4(3) + 2(–6) = 0, scalar product is zeroAC is perpendicular to BD AGN0

(b) (i)correct “position” vector for u; “direction” vector for vA1A1N2

e.g. u =

1

2,

2

4;

0

1,

2

5vvu

accept in equation e.g.

2

4

2

5t

(ii) any correct equation in the form r = a + tb, where b = BD

e.g. r =

2

1

1

4,

6

3

5

1t

y

xt A2N2

(c) METHOD 1

substitute (3, k) into equation for (AC) or (BD) (M1)e.g. 3 = 1 + 4s, 3 = 1 + 3t

value of t or s A1

e.g. s =3

1,

3

2,

2

1,

2

1 t ,

substituting A1

e.g. k = 0 + )2(2

1,

k = 1 AGN0

METHOD 2

setting up two equations (M1)e.g. 1 + 4s = 4 + 3t, 2s = –1 – 6t; setting vector equations of lines equal

value of t or s A1

e.g. s =3

1,

3

2,

2

1,

2

1 t

substituting A1

e.g. r =

6

3

3

1

1

4,

k = 1 AGN0

(d)

2

1PD (A1)

)5(12PD 22 (A1)

)20(24AC 22 (A1)

area =

520

2

1PDAC

2

1M1

= 5 A1N4[17]

9.) correct substitutions for v • w; │v│; │w│ (A1)(A1)(A1)

e.g. 2k + (– 3) × (–2) + 6 × 4, 2k + 30; 20,4)2(;49,6)3(2 2222222 kk

evidence of substituting into the formula for scalar product (M1)

e.g.207

3022

k

k

correct substitution A1

e.g.207

302

3

πcos

2

k

k

k = 18.8 A2 N5[7]

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

e.g. ,OQPOPQ Q – P

1

2

1

PQ A1 N2

(ii)

4

2

2

PR A1N1

(b) METHOD 1

choosing correct vectors PRandPQ (A1)(A1)

finding PR,PQ,PRPQ (A1) (A1)(A1)

PRPQ = –2 + 4 + 4 (= 6)

)24(422PR),6(12)1(PQ 222222

substituting into formula for angle between two vectors M1

e.g.246

6QPRcos

simplifying to expression clearly leading to2

1A1

e.g.12

6,

144

6,

626

6

2

1QPRcos AGN0

METHOD 2

evidence of choosing cosine rule (seen anywhere) (M1)

3

0

3

QR A1

24PRand6PQ,18QR (A1)(A1)(A1)

2462

)18()24()6(QPRcos

222

A1

24

12

24

18–246QPRcos A1

2

1QPRcos AGN0

(c) (i) METHOD 1

evidence of appropriate approach (M1)e.g. using 1QPRcosQPRsin 22 , diagramsubstituting correctly (A1)

e.g.2

2

1–1QPRsin

2

3

4

3QPRsin A1N3

METHOD 2

since 60P,2

1Pcos (A1)

evidence of approache.g. drawing a right triangle, finding the missing side (A1)

2

3Psin A1N3

(ii) evidence of appropriate approach (M1)

e.g. attempt to substitute into2

1ab sin C

correct substitution

e.g. area =2

3246

2

1 A1

area = 33 A1N2[16]

11.) finding scalar product and magnitudes (A1)(A1)(A1)scalar product = 12 – 20 – 15 (= 23)magnitudes = )50,50()3()5(4,543 222222 substitution into formula M1

e.g. cos θ =))3()5(4()543(

152012222222

cos θ =50

23 (= –0.46) A2 N4

[6]

12.) (a) L1: r =

1

0

0

0

1

8

t A2 N2

(b) evidence of equating r and OA (M1)

e.g.

5

1

2

1

4

2

9

2

6

s , A = r

one correct equation A1e.g. 6 = 2 + 2s, 2 = 4 – s, 9 = –1 + 5s, s=2 A1

evidence of confirming for other two equations A1e.g. 6 = 2 + 4, 2 = 4 – 2, 9 = –1 + 10so A lies on L2 AGN0

(c) (i) evidence of approach M1

e.g.

1

0

0

0

1

8

5

1

2

1

4

2

ts L1 = L2

one correct equation A1e.g. 2 + 2s = 8, 4 – s = 1, –1 + 5s = t

attempt to solve (M1)finding s = 3 A1

substituting M1

e.g.

5

1

2

3

1

4

2

OB

14

1

8

OB AGN0

(ii) evidence of appropriate approach (M1)

e.g. OAOBAB,OBAOAB

5

1

2

AB A1N2

(d) evidence of appropriate approach (M1)e.g. DCAB

correct values A1

e.g.

z

y

x

z

y

x

4

1

2

5

1

2

,

4

1

2

5

1

2

,

4

1

2

5

1

2

OD

9

2

0

OD A1N2

[16]

13.) evidence of appropriate approach (M1)

e.g.

1

5

3

2

2

9

2

3

5

1

3

2

ts

two correct equations A1A1e.g. 2 + 5s = 9 – 3t, 3 – 3s = 2 + 5t, –1 + 2s = 2 – t

attempting to solve the equations (M1)one correct parameter s = 2, t = –1 A1

P is (12, –3, 3)

3

3-

12

accept A1 N3

[6]

14.) (a) evidence of equating scalar product to 0 (M1)2 × 3 + 3 × (–1) + (–1) × p = 0 (6 – 3 – p = 0, 3 – p = 0)A1p = 3 A1 N2

(b) evidence of substituting into magnitude formula (M1)

e.g. 251 2 q , 1 + q2 + 25

setting up a correct equation A1

e.g. 42251 2 q , 1 + q2 + 25 = 42, q2 = 16

q = ±4 A1N2[6]

15.) (a) evidence of correct approach A1

e.g.

5

1

2

8

3

3

,OPOQPQ

3

2

1

PQ AGN0

(b) (i) correct description R1 N1

e.g. reference to

8

3

3

being the position vector of a point on the line,

a vector to the line, a point on the line.

(ii) any correct expression in the form r = a + tb A2N2

where a is

8

3

3

, and b is a scalar multiple of

3

2

1

e.g. r =

s

s

s

t

68

43

23

,

3

2

1

8

3

3

r

(c) one correct equation (A1)e.g. 3 + s = –1, –3 – 2s = 5

s = –4 A1p = –4 A1N2

(d) one correct equation A1e.g. –3 + t = –1, 9 – 2t = 5

t = 2 A1substituting t = 2e.g. 2 + 2q = –4, 2q = –6 A1

q = –3 AGN0

(e) choosing correct direction vectors

3

2

1

and

3

2

1

(A1)(A1)

finding correct scalar product and magnitudes (A1)(A1)(A1)

scalar product (1)(1) + (–2)(–2) + (–3)(3) (= –4)magnitudes 14)3()2(1,143)2(1 232222 evidence of substituting into scalar product M1

e.g. cos θ =...741.3...741.3

4

θ = 1.86 radians (or 107°) A1N4[17]

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

e.g. 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

5

2

2

4

2

..

kge

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) AD =

2

2

2

(A1)

BC =

1

1

1

A1

evidence of correct approach (M1)

e.g.

1

1

1

2

1

3

,

1

1

1

2

1

3

,BCOBOC

z

y

x

OC =

1

2

4

A1 N3

(d) METHOD 1

choosing appropriate vectors,

BC,BA (A1)

finding the scalar product M1e.g. −2(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]

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

attempt to find v + pw (M1)e.g. 3i + 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)e.g. 1(3 + p) + 2(4 + 2p) − 3(1 − 3p)

setting their dot product equal to 0 (M1)e.g. 1(3 + p) + 2(4 + 2p) − 3(1 − 3p) = 0

simplifying A1e.g. 3 + p + 8 + 4p − 3 + 9p = 0, 14p + 8 = 0

P = − 0.571

14

8A1 N3

[7]

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

e.g.

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

p- 0.799. If they then say OAB ˆ= 0.799, this is a correct solution.

calculating

AO • AB ,

AB,AO (A1)(A1)(A1)

e.g. d1•d2 = (−1)(−4) + (2)(6) + (−3)(−1) (= 19)

,14321 2221 d

53164 2222 d

evidence of using the formula to find the angle M1

e.g. cosq =

,

164321

136241222222

...69751.0,5314

19

OAB = 0.799 radians (accept 45.8) A1 N3

(b) two correct answers A1A1e.g. (1, − 2, 3), (−3, 4, 2), (−7, 10, 1), (−11, 16, 0) N2

(c) (i) r =

2

4

3

3

2

1

t A2 N2

(ii) C on L2, so

2

4

3

3

2

1

5

tk

k

(M1)

evidence of equating components (A1)

e.g. 1 − 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)

e.g. 3 + p = −2, −8 − 2p = 2, − p = 5

p = − 5 A1 N2[18]

19.) evidence of equating vectors (M1)e.g. L1 = L2

for any two correct equations A1A1e.g. 2 + 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]

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

e.g.

1015

26

69

A,–B,OBAO

5

4

3

AB (accept (3, 4, 5)) A1 N2

(ii) evidence of finding the magnitude of the velocity vector M1

e.g. speed = 222 543

speed = )25(50 A1 N1

(b) correct equation (accept Cartesian and parametric forms) A2 N2

e.g. r =

5

4

3

15

6

9

,

5

4

3

10

2

6

tt r

[6]

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

e.g. B – A,

3

2

1

4

4

6

,OBAO

1

2

5

AB AG N0

(ii) evidence of approach (M1)

e.g. D – A,

3

2

1

5

5

2

,ODAO

2

3

1

AD A1 N2

(iii) evidence of approach (M1)e.g. ADABAC correct substitution A1

e.g.

2

3

1

1

2

5

AC

3

5

6

AC AG N0

(b) evidence of combining vectors (there are at least 5 ways) (M1)e.g. ODOCAB,ADOBOC,ACOAOC correct substitution A1

e.g.

6

7

7

3

5

6

3

2

1

OC

coordinates of C are (7, 7, 6) A1 N1

(c) (i) evidence of using scalar product on ADandAB (M1)

e.g. ADAB = 5(1) + 2(3) + 1(2)

ADAB = 13 A1 N2

(ii) AB = 5.477..., AD = 3.741... (A1)(A1)

evidence of using cos A =ADAB

ADAB (M1)

correct substitution A1

e.g. cos A =...493.20

13

A = 0.884 (50.6°) A1 N3

(d) METHOD 1

evidence of using area =

BADsinABAD2

12 (M1)

correct substitution A1

e.g. area = 2

...883.0sin...)477.5...)(741.3(2

1

area = 15.8 A1 N2

METHOD 2

evidence of using area = b × h (M1)finding height of parallelogram A1

e.g. h = 3.741... × sin 0.883... (= 2.892...), h = 5.477... × sin 0.883... (= 4.234...)area = 15.8 A1 N2

[18]

22.) (a) (i) evidence of combining vectors (M1)e.g. OAOBAB

3

2AB A1 N2

(ii)

2

3AC A1N1

(b) (i) ACAB = (–2)(3) + (–3)(–2) = 0 A1 N1

(ii) scalar product 0 = perpendicular, θ = 90° (R1)sin θ = 1 A1N2

[6]

23.) (a) Using direction vectors u =

2

10

6

and

10

6

2

v (M1)

140410036,140100364 vu A1A1

u • v = 12 + 60 – 20 = 52 A1

cos θ =140140

52A1

=140

52AG N0

(b) (i) For substituting s = 1 (M1)Correct calculations (A1)9 + 1(–2) = 7, 4 + 1(6) = 10, –6 + 1(10) = 4

position vector of P is

4

10

7

A1 N3

(ii) For substituting into the equation

2

10

6

2

20

1

4

10

7

t (M1)

For one correct equation A1e.g. 7 = 1 – 6tSolving gives t = –1 A1

verify for second coordinate, 10 = 20 + (–1)(10) A1verify for third coordinate, 4 = 2 + (–1)(–2) A1Thus, P is also on L2. AGN0

(c) k

30

6

10

6

2

x (M1)

–2k = 6

k = –3 A1x = –3 × 6 = –18 A1N2

[16]

24.) (a) Finding correct vectors,

1

3AC

3

4AB A1A1

Substituting correctly in the scalar product ACAB = 4(–3) + 3(1) A1= –9 AG N0

(b) 10AC5AB (A1)(A1)

Evidence of using scalar product formula M1

e.g.105

9CABcos

= –0.569 (3 s.f.)

CAB = 2.47 (radians), 125° A1N3[7]

25.) (a) Attempting to find unit vector (eb) in the direction of b (M1)

Correct values =

0

4

3

043

1222

A1

=

0

8.0

6.0

A1

Finding direction vector for b, vb = 18 × eb (M1)

b =

0

4.14

8.10

A1

Using vector representation b = b0 + tvb (M1)

=

0

4.14

8.10

5

0

0

t AG N0

(b) (i) t = 0 (49, 32, 0) A1 N1

(ii) Finding magnitude of velocity vector (M1)

Substituting correctly vh = 222 6)24()48( A1

= 54 (km h–1) A1N2

(c) (i) At R,

t

t

t

t

t

6

2432

4849

5

4.14

8.10

A1

t =6

5(= 0.833)(hours) A1 N1

(ii) For substituting t =6

5 into expression for b or h (M1)

(9, 12, 5) A2N3

[15]

26.) (a) u • v = 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 q A1

q = 74.314 A1 N2[6]

27.) Note: In this question, accept any correct vector notation, including rowvectors eg (1,- 2, 3).

(a) (i) PQ = OQ − OP (M1)

= i − 2j + 3k A1 N2

(ii) r =

OP + s PQ (M1)

= −5i + 11j −8k + s(i − 2j + 3k) A1

= (−5 + s) i + (11 − 2s)j + (−8 + 3s) k AG N0

(b) If (2, y1, z1) lies on L1 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 A1A1eg −5 + 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 for L1 is d1 = i − 2j + 3k (A1)

Note: Award A1FT for their vector from (a)(i).

Direction vector for L2 is d2 = i − 2j + 3k (A1)

d1 • d2 = 6, 1d = ,14 2d = ,14 (A1)(A1)(A1)

7

3

14

6

1414

6cosθ A1

q = 64.6 (= 1.13radians) A1 N4

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

d1 • d2 =-6, leading toq = 115(= 2.01 radians).

[22]

28.) (a) speed = 222 1043 (M1)

= 125 = 55 , 11.2, (metres per minute) A1 N2

(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

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

t = 8 (A1)

Correct answer A1

eg after 8 minutes, 13:08 N3

(ii) Substituting for t (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

(d) For choosing both direction vectors d1 =

10

4

3

and d2 =

8

3

4

(A1)

d1 d2 = 104, 1d = 125 , 2d = 89 (A1)(A1)(A1)

cosq =89125

104 = 0.98601... A1

q = 0.167 (radians) (acceptq = 9.59) A1 N3[17]

29.) (a) (i) Evidence of approach

egJQ =

OQJOJQ,

0

0

6

10

7

0

M1

10

7

6

JQ AG N0

(ii)

10

7

6

MK A1 N1

(b) (i) r =

0

0

6

+ t

10

7

6

or r =

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

eg cosq =

0811.0

185

15

185185

15

185185

15accept

q= 1.49 (radians), 85.3 A1 N4

(c) METHOD 1

Geometric approach (M1)

Valid reasoning A2

eg diagonals bisect each other,

MK2

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

METHOD 2

Correct approach (M1)

eg

0

0

6

+ t

10

7

6

=

10

7

0

+ s

10

7

6

Two correct equations A1

eg 6 − 6t = 6s,7t = 7 − 7s, 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

METHOD 3

Correct approach (M1)

eg

10

7

0

+ t

10

7

6

=

0

7

0

+ s

10

7

6

Two correct equations A1

eg − 6t = 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]

30.) (a)

PQ =

3

5A1A1 N2

(b) Using r = a + tb

3

5

6

1t

y

xA2A1A1 N4

[6]

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

eg kjikji

322154OAOBAB

= 2i −8j + 20k AG N0

(ii) AB = 6.2111721364682082 222 (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 relationship betweenthe scalar product and perpendicularity, seenanywhere in the solution.

Finding an appropriate scalar product

OAABorOAu M1

eg 1468

203

468

82

468

2OA

u

468

20244

1203822OAAB

0OAABor0OA

u A1 N0

(b) (i) EITHER

2

211,

2

53,

2

42S (M1)(A1)

Therefore,

OS = 3i − j + 11k (accept (3, −1, 11)) A1 N3

OR

AB

2

1OAOS (M1)

= (2i + 3j + k) +21

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

OS = 3i − j + 11k A1 N3

(ii) L1 : r = (3i − j + 11k) + t (2i + 3j + 1k) A1 N1

(c) Using direction vectors (eg 2i + 3j + 1k and −2i + 5j − 3k) (M1)Valid explanation of why L1 is not parallel to L2 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 d1d2.

Note: Award R0 for “direction vectors are not equal”.

(d) Setting up any two of the three equations (M1)For each correct equation A1A1eg 3 + 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 A2 N4

Notes: Award (M1)A2 if the same parameter is usedfor both lines in the initial correct equations.Award no further marks.

[19]

32.) (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)) A1 N1

(ii) 0ADAB

((10)(−6) + 5(6) + 10(3)) A1 N1

(iii) 0AEAB

((10)(−2) + 5(−4) + 10(4)) A1 N1

(iv) 90

2or A1 N1

(c) Volume = AB

AD

AE (A1)

= 15 9 6

= 810 (cubic units) A1 N2

(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) A1 N3

(e)

3

3

18

HB A1

Attempting to use formula cos

HBAG

HBAGP

(M1)

=

342342

108

33181772

31737182222222

A1

= 0.31578... (A1)

6.71P (= 1.25 radians) A1 N3[19]

33.) (a)

1

2

3

3

5

1

OAOBAB (M1)

2

3

2

AB A2 N3

(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

A1A1A1 N3

[6]

34.) (a)

x

x

33OR,

3

1AB A1A1 N2

(b) xx 333ORAB

A1

09100ORAB

x M1

R is

10

3,

10

9A1A1 N2

[6]

35.) (a) OG = 5i + 5j – 5k A2 2

(b) BD = 5i + 5k A2 2

(c) EB = 5i + 5j – 5k A2 2

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

36.) (a) Finding correct vectors, AB =

3

4AC =

1

3A1A1

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 =1059

M1

= –0.569 (3 s.f) AG 3[6]

37.) (a) Attempting to find unit vector (eb) in the direction of b (M1)

Correct values =

0

4

3

043

1222

A1

=

0

8.0

6.0

A1

Finding direction vector for b, vb = 18 × eb (M1)

b =

0

4.14

8.10

A1

Using vector representation b = b0 + tvb (M1)

=

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 h–1) 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 for b or h M1

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

38.) METHOD 1

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

cos1

2–4

3

1

2–4

3

(A1)

Correct value of scalar product 2–142–31

2–4

3

(A1)

Correct magnitudes ( )3 225 5 , 5

4 1

− = = =

(A1)(A1)

2cos

125q

−= (A1) (C6)

METHOD 2

325

4

=

(A1)

25

1

− =

(A1)

534

3

=

(A1)

Using cosine rule (M1)

34 25 5 25 5 cosq= + − (A1)

2cos

125q= − (A1) (C6)

[6]

39.) (a) (i)200 600

AB400 200

→ − = − − (A1)

800

600

=

(A1) (N2)

(ii) 2 2AB 800 600 1000→

= + = (must be seen) (M1)

unit vector8001

6001000

=

(A1)

0.8

0.6

=

(AG) (N0)4

Note: A reverse method is not acceptable in “show that”questions.

(b) (i)0.8

2500.6

=

v (M1)

200

150

=

(AG) (N0)

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

(ii) at 13:00, t = 1

600 2001

200 150

x

y

− = + − (M1)

400

50

− = − (A1) (N1)

(iii) AB 1000→

=

Time1000

4 (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, withthe mark allocation. Use discretion, following this allocation as closely aspossible.

Time for A to B to C= 9 hours

Distance from A to B toC

= 2250 km

Fuel used from A to B= 1800 4 7200× = litres

(A1)

Light goes on after16000 litres

Light goes on after16000 litres

Fuel remaining= 9800 litres (A1)

Time for 16 000litres

Distance on 16000 litres

2501800

16000

Hours before light8800

1800

( )84 4.889

9= =

(A1)(A1)

)889.8(9

88

1800

16000

Time remaining is

= )111.0(9

1 hour

)22.2222(9

22222 km

Time remaining is

( )10.111

9= = hour

(A1)

Distance1

2509

= ×

= 27.8 km

Distance to C= 2250 – 2222.22

= 27.8 km

Distance1

2509

= ×

= 27.8 km (A2) (N4) 7

[17]

40.) (a) 916 = 25 = 5 (M1)(A1) (C2)

(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, ie notan equation.

[6]

41.) (a) DE =

511

412 =

6

8(M1)(A1) (N2)

(b)

DE = 22 68 3664 (M1)

= 10 (A1) (N2)

(c) Vector geometry approach

Using DG = 10 (M1)

(x − 4)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 =86

(A1)

gradient of DG =68 (A1)

equation of line DG is y − 5 = 4)(3

4 x (A1)

Using DG = 10 (M1)

(x − 4)2 + (y − 5)2 = 100 (A1)

Solving simultaneous equation (M1)

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

Note: Award full marks for an appropriatelylabelled diagram (eg showing that DG =10 ,displacements of 6 and 8), or an accuratediagram leading to the correct answers.

[12]

42.) (a) p = 2

3

52

12

0(A1)

=

6

10(accept any other vector notation, including (10, 6) ) (A1) (N2)

(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) (accept x = 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 3x − y = 42 (A1)

Solving simultaneously gives x = 15, y = 3

Substituting to find p 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) (accept x = 15, y = 3 seen above) (A1)(A1)(N1)(N1)[8]

43.) Direction vectors are a = i – 3j and b = i – j. (A2)a b = (1 + 3) (A1)a = 10 , b = 2 (A1)

cos θ =

210

4

ba

ba(M1)

cos θ =20

4(A1) (C6)

[6]

44.) (a) (i)

2

2

1

3OAOBAB (M1)

=

1

5 (A1) (N2) 2

(ii) 125AB (M1)

= 26 (= 5.10 to 3 sf) (A1) (N2) 2Note: An answer of 5.1 is subject to AP.

(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

5

1ABofGradient

d

(A1)

5

1

2

25

d = –1 (A1)

d = 7 (AG)

(ii)

23

7OD (correct answer only) (A1) 3

(d) BCAD (M1)

25

5BC (A1)

BCOBOC (M1)

25

5

1

3OC

=

24

2 (A1) (N3) 4

Note: Many other methods, including scale drawing, are

acceptable.

(e) 650255BCorAD 22

(A1)

Area = 65026 =( 5.099 × 25.5)

= 130 (A1) 2[15]

45.) (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

(b) BD OD OB→ → →

= −3 3= − +i j (A1)

AC OC OA→ → →

= −9 7= − −i j (A1)

Let q be the angle between BD→

and AC→

jiji

jiji

79)33(

)79()33(cosθ (M1)

numerator = + 27 – 21 (= 6) (A1)

denominator ( )18 130 2340= = (A1)

therefore,6

cos2340

q=

82.9q= ° (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 is 15 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]

46.) METHOD 1

At point of intersection:

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

Attempting to solve the linear system (M1)λ = –l (or t = 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) for the point P(2, 3).

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

[2]

48.) (a) PQOR = q – p

=

3

7–

1

10(A1)(A1)

=

2–

3(A1) 3

(b) cosPQPO

PQPOQPO

(A1)

22 3–7–PO = 58 , 22 2–3PQ = 13 (A1)(A1)

PQPO = –21 + 6 = –15 (A1)

cos754

15–1358

15–QPO (AG) 4

(c) (i) Since QPO + RQP = 180° (R1)

cos RQP = –cos QPO

754

15(AG)

(ii) sin RQP =2

754

15–1

(M1)

=754

529(A1)

=754

23(AG)

OR

cosq=754

15

15754

Px

(M1)

therefore x2 = 754 – 225 = 529 x = 23 (A1)

sinq=754

23(AG)

Note: Award (A1)(A0) for the following solution.

cosq=754

15q= 56.89°

sinq= 0.8376

754

23= 0.8376 sinq=

754

23

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

= 2 × RQPsinQRPQ21

(A1)

= 2 ×754

235813

21

(A1)

= 23 sq units. (A1)ORArea of OPQR = 2 (area of triangle OPQ) (M1)

= 2 103–1721

(A1)(A1)

= 23 sq units. (A1) 7

Notes: Other valid methods can be used.Award final (A1) for the integer answer.

[14]

49.) B, or r =

2

6

4

4t (C3)

D, or r =

1

3

5

7t (C3)

Note: Award C4 for B, D and one incorrect,C3 for one correct and nothing else, C1 for one correct and oneincorrect, C0 for anything else.

[6]

50.) (a)

40

30–25

60 = 60 × (–30) + 25 × 40 (M1)

= –800 (A1) (C2)

(b) cos θ = 2222 40302560

800

(M1)(A1)

Note: Trig solutions:Award M1 for attempt to use a correct strategy, A1 for correctvalues.

cos θ = –0.246... (A1)θ = 104.25...° (or 255.75...°) (A1) (C4)She turns through 104° (or 256°)

Note: Accept answers in radians ie 1.82 or 4.46.[6]

51.) (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

1–or

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 =3

2(A1)

Note: The equation

1

4

7

1t

y

xleads to t = 2.

when t =3

2, y = 7 +

3

2(–3) (M1)

= 7 – 2 = 5 (A1)ie P on line (AG)

OR

5 – 7 = –3t = –2

t =3

2(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]

52.) x = l – 2t (A1)y = 2 + 3t (A1)

3

2–2–1– yx (M1)

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

53.)

T

S

U

V

y

x

(a) ST = t – s (M1)

=

2–2–

–7

7

=

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

(c)

11

1is on the line because it gives the same value ofl, for both the x

and y coordinates. (R1)

For example, 1 = 5 + 9l l = –9

4

11 = 15 + 9l l = –9

4(A1) 2

(d) (i)

11

1–

17EW

a(M1)

=

6

1–a(A1)

EW = 2 13 361– 2 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

24–24–(A1)

= –13

12

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

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

Direction vectors are

3

4and

1–

1. (A1)

3

4.

1–

1 =

3

4

1–

1cosq (M1)

4(1) + 3(–1) =

2222 1–134 cosq(A1)

cosq =25

1 = 0.1414 (A1)

q= 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

2cosq (M1)

4(2) + (–1) 4 = 2222 421–4

cosq (A1)

2017

4 = cosq = 0.2169 (A1)

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

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

(ii) b= 22 86 = 10 (A1)

=> unit vector in direction of b =101

(6i + 8j) (A1)

= 0.6i + 0.8j

(iii) a . b = ab cosq (M1)

=> cos θ = 1013

85612 (A1)

=6556

130112 (A1) 6

[6]

56.) cos θ = baa.b

(M1)

=5020

144 (A1)

=1010

10

=10

1 (= 0.3162) (A1)

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

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

57.) (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 s–1 (A1) 2

(c) x = 2 + 0.7 t and y = 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

29

160and

29

170or yx (A1)(A1) 3

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

1

7.0

0

2

52.5

86.5t for t (M1)

t = 5.52 s (A1)[ie collision occurred 5.52 seconds after the vehicles set out].Distance d travelled 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 s–1 (A1) 5[14]

58.) 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]

59.) (a)

5

1

3

2 x

x

x = 0 (M1)(M1)

2x(x + 1) + (x – 3)(5) = 0 (A1)

2x2 + 7x – 15 = 0 (C3)

(b) METHOD 1

2x2 + 7x – 15 = (2x – 3)(x + 5) = 0

x =2

3 or x = –5 (A1) (C1)

METHOD 2

x =)2(2

)15)(2(477 2

x =2

3 or x = –5 (A1) (C1)

[4]

60.) (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 θ =)300)(250(

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

ABOA

ABOA

(M1)

= 0.936 (A1) θ = 20.6° (A1) 4

(c) (i)

168

99

70238

240339AX (A1)

(ii)

180

240

4

3 = –720 + 720 = 0 (M1)(A1)

n ⊥ AB (AG)

(iii) Projection of AX in the direction of n 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]

61.)

8

6.

2

1 = 6 – 16 = –10 (A1)

100868

6,521

2

1 2222

= 10 (A1)

8

6

2

1

8

6.

2

1cosq

–10 = 5 × 10 cosq cosq =5

1

510

10

q = arccos

5

1(M1)

q 117° (A1)[4]

62.)

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 is y + 1 = –3

2(x – 4) (A1)

3y + 3 = –2x + 8 2x + 3y = 5 (A1)

[4]

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

20

6

8

61

28

0

y

x(A1) 2

(b) (i) Velocity vector:01

tty

x

y

x(M1)

=

8

6

28

0

20

6 (km h–1)(A1)

(ii) Speed = ))8(6( 22 ; (M1)

= 10; 10 km h–1 (A1) 4

(c) EITHER

ty

tx

828

6(M1)

Note: Award (M1) for both equations.

y = 28 – 8

6

x(M1)(A1)

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

4x + 3y = 84 (a1) 4

OR

8

6.

28

0

8

6.

y

x(M1)

6

8.

28

0

6

8.

y

x(M1)(A1)

4x + 3y = 84 (A1) 4

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

OR

8

6

28

0

4

18t for some t,

t

t

8–284

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]

64.) 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]

65.) Required vector will be parallel to

4

1

1

3(M1)

=

5

4(A1)

Hence required equation is r =

5

4

4

1t (A1)(A1) (C4)

Note: Accept alternative answers, eg

5

4

1

3s .

[4]

66.) (a)

24

18 = 30 km h–1 (A1)

22 )16(3616

36

= 39.4 (A1) 2

(b) (i) After ½ hour, position vectors are

12

9 and

8

18(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

(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]

67.) Vector equation of a line r = a +lt (M1)

a =

0

0, t =

3

2(M1)(M1)

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

68.) (a)

A

B

C4

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]

69.) (a) (i) r1 =

5

12

12

16t

t = 0 r1 =

12

16(M1)

|r1| = )1216( 22 = 20 (A1)

(ii) Velocity vector =

5

12

speed = ))5(12( 22 (M1)

= 13 (A1) 4

(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 – 5t t =5

12 y(M1)

x = 16 + 12

5

12 y(A1)

5x = 80 + 144 – 12 y 5x + 12y = 224 (A1)(AG) 3

(c) v1 =

5

12v2 =

6

5.2(M1)

v1.v2 =

6

5.2.

5

12(M1)

= 30 – 30 v1.v2 = 0 (A1)q = 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)

(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 + 2.5t 9.5t = 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]

70.) (a) OCODCD (A1) (C1)

(b) CDOA2

1

= )(2

1OCOD (A1) (C1)

(c) OAODAD

= )(2

1OCODOD (A1)

= OCOD2

1

2

1 (A1) (C2)

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

[4]

71.) (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]

72.) (a) OA = 6 A is on the circle (A1)

OB = 6 B is on the circle. (A1)

11

5OC

= 1125 = 6 C is on the circle. (A1) 3

(b) OAOCAC

=

0

6

11

5(M1)

=

11

1(A1) 2

(c)ACAO

ACAOCAO

ˆcos (M1)

=

1116

11

1.

0

6

=126

6(A1)

=6

3

32

1 (A1)

OR1262

6)12(6ˆcos222

CAO (M1)(A1)

=12

1 as before (A1)

OR using the triangle formed by AC and its horizontal andvertical components:

12AC (A1)

12

1ˆcos CAO (M1)(A1) 3

Note: The answer is 0.289 to 3 sf

(d) A number of possible methods here

OBOCBC

=

0

6

11

5(A1)

=

11

11(A1)

BC = 132

ABC = 121322

1 (A1)

= 116 (A1)

OR ∆ABC has base AB = 12 (A1)and height = 11 (A1)

area = 11122

1 (A1)

= 116 (A1)

OR Given6

3ˆcos CAB

6

331212

2

1

6

33ˆsin ABCCAB (A1)(A1)(A1)

= 116 (A1) 4[12]

73.) (a)

5

10OB (A1) (C1)

6

3AC (A1) (C1)

(b) ACOB = (10 × (–3)) + (5 × 6) = 0 (M1)

Angle = 90° (A1) (C2)[4]