vectors sl marks
-
Upload
uelle-koduste -
Category
Documents
-
view
228 -
download
0
Transcript of 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
-
8/7/2019 vectors sl marks
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
-
8/7/2019 vectors sl marks
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
-
8/7/2019 vectors sl marks
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
-
8/7/2019 vectors sl marks
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
-
8/7/2019 vectors sl marks
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
-
8/7/2019 vectors sl marks
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
-
8/7/2019 vectors sl marks
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
-
8/7/2019 vectors sl marks
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
-
8/7/2019 vectors sl marks
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
-
8/7/2019 vectors sl marks
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
-
8/7/2019 vectors sl marks
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
-
8/7/2019 vectors sl marks
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
-
8/7/2019 vectors sl marks
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
-
8/7/2019 vectors sl marks
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
-
8/7/2019 vectors sl marks
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
-
8/7/2019 vectors sl marks
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
-
8/7/2019 vectors sl marks
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
-
8/7/2019 vectors sl marks
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
-
8/7/2019 vectors sl marks
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
-
8/7/2019 vectors sl marks
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
-
8/7/2019 vectors sl marks
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
-
8/7/2019 vectors sl marks
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
-
8/7/2019 vectors sl marks
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
-
8/7/2019 vectors sl marks
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
-
8/7/2019 vectors sl marks
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
-
8/7/2019 vectors sl marks
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
-
8/7/2019 vectors sl marks
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
-
8/7/2019 vectors sl marks
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
-
8/7/2019 vectors sl marks
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
-
8/7/2019 vectors sl marks
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
-
8/7/2019 vectors sl marks
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
-
8/7/2019 vectors sl marks
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
-
8/7/2019 vectors sl marks
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
-
8/7/2019 vectors sl marks
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
-
8/7/2019 vectors sl marks
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
-
8/7/2019 vectors sl marks
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
-
8/7/2019 vectors sl marks
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
-
8/7/2019 vectors sl marks
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
-
8/7/2019 vectors sl marks
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
-
8/7/2019 vectors sl marks
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
-
8/7/2019 vectors sl marks
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
-
8/7/2019 vectors sl marks
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
-
8/7/2019 vectors sl marks
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
-
8/7/2019 vectors sl marks
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
-
8/7/2019 vectors sl marks
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
-
8/7/2019 vectors sl marks
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]