1. (a)OC OD CD (A1)(C1) (b)CD OA21= ) (21OC OD (A1)(C1) (c)OA OD AD = ) (21OC OD OD (A1)= OC OD2121+(A1)(C2)Note: Deduct [1 mark] (once only) if appropriate vector notation is omitted.[4] 2. (a)j i u 2 + j i v 5 3 + j i v u 12 5 2 + +(A1)(C1)(b)2 212 5 2 + + v u = 13 (A1)Vector ) 12 5 (1326j i w + (A1)= j i 24 10 +(A1) (C3)[4] 3. (a)OA = 6 A is on the circle (A1)OB = 6 B is on the circle. (A1)

,_

115OC= 11 25 += 6 C is on the circle. (A1) 31(b)OA OC AC =

,_

,_

06115(M1)=

,_

111(A1) 2 (c)AC AOAC AOC A Ocos(M1)= 11 1 6111.06+

,_

,_

= 12 66(A1)= 633 21(A1) OR 12 6 26 ) 12 ( 6cos2 2 2 + C A O(M1)(A1)= 121 as before (A1)OR using the triangle formed by AC and its horizontal andvertical components:12 AC(A1)121cos C A O(M1)(A1)3Note: The answer is 0.289 to 3 sf2(d) A number of possible methods hereOB OC BC =

,_

,_

06115(A1)=

,_

1111(A1) BC = 132 ABC= 12 13221 (A1)= 11 6(A1) OR ABC has base AB = 12 (A1)and height = 11(A1)area = 11 1221 (A1)= 11 6(A1) OR Given 63cos C A B63312 1221633sin ABC C A B(A1)(A1)(A1)= 11 6(A1) 4[12] 4. (a)

,_

510OB(A1)(C1)

,_

63AC(A1) (C1) (b)AC OB= (10 (3)) + (5 6) = 0 (M1)Angle = 90 (A1) (C2)[4]35. 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

,_

,_

4113(M1)=

,_

54(A1)Hence required equation is r =

,_

+

,_

5441t(A1)(A1) (C4)Note: Accept alternative answers, eg

,_

+

,_

5413s.[4] 7. (a)

,_

2418 = 30 km h1(A1)2 2) 16 ( 361636 +

,_

= 39.4(A1) 2 (b) (i) After hour, position vectors are

,_

129 and

,_

818(A1)(A1) (ii) At 6.30 am, vector joining their positions is

,_

,_

,_

209818129 (or

,_

209) (M1)

,_

209(M1)= 481 (= 21.9 km to 3 sf) (A1)54(c) The Toyundai must continue until its position vector is

,_

k18(M1)Clearly k = 24, ie position vector

,_

2418. (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 kmTheir 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 . 20186 . 3 2418(M1)(A1)Distance to base camp =

,_

4 . 2018(A1) = 27.2 kmTime to cover this distance = 302 . 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 =

,_

00, t =

,_

32(M1)(M1)r = (2i + 3j) (A1) (C4)[4]59. (a)ABC43210 1 2 31 2 3 4 5xy(A3)(C3)Note: Award (A1) for B at (5, 1); (A1) for BC perpendicular to AB; (A1) for AC parallel to the y-axis. (b)

,_

25 . 32OC(A1)(C1)Note: Accept correct readings from diagram (allow 0.1).[4] 10. (a) (i) r1 = 1]1

+1]1

5121216tt = 0 r1 = 1]1

1216(M1) r1= ) 12 16 (2 2+ = 20 (A1) (ii) Velocity vector = 1]1

512speed = ) ) 5 ( 12 (2 2 +(M1)= 13 (A1) 46(b)1]1

+1]1

1]1

5121216tyx1]1

1]1

+1]1

1]1

1]1

1]1

512.125126.125.125tyx(M1)5x + 12y = 80 + 144 (A1)5x + 12y = 224 (A1)(AG) OR5121216 y x(M1)5x 80 = 144 12y (A1)5x + 12y = 224 (A1)(AG) ORx = 16 + 12t, y = 12 5t t = 512 y (M1)x = 16 + 12

,_

512 y(A1)5x = 80 + 144 12 y5x + 12y = 224 (A1)(AG)3 (c) v1 = 1]1

512v2 = 1]1

65 . 2(M1)v1.v2 = 1]1

1]1

65 . 2.512(M1)= 30 30v1.v2 = 0 (A1)= 90 (A1) 4 (d) (i)1]1

1]1

1]1

1]1

512.523512.yx(M1)12x 5y = 23 12 + 25 = 301 (A1)OR655 . 223 + y x6x 138 = 2.5y + 12.5 (M1)12x 276 = 5y + 2512x 5y = 301 (A1) 7(ii); +; +3612 60 1441120 60 25301 5 12224 12 5y xy xy xy x(M1)169x = 4732 x = 28, y = (12 28 301) 5 = 7 (28, 7) (A1)(A1) 5Note: 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)11175 . 97(A1)planes cannot be at the same place at the same time (R1) ORr1 = 1]1

+1]1

1]1

1]1

5121216728728t(M1) ' 5 512 12tt t = 1 (A1)When t = 1 r2 = 1]1

1]1

1]1

+1]1

72815 . 2565 . 2523(A1)(R1) OR r2 = 1]1

+1]1

1]1

1]1

65 . 2523728728t(M1) t = 2 (A1) 4[20]811.

,_

,_

86.21 = 6 16 = 10 (A1)100 8 686, 5 2 1212 2 2 2 +

,_

+

,_

= 10 (A1)

,_

,_

,_

,_

862186.21cos10 = 5 10 coscos= 515 1010 = arccos51 (M1)117 (A1)[4] 12.

,_

+

,_

14.32yx(M1) (M1)Notes: Award (M1) for using scalar product.Award (M1) for

,_

+14yx.2(x 4) + 3(y + 1) = 0 (A1)2x 8 + 3y + 3 = 02x + 3y = 5 (A1) ORGradient of a line parallel to the vector

,_

32 is 23(M1)Gradient of a line perpendicular to this line is 32(M1)So the equation is y + 1 = 32(x 4) (A1)3y + 3 = 2x + 82x + 3y = 5 (A1)[4] 13. (a) At 13:00, t = 1 (M1)

,_

,_

+

,_

,_

206861280yx(A1)2 9(b) (i) Velocity vector: 0 1

,_

,_

t tyxyx(M1)=

,_

,_

,_

86280206 (km h1) (A1) (ii) Speed = ) ) 8 ( 6 (2 2 +; (M1)= 10; 10 km h1(A1)4 (c) EITHER ; t yt x8 286(M1)Note: Award (M1) for both equations.y = 28 8

,_

6x(M1)(A1)Note: Award (M1) for elimination, award (A1) for equation in x, y.4x + 3y = 84 (a1)4 OR

,_

,_

,_

,_

86.28086.yx(M1)

,_

,_

,_

,_

68.28068.yx(M1)(A1) 4x + 3y = 84 (A1)410(d) They collide if

,_

418 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)4OR

,_

+

,_

,_

86280418t for some t, ;'tt8 28 46 18and(A1) ;'24 83ttand(A1) ;'33ttandThey collide at 15:00 (A1)4 (e)

,_

+

,_

,_

125) 1 (418tyx(M1)=

,_

+ +12 12 45 5 18tt(M1)2=

,_

+

,_

125813t(AG) (f) At t = 3, (M1)

,_

,_

+ +

,_

282812 3 85 3 13yx(A1)

,_

,_

,_

24104182828(A1)) 676 ( ) 24 10 (2 2 + = 2626 km apart (A1) 4[20]1114. cos = b aa.b(M1)= 50 2014 4 + (A1)= 10 1010= 101 (= 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,

,_

,_

+

,_

24 . 317 . 0202(M1)Distance from (0, 0) = 2 22 4 . 3 + = 3.94 m (A1)2 (b)2 21 7 . 017 . 0+

,_

(M1) = 1.22 m s1(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

,_

29160and29170ory x(A1)(A1)312(e) The time of the collision may be found by solving

,_

+

,_

,_

17 . 00252 . 586 . 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 byd = 2 2) 52 . 3 ( ) 86 . 5 (2052 . 586 . 5+

,_

,_

(M1)= 73 . 46= 6.84 m (A1)Speed of the motorcycle = 52 . 584 . 6td= 1.24 m s1(A1) 5[14] 16. Direction vector =

,_

,_

3156(M1)=

,_

25(A1)

,_

+

,_

,_

2531tyx(A2) OR

,_

+

,_

,_

2556tyx(A2) (C4)[4] 17. (a)

,_

+

,_

5132 xxx = 0 (M1)(M1)2x(x + 1) + (x 3)(5) = 0 (A1)2x2 + 7x 15 = 0(C3)13(b) METHOD 12x2 + 7x 15 = (2x 3)(x + 5) = 0x = 23 or x = 5 (A1)(C1)METHOD 2x = ) 2 ( 2) 15 )( 2 ( 4 7 72 t x = 23 or x = 5 (A1)(C1)[4] 18. (a) (i)

,_

70240OA OA = 2 270 240 + = 250 (A1)unit vector =

,_

,_

28 . 096 . 0702402501(M1)(AG) (ii)

,_

,_

8428828 . 096 . 0300 v(M1)(A1)(iii) t = 65288240hr (= 50 min) (A1)5 (b)

,_

,_

18024070 250240 480AB(A1)AB = 2 2180 240 += 300cos = ) 300 )( 250 () 180 )( 70 ( ) 240 )( 240 (AB OAAB OA +(M1)= 0.936 (A1) = 20.6 (A1) 4 (c) (i)

,_

,_

1689970 238240 339AX(A1)(ii)

,_

,_

18024043 = 720 + 720 = 0 (M1)(A1)n AB(AG)14 (iii) Projection of AX in the direction of n isXY = 5672 297431689951 +

,_

,_

= 75 (M1)(A1)(A1)6 (d) AX = 2 2168 99 + = 195 (A1)AY = 2 275 195 = 180 km (M1)(A1)3[18] 19. x = l 2t (A1)y = 2 + 3t (A1)32 2 1 y x(M1)3x + 2y = 7 (A1)(A1)(A1) (C6)[6]1520.TSUVyx(a)ST = t s (M1)=

,_

,_

2 2 77=

,_

99(A1)VU = ST(M1)u v =

,_

99v = u

,_

99=

,_

,_

,_

64 99155(A1)V(4, 6) (A1)5 (b) Equation of (UV): direction is =

,_

,_

,_

11or 99k(A1)r =

,_

+

,_

99155 or

,_

+

,_

11155(A1)ORr =

,_

+

,_

9964 or

,_

+

,_

1164 (A1)216(c)

,_

111is on the line because it gives the same value of , for both the x and y coordinates. (R1)For example, 1 = 5 + 9= 9411 = 15 + 9= 94(A1) 2 (d) (i)

,_

,_

11117EWa(M1) =

,_

61 a(A1)|EW |= 213 ( ) 36 1 2+ a = 213 (or (a 1)2 + 36 = 52) (M1)a2 2a + 1 +36 = 52a2 2a 15 = 0 (A1)a = 5 or a =3 (A1)(AG) (ii) For a = 3EW=

,_

64

ET = t e =

,_

4 6(A1)(A1)cos T EW= ET EWET EW(M1)= 52 5224 24 (A1)= 1312Therefore, T EW = 157 (3 sf) (A1)10[19]1721. Angle between lines = angle between direction vectors. (M1)Direction vectors are

,_

34 and

,_

1 1. (A1)

,_

34 .

,_

1 1 =

,_

34

,_

1 1cos (M1)4(1) + 3(1) = ( ) ( ) ,_

+ +2 2 2 21 1 3 4cos (A1)cos = 2 51 = 0.1414 (A1) = 81.9 (3 sf), (1.43 radians) (A1) (C6) Note: If candidates find the angle between the vectors

,_

1 4and

,_

42, award marks as below:Angle required is between

,_

1 4 and

,_

42(M0)(A0)

,_

1 4 .

,_

42=

,_

1 4

,_

42cos (M1)4(2) + (1) 4 = ( ) ( )2 2 2 24 2 1 4 + ,_

+cos (A1)20 174 = cos = 0.2169 (A1)= 77.5 (3sf), (1.35 radians) (A1) (C4)[6] 22. (i) |a|= 2 25 12 += 13 (A1)(ii) |b|= 2 28 6 += 10 (A1)=> unit vector in direction of b = 101(6i + 8j) (A1) = 0.6i + 0.8j18(iii) a . b =a1b1cos (M1)=> cos = ( ) ( )( ) 10 138 5 6 12 +(A1) = 6556130112(A1)6[6] 23. METHOD 1At point of intersection:5 + 3 = 2 + 4t (M1)l 2 = 2 + t (M1)Attempting to solve the linear system (M1) = l (or t = 1) (A1)

,_

32OP(A1)(A1) (C6) METHOD 2(changing to Cartesian coordinates)2x + 3y = 13, x 4y = 10 (M1)(A1)(A1)Attempt to solve the system (M1)

,_

32OP(A1)(A1) (C6)Note: Award (C5) for the point P(2, 3). 24. (a) c d = 3 5 + 4 (12) (M1) = 33 (A1) (C2)[2]1925. (a)PQ OR = q p=

,_

,_

37110(A1)(A1)=

,_

2 3(A1)3 (b) cos PQ POPQ POQ PO(A1)( ) ( )2 23 7 PO + = 58, ( )2 22 3 PQ + = 13(A1)(A1)PQ PO = 21 + 6 = 15 (A1)cos 75415 13 5815 Q PO (AG)4 (c) (i) Since Q PO + R QP = 180 (R1)cos R QP = cos Q PO

,_

75415(AG)20(ii) sin R QP= 275415 1 ,_

(M1) = 754529(A1) = 75423(AG)ORcos = 754151 57 5 4Px(M1)therefore x2 = 754 225 = 529 x = 23 (A1)sin = 75423(AG)Note: Award (A1)(A0) for the following solution.cos = 75415 = 56.89sin = 0.837675423 = 0.8376 sin = 75423 (iii) Area of OPQR = 2 (area of triangle PQR) (M1)= 2 R QP sin QR PQ21 (A1)= 2 7542358 1321(A1)= 23 sq units. (A1)ORArea of OPQR = 2 (area of triangle OPQ) (M1) = 2 ( ) 10 3 1 721

,_

(A1)(A1) = 23 sq units. (A1)7Notes:Other valid methods can be used.Award final (A1) for the integer answer.[14]21 26. B, or r =

,_

+

,_

2644t(C3)D, or r =

,_

+

,_

1357t(C3)Note: Award C4 for B, D and one incorrect,C3 for one correct and nothing else, C1 for one correct and one incorrect, C0 for anything else.[6] 27. (a)

,_

,_

4030 2560 = 60 (30) + 25 40 (M1) = 800 (A1) (C2) (b) cos = ( )2 2 2 240 30 25 60800+ +(M1)(A1)Note: Trig solutions:Award M1 for attempt to use a correct strategy, A1 for correct values.cos = 0.246... (A1) = 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] t28. (a)

,_

71OB

,_

98OC(A1)(A1)2 (b)OB OC BC AD (M1)=

,_

,_

,_

297198(A1)

,_

,_

,_

+

,_

,_

,_

+

,_

+ 4115398or 4112922AD OA ODd = 11

,_

,_

411accept (A1)322(c)

,_

,_

,_

31271411BD(A1)1 (d) (i) l :

,_

,_

+

,_

,_

+

,_

,_

1471 or31271t tyx(A2)(ii) At B, t = 0 by observation (A1)OR

,_

+

,_

,_

3127171tt = 0 (A1) 3 (e)

,_

+

,_

,_

3127157t 7 + 1 = 12t = 8t = 32(A1)Note: The equation

,_

+

,_

,_

1471tyx leads to t = 2.when t = 32, y = 7 +

,_

32(3) (M1) = 7 2 = 5 (A1)ie P on line (AG) OR5 7 = 3t = 2t = 32(A1)when t = 32, x = 1 + 32 12 (M1) = 1 + 8 = 7 (A1)ie P on line (AG) 323(f)

,_

,_

,_

419857CP(A1)

,_

,_

31241 = 12 +12 = 0 (M1)(A1)Scalar product of non-zero vectors = 0 are perpendicular (R1)(AG)ORGeometric approachCP: m = 4 (A1)BD: m1 = 41 (A1)mm1 = 4

,_

41 = 1 (A1)Product of gradients is 1 lines (vectors) are perpendicular (R1)(AG)4[16] 29. Direction vectors are a = i 3j and b = i j. (A2)a b = (1 + 3) (A1) a= 10, b= 2(A1)cos =

,_

2 104b ab a(M1)cos = 204(A1) (C6)[6] 30. (a) (i)

,_

,_

2213OA OB AB(M1)=

,_

15(A1) (N2)2 (ii)1 25 AB + (M1)= 26 (= 5.10 to 3 sf) (A1) (N2)2Note: An answer of 5.1 is subject to AP.24(b)OA OD AD =

,_

,_

2223d=

,_

252 d(A1)(A1) 2 (c) (i) EITHERAD AB 90 D AB = 0 or mention of scalar (dot) product. (M1)

,_

,_

25215 d = 05d + 10 + 25 = 0 (A1)d = 7 (AG) OR; 225AD of Gradient 51AB of Gradient d(A1)

,_

,_

51225d = 1 (A1)d = 7 (AG) (ii)

,_

237OD (correct answer only) (A1)3 (d)BC AD(M1)

,_

255BC(A1)BC OB OC + (M1)

,_

+

,_

25513OC=

,_

242 (A1) (N3)4Note: Many other methods, including scale drawing, are acceptable.25(e)650 25 5 BC or AD2 2 + ,_

(A1)Area = 650 26 =( 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 (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

,_

+ + j i j ij i j i7 9 ) 3 3 () 7 9 ( ) 3 3 (cos (M1)numerator = + 27 21 (= 6) (A1)denominator ( )18 130 2340 (A1)therefore,6cos2340 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) 126(d) EITHER) 7 2 ( 3 ) 4 ( 2 4 j i j i j i j i + + + + + t s(may be implied) (M1)4 1 22 4 3 7s ts t+ + ;+ +(A1)7 and/or 11 t s (A1)Position vector of P is 15 46 + i j(A1)(N2)OR7 2 13 or equivalent x y (A1)4 14 or equivalent x y (A1)15 , 46 x y (A1)Position vector of P is 15 46 + i j(A1)(N2) 4[15] 32. (a)OG= 5i + 5j 5k A22(b)BD= 5i + 5k A22(c)EB= 5i + 5j 5k A22Note: Award A0(A2)(A2) if the 5 is consistently omitted.[6] 33. (a) Finding correct vectors, AB =

,_

34AC =

,_

13A1A1Substituting correctly in the scalar productAC AB = 4(3) + 3(1) A1= 9 AG 3(b) |AB| = 5 |AC| = 10(A1)(A1)Attempting to use scalar product formula cos BAC = 10 59 M1= 0.569 (3 s.f) AG3[6]2734. (a) Attempting to find unit vector (eb) in the direction of b (M1)Correct values =

,_

+ +0430 4 312 2 2A1 =

,_

08 . 06 . 0A1 Finding direction vector for b, vb = 18 eb(M1)b =

,_

04 . 148 . 10A1Using vector representation b = b0 + tvb(M1)=

,_

+

,_

04 . 148 . 10500tAG 6 (b) (i) t = 0 (49, 32, 0) A11(ii) Finding magnitude of velocity vector (M1)Substituting correctly vh = 2 2 26 ) 24 ( ) 48 ( + + A1 = 54(km h1) A1 3 (c) (i) At R,

,_

,_

ttttt624 3248 4954 . 148 . 10A1t = 65 (= 0.833) (hours) A12(ii) For substituting t = 65 into expression for b or h M1(9,12,5) A2 3[15]2835. METHOD 1Using a b = ab cos (may be implied) (M1) cos12 4312 43

,_

,_

,_

,_

(A1)Correct value of scalar product ( ) ( ) 2 1 4 2 312 43 +

,_

,_

(A1)Correct magnitudes ( )3 225 5 , 54 1 _ _ , ,(A1)(A1)2cos125(A1) (C6) METHOD 23254 _ ,(A1)251 _ ,(A1)5343 _ ,(A1)Using cosine rule (M1)34 25 5 25 5cos + (A1)2cos125 (A1) (C6)[6] 36. (a) (i)200 600AB400 200 _ _ , ,(A1)800600 _ ,(A1) (N2)29(ii)2 2AB 800 600 1000 + (must be seen) (M1)unit vector 8001600 1000 _ ,(A1)0.80.6 _ ,(AG) (N0) 4Note: A reverse method is not acceptable in show that questions. (b) (i)0.82500.6 _ ,v(M1)200150 _ ,(AG) (N0)Note: A correct alternative method is using the given vector equation with t = 4. (ii) at 13:00, t = 1600 2001200 150xy _ _ _ + , , ,(M1)40050 _ ,(A1) (N1) (iii)AB 1000Time 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) 630(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 hoursDistance from A to B to C= 2250 kmFuel used from A to B= 1800 4 7200 litres(A1)Light goes on after16000 litresLight goes on after16000 litresFuel remaining= 9800 litres (A1)Time for 16 000 litres) 889 . 8 (988180016000 Time remaining is=) 111 . 0 (91hourDistance on 16000 litres250180016000 ) 22 . 2222 (922222 kmHours before light 88001800( )84 4.8899 Time remaining is( )10.1119 hour(A1)(A1)(A1)Distance 12509 = 27.8 kmDistance to C= 2250 2222.22= 27.8 kmDistance 12509 = 27.8 km (A2) (N4)7[17] 37. (a)9 16+ = 25 = 5 (M1)(A1)(C2) (b)

,_

,_

+

,_

7634212(so B is (6, 7) ) (M1)(A1)(C2) (c) r =

,_

+

,_

3412t(not unique) (A2)(C2)Note: Award (A1) if r = is omitted, ie notan equation.[6] 3138. (a)DE =

,_

5 114 12 =

,_

68(M1)(A1)(N2) (b) DE = 2 26 8 + ( ) 36 64+ (M1)= 10 (A1) (N2) (c) Vector geometry approachUsing DG = 10 (M1)(x 4)2 + (y 5)2 = 100 (A1)Using (DG) perpendicular to (DE) (M1)Leading to DG =

,_

86, DG =

,_

86(A1)(A1)Using DG = +OG DO(O is the origin) (M1)G (2, 13), G (10, 3) (accept position vectors) (A1)(A1)Algebraic approachgradient of DE = 86(A1)gradient of DG = 68(A1)equation of line DG is y 5 = 4) (34 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] 39. (a) p = 2

,_

+

,_

352120(A1)=

,_

610(accept any other vector notation, including (10, 6) ) (A1)(N2)32(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 + qy =12 3p y = 3qgiving 3x + 5y = 60 3x y = 42 (A1)Solving simultaneously gives x = 15, y = 3Substituting to find p and q,31014315,35120315

,_

+

,_

,_

,_

+

,_

,_

q pp = 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] 40. (a)PQ =

,_

35A1A1N2 (b) Using r = a + tb

,_

+

,_

,_

3561tyxA2A1A1 N4[6]3341. (a) (i) Evidence of subtracting all three components in the correct order M1eg ( ) ( ) k j i k j i + + + 3 2 21 5 4 OA OB AB= 2i 8j + 20k AGN0(ii) AB= ( ) ( ) 6 . 21 117 2 13 6 468 20 8 22 2 2 + +(A1)u = ( ) k j i 20 8 24681+ A1 N2

,_

+ + etc. , 925 . 0 370 . 0 0925 . 0 ,4682046884682k j i k j i(iii) If the scalar product is zero, the vectors are perpendicular. R1Note: Award R1 for stating the relationship betweenthe scalar product and perpendicularity, seenanywhere in the solution.Finding an appropriate scalar product

,_

OA AB or OA uM1eg 1468203468824682OA

,_

+

,_

+

,_

u

,_

+ 46820 24 4( ) 1 20 3 8 2 2 OA AB + + 0 OA AB or 0 OA uA1 N034(b) (i) EITHER

,_

+ +221 1,25 3,24 2S(M1)(A1)Therefore, OS = 3i j + 11k (accept (3, 1, 11)) A1 N3OR + AB21OA OS(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) A1N1 (c) Using direction vectors (eg 2i + 3j + 1k and 2i + 5j 3k) (M1)Valid explanation of why L1 is not parallel to L2R1N2eg. Direction vectors are not scalar multiples of each other.Angle between the direction vectors is not zero or 180.Finding the angled1 d2 d1d2 .Note: Award R0 for direction vectors are not equal. (d) Setting up any two of the three equations (M1)For each correct equationA1A1eg 3 + 2t = 5 2s, 1 + 3t = 10 + 5s, 11 + t = 10 3sAttempt to solve these equations (M1)Finding one correct parameter (s = 1, t = 2) (A1)P has position vector 7i + 5j + 13k A2N4Notes: Award (M1)A2 if the same parameter is usedfor both lines in the initial correctequations.Award no further marks.[19]3542. (a) (i) OA OB AB(A1)=

,_

,_

5375217=

,_

10510A1 N2(ii)2 2 210 5 10 AB + + (M1)= 15 A1 N2 (b) Evidence of correct calculation of scalar product (may be in (i), (ii)or (iii)) A1(i)0 AE AB ((6)(2) + 6(4) + 3(4)) A1N1(ii)0 AD AB ((10)(6) + 5(6) + 10(3)) A1N1(iii)0 AE AB ((10)(2) + 5(4) + 10(4)) A1N1(iv) 90

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2orA1 N1 (c) Volume = ABAD AE(A1)= 15 9 6= 810 (cubic units) A1 N236(d) Setting up a valid equation involving H. There are many possibilities.eg

,_

,_

+ + + 105101249, EH AE OA OH , GH OG OHzyx(M1)Using equal vectors (M1)eg AD EH , AB GH

,_

,_

+

,_

+

,_

,_

,_

,_

211366442537OH ,211105101249OHcoordinates of H are (1, 1, 2) A1N3 (e)

,_

3318HBA1Attempting to use formula cos HB AGHB AGP(M1)=

,_

+ + + + + + 342 3421083 3 18 17 7 23 17 3 7 18 22 2 2 2 2 2A1= 0.31578... (A1) 6 . 71P(= 1.25 radians) A1N3[19] 43. (a)

,_

,_

,_

123351OA OB AB(M1)

,_

232ABA2 N337(b) Using r = a + tb

,_

+

,_

,_

,_

+

,_

,_

232351or232123tzyxtzyxA1A1A1N3[6] 44. (a)

,_

,_

xx3 3OR ,31ABA1A1N2(b)( ) x x 3 3 3 OR AB A1( ) 0 9 10 0 OR AB xM1R is

,_

103,109A1A1 N2[6] 45. (a) u v = 8 + 3 + p (A1)For equating scalar product equal to zero (M1)8 + 3 + p = 0p = 11 A1 N3 (b)u = ( ) 74 . 3 , 14 1 3 22 2 2 + +(M1)14 14 qA1q = ( ) 74 . 3 14 A1 N2[6]3846. 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 + sPQ(M1)= 5i + 11j 8k + s(i 2j + 3k) A1= (5 + s) i + (11 2s)j + (8 + 3s) k AGN0 (b) If (2, y1, z1) lies on L1 then 5 + s = 2 (M1)s = 7 A1y1 = 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 + 3tAttempting to solve their equations (M1)One correct parameter (s = 4, t = 3) A1OT = 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)

,_

7314614 146cosA1= 64.6 (= 1.13radians) A1N4Note: Award marks as per the markschemeif their (correct) direction vectors gived1 d2 = 6, leading to = 115(= 2.01 radians).[22]3947. (a) speed = 2 2 210 4 3 + +(M1)= 125 = 5 5, 11.2, (metres per minute) A1N2 (b) Let the velocity vector be

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cbaFinding a velocity vector A2eg

,_

39163 =

,_

23105 + 2

,_

cba,

,_

39163

,_

23105Dividing by 2 to give

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834A1

,_

zyx =

,_

23105 + t

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834AGN0 (c) (i) At Q,

,_

723 + t

,_

1043 =

,_

23105 + t

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834(M1)Setting up one correct equation A1eg 3 + 3t = 5 + 4t, 2 + 4t = 10 + 3t, 7 + 10t = 23 + 8tt = 8 (A1)Correct answer A1eg after 8 minutes, 13:08 N3(ii) Substituting for t (M1)

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zyx =

,_

723 + 8

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1043, or

,_

zyx =

,_

23105 + 8

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834x = 27, y = 34, z = 87 or (27, 34, 87), or

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873427A1N2 40(d) For choosing both direction vectors d1 =

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1043 and d2 =

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834(A1)d1 d2 = 104, 1d = 125, 2d = 89(A1)(A1)(A1)cos = 89 125104 = 0.98601... A1= 0.167 (radians)(accept = 9.59) A1 N3[17] 48. (a) (i) Evidence of approacheg JQ = +

,_

,_

OQ JO JQ ,0061070M1

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1076JQAG N0(ii)

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1076MKA1 N141(b) (i) r =

,_

006 + t

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1076 or r =

,_

1070 + t

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1076A2N2Note: Award A1 if r = is missing.(ii) Evidence of choosing correct vectors

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1076,

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1076(A1)(A1)Evidence of calculating magnitudes (A1)(A1)eg ( ) ( )2 2 2 2 2 210 7 6 185 10 7 6 + + + + 185

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1076

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1076 = 36 49 + 100 (= 15) (accept 15) (A1)For evidence of substitution into the correct formula M1eg cos =

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0811 . 018515185 18515

,_

185 18515accept= 1.49 (radians), 85.3 A1 N4 (c) METHOD 1Geometric approach (M1)Valid reasoning A2eg diagonals bisect each other, + MK21OM ODCalculation of mid point (A1)eg

,_

+ + +210 0,27 0,20 6

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55 . 33OD(accept (3,3.5,5)) A1N342METHOD 2Correct approach (M1)eg

,_

006 + t

,_

1076 =

,_

1070 + s

,_

1076Two correct equations A1eg 6 6t = 6s,7t = 7 7s, 10t = 10sAttempt to solve (M1)One correct parameters = 0.5 t = 0.5 A1

,_

55 . 33OD(accept (3, 3.5, 5)) A1N3 METHOD 3Correct approach (M1)eg

,_

1070 + t

,_

1076 =

,_

070 + s

,_

1076Two correct equations A1eg 6t = 6s, 7 + 7t = 7 7s, 10 + 10t = 10sAttempt to solve (M1)One correct parameters = 0.5 t = 0.5 A1

,_

55 . 33OD(accept (3, 3.5, 5)) A1N3[16]4349. (a) (i) evidence of combining vectors (M1)eg AB = OB OA (or AD = AO + OD in part (ii))AB =

,_

242A1 N2(ii)AD =

,_

252kA1 N1 (b) evidence of using perpendicularity scalar product = 0 (M1)0252242. .

,_

,_

k g e4 4(k 5) + 4 = 0 A14k + 28 = 0 (accept any correct equation clearly leading to k = 7) A1k = 7 AGN0 (c)AD =

,_

222(A1)BC =

,_

111A1evidence of correct approach (M1)eg

,_

,_

,_

+

,_

+ 111213,111213, BC OB OCzyxOC =

,_

124A1 N344(d) METHOD 1choosing appropriate vectors, BC , BA(A1)finding the scalar productM1eg 2(1) + 4(1) + 2(1), 2(1) + (4)(1) + (2)(1)cos C BA = 0 A1 N1METHOD 2BC parallel to AD (may show this on a diagram with points labelled) R1BC AB (may show this on a diagram with points labelled) R1C BA = 90cos C BA = 0 A1 N1[13] 50. pw = pi + 2pj 3pk (seen anywhere) (A1)attempt to find v + pw (M1)eg 3i + 4j + k + p(i + 2j 3k)collecting terms (3 + p)i + (4 + 2p)j + (1 3p) k A1attempt to find the dot product (M1)eg 1(3 + p) + 2(4 + 2p) 3(1 3p)setting their dot product equal to 0 (M1)eg 1(3 + p) + 2(4 + 2p) 3(1 3p) = 0simplifying A1eg 3 + p + 8 + 4p 3 + 9p = 0, 14p + 8 = 0P = 0.571

,_

148A1N3[7]4551. (a) (i) evidence of approach M1eg AO + OB = AB, B AAB =

,_

164AG N0(ii) for choosing correct vectors, (AOwith AB, or OA withBA) (A1)(A1)Note: Using AO with BA will lead to0.799. If they then say BAO= 0.799, this is a correct solution.calculating AOAB, AB , AO(A1)(A1)(A1)eg d1d2 = (1)(4) + (2)(6) + (3)(1) (= 19)( ) ( ) ( ) , 14 3 2 1 d2 2 21 + + ( ) ( ) ( ) 53 1 6 4 d2 2 22 + + evidence of using the formula to find the angle M1eg cos = ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ),1 6 4 3 2 11 3 6 2 4 12 2 2 2 2 2 + + + + + + ... 69751 . 0 ,53 1419O AB = 0.799 radians (accept 45.8) A1N3 (b) two correct answers A1A1eg (1, 2, 3), (3, 4, 2), (7, 10, 1), (11, 16, 0)N246(c) (i) r =

,_

+

,_

243321tA2N2(ii) C on L2, so

,_

+

,_

,_

2433215t kk(M1)evidence of equating components (A1)eg 1 3t = k, 2 + 4t = k, 5 = 3 + 2tone 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

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+

,_

,_

121083522p(M1)eg 3 + p = 2, 8 2p = 2, p = 5p = 5 A1 N2[18] 52. evidence of equating vectors (M1)eg L1 = L2for any two correct equations A1A1eg 2 + s = 3 t, 5 + 2s = 3 + 3t, 3 + 3s = 8 4tattempting to solve the equations (M1)finding one correct parameter (s = 1, t = 2) A1the coordinates of T are (1, 3, 0) A1N3[6]47