Vectors Practice MS

7/17/2019 Vectors Practice MS

http://slidepdf.com/reader/full/vectors-practice-ms 1/43

1. (a) = b – a A1AB

= a + b A1CB

(b) = ( b – a)•( b + a) M1CBAB

= │ b│2

– │ a│2

A1 = 0 since │ b│=│ a│ R1

Note: Only award the A1 and R1 if working indicates that they

understand that they are working with vectors.

so is perpendicular to i.e. is a right angle AGAB CB CBA

[5]

2. (a) A1A1

1

3

4

AC,

3

1

4

AB

Note: Accept row vectors.

(b) M1A1

16

16

8

134

314ACAB

k ji

normal n = so r • (M1)

2

2

1

2

2

1

1

2

1

2

2

1

x + 2 y + 2 z = 7 A1

Note: If attempt to solve by a system of equations:

Award A1 for 3 correct equations, A1 for eliminating a variable

and A2 for the correct answer.



(c) r = (or equivalent) A1

2

2

1

7

3

5

1(5 + λ) + 2(3 + 2 λ) + 2(7 + 2 λ) = 7 M1

9 λ = –18 λ = –2 A1

Note: λ = –

is used.

16

16

8

if 4

1

distance = (M1)222 2212

= 6 A1

(d) (i) area = (M1)222 161682

1ACAB2

1

= 12 (accept ) A15762

1

(ii) EITHER

volume = × area × height (M1)3

1

= × 12 × 6 = 24 A13

1

OR

volume = M1 )ACAB(AD6

1

= 24 A1



(e) 222 16168ACAB

M1

614

134ADAC

k ji

= │ –19i – 20 j + 16 k│ A1

EITHER

M1222222 161682

1162019

2

1

therefore since area of ACD bigger than area ABC implies that

B is closer to opposite face than D R1

OR

correct calculation of second distance as A1222 162019

144

which is smaller than 6 R1

Note: Only award final R1 in each case if the calculations are correct.[19]

3. (a) = b – c, = b + c A1A1CB AC

Note: Condone absence of vector notation in (a).

(b) = ( b + c) • ( b – c) M1CBAC

= │ b│2 – │ c│2

A1

= 0 since │ b│=│ c│ R1

Note: Only award the A1 and R1 if working indicates that they understand

that they are working with vectors.

so is perpendicular to i.e. is a right angle AGAC CB BCA

[5]



4. METHOD 1

equation of journey of ship S 1

r1 =

20

10t

equation of journey of speedboat S 2 ,setting off k minutes later

r2 = M1A1A1

30

60)(

30

70k t

Note: Award M1 for perpendicular direction, A1 for speed, A1 for

change in parameter (e.g. by using t – k or T , k being the time

difference between the departure of the ships).

solve (M1)

30

60)(

30

70

20

10k t t

Note: M mark is for equating their two expressions.

10t = 70 – 60t + 60k

20t = 30 + 30t – 30k M1

Note: M mark is for obtaining two equations involving two different parameters.

7t – 6k = 7

– t + 3k = 3

k = A115

28

latest time is 11:52 A1



METHOD 2

x

y

time takent - k

time

taken t

S (26,52)

22 526 5

10 58

(A)

O

B (70,30)

SB = 22 M1A15

(by perpendicular distance)

SA = 26 M1A15

(by Pythagoras or coordinates)

t = A1510526

t – k = A1530

522

k = leading to latest time 11:52 A115

28

[7]

5. (a)

k z

y

x

1

3

212

311

120

= 0 – 2( – 2 + 6) + ( – 1 + 2) = –7 M1A1

212

311

120

since determinant ≠ 0 unique solution to the system planes R1

intersect in a point AG

Note: For any method, including row reduction, leading to the explicit

solution , award M1 for an attempt at

7

21,

7

10,

7

56 k k k

a correct method, A1 for two correct coordinates and A1 for

a third correct coordinate.



(b)

212

311

12

a

a

a

= a((a + 1)(a + 2) – 3) – 2(–1(a + 2) + 6) + (–1 + 2(a + 1)) M1(A1)

planes not meeting in a point no unique solution i.e. determinant = 0(M1)a(a

2 + 3a – 1) + (2a – 8) + (2a + 1) = 0

a3 + 3a2 + 3a – 7 = 0 A1

a = 1 A1[5]

(c) M1

21

312

4440

3121 r r

k

(A1)312

6550

44403121 r r

k

(A1)

23 54

44000

4440

3121 r r

k

for an infinite number of solutions to exist, 4 + 4k = 0 k = –1 A1

x + 2 y + z = 3

y + z = 1 M1

A1

1

1

1

0

1

1

z

y

x

Note: Accept methods involving elimination.

Note: Accept any equivalent form e.g. .

1

1

1

1

2

0

or

1

1

1

1

0

2

z

y

x

z

y

x

Award A0 if or r = is absent.

z

y

x

[14]

6. (a) x3 + 1 =

1

13 x

(–1.26, –1) (= ( , –1)) A13 2



(b) f ′(–1.259...) = 4.762... (3 × ) A13

2

2

g ′(–1.259...) = –4.762... (–3 × ) A13

2

2

required angle = 2arctan

M1

...762.4

1

= 0.414 (accept 23.7°) A1

Note: Accept alternative methods including finding the obtuse angle first.[5]

7. (a) (M1)

z

y

x

1

5

0

SR ,

3

1

1

PQ

point S = (1, 6, –2) A1

(b) A1

1

4

2

PS

3

1

1

PQ

2

7

13

PSPQ

m = – 2 A1

(c) area of parallelogram PQRS = M1222 )2(7)13(PSPQ

= = 14.9 A1222

(d) equation of plane is –13 x + 7 y – 2 z = d M1A1

substituting any of the points given gives d = 33

–13 x + 7 y – 2 z = 33 A1

(e) equation of line is r = A1

2

7

13

0

0

0

Note: To get the A1 must have r = or equivalent.



(f) 169 λ + 49 λ + 4 λ = 33 M1

λ = (= 0.149...) A1222

33

closest point is (= (–1.93, 1.04, –0.297)) A1

37

11,

74

77,

74

143

(g) angle between planes is the same as the angle between the normals (R1)

cos θ = M1A16222

1227113

θ = 143° (accept θ = 37.4° or 2.49 radians or 0.652 radians) A1[17]

8. (a) for using normal vectors (M1)

= 1 – 1 = 0 M1A1

1

0

1

1

2

1

hence the two planes are perpendicular AG

(b) METHOD 1

EITHER

= – 2i – 2 j – 2 k M1A1

101

121

k ji

OR

if is normal to π 3, then

c

b

a

a + 2b – c = 0 and a + c = 0 M1

a solution is a = 1, b = –1, c = – 1 A1

THEN

π 3 has equation x – y – z = d (M1)

as it goes through the origin, d = 0 so π 3 has equation x – y – z = 0 A1

Note: The final (M1)A1 are independent of previous working.



METHOD 2

r = A1(A1)A1A1

1

0

1

1

2

1

0

0

0

t s

[7]

9. ( a + b)•( a – b) = a • a + b • a – a • b – b • b M1

= a • a – b • b A1

= | a|2 – | b|2 = 0 since | a| = | b| A1

the diagonals are perpendicular R1

Note: Accept geometric proof, awarding M1 for recognizing OACB is a

rhombus, R1 for a clear indication that ( a + b) and ( a – b) are the

diagonals, A1 for stating that diagonals cross at right angles and

A1 for “hence dot product is zero”.

Accept solutions using components in 2 or 3 dimensions.[4]

10. (a) 2 y + 8 x = 4 M1

–3 x + 2 y = – 7 A1

2 x + 6 – 2 x = 6

Note: Award M1 for attempt at components, A1 for two correct equations.

No penalty for not checking the third equation.

solving : x = 1, y = – 2 A1

(b) │ a + 2 b│=

2

2

4

2

2

3

4

=

6

7

4

(M1)222 6)7(42 b a

= A1101

[5]



11. (a) (i) use of a • b = │ a││ b│cos θ (M1)

a • b = –1 (A1)

│ a│ = 7, │ b│ = 5 (A1)

cos θ = A135

1

(ii) the required cross product is

= 18i – 24 j – 18 k M1A1

430

236

k ji

(iii) using r • n = p • n the equation of the plane is (M1)

18 x – 24 y – 18 z = 12 (3 x – 4 y – 3 z = 2) A1

(iv) recognizing that z = 0 (M1)

x-intercept = , y-intercept = (A1)3

2

2

1

area = A16

1

2

1

2

1

3

2

(b) (i) p • p = │ p││ p│cos 0 M1A1

= │ p│2 AG

(ii) consider the LHS, and use of result from part (i)

│ p + q│2 = ( p + q)•( p + q) M1

= p • p + p • q + q • p + q • q (A1)

= p • p + 2 p • q + q • q A1

= │ p│2 + 2 p • q + │q│2

AG



(iii) EITHER

use of p • q ≤ │ p││q│ M1

so 0 ≤ | p + q|2 = │ p│2 + 2 p • q + │q│2 ≤ │ p│2 + 2│ p││q│+│q│2 A1

take square root (of these positive quantities) to establish A1

│ p + q│≤│ p│+│q│ AG

OR

M1M1

Note: Award M1 for correct diagram and M1 for correct labelling

of vectors including arrows.

since the sum of any two sides of a triangle is greater than the third side,

│ p│ + │q│ > │ p + q│ A1

when p and q are collinear │ p│ + │q│ = │ p + q│ │ p + q│ ≤ │ p│ + │q│ AG

[19]



12. EITHER

using row reduction (or attempting to eliminate a variable) M1

132

13222

2

21

213

312

R R

R R

ba

A15/2

22

10

2

3230

550

312

R

ba

Note: For an algebraic solution award A1 for two correct equations in two variables.

23322

2

2

3230

110

312

R Rba

82

2

2

6200

110

312

ba

Note: Accept alternative correct row reductions.

recognition of the need for 4 zeroes M1

so for multiple solutions a = – 3 and b = – 4 A1A1

OR

M10

21

213

312

a

2(a – 4) + (3a + 2) + 3(6 + 1) = 0

5a + 15 = 0

a = –3 A1

M10

21

213

212

b

2(b + 4) + (3b – 2) + 2(6 + 1) = 0 A1

5b + 20 = 0

b = –4 A1

[5]



13. (a) EITHER

normal to plane given by

M1A1

236

232

k ji

= 12i + 8 j – 24 k A1

equation of π is 3 x + 2 y – 6 z = d (M1)

as goes through (–2, 3, –2) so d = 12 M1A1

π :3 x + 2 y – 6 z = 12 AG

OR

x = – 2 + 2 λ + 6 μ

y = 3 + 3 λ – 3 μ

z = – 2 + 2 λ + 2 μ

eliminating μ

x + 2 y = 4 + 8 λ2 y + 3 z = 12 λ M1A1A1

eliminating λ

3( x + 2 y) – 2(2 y + 3 z ) = 12 M1A1A1

π : 3 x + 2 y – 6 z = 12 AG

(b) therefore A(4, 0, 0), B(0, 6, 0) and C(0, 0, –2) A1A1A1

Note: Award A1A1A0 if position vectors given instead of coordinates.

(c) area of base OAB = = 12 M1642

1

V = = 8 M1A12123

1

(d) = 3 = 7 × 1 × cos M1A1

0

0

1

6

2

3

7

3arccos

so θ = 90 – arccos

= 25.4° (accept 0.443 radians) M1A17

3

(e) d = 4 sin θ = (= 1.71) (M1)A17

12



(f) 8 = area = 14 M1A1 area7

12

3

1

Note: If answer to part (f) is found in an earlier part, award M1A1,

regardless of the fact that it has not come from their answers

to part (c) and part (e).[20]

14. (a) use GDC or manual method to find a, b and c (M1)

obtain a = 2, b = – 1, c = 3 (in any identifiable form) A1

(b) use GDC or manual method to solve second set of equations (M1)

obtain x = , z = t (or equivalent) (A1)2

7;

2

114 t y

t

r = (accept equivalent vector forms) M1A1

1

5.3

5.5

0

0

2

t

Note: Final A1 requires r = or equivalent.[6]

15. (a) a = is parallel to the line (A1)(A1)

k

1

2

planetothe

1

2

4

e

Note: Award A1 for each correct vector written down, even if not identified.

line plane e parallel to a

since (M1)A12

11

2

1

2

4

k

k

t



(b) 4(3 – 2 λ) – 2 λ – = 1 (M1)(A1)

2

11

Note: FT their value of k as far as possible.

λ = A17

8

A1

7

3,

7

8,

7

5P

[8]

16. (a) cos θ = M1A1

2

13sin

22

12cossincos2sin

b a

ab

(b) a b cos θ = 0 M1

sin 2α cos α + sin α cos 2α – 1 = 0

α = 0.524 A1

6

π

(c) METHOD 1

(M1)1sincos

12cos2sin

k ji

assuming α =6

π7

Note: Allow substitution at any stage.

A1

121

23

12

1

2

3

k ji

=

2

3

2

1

2

1

2

3

2

3

2

3

2

1

2

1 k ji

= 0 A1

a and b are parallel R1

Note: Accept decimal equivalents.



METHOD 2

from (a) cos θ = –1 (and sin θ = 0) M1A1

a × b = 0 A1

a and b are parallel R1[8]

17. (a) A1A1A1

1

2

0

OP and

2

1

0

ON,

2

2

1

OM

(b) A1A1 ,

0

1

1

MNand

1

0

1

MP

(M1)A1

1

1

1

011

101MNMP

k ji

(c) (i) area of MNP = M1MNMP2

1

=

1

1

1

21

= A12

3

(ii)

2

2

0

OG,

0

0

2

OA

A1

2

2

2

AG

since AG is perpendicular to MNP R1)MNMP(2AG



(iii) r M1A1

1

1

1

2

2

1

1

1

1

r = 3 (accept – x + y + z = 3) A1

1

1

1

(d) r = A1

2

2

2

0

0

2

= 3 M1A1

1

1

1

2

2

22

–2 + 2 λ + 2 λ + 2 λ = 3

λ = A16

5

r = M1

2

2

2

6

5

0

0

2

coordinates of point A1

3

5,

3

5,

3

1

[20]



18. METHOD 1

for finding two of the following three vectors (or their negatives)

(A1)(A1)

1

0

2

BC,

2

2

2

AC,

1

2

0

AB

and calculating

EITHER

M1A1ACAB

4

2

2

222

120

k ji

area ∆ABC = M1ACAB2

1

OR

M1A1BCBA

4

2

2

102

120

k ji

area ∆ABC = M1BCBA2

1

OR

M1A1CBCA

4

2

2

102

222

k ji

area ∆ABC = M1CBCA2

1

THEN

area ∆ABC = A12

24

= AG N06



METHOD 2


(A1)(A1)

1

0

2

BC,

2

2

2

AC,

1

2

0

AB

EITHER

cos A = M1ACAB

ACAB

=

15

3or

60

6

125

6

sin A = A15

2

area ∆ABC = M1 AsinACAB2

1

=5

2125

2

1

= A1242

1

= AG N06

OR

cos B = M1BCBA

BCBA

=5

1

55

1

sin B = A1

5

24or

25

24

area ∆ABC = M1 BsinBCBA2

1

=25

2455

2

1

= A1242

1

= AG N06



OR

cos C = M1CBCA

CBCA

=

153or

606

5126

sin C = A15

2

area ∆ABC = M1C sinCBCA2

1

=5

2512

2

1

= A124

2

1

= AG N06

METHOD 3


(A1)(A1)

1

0

2

BC,

2

2

2

AC,

1

2

0

AB

AB = M1A1abc 5BC,3212AC,5

s = M1532

5325

area ∆ABC = ))()(( c sb sa s s

= )3)(35)(3)(53(

= A1)35(3

= AG N06



METHOD 4


(A1)(A1)

1

0

2

BC,

2

2

2

AC,

1

2

0

AB

AB = BC = and AC = M1A15 3212

∆ABC is isosceles

let M be the midpoint of [AC], the height BM = M1235

area ∆ABC = A12

232

= AG N06

[6]

19. (a) identifies a direction vector e.g. A1

1

1

2

BAor

1

1

2

AB

identifies the point (1, –1, 2) A1

line l 1: AG1

2

1

1

2

1

z y x

(b) r =

1

2

1

3

2

1

1

1

2

2

1

1

r

1 + 2 λ = 1 + μ, –1 + λ = 2 + 2 μ, 2 + λ = 3 + μ (M1)

equating two of the three equations gives λ = –1 and μ = – 2 A1A1

check in the third equation

satisfies third equation therefore the lines intersect R1

therefore coordinates of intersection are (–1, –2, 1) A1



(c) d 1 = 2i + j + k, d 2 = i + 2 j + k A1

d 1 × d 2 = = – i – j + 3 k M1A1

121

112

k ji

Note: Accept scalar multiples of above vectors.

(d) equation of plane is – x – y + 3 z = k M1A1

contains (1, 2, 3) (or (–1, –2, 1) or (1, –1, 2)) k = –1 – 2 + 3 × 3 = 6 A1

– x – y + 3 z = 6 AG

(e) direction vector of the perpendicular line is (M1)

3

1

1

r = A1

3

11

4

13

m

Note: Award A0 if r omitted.

(f) (i) find point where line meets plane

–(3 – m) – (1 – m) + 3( – 4 + 3m) = 6 M1

m = 2 A1

point of intersection is (1, –1, 2) A1

(ii) for T′, m = 4 (M1)

so T′ = (–1, –3, 8) A1

(iii) (M1)222 8)4(3)(11)(3TT

= A1)114(176

[22]



20. consider a vector parallel to each line,

e.g. u = A1A1

1

3

3

and

1

2

4

v

let θ be the angle between the lines

cos θ = M1A11921

1612

vu

vu

= = 0.350... (A1)1921

7

so 0 = 69.5° A1 N4

1921

7arccosorrad1.21or

Note: Allow FT from incorrect reasonable vectors.

[6]

21. (a) let A = (M1)

5

2

2

and,

415

312

211

B X

z

y

x

point of intersection is (or (0.917, 0.583, 0.25)) A1

4

1,

12

7,

12

11

(b) METHOD 1

(i) det

M10

15

312

211

a

– 3a + 24 = 0 (A1)

a = 8 A1 N1

(ii) consider the augmented matrix M1

5

2

2

815

312

211

use row reduction to obtain

1

0

0

000

10

01

or

1

2

2

000

130

211

31

35

(or equivalent) A1

any valid reason R1

(e.g. as the last row is not all zeros, the planes do not meet) N0



METHOD 2

use of row reduction (or equivalent manipulation of equations) M1

e.g. A1A1

5

2

2

1060

130

211

5

2

2

15

312

211

aa

Note: Award an A1 for each correctly reduced row.

(i) a – 10 = – 2 a = 8 M1A1 N1

(ii) when a = 8, row 3 ≠ 2 × row 2 R1 N0[8]

22. (a) = i + 2 j – k (M1)OP

the coordinates of P are (1, 2, –1) A1

(b) EITHER

x = 1 + t , y = 2 – 2t , z = 3t – 1 M1

x – 1 = t , A1t z

t y

3

1,

2

2

x – 1 = AG N03

1

2

2

z y

OR

M1A1

3

2

1

1

2

1

t

z

y

x

x – 1 = AG3

1

2

2

z y

(c) (i) 2(1 + t ) + (2 – 2t ) + (3t – 1) = 6 t = 1 M1A1 N1

(ii) coordinates are (2, 0, 2) A1

Note: Award A0 for position vector.

(iii) distance travelled is the distance between the two points (M1)

(= 3.74) (M1)A114)12()20()12( 222



(d) (i) distance from Q to the origin is given by

d (t ) = (or equivalent) M1A12224 )1()1( t t t

e.g. for labelled sketch of graph of d or d 2 (M1)(A1)

the minimum value is obtained for t = 0.761 A1 N3

(ii) the coordinates are (0.579, 0.239, 0.421) A1

Note: Accept answers given as a position vector.

(e) (i) (M1)A1

3

1

4

and

0

0

1

,

1

1

0

c b a

substituting in the equation a – b = k ( b – c), we have (M1)

A1

3

1

3

1

1

1

3

1

4

0

0

1

0

0

1

1

1

0

k k

which is impossible

3

1 and1 k k

so there is no solution for k R1

(ii) are not parallel R2CBandBA

(hence A, B, and C cannot be collinear)

Note: Only accept answers that follow from part (i).

[23]



23. direction vector for line = or any multiple A1

1

1

M101

1

sin1

sin2

2 sin θ – 1 + sin θ = 0 A1

Note: Allow FT on candidate’s direction vector just for line above only.

3 sin θ = 1

sin θ = A13

1

θ = 0.340 or 19.5° A1

Note: A coordinate geometry method using perpendicular gradients is acceptable.[5]

24. EITHER

l goes through the point (1, 3, 6), and the plane contains A(4, –2, 5)

the vector containing these two points is on the plane, i.e.

(M1)A1

1

5

3

5

2

4

6

3

1

= 7i + 4 j + k M1A1

153

121

1

5

3

1

2

1

k ji

(M1)25

1

4

7

5

2

4

hence, Cartesian equation of the plane is 7 x + 4 y + z = 25 A1



OR

finding a third point M1

e.g. (0, 5, 5) A1

three points are (1, 3, 6), (4, –2, 5), (0, 5, 5)

equation is ax + by + cz = 1

system of equationsa + 3b + 6c = 1 M1

4a – 2b + 5c = 1

5b + 5c = 1

a = , from GDC M1A125

1,

25

4,

25

7 cb

so A1125

1

25

4

25

7 z y x

or 7 x + 4 y + z = 25[6]

25. (a) on l 1 A(–3 + 3 λ, –4 + 2 λ, 6 – 2 λ) A1

on l 2 l 2 : r = (M1)

1

4

3

3

7

4

B(4 – 3 μ, –7 + 4 μ, – 3 – μ) A1

(M1)A1BA

92

342

733

b a



EITHER

BA

0

2

2

3

BA1

l

3(3 λ + 3 μ – 7) + 2(2 λ – 4 μ + 3) – 2(–2 λ + μ + 9) = 0 M1

17 λ – μ = 33 A1

BA

0

1

4

3

BA2

l

–3(3 λ + 3 μ – 7) + 4(2 λ – 4 μ + 3) – 2(–2 λ + μ + 9) = 0 M1

λ – 26 μ = –24 A1

solving both equations above simultaneously gives

λ = 2; μ = 1 A(3, 0, 2), B(1, –3, –4) A1A1A1A1

OR

= 6i + 9 j + 18 k M1A1

143

223

k ji

so M1A1AB

92

342

733

6

3

2

p

3 λ + 3 μ – 2 p = 7

2 λ – 4 μ – 3 p = – 3

–2 λ + μ – 6 p = –9

λ = 2, μ = 1, p = 1 A1A1

A(–3 + 6, –4 + 4, 6 – 4) = (3, 0, 2) A1

B(4 – 3, –7 + 4, –3 – 1) = (1, –3, –4) A1

(b) AB = (A1)

6

3

2

2

0

3

4

3

1

|AB| = = 7 M1A149)6()3()2( 222



(c) from (b) 2i + 3 j + 6 k is normal to both lines

l 1 goes through (–3, –4, 6) = 18 M1A1

6

3

2

6

4

3

hence, the Cartesian equation of the plane through l 1, but not l 2,is 2 x + 3 y + 6 z = 18 A1

[19]

26. (a) (i) METHOD 1

= b a = (A1)

AB

110

211

321

= c a =

(A1)AC

112

211

103

= M1

AB

AC112110

k ji

= i (1 + 1) j(0 2) + k (0 2) (A1)

= 2 j 2 k A1

Area of triangle ABC = sq. units M1A1 28

2

122

2

1 k j

Note: Allow FT on final A1.



METHOD 2

A1A1A16AC,12BC,2AB

Using cosine rule, e.g. on M1C ˆ

cos C = A13

22

722

2126

M1C ab sin

2

1ABCArea

=

3

22arccossin612

2

1

= A1 2

3

22arccossin23

Note: Allow FT on final A1.

(ii) AB = A12

= equals the shortest distance (M1)2 hhh ,22

1AB

2

1

h = 2 A1

(iii) METHOD 1

has form r •

(M1)d

2

2

0

Since (1, 1, 2) is on the plane

d = •

M1A1

2

1

1

242

2

2

1

Hence r •

= 2

2

2

0

2 y 2 z = 2 (or y z = 1) A1



METHOD 2

r = (M1)

1

1

2

1

1

0

2

1

1

μ λ

x = 1 + 2 (i)

y = 1 + (ii)

z = 2 + (iii) A1

Note: Award A1 for all three correct, A0 otherwise.

From (i) =2

1 x

substitute in (ii) y = 1 +

2

1 x

= y 1 +

2

1 x

substitute and in (iii) M1

z = 2 + y 1 +

2

1

2

1 x x

y z = 1 A1

(b) (i) The equation of OD is

r =

M1

1

1

0

or ,

2

2

0

λ r

This meets where

2 + 2 = 1 (M1)

= A14

1

Coordinates of D are A1

2

1,

2

1,0

(ii) (M1)A12

1

2

1

2

10OD

22

[20]



27. METHOD 1

Use of | a b | = | a | | b | sin (M1)

| a b |2 = | a |2 | b |2 sin2 (A1)

Note: Only one of the first two marks can be implied.

= | a |2

| b |2

(1 cos2

) A1

= | a |2 | b |

2 | a |

2 | b |

2 cos

2 (A1)

= | a |2 | b |

2 (| a | | b | cos )

2 (A1)

Note: Only one of the above two A1 marks can be implied.

= | a |2 | b |2 ( a • b)2 A1

Hence LHS = RHS AG N0

METHOD 2

Use of a • b = | a | | b | cos (M1)

| a |2 | b |2 ( a • b)2 = | a |2 | b |2 (| a | | b | cos )2 (A1)

= | a |2 | b |2 | a |2 | b |2 cos2 (A1)

Note: Only one of the above two A1 marks can be implied.

= | a |2 | b |2 (1 cos2 ) A1

= | a |2 | b |2 sin2 A1

= | a b |2 A1

Hence LHS = RHS AG N0

Notes: Candidates who independently correctlysimplify both sides

and show that LHS = RHS should be awarded full marks.

If the candidate starts off with expression that they are trying to

prove and concludes that sin2 = (1 cos

2) award M1A1A1A1A0A0.

If the candidate uses two general 3D vectors and explicitly finds

the expressions correctly award full marks. Use of 2D vectors

gains a maximum of 2 marks.

If two specific vectors are used no marks are gained.[6]



28. (a) Use of cos = (M1)

ABOA

ABOA

= i j + k A1

AB

= and = A1

AB 3

OA 23

= 6 A1

ABOA

substituting gives cos = or equivalent M1 N1

3

6

6

2

(b) L1: r = + s or equivalent (M1)

OA

AB

L1: r = i j + 4 k + s(i j + k) or equivalent A1

Note: Award (M1)A0 for omitting “r =” in the final answer.

(c) Equating components and forming equations involving s and t (M1)

1 + s = 2 + 2t , 1 s = 4 + t , 4 + s = 7 + 3t

Having two of the above three equations A1A1

Attempting to solve for s or t (M1)

Finding either s = 3 or t = 2 A1Explicitly showing that these values satisfy the third equation R1

Point of intersection is (2, 2, 1) A1 N1

Note: Position vector is not acceptable for final A1.

(d) METHOD 1

r = (A1)

3

3

3

3

1

2

4

1

1

x = 1 + 2 3 , y = 1 + + 3 and z = 4 + 3 3 M1A1

Elimination of the parameters M1

x + y = 3 so 4( x + y) = 12 and y + z = 4 + 3

so 3( y + z ) = 12 + 9

3( y + z ) = 4( x + y) + 9 A1

Cartesian equation of plane is 4 x + y 3 z = 9 (or equivalent) A1 N1



METHOD 2

EITHER

The point (2, 4, 7) lies on the plane.

The vector joining (2, 4, 7) and (1, 1, 4) and 2i + j + 3 k

are parallel to the plane. So they are perpendicular to thenormal to the plane.

(i j + 4 k) (2i + 4 j + 7 k) = i 5 j 3 k (A1)

M1

312

351

k ji

n

= 12i 3 j + 9 k or equivalent parallel vector A1

OR

L1 and L2 intersect at D (2, 2,1)

= (2i + 2 j + k) (i j + 4 k) = 3i + 3 j 3 k (A1)

AD

M1

333

312

k ji

n

= 12i 3 j + 9 k or equivalent parallel vector A1

THEN

r • n = (i j + 4 k) • (12i 3 j + 9 k) M1

= 27 A1

Cartesian equation of plane is 4 x + y 3 z = 9 (or equivalent) A1 N1[20]



29. The normal vector to the plane is (A1).

2

3

1

EITHER

is the angle between the line and the normal to the plane.

(M1)A1A1

67

3

2114

3

2114

2

3

1

2

1

4

cosθ

= 79.9 (= 1.394 ...) A1

The required angle is 10.1 (= 0.176) A1

OR

is the angle between the line and the plane.

(M1)A1A12114

3

2114

2

3

1

2

1

4

sin

= 10.1 (= 0.176) A2[6]

30. METHOD 1

(from GDC)

(M1)

0

6

112

1

000

3

210

6

101

A1

12

1

6

1 λ x

A1

6

1

3

2 λ y

r = A1A1A1 N3

k ji ji

3

2

6

1

6

1

12

1



METHOD 2

(Elimination method either for equations or row reduction of matrix)

Eliminating one of the variables M1A1

Finding a point on the line (M1)A1

Finding the direction of the line M1

The vector equation of the line A1 N3[6]

31. = c – bBC

= a – cCA

a • ( c – b) = 0 M1

and b • ( a – c) = 0 M1

a • c = a • b A1

and a • b = b • c A1

a • c = b • c M1

b • c – a • c = 0

c • ( b – a) = 0 A1

is perpendicular to , as b ≠ a. AGOC AB

[6]

32. a • b = │ a││ b│cos θ (M1)

a • b = = 7 + 3m A1

m

2

3

3

2

1

A121314 m b a

30cos1314cos 2m b a

7 + 3m = cos 30° A121314 m

m = 2.27, m = 25.7 A1A1[6]



33. (a) M1

5705

4213

321 k

R1 – 2 R2

(A1)

5705

42138705 k

R1 + R3

(A1)

5705

4213

3000 k

Hence no solutions if k , k ≠ 3 A1

(b) Two planes meet in a line and the third plane is parallel to that line. A1[5]

34. (a) x = 3 + 2m

y = 2 – m

z = 7 + 2m A1

x = 1 + 4n

y = 4 – n

z = 2 + n A1

(b) 3 + 2m = 1 + 4n 2m – 4n = – 2(i)

2 – m = 4 – n m – n = –2(ii) M1

7 + 2m = 2 + n 2m – n = – 5(iii)

(iii) – (ii) m = –3 A1

n = –1 A1

Substitute in (i), – 6 + 4 = – 2. Hence lines intersect. R1

Point of intersection A is (–3, 5, 1) A1



(c) M1A1

2

6

1

114

212

k ji

r • (M1)

2

61

7

23

2

61

r • = 29

2

6

1

x + 6 y + 2 z = 29 A1

Note: Award M1A0 if answer is not in Cartesian form.

(d) x = –8 + 3 λ

y = – 3 + 8 λ (M1)

z = 2 λ

Substitute in equation of plane.

–8 + 3 λ – 18 + 48 λ + 4 λ = 29 M1

55 λ = 55

λ = 1 A1

Coordinates of B are (–5, 5, 2) A1

(e) Coordinates of C are (A1)

2

3,5,4

r = M1A1

2

6

1

2

3

5

4

Note: Award M1A0 unless candidate writes r = or

z

y

x

[18]



35. EITHER

Let s be the distance from the origin to a point on the line, then

s2 = (1 – λ)2 + (2 – 3 λ)2 + 4 (M1)

= 10 λ2 – 14 λ + 9 A1

= 20 λ – 14 A1

d)d(

2

s

For minimum A110

7,0

d

)d( 2

s

OR

The position vector for the point nearest to the origin is perpendicular to

the direction of the line. At that point:

= 0 (M1)A1

03

1

232

1

Therefore, 10 λ – 7 = 0 A1

Therefore, λ = A110

7

THEN

x = (A1)(A1)10

1,

10

3 y

The point is . N3

2,

10

1,

10

3

[6]



36. (a)

M1

(M1)

M1

When a = – 1 the augmented matrix is

A1Hence the system is inconsistent a ≠ – 1 R1

(b) When a ≠ – 1, ( –a – 1) z = 9 – a2

(a + 1) z = a2 – 9

M1A1

1

92

a

a z

2 y – z = 0 M1A1)1(2

9

2

1 2

a

a z y

x = – 3 y + z = M1A1)1(2

9

)1(2

)9(2

)1(2

)9(3 222

a

a

a

a

a

a

The unique solution is when a ≠ –1

1

9,

)1(2

9,

)1(2

9 222

a

a

a

a

a

a

(c) 2 – a = 1 a = 1 M1

or (2, –2, –4) A1

2

8,

4

8,

4

8issolutionThe

[13]

37. (a) = – i – 3 j + k, = i + j A1A1AB BC



(b) M1

011

131BCAB

k ji

= – i + j + 2 k A1

(c) Area of ∆ABC = │ – i + j + 2 k│ M1A12

1

= 4112

1

= A12

6

(d) A normal to the plane is given by n = = – i + j + 2 k (M1)BCABTherefore, the equation of the plane is of the form – x + y + 2 z = g

and since the plane contains A, then –1 + 2 + 2 = g g = 3. M1

Hence, an equation of the plane is – x + y + 2 z = 3. A1

(e) Vector n above is parallel to the required line.

Therefore, x = 2 – t A1

y = – 1 + t A1

z = – 6 + 2t A1

(f) x = 2 – t

y = – 1 + t

z = – 6 + 2t

– x + y + 2 z = 3

–2 + t – 1 + t – 12 + 4t = 3 M1A1

–15 + 6t = 3

6t = 18

t = 3 A1

Point of intersection (–1, 2, 0) A1

(g) Distance = (M1)A154633 222

(h) Unit vector in the direction of n is e = (M1) n n

1

= (– i + j + 2 k) A16

1

Note: –e is also acceptable.



(i) Point of intersection of L and P is (–1, 2, 0).

(M1)A1

6

3

3

DE

M1

6

33

EF

coordinates of F are (–4, 5, 6) A1

[25]

38. (a) L1 : x = 2 + λ; y = 2 + 3 λ; z = 3 + λ (A1)

L2 : x = 2 + µ; y = 3 + 4 µ; z = 4 + 2 µ (A1)

At the point of intersection (M1)2 + λ = 2 + µ (1)

2 + 3 λ = 3 + 4 µ (2)

3 + λ = 4 + 2 µ (3)

From (1), λ = µ A1

Substituting in (2), 2 + 3 λ = 3 + 4 λ

λ = µ = –1 A1

We need to show that these values satisfy (3). (M1)

They do because LHS = RHS = 2; therefore the lines intersect. R1

So P is (1, –1, 2). A1 N3

(b) The normal to Π is normal to both lines. It is therefore given bythe vector product of the two direction vectors.

Therefore, normal vector is given by M1A1

241

131

k ji

= 2i – j + k A2

The Cartesian equation of Π is 2 x – y + z = 2 + 1 + 2 (M1)

i.e. 2 x – y + z = 5 A1 N2



(c) The midpoint M of [PQ] is . M1A1

2

5,

2

3,2

The direction of is the same as the normal to Π , i.e. 2i – j + k (R1)MS

The coordinates of a general point R on are thereforeMS

(M1)

2

5,

2

3,22

It follows that = (1 + 2 λ)i + A1A1A1PR k j

2

1

2

5

At S, length of is 3, i.e. (M1)PR

(1 + 2 λ)2 + A19

2

1

2

522

1 + 4 λ + 4 λ2 + – 5 λ + λ2 + + λ + λ2 = 9 (A1)

4

25

4

1

6 λ2 = A1

4

6

λ = A12

1

Substituting these values, (M1)

the possible positions of S are (3, 1, 3) and (1, 2, 2) A1A1 N2[29]

39. (a) Finding correct vectors A1A1

1

1

3

AC

1

3

4

AB

Substituting correctly in scalar product = 4(–3) + 3(1) – 1(1) A1ACAB

= –10 AG N0

(b) (A1)(A1)11AC26AB

Attempting to use scalar product formula, M11126

10

CAˆ

Bcos

= –0.591 (to 3 s.f.) A1

= 126° A1 N3CAB

[8]

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