Date post: | 09-Jan-2016 |
Category: |
Documents |
Upload: | raahish-kalaria |
View: | 14 times |
Download: | 1 times |
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.
7/17/2019 Vectors Practice MS
http://slidepdf.com/reader/full/vectors-practice-ms 2/43
(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
7/17/2019 Vectors Practice MS
http://slidepdf.com/reader/full/vectors-practice-ms 3/43
(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]
7/17/2019 Vectors Practice MS
http://slidepdf.com/reader/full/vectors-practice-ms 4/43
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
7/17/2019 Vectors Practice MS
http://slidepdf.com/reader/full/vectors-practice-ms 5/43
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.
7/17/2019 Vectors Practice MS
http://slidepdf.com/reader/full/vectors-practice-ms 6/43
(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
7/17/2019 Vectors Practice MS
http://slidepdf.com/reader/full/vectors-practice-ms 7/43
(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.
7/17/2019 Vectors Practice MS
http://slidepdf.com/reader/full/vectors-practice-ms 8/43
(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.
7/17/2019 Vectors Practice MS
http://slidepdf.com/reader/full/vectors-practice-ms 9/43
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]
7/17/2019 Vectors Practice MS
http://slidepdf.com/reader/full/vectors-practice-ms 10/43
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
7/17/2019 Vectors Practice MS
http://slidepdf.com/reader/full/vectors-practice-ms 11/43
(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]
7/17/2019 Vectors Practice MS
http://slidepdf.com/reader/full/vectors-practice-ms 12/43
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]
7/17/2019 Vectors Practice MS
http://slidepdf.com/reader/full/vectors-practice-ms 13/43
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
7/17/2019 Vectors Practice MS
http://slidepdf.com/reader/full/vectors-practice-ms 14/43
(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
7/17/2019 Vectors Practice MS
http://slidepdf.com/reader/full/vectors-practice-ms 15/43
(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.
7/17/2019 Vectors Practice MS
http://slidepdf.com/reader/full/vectors-practice-ms 16/43
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
7/17/2019 Vectors Practice MS
http://slidepdf.com/reader/full/vectors-practice-ms 17/43
(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]
7/17/2019 Vectors Practice MS
http://slidepdf.com/reader/full/vectors-practice-ms 18/43
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
7/17/2019 Vectors Practice MS
http://slidepdf.com/reader/full/vectors-practice-ms 19/43
METHOD 2
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
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
7/17/2019 Vectors Practice MS
http://slidepdf.com/reader/full/vectors-practice-ms 20/43
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
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
AB = M1A1abc 5BC,3212AC,5
s = M1532
5325
area ∆ABC = ))()(( c sb sa s s
= )3)(35)(3)(53(
= A1)35(3
= AG N06
7/17/2019 Vectors Practice MS
http://slidepdf.com/reader/full/vectors-practice-ms 21/43
METHOD 4
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
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
7/17/2019 Vectors Practice MS
http://slidepdf.com/reader/full/vectors-practice-ms 22/43
(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]
7/17/2019 Vectors Practice MS
http://slidepdf.com/reader/full/vectors-practice-ms 23/43
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
7/17/2019 Vectors Practice MS
http://slidepdf.com/reader/full/vectors-practice-ms 24/43
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
7/17/2019 Vectors Practice MS
http://slidepdf.com/reader/full/vectors-practice-ms 25/43
(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]
7/17/2019 Vectors Practice MS
http://slidepdf.com/reader/full/vectors-practice-ms 26/43
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
7/17/2019 Vectors Practice MS
http://slidepdf.com/reader/full/vectors-practice-ms 27/43
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
7/17/2019 Vectors Practice MS
http://slidepdf.com/reader/full/vectors-practice-ms 28/43
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
7/17/2019 Vectors Practice MS
http://slidepdf.com/reader/full/vectors-practice-ms 29/43
(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.
7/17/2019 Vectors Practice MS
http://slidepdf.com/reader/full/vectors-practice-ms 30/43
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
7/17/2019 Vectors Practice MS
http://slidepdf.com/reader/full/vectors-practice-ms 31/43
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]
7/17/2019 Vectors Practice MS
http://slidepdf.com/reader/full/vectors-practice-ms 32/43
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]
7/17/2019 Vectors Practice MS
http://slidepdf.com/reader/full/vectors-practice-ms 33/43
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
7/17/2019 Vectors Practice MS
http://slidepdf.com/reader/full/vectors-practice-ms 34/43
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]
7/17/2019 Vectors Practice MS
http://slidepdf.com/reader/full/vectors-practice-ms 35/43
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
7/17/2019 Vectors Practice MS
http://slidepdf.com/reader/full/vectors-practice-ms 36/43
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]
7/17/2019 Vectors Practice MS
http://slidepdf.com/reader/full/vectors-practice-ms 37/43
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
7/17/2019 Vectors Practice MS
http://slidepdf.com/reader/full/vectors-practice-ms 38/43
(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]
7/17/2019 Vectors Practice MS
http://slidepdf.com/reader/full/vectors-practice-ms 39/43
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]
7/17/2019 Vectors Practice MS
http://slidepdf.com/reader/full/vectors-practice-ms 40/43
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
7/17/2019 Vectors Practice MS
http://slidepdf.com/reader/full/vectors-practice-ms 41/43
(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.
7/17/2019 Vectors Practice MS
http://slidepdf.com/reader/full/vectors-practice-ms 42/43
(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
7/17/2019 Vectors Practice MS
http://slidepdf.com/reader/full/vectors-practice-ms 43/43
(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]