H2 Revision: Vectors 2010 Mathematics Department
Page 1 of 23
Vectors
Qn Solution 1 AJC/2009 Prelim/I/13 (i) 0 2
1 , 0 0 02
1 1
xz y r λ
� � � �� � � �= − − = � = +� � � �� � � �− −� � � �
�
Vector parallel to 2
1 0 11 0 1
0 1 1π
� � � � � �� � � � � �= − − = −� � � � � �� � � � � �−� � � � � �
1
1 2 11 0 3
1 1 2
1 0 1 1: 3 0 3 3 2
2 1 2 2
n
r rπ
� � � � � �� � � � � �= − × =� � � � � �� � � � � �−� � � � � �� � � � � � � �� � � � � � � �• = • � • = −� � � � � � � �� � � � � � � �−� � � � � � � �
�
(ii) Let foot of perpendicular from Q to 2π be N.
3 113 36 2
ON λ� � � �� � � �= +� � � �� � � �� � � �
����
3 113 3 3 26 2 2
4
λλλ
λ
+� � � �� � � �+ • =� � � �� � � �+� � � �� = −
112
ON
−� �� �= � �� �−� �
����
(iii) 1 3 5'
' 2 1 13 112
2 6 10
OQ OQON OQ
− −� � � � � �+ � � � � � �= � = − = −� � � � � �
� � � � � �− −� � � � � �
���� ��������� �����
2 2
5 1 6 3' 11 1 10 2 5
10 0 10 5
1 3 5 3: 1 5 OR : 11 5
0 5 10 5
PQ
l r l rλ λ
− −� � � � � � � �� � � � � � � �= − − − = − = −� � � � � � � �� � � � � � � �− −� � � � � � � �
−� � � � � � � �� � � � � � � �∴ = − + ∴ = − +� � � � � � � �� � � � � � � �−� � � � � � � �
�����
� �
H2 Revision: Vectors 2010 Mathematics Department
Page 2 of 23
(iv) 132
� �� �� �� �� �
= 6 24
a
k a� �� �
� =� �� �� �
Method 1:
1 2
1 1: 3 2 : 3
22 2
br rπ π� � � �� � � �• = − • =� � � �� � � �� � � �
Distance between the 2 planes = ( )2
2 2241 9 4
b − −=
+ + 108 or -116b� =
Method 2:
Distance QN = ( )( ) ( ) ( )( )2 223 1 13 1 6 2 224− − + − + − − = 2 3 2 2 5 26 13 6 108 OR ' 6 11 6 1164 6 4 4 10 4
b OQ b OQ
−� � � � � � � � � � � �� � � � � � � � � � � �= • = • = = • = − • = −� � � � � � � � � � � �� � � � � � � � � � � �−� � � � � � � � � � � �
���� �����
Method 3
1
1 11 2
: 3 2 314 14
2 2r rπ� � � �� � � �• = − � • = −� � � �� � � �� � � �
2
11 2
: 3 22414 14
2
13 2 56 54 or 58 108 or 1162
r
r b
π� �� �• = − ±� �� �� �
� �� �� • = − ± = − � = −� �� �� �
(v) 2,a b≠ ∈�
Qn Solution 2 AJC/2009 Prelim/II/2 (i)
' 4tan ' 3 4 ' 4
3 3EE
EE E E kθ= � = × = ∴ =�����
�
' ' 3 4 10OF OE E E EF j k i= + + = + +���� ����� ����� ����
� ��
(Shown)
H2 Revision: Vectors 2010 Mathematics Department
Page 3 of 23
(ii) 10 06 3
0 4DB DE
� � � �� � � �= − = −� � � �� � � �� � � �
���� ����
10 0 24 126 3 40 2 20
0 4 30 15n
−� � � � � � � �� � � � � � � �= − × − = − = −� � � � � � � �� � � � � � � �−� � � � � � � �
�
12 1020 315 4
sin769 125
50.7228.. 50.7
θ
θ
� � � �� � � �•� � � �� � � �� � � �=
= =� �
Qn Solution 3 ACJC/Prelim 2009/I/12 (i) Let N be the foot of perpendicular from A to the plane 1p .
111
AN λ� �� �= � �� �� �
����
1 1 12 1 24 1 4
ON OA AN
λλ λ
λ
− − +� � � � � �� � � � � �= + = + = +� � � � � �� � � � � �+� � � � � �
���� ���� ����
Since N is a point on the plane 1p ,
1. 1 7
1ON
� �� � =� �� �� �
����
1 12 . 1 74 1
λλλ
− +� � � �� � � �+ =� � � �� � � �+� � � �
( 1 ) (2 ) (4 ) 7λ λ λ− + + + + + = 2λ =
21
31 12 1
2 2 83 3
4 1424
3
ON
λλλ
� �− +� �− + −� � � �� �� � � �� �= + = + =� � � �� �� � � �+ � �� � � �
� �+� �� �
����
H2 Revision: Vectors 2010 Mathematics Department
Page 4 of 23
(ii) 4 13 . 1 4 3 0 70 1
� � � �� � � � = + + =� � � �� � � �� � � �
(satisfies equation of plane 1p )
4 13 . 1 4 3 0 10 a
� � � �� � � �− = − + =� � � �� � � �� � � �
(satisfies equation of plane 2p )
Hence the point (4, 3, 0) is on both planes 1p and 2p . (4, 3, 0) is then a point on the line � where the 2 planes intersect.
1 1 11 1 11 2
a
a
a
+� � � � � �� � � � � �× − = −� � � � � �� � � � � �−� � � � � �
is the direction vector of the line � ,
Hence r 4 13 10 2
a
aλ+� � � �
� � � �= + −� � � �� � � �−� � � �
is the equation of line � .
(iii)
If the planes 3p with equation r1
. 23
b� �� � =� �� �� �
intersects with 1p and 2p at line �
, then
1 12 . 1 03 2
a
a
+� � � �� � � �− =� � � �� � � �−� � � �
(1 ) 2(1 ) 6 0a a+ + − − = 3a = −
The line � lies in the plane 3p and thus
430
� �� �� �� �� �
lies in the plane 3p also.
4 13 . 2 4 6 0 100 3
b b� � � �� � � � = � = + + =� � � �� � � �� � � �
(iv) 0 1 1
2 2 45 4 1
AB OB OA
−� � � � � �� � � � � �= − = − − = −� � � � � �� � � � � �� � � � � �
����.
Given the length of projection of line segment AB on the line � is 16
.
H2 Revision: Vectors 2010 Mathematics Department
Page 5 of 23
1
. 1
2 11 61
2
a
AB a
a
a
+� �� �−� �� �−� � =+� �
� �−� �� �−� �
����
2 2 2
(1 ) 4(1 ) 2 16(1 ) (1 ) ( 2)
a a
a a
+ − − −=
+ + − + −
2
5( 1) 13( 3)
a
a
− =+
2
2
25( 1) 1( 3) 3
aa
− =+
2 275( 1) 3a a− = + 274 150 72 0a a− + = 150 1188 75 247
2(74) 74a
± ±= =
0.781a = or 1.25a =
Qn Solution 4 ACJC/Prelim 2009/II/1
Equation of line AB: r1 32 41 1
λ� � � �� � � �= + −� � � �� � � �� � � �
or r4 32 4
2 1λ
� � � �� � � �= − + −� � � �� � � �� � � �
C is on the line AB:
1 32 41
OC
λλ
λ
+� �� �= −� �� �+� �
����
Let AOC BOCθ = =� � . .
cosOAOC OB OC
OA OC OB OCθ = =
�������� ��������
���� ���� ���� ����
Hence, . .OAOC OB OC
OA OB=
�������� ��������
���� ����
H2 Revision: Vectors 2010 Mathematics Department
Page 6 of 23
1 1 3 4 1 32 . 2 4 2 . 2 41 1 2 1
1 12 2 21 1
λ λλ λ
λ λ
+ +� � � � � � � �� � � � � � � �− − −� � � � � � � �� � � � � � � �+ +� � � � � � � �=
� � � �� � � �� � � �� � � �� � � �
[ ]2 2(1 3 ) (2 4 ) (1 )(1 3 ) 2(2 4 ) (1 )
6 2 6
λ λ λλ λ λ + − − + ++ + − + + =
(3 8 1) (1 4 1) (6 4 1) (2 2 1)λ λ− + + + + = + + + − + 15 5λ =
13
λ =
32
13
2OC
� �� �= � �� �� �
����
3
14
31
AC OC OA� �� �= − = −� �� �� �
���� ����
32
43
1CB OB OC
� �� �= − = −� �� �� �
���� ����
12
ACCB
=
: 1: 2AC CB =
Qn Solution 9 HCI/Prelim 2009/II/2 1 3
Given 2 , 23 1
OA OB
−� � � �� � � �= =� � � �� � � �−� � � �
��� ����
Equation of l:
32 , 1
m m
−� �� �= ∈� �� �� �
r � .
Method 1:
Let F be the foot of perpendicular from C to the line l,
32 01
CF
−� �� �⋅ =� �� �� �
����.
H2 Revision: Vectors 2010 Mathematics Department
Page 7 of 23
Thus,
3( ) 2 0
1OF OC
−� �� �− ⋅ =� �� �� �
���� ����3 3
2 2 . 2 01
m k
m k
m k
− − −� � � �� � � �
� + =� � � �� � � �−� � � �
39 3 4 4 0
7m k m k m k m k� + + + + − = � = −
Since F is the mid-point of C and 'C , ' 2OC OF OC= −���� ����� ����
=
111
27
13k� �� �� �� �−� �
(Shown)
Method 2:
Let OF����
be the projection vector of OC����
on the line l.
Then
1 3 32 . 2 2
1 1 1
14 14
k
OB OBOF OC
OB OB
− −� � � � � �� � � � � �−� � � � � �� � � � � �� � � � � � � �= ⋅ =� �� �
� �
���� �������� ����
=
33
27
1k
−� �� �− � �� �� �
Since F is the mid-point of C and 'C , using Ratio Theorem Since F is the mid-point of C and 'C ,
' 2OC OF OC= −���� ����� ����
=
3 1 116 1
2 2 27 7
1 1 13k k k
−� � � � � �� � � � � �− − − =� � � � � �� � � � � �−� � � � � �
(shown)
When BC is perpendicular to OA, 0
3 12 2 . 2 0
1 3
3 12 2 . 2 0
1 3
3 4 4 3 3 013
BC OA
k
k
k
k
k
k
k k k
k
⋅ =
� −� � � � � �
�� � � � � �
� − − = �� � � � � �� � � � � � � −� � � � � ��
+� � � �� � � �
� − − =� � � �� � � �− −� � � �
� + − − − + =
� =
���� ���
When k = 13
,
13 11
12 and ' 23 21131
3
OC OC
� �� �� �� �� �= − = � �� �� �−� � � �� �
� �
���� ����.
H2 Revision: Vectors 2010 Mathematics Department
Page 8 of 23
Hence,
131 11
2LHS = 6 15 6 2 15 233 131
311
RHS = 21 ' 2 LHS13
OA OC
OC
� �� � � �� �� � � �� �+ = + − =� � � �� �� � � �− −� �� � � �� �
� �
� �� �= =� �� �−� �
��� ����
����
(Verified)
Method 1:
6 15 21 '
6 6 ' 15 15 '
6( ') 15( ')
6 ' 15 '
OA OC OC
OA OC OC OC
OA OC OC OC
C A CC
+ − =
� − + − =
� − + − =
� =
0
0
0
��� ���� ����
��� ���� ���� ����
��� ���� ���� ����
����� ����
Thus, 'C A�����
is parallel to 'CC����
. Since 'C is the common point, the points A, C and 'C are collinear.
Method 2:
6 15 21 '
6 15'
21
OA OC OC
OA OCOC
+ =
+� =
��� ���� ����
��� ��������
Hence by Ratio Theorem, we can deduce that A, C and 'C are collinear.
Qn Solution 11 IJC/Prelim 2009/II/2 (a)
shortest distance from the origin O to the plane � 14 1276
• −= =� �� �� �
r n= unitsn
(b) Let F be the foot of perpendicular from O to AB. ( )OF λ= + −a b a
����
Since OF is perpendicular to the line AB,
H2 Revision: Vectors 2010 Mathematics Department
Page 9 of 23
[ ]
( )( )2 2 2
2 2 2
2
2 2
( ) 0
( ) ( ) 0
( ) ( ) ( ) 0
2 0
0
OF
λλ
λ
λ
λ
• − =+ − • − =
• − + − • − =
• − + − • + =
− + + =
=+
b aa b a b a
a b a b a b a
a b a b a b a
a b a
aa b
����
( )2
2 2
( )OF λ= + −
= + −+
a b a
aa b a
a b
����
Qn Solution 13 JJC/Prelim 2009/II/1 (i)
1
1 5: 0 8
3 1l λ
� � � �� � � �= +� � � �� � � �� � � �
r 2
1 3: 0 1
3 0l α
� � � �� � � �= +� � � �� � � �� � � �
r
2 2 2
6 1 1 38 0 0 1 for some 4 3 3 0
35 18
01
5 8 1 90
OB OA OC
AB OB OA AC OC OA
AB AC
+
� � � � � � � �� � � � � � � �= = = + ∈� � � � � � � �� � � � � � � �� � � � � � � �
� �� �� �� �= − = = − = � �� �� �� �� �� �
= + + = =
���� ���� �����
���� ���� ���� ���� ���� ����
���� ����
α α
α
2 2 2
2
2
(3 ) 10
Since ,
10 90
9
At point , 3 (since > 0.)
1 3 10
0 3 1 3 (Shown)3 0 3
AB AC
C
OC
+ =
=
==
=
� � � � � �� � � � � �= + =� � � � � �� � � � � �� � � � � �
���� ����
����
α α α
αα
α α
(ii)
1 5 9 150 8 3 113 1 0 4
OD OA AB BD
OA AB BD
= + +
= + +
� � � � � � � �� � � � � � � �= + + =� � � � � � � �� � � � � � � �� � � � � � � �
���� ���� ���� ����
���� ���� ���� OR
10 1 6 153 0 8 113 3 4 4
BD AC
OD OC OA OB
=
= − +
� � � � � � � �� � � � � � � �= − + =� � � � � � � �� � � � � � � �� � � � � � � �
���� ����
����
H2 Revision: Vectors 2010 Mathematics Department
Page 10 of 23
2 2 2
5 9 38 3 91 0 57 ( 3) 9 (57) 371ˆsin
90 90 3090 90
Area of parallelogram
ˆsin
37190. 90. 3 371
30
AB ACBAC
AB AC
ABDC AC AB
AB AC BAC
−� � � � � �� � � � � �×� � � � � �� � � � � �× − − + +� � � � � �= = = = =
= ×
=
= =
���� ����
������������������
��������
��������
Qn Solution 14 MI/Prelim 2009/I/7 (i)
���
�
�
���
�
�
−−−
+���
�
�
���
�
�
=372411
321
: λrl
���
�
�
���
�
�
+���
�
�
���
�
�
=���
�
�
���
�
�
+���
�
�
���
�
�
=210
321
420
321
: µλrl( ) 222222 210143
210
14
3
sin++⋅+−+
���
�
�
���
�
�
⋅���
�
�
���
�
�
−
=θ
���
����
�
⋅+−= −
526240
sin 1θ
rad176.0=θ or 10.1 (ii) Let the l1 represent the line containing point A, perpendicular to the plane.
Let F represent the foot of the perpendicular.
���
�
�
���
�
�
−+���
�
�
���
�
�
=14
3
321
:2 αrl
21
132611316893
1114
3
14
3
321
=
==+++−+
=���
�
�
���
�
�
−⋅���
�
�
���
�
�
���
�
�
−+���
�
�
���
�
�
α
αααα
α
���
�
�
���
�
�
=
���
�
�
���
�
�
−+���
�
�
���
�
�
=
21
21
30
2
14
3
21
321
OF
OF
H2 Revision: Vectors 2010 Mathematics Department
Page 11 of 23
(iii)
zz
zy
zx
=−−=
−=
72
78
75
715
Line of intersection l:
���
�
�
���
�
�
−+
���
�
�
���
�
�
−=10
7275
78
715
λr
Qn Solution 20 NJC/Prelim 2009/I/10 (i)
2 1 3 11 2 3 3 1
7 2 9 3AB
−� � � � � � � �� � � � � � � �= − − = − = −� � � � � � � �� � � � � � � �−� � � � � � � �
����
normal vector of 2∏ is 1 1 3 11 1 3 3 1
3 0 0 0
� � � � � � � �� � � � � � � �− × − = =� � � � � � � �� � � � � � � �� � � � � � � �
Therefore the equation of 2∏ is 1 1 11 1 2 10 0 2
r
−� � � � � �� � � � � �= =� � � � � �� � � � � �−� � � � � �
�
(ii)
:ANl
1 12 12 0
r λ−� � � �� � � �= + −� � � �� � � �−� � � �
�
Since N lies on the plane, then we have
1 12 1 5 1 2 5 4
2 0
λλ λ λ λ
− +� � � �� � � �− − = � − + − + = � =� � � �� � � �−� � � �
Therefore
322
ON� �� �= −� �� �−� �
����
(iii) 1 1 0 01 1 0 2 00 0 2 1
� � � � � � � �� � � � � � � �× − = = −� � � � � � � �� � � � � � � �−� � � � � � � �
Thus the equation of line of intersection is
3 02 02 1
r λ� � � �� � � �= − +� � � �� � � �−� � � �
�
H2 Revision: Vectors 2010 Mathematics Department
Page 12 of 23
(iv)
3
4: 2
1r D� �� �∏ =� �� �−� �
�
Since
4 02 0 01 1
� � � �� � � � ≠� � � �� � � �−� � � �
, then the line of intersection is not parallel to 3∏ .
Therefore the 3 planes meet at a common point.
(v)
Let 4
00
D
OC
� �� �� �=� �� �� �
���� where C lies on the plane 3∏ .
Then we have
( )
41
2 2121
1
2 441 2 217 1
1 21
1 21 or 1 21
20 or 22 NA
BC
D
D
D D
D D
� �� � =� �� �−� �
� �− � �� �� �� � =� �� � � �− −� � � �
� �
+ =+ = + = −= = −
����
OR Let N be the foot of perpendicular from B to 3∏ .
4 421 2 2
1 1BN
� � � �� � � �= ± = ±� � � �� � � �− −� � � �
����.
Therefore
616
ON OB BN� �� �= + = � �� �� �
���� ���� ���� or
23
8
−� �� �−� �� �� �
.
Hence 6 41 2 216 1
� � � �� � � � =� � � �� � � �−� � � �
or
2 43 2 22
8 1
−� � � �� � � �− = −� � � �� � � �−� � � �
H2 Revision: Vectors 2010 Mathematics Department
Page 13 of 23
Qn Solution 21 PJC/Prelim 2009/I/4 (i) 7 3
2 6 3 41 5
� � � �� � � �+ =� � � �� � � �−� � � �
p
138
� �� �= � �� �� �
p
(ii) cosAP CP AP CP θ=
���� �������� ����
0
6 33 2 126 14 cos
9 1
21cos
126 1460
θ
θ
θ
− −� � � �� � � �−� � � �� � � �� � � �
=
=
(iii) 1Area = sin
2AP CP ��� ����
1 3126 14
2 221 3
2
=
= OR 1Area =
2AP CP���� ����
=
6 31
3 22
9 1
− −� � � �� � � �− ×� � � �� � � �� � � �
= 21 3
2
Qn Solution 23 RI(JC)/Prelim 2009/I/6 (i)
Since 3252
3
MN
� �� �� �=� �� �� �
� and
435
BH−� �� �=� �� �� �
�
the acute angle between the lines 1l and 2l
A ( )7,6, 1−
B(3,4,7)
P
C(4,1,7)
θ
H2 Revision: Vectors 2010 Mathematics Department
Page 14 of 23
325 9 2521 1 2 2
3252
3 435 12
cos cos 80.350 17.53 4
35
− −
� �� �� �−� �� �� �� �� �•� �� �� � � �� �� �� � � �− + +� �� � � �= = = °� � � �� �� � � �−� �� �� �� �� �� �� �� � � �� �� �� �� �� �� �
(ii) A vector normal to the plane EFGH is k
The acute angle between the plane EFGH and a line parallel to AP�
1 1
12 01
21 010
35 1 3.5sin sin 55.4
12 0 18.11
21 0100
35 1
− −
� �� � � �� �� � � �•� �� � � �� � � �� � � �� � � �= = = °� � � �
� � � �� � � �� � � �� �� � � �� �� � � �� �� � � �� �
Qn Solution 25 SAJC/Prelim 2009/I/6 (i) Let
5 0 13 , 1 1
0 1 1OA OP and OB
� � � � � �� � � � � �= − = − = −� � � � � �� � � � � �� � � � � �
���� ���� ����
.
We have 5 4
2 , 21 1
AP AB
− −� � � �� � � �= =� � � �� � � �� � � �
���� ����
H2 Revision: Vectors 2010 Mathematics Department
Page 15 of 23
1 1
2 2
2 3 11 3 3 1
1 9 3
1Take normal vector for plane , 1
3
0 3 31 4 1 4
1 4 4
3Take normal vector for plane , 4
4
AP
n
AB
n
π
π
−� � � � � �� � � � � �× = − = −� � � � � �� � � � � �− −� � � � � �
� �� �= � �� �� �
−� � � � � �� � � � � �× = − = −� � � � � �� � � � � �− −� � � � � �
� �� �= ��� �
����
����
��
Hence the direction vector of � 3 has direction vector
1 3 81 4 53 4 1
−� � � � � �� � � � � �× =� � � � � �� � � � � �� � � � � �
Vector equation of � 3 is
r = 5 8
3 5 , where0 1
γ γ−� � � �
� � � �− + ∈� � � �� � � �� � � �
� (Ans)
(ii)
For 3π , 3
2 0 01 1 2 1
1 1 1n
� � � � � �� � � � � �= × =� � � � � �� � � � � �− −� � � � � �
Therefore, angle between � 3 and the plane 3π is given by
0
8 05 11 1
sin 26.6 (1decimal place)90 2
θ θ
−� � � �� � � �•� � � �� � � �� � � �= � =
Qn Solution 26 SRJC/Prelim 2009/II/5 (i) 4 2 2
1 1 2
3 3 6
AB� � � � � �� � � � � �= − − =� � � � � �� � � � � �− −� � � � � �
����
H2 Revision: Vectors 2010 Mathematics Department
Page 16 of 23
Length of projection =
2 2
2 16 5
21
5
−� � � �� � � �• −� � � �� � � �−� � � �
−� �� �−� �� �� �
36 6
30530
= =
(ii) Method 1: Using GC, the vector equation of the line l is
5 6
, .: 7 7
0 1
l λ λ� � � �� � � �= + ∈� � � �� � � �� � � �
−− �r
Method 2:
Direction vector of l = 2 1 6
1 1 7
5 1 1
� � � � � �� � � � � �× =� � � � � �� � � � � �� � � � � �
−− − .
2 5 3 (1)2 (2)
x y z
x y z
− − + = − − − −+ + = − − − − − −
Let 0x =
(1) + (2): 6 5
56
z
z
=
=
,
06
7: 7
615
6
.l λ λ
� �� �
� �� �� �� �= + ∈� �� � � �� � � �
� �� �
−r �
Method 3:
Direction vector of l =
2 1 61 1 7
5 1 1
� � � � � �� � � � � �× =� � � � � �� � � � � �� � � � � �
−− −
2 5 3 (1)2 (2)
x y z
x y z
− − + = − − − −+ + = − − − − − −
Let 0y = (1) + (2): 7 7
1z
z=
= 1x =
H2 Revision: Vectors 2010 Mathematics Department
Page 17 of 23
45
y = 1 6
: 0 71 1
, .l λ λ−� � � �� � � �= + ∈� � � �� � � �� � � �
r �
(iii)
A vector parallel to plane 3π = 1 52 73 0
−� � � �� � � �−� � � �� � � �� � � �
=
6
35
� �� �−� �� �� �
Normal vector of 3π = 6 6 16
7 12 43 12
45 3
1 3
� � � � � � � �� � � � � � � �× − = =� � � � � � � �� � � � � � � �−� �
−
� −� � � � �
.
Thus 3
4 1 43 33 3 3
: 2 1π� � � � � �� � �
− −
� � �= =� � � � � �� � � � � �� � � � � �
r .
and Cartesian equation of 3π is 4 3 3 1x y z+ − = .
3
4 5 43 7 33 0
:3
1π� � � � � �� � � � � �= =� � � � � �� � � � � �� � � �− −� �
−r
For Method 2:
,
06
7: 7
615
6
.l λ λ
� �� �
� �� �� �� �= + ∈� �� � � �� � � �
� �� �
−r �
A vector parallel to plane 3π =
07
26
3 5
1
6
� �� �
� � � �� � � �−� � � �� � � �� �
� �� �
=
156
136
� �� �� �� �� �� �� �� �
Normal vector of 3π = 5
7 12
16 16 4
3
1 313
46
12
6
� �� �
� � � � � �� �� � � � � �� �× − = − = −� � � � � �� � � � � � � �� � � � � � � �
� ��
−
�
−.
H2 Revision: Vectors 2010 Mathematics Department
Page 18 of 23
Thus 3
4 1 43 33 3 3
: 2 1π� � � � � �� � �
− −
� � �= =� � � � � �� � � � � �� � � � � �
r .
Or 3
04 4
73 3
63
: 1
356
π
� �� �
� � � �� �� � � ��
− −
�= =� � � �� �� � � �� �� � � �� �� �
r
and Cartesian equation of 3π is 4 3 3 1x y z+ − = .
For Method 3: 1 6
: 0 71 1
, .l λ λ−� � � �� � � �= + ∈� � � �� � � �� � � �
r �
A vector parallel to plane 3π = 1 12 03 1
� � � �� � � �−� � � �� � � �� � � �
=
022
� �� �� �� �� �
Normal vector of 3π = 0 6 16 4
32 12 7 12 4
1 32
� � � � � � � �� � � � � � � �× − = − = −� � � � � � � �� � � � � � � �� � � �
−
−� � � �
.
Thus 3
4 1 43 33 3 3
: 2 1π� � � � � �� � �
− −
� � �= =� � � � � �� � � � � �� � � � � �
r .
Or 3
4 1 43 0 33 1 3
: 1π� � � � � �� � �
− −
� � �= =� � � � � �� � � � � �� � � � � �
r
and Cartesian equation of 3π is 4 3 3 1x y z+ − = . (iv) The system of linear equations represents the intersection between
1 2 3, a .nd π π π
(Keywords: Intersection of 3 planes) Since l is a common line to 1 2 3, a ,nd π π π thus the system has an infinite
number of solutions.
(Keywords: 3 Planes intersect along l)
(v) Since 4π is parallel to 1π , the planes 2 3 4, and π π π will form an infinite triangular prism
H2 Revision: Vectors 2010 Mathematics Department
Page 19 of 23
Qn Solution 28 TJC/Prelim 2009/I/6 (i) Method 1: Using Ratio Theorem
OP����
= 21
( OA����
+ OB����
)1 3
13 2
22 1
� � � � �
�� � � �= + �� � � �� � � � �−� � � ��
4 21
5 2.52
1 0.5
� � � �� � � �= =� � � �� � � �� � � �
Method 2: Law of Vector Addition 2
13
AB OB OA� �� �= − = −� �� �−� �
���� ���� ����
1 2 21 1
3 1 2.52 2
2 3 0.5OM OA AM OA AB
� � � � � �� � � � � �= + = + = + − =� � � � � �� � � � � �−� � � � � �
����� ���� ����� ���� ����
(ii) 1 33 2
2 1 3 6 2 1cos
14 2| || | 14 14
.OA OB
AOBOA OB
� � � �� � � �� � � �� � � �− + −� � � �= = = =
���� ����
���� �����
(iii) Area of OAXB =
12 ( )( ) sin
2OA OB AOB× × �
= 314 14 147 7 32
� �= =� �� �
� � units2
OAXB is a rhombus since OA = OB and 90AOB ≠ �� .
Qn Solution 29 TJC/Prelim 2009/I/12 (i)
���
�
�
���
�
�
+−
+−=
���
�
�
���
�
�
+−−
+−
µµµ
λλ
λ
2337
2
24
15
8,2152
3313
==���
��
�
=−=+−
=−� µλ
µλµλ
µλ
O
A B M
H2 Revision: Vectors 2010 Mathematics Department
Page 20 of 23
Position vector of pt of intersection = ���
�
�
���
�
�
−19
176
(ii) 15 24 0 20
2 1.
λλ
λ
− +� � � �� � � �− =� � � �� � � �− + −� � � �
48202230 =�=−++−� λλλ
OA����
= ���
�
�
���
�
�
−=���
�
�
���
�
�
+−−
+−
4644
33
482484
4815
(iii) 1 23 0 0 //
2 1. m �
� � � �� � � �− = �� � � �� � � �−� � � �
2 27 0 7 203 1
.−� � � �� � � � = − ≠ �� � � �� � � �−� � � �
a point on m does not lie on Π.
m∴ does not intersect Π.
Alternative:
2 27 3 0 203 2 1
4 2 3 2 20 (no solution)
.µµµµ µ
− +� � � �� � � �− =� � � �� � � �+ −� � � �− + − − =
Every point on m does not lie in Π. Thus m does not intersect Π.
(iv)
A vector // to Рis ���
�
�
���
�
�
−
−=
���
�
�
���
�
�
−−���
�
�
���
�
�−
4351
35
4644
33
37
2
A normal is ���
�
�
���
�
�−=
���
�
�
���
�
�−=
���
�
�
���
�
�
−���
�
�
���
�
�
−
−
21
127
5427
27
23
1x
4351
35
∴equation of plane required is
r.1 2 1
1 7 12 3 2
.− − −� � � � � �� � � � � �=� � � � � �� � � � � �� � � � � �
i.e. r. 1521
1=
���
�
�
���
�
�−
H2 Revision: Vectors 2010 Mathematics Department
Page 21 of 23
Qn Solution 30 VJC/Prelim 2009/I/8 (i)
(ii)
Qn Solution 31 VJC/Prelim 2009/II/4 (i) 1 3
2 41 2 3
cos� � 103.1466 29 74
� � � �� � � �⋅ −� � � �� � � �− − −� � � �= = � = °
acute angle between normal and line
76.854� 90 76.854 13.146 13.1 (3 sf)
= °∴ = ° − ° = ° = °
(ii) 3 1 3
4 1 4 3 4 4 3
2 2 2
lies on �.
OS
S
� � � � � �� � � � � �⋅ − = − ⋅ − = + − =� � � � � �� � � � � �− −� � � � � �
∴
���
(iii) 1
11
Since ,
1 1
1 2 01 1
2 2
a
ST b
ST l
a
b
a b
−� �� �= +� �� �−� �
⊥−� � � �
� � � �+ ⋅ =� � � �� � � �− −� � � �� + = −
���
H2 Revision: Vectors 2010 Mathematics Department
Page 22 of 23
lies on � 3 4 2 33 4 5
1 11Solving, ,
5 10
T a b
a b
a b
� − − =− =
= = −
(iv) contains and is parallel to
451 11
: 1 � 2 � , �,�10
2 1 1
p l ST
p r
� �−� �� � � � � �� � � � � �∴ = − + + − ∈� � � � � �� � � �− � �� � � � −� �� �
� �
��
Qn Solution 33 YJC/Prelim 2009/II/4
(a)(i)
Consider ���
�
�
���
�
�
=���
�
�
���
�
�
=���
�
�
���
�
�
−×���
�
�
���
�
�
−310
1648160
391
13
5
A vector which is parallel to both planes 1p and 2p is ���
�
�
���
�
�
310
.
(b)(ii) Let 0=y , 25 =+ zx 63 −=− zx Solving, 2,0 == zx . Therefore, ( )2,0,0 is common to both planes.
1l : r = ���
�
�
���
�
�
+���
�
�
���
�
�
310
200
λ where λ is a real parameter.
1l : 0,32 =−= xzy
(b)(i) If 3 planes, 1p , 2p and 3p intercept in a 1l � 1l lies on 3p .
3p must contain ���
�
�
���
�
�
200
: 61
200
=���
�
�
���
�
�−•���
�
�
���
�
�
b
a � 3=b
Normal vector of 3p is perpendicular to 1l : 01
310
=���
�
�
���
�
�−•���
�
�
���
�
�
b
a � 03 =+ ba
If 3=b , then 9−=a .
H2 Revision: Vectors 2010 Mathematics Department
Page 23 of 23
(b)(ii) Consider
222222 1
6
135
2
ba ++=
++
� 221
6353
6
ba ++=
� 221353 ba ++= � 221315 ba ++= � 31422 =+ ba
Top Related