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  • 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++⋅+−+

    ���

    ���

    ⋅���

    ���

    ���

    ����

    ⋅+−= −

    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

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    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