Vectors Qn Solutiona- H2 Revision: Vectors 2010 Mathematics Department Page 3 of 23 (ii) 10 0 6 3 0

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Transcript of Vectors Qn Solutiona- H2 Revision: Vectors 2010 Mathematics Department Page 3 of 23 (ii) 10 0 6 3 0

  • 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 0 2

    1 1

    x z y r λ

    � � � � � � � �= − − = � = +� � � � � � � �− −� � � �

    Vector parallel to 2

    1 0 1 1 0 1

    0 1 1 π

    � � � � � � � � � � � �= − − = −� � � � � � � � � � � �−� � � � � �

    1

    1 2 1 1 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 1 13 3 6 2

    ON λ � � � � � � � �= +� � � � � � � � � � � �

    ����

    3 1 13 3 3 2 6 2 2

    4

    λ λ λ

    λ

    +� � � � � � � �+ • =� � � � � � � �+� � � � � = −

    1 1 2

    ON

    −� � � �= � � � �−� �

    ����

    (iii) 1 3 5 '

    ' 2 1 13 11 2

    2 6 10

    OQ OQ ON 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) 1 3 2

    � � � � � � � � � �

    = 6 2 4

    a

    k a � � � �

    � =� � � � � �

    Method 1:

    1 2

    1 1 : 3 2 : 3

    2 2 2

    b r rπ π � � � � � � � �• = − • =� � � � � � � � � � � �

    Distance between the 2 planes = ( )2

    2 224 1 9 4

    b − − =

    + + 108 or -116b� =

    Method 2:

    Distance QN = ( )( ) ( ) ( )( )2 223 1 13 1 6 2 224− − + − + − − = 2 3 2 2 5 2 6 13 6 108 OR ' 6 11 6 116 4 6 4 4 10 4

    b OQ b OQ

    −� � � � � � � � � � � � � � � � � � � � � � � �= • = • = = • = − • = −� � � � � � � � � � � � � � � � � � � � � � � �−� � � � � � � � � � � �

    ���� �����

    Method 3

    1

    1 1 1 2

    : 3 2 3 14 14

    2 2 r rπ � � � � � � � �• = − � • = −� � � � � � � � � � � �

    2

    1 1 2

    : 3 224 14 14

    2

    1 3 2 56 54 or 58 108 or 116 2

    r

    r b

    π � � � �• = − ±� � � � � �

    � � � �� • = − ± = − � = −� � � � � �

    (v) 2,a b≠ ∈�

    Qn Solution 2 AJC/2009 Prelim/II/2 (i)

    ' 4 tan ' 3 4 ' 4

    3 3 EE

    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 0 6 3

    0 4 DB DE

    � � � � � � � �= − = −� � � � � � � � � � � �

    ���� ����

    10 0 24 12 6 3 40 2 20

    0 4 30 15 n

    −� � � � � � � � � � � � � � � �= − × − = − = −� � � � � � � � � � � � � � � �−� � � � � � � �

    12 10 20 3 15 4

    sin 769 125

    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 .

    1 1 1

    AN λ � � � �= � � � � � �

    ����

    1 1 1 2 1 2 4 1 4

    ON OA AN

    λ λ λ

    λ

    − − +� � � � � � � � � � � �= + = + = +� � � � � � � � � � � �+� � � � � �

    ���� ���� ����

    Since N is a point on the plane 1p ,

    1 . 1 7

    1 ON

    � � � � =� � � � � �

    ����

    1 1 2 . 1 7 4 1

    λ λ λ

    − +� � � � � � � �+ =� � � � � � � �+� � � �

    ( 1 ) (2 ) (4 ) 7λ λ λ− + + + + + = 2λ =

    2 1

    31 1 2 1

    2 2 8 3 3

    4 142 4

    3

    ON

    λ λ λ

    � �− +� � − + −� � � �� � � � � �� �= + = + =� � � �� � � � � �+ � �� � � �

    � �+� � � �

    ����

  • H2 Revision: Vectors 2010 Mathematics Department

    Page 4 of 23

    (ii) 4 1 3 . 1 4 3 0 7 0 1

    � � � � � � � � = + + =� � � � � � � � � � � �

    (satisfies equation of plane 1p )

    4 1 3 . 1 4 3 0 1 0 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 1 1 1 1 1 2

    a

    a

    a

    +� � � � � � � � � � � �× − = −� � � � � � � � � � � �−� � � � � �

    is the direction vector of the line � ,

    Hence r 4 1 3 1 0 2

    a

    aλ +� � � �

    � � � �= + −� � � � � � � �−� � � �

    is the equation of line � .

    (iii)

    If the planes 3p with equation r 1

    . 2 3

    b � � � � =� � � � � �

    intersects with 1p and 2p at line �

    , then

    1 1 2 . 1 0 3 2

    a

    a

    +� � � � � � � �− =� � � � � � � �−� � � �

    (1 ) 2(1 ) 6 0a a+ + − − = 3a = −

    The line � lies in the plane 3p and thus

    4 3 0

    � � � � � � � � � �

    lies in the plane 3p also.

    4 1 3 . 2 4 6 0 10 0 3

    b b � � � � � � � � = � = + + =� � � � � � � � � � � �

    (iv) 0 1 1

    2 2 4 5 4 1

    AB OB OA

    −� � � � � � � � � � � �= − = − − = −� � � � � � � � � � � � � � � � � �

    ���� .

    Given the length of projection of line segment AB on the line � is 1 6

    .

  • H2 Revision: Vectors 2010 Mathematics Department

    Page 5 of 23

    1

    . 1

    2 1 1 6 1

    2

    a

    AB a

    a

    a

    +� � � �−� � � �−� � = +� �

    � �−� � � �−� �

    ����

    2 2 2

    (1 ) 4(1 ) 2 1 6(1 ) (1 ) ( 2)

    a a

    a a

    + − − − =

    + + − + −

    2

    5( 1) 1 3( 3)

    a

    a

    − = +

    2

    2

    25( 1) 1 ( 3) 3

    a a

    − = +

    2 275( 1) 3a a− = + 274 150 72 0a a− + = 150 1188 75 247

    2(74) 74 a

    ± ±= =

    0.781a = or 1.25a =

    Qn Solution 4 ACJC/Prelim 2009/II/1

    Equation of line AB: r 1 3 2 4 1 1

    λ � � � � � � � �= + −� � � � � � � � � � � �

    or r 4 3 2 4

    2 1 λ

    � � � � � � � �= − + −� � � � � � � � � � � �

    C is on the line AB:

    1 3 2 4 1

    OC

    λ λ

    λ

    +� � � �= −� � � �+� �

    ����

    Let AOC BOCθ = =� � . .

    cos OAOC OB OC

    OA OC OB OC θ = =

    �������� ��������

    ���� ���� ���� ����

    Hence, . .OAOC OB OC

    OA OB =

    �������� ��������

    ���� ����

  • H2 Revision: Vectors 2010 Mathematics Department

    Page 6 of 23

    1 1 3 4 1 3 2 . 2 4 2 . 2 4 1 1 2 1

    1 1 2 2 2 1 1