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  • 7/29/2019 2012 Vectors (Basic) Tutorial Solutions Barely Passed

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    CJC MATHEMATICS DEPARTMENT2012 JC1 H2 MATHEMATICSTOPIC: VECTORS (BASIC)

    Unit Vectors, Parallel Vectors, Collinear Points,Properties of a Parallelogram

    1.

    =

    =

    =

    4

    0

    ,

    2

    5

    3

    ,

    2

    1

    1

    pOCOBOA

    (i)

    =

    =

    4

    6

    2

    2

    5

    3

    2

    1

    1

    BA

    Unit vector in the direction ofBA

    =

    =

    2

    3

    1

    14

    14

    4

    6

    2

    56

    1

    (ii)

    =

    4

    6

    2

    BA

    =

    ==

    2

    1

    1

    4

    0

    2

    1

    1

    ppOCOACA

    ForA,B, Cto be collinear, CAkBA =

    4

    )1(26

    2n,observatioBy

    21

    1

    46

    2

    =

    =

    =

    =

    p

    p

    k

    pk

    (iii) Ifp = 1,

    =

    4

    1

    0

    OC

    =

    4

    6

    2

    BA

    ForABCD to be a parallelogram,

    =

    =

    =

    8

    7

    2

    41

    0

    46

    2

    OD

    OD

    CDBA

    (iv) Suppose 3|| =AC 3

    2

    1

    1

    =

    +

    p

    1or3

    0)1)(3(

    032

    9215

    32)1(1

    2

    2

    22

    ==

    =+

    =+

    =+++

    =+++

    pp

    pp

    pp

    pp

    p

    Scalar Product (Dot Product)

    2(a) Length of projection ofa on b = |a ^b |

    ++

    =

    8

    3

    1

    831

    1

    5

    4

    2

    222

    =1

    74| 2 12 + 40| =

    30

    74

    (b) Length of projection ofa on b = |a ^b |

    ++

    =

    4

    2

    5

    425

    1

    2

    7

    3

    222

    =1

    45|15 14 8| =

    7

    45

    3

    Given

    =

    =

    =

    11

    1

    2

    ,

    7

    5

    2

    ,

    6

    8

    2

    OROQOP

    By Ratio Theorem,

    =

    3

    4

    1

    OA

    =

    +

    =+=

    8

    4

    2

    411

    1

    2

    7

    5

    2

    3

    43 OROQOB

    (i) Shortest distance fromB to line OP

    =|| OP

    OPOB

    =

    6

    8

    2

    104

    1

    8

    4

    2

    C

    A B

    D

    O P

    QR

    A

    B

    C

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    2

    13

    370

    104

    2960

    24

    28

    40

    104

    1==

    = (shown)

    (ii)

    =

    =

    5

    0

    3

    8

    4

    2

    3

    4

    1

    BA

    =

    =

    3

    3

    0

    8

    4

    2

    11

    1

    2

    BR

    25999

    5

    0

    3

    3

    3

    0

    ||||cos

    ++

    =

    =

    BABR

    BABR

    0

    3.127=

    Vector Product (Cross Product)

    4.

    AB =

    OB

    OA =

    32

    1

    01

    2=

    31

    1

    AC=

    OC

    OA =

    1

    1

    0

    0

    1

    2

    =

    1

    2

    2

    n =

    31

    1

    12

    2=

    45

    7

    Unit vector =1

    42+5

    2+7

    2

    45

    7=

    1

    90

    45

    7

    5. ab = ac

    (ab) (ac) = 0 a (b c) = 0

    a is parallel to (b c)Hence b c = ka, where kis a scalar (shown)

    6. [10/NJC/I/3]

    (i) ( )AB OP = b auuuv uuuv

    p

    = b p a p

    = a p a p (since b p = a p)= 0

    Hence,AB is perpendicular to OP.

    OR

    b p = a pb p a p = 0

    ( )b a p = 0

    0AB OP =uuuv uuuv

    Hence,AB is perpendicular to OP.

    (ii) Since =a b , then P must be the midpoint ofAB.

    Using ratio theorem, ( )12

    OP = +a buuur

    Thus, 2OD OP=uuuv uuuv

    ( )1

    22

    = +

    a b = a + b

    (iii)

    a b represents the

    area of rhombus OADB or OBDAOR

    magnitude of a vector which is perpendicular to

    a and b

    A B

    O

    D

    P