6
VECTOR OPERATIONS: The possible vector operations are: a. Addition or subtraction of vectors b. Multiplication of vectors 1. 6VECTOR ADDITION a) Law of Triangle : If two sides of triangle are shown by two continuous vectors then third side of triangle in opposite direction shown resultant of two vectors ( ) 7. COMPONENT 2. OF A VECTOR ALONG ANOTHER VECTOR Two vector and there as shown below to get component of along shift || to itself and made it coinitial with , draw a perpendicular from B to OA let is meets at Mθ is angle in between and . Component of , along = OB cos θ = OB cosθ Note: (i) θ = 0, OM maximum (ii) θ = 90 O , OM = zero A B B A B A A A B B OM A ) B and A ( C B A C O A B O A B C C A AC OA OC B A C B A B A M B

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  • VECTOR OPERATIONS:The possible vector operations are:

    a. Addition or subtraction of vectorsb. Multiplication of vectors

    1. 6VECTOR ADDITIONa) Law of Triangle : If two sides of triangle are shown by two

    continuous vectors then third side of triangle in oppositedirection shown resultant of two vectors ( )

    7. COMPONENT

    2. OF A VECTOR ALONG ANOTHER VECTORTwo vector and there as shown below to get component ofalong shift || to itself and made it coinitial with , draw aperpendicular from B to OA let is meets at M is angle in betweenand .Component of , along = OB cos = OB cos

    Note: (i) = 0, OM maximum (ii) = 90O, OM = zero

    A B BA B A

    AA

    BB OMA

    )BandA( C

    BAC

    O A

    B

    O AB

    CC

    A ACOAOC BAC

    B

    A

    B

    A

    M

    B

  • 3. SUBTRACTION OF VECTORSVector which is want to subtracted just change direction of that vectorand then add.

    4. PARALLELOGRAM LAW OF VECTORTo determine magnitude & direction of resultant vector, when twovectors act at an angle .According to this law if two vectors and are represented by twoadjacent sides of a parallelogram both pointing outwards as shown infig. The diagonal drawn through the intersection of the two vectorsrepresents the resultant .

    From triangle OCMOC2 = OM2 + CM2

    = (P + Q cos)2 + (Q sin )2= P2 + Q2 cos2 + 2PQ cos + Q2 sin2

    Since Q2 (cos2 + sin2) = Q2R2 = P2 + Q2 + 2PQ cos

    )B(ABA

    P Q

    R

    QPR

  • 5. SCALAR PRODUCT OR DOT PRODUCT OF TWO VECTORSIf is the angle between and .

    Then A (B cos) = , A and B are the magnitude of andThe quantity AB cos is a scalar quantity.

    B cos is the component of vector in the direction of .Hence, the scalar product of two vectors is equal to the product of themagnitude of one vector and the component of the second vector in thedirection of the first vector.

    A B B.A A B

    B A

  • 6. CROSS PRODUCT OR VECTOR PRODUCT OF TWO VECTORSCross product of and inclined to each other at an angle q is

    defined as:to plane of and

    A B

    BAnsinBA n A B

  • Direction of is given by right hand thumb rule. Curl the fingers of yourright hand from to . Then the direction of the erect thumb willpoint in the direction

    .

    Properties of vector producta) The vector product is not commutative i.e

    b) The vector product is distributive i.e.

    c) The magnitude of the vector product of two vectors mutually atright angles is equal to the product of the magnitudes of thevectors.

    (if q = 90)d) The vector product of two parallel vectors is a null vector (or zero

    vector)

    e) The vector product of a vector by itself is a null vector (zerovector)

    nA B

    BA

    ABBA ABBA

    CABA)CB(A

    AB|BA|,nABn90sinABBA

    0or0n)0(sinABBA

  • f) The vector product of unit orthogonal vectors have thefollowing relations in the right handed coordinate system

    I.

    II.

    The magnitude of each of the vectors and is 1 and the anglebetween any two of them is 90.

    Therefore, we write (1) sin 90 , where is a unit avector perpendicular to the plane of and i.e. it is just the thirdunit vector .

    g) The vector product of two vectors in terms of their x, y & zcomponents can be expressed as a determinant. Let and betwo vectors. Let us write their rectangular components :

    X =

    0or0n)0(sinAABA k,j,i

    kj

    i

    + ivejkijikijkikjkijkji

    kj

    -ivei

    0kk0jj0ii

    j,i k

    )1(ji nn ni j

    k

    A B

    kAjAiAA zyx

    kBjBiBB zyx

    )kAjAiA(BA zyx

    )kBjBiB( zyx k)ABBA(j)ABBA(i)ABBA( zxyxzxzxzyzy

    zyxzyx

    BBBAAAkji

    BA