On the diagram below the displacement is represented by the directed line segment OA.

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Vectors Sclrs nd Vectors A vector is quntity tht hs mgnitude nd direction. One exmple of vector is velocity. The velocity of n oject is determined y the mgnitude(speed) nd direction of trvel.

1 Vectors Sclrs nd Vectors A vector is quntity tht hs mgnitude nd direction. One exmple of vector is velocity. The velocity of n oject is determined y the mgnitude(speed) nd direction of trvel. Other exmples of vectors re force, displcement nd ccelertion. A sclr is quntity tht hs mgnitude only. Mss, time nd volume re ll exmples of sclr quntities. Exmple 1. If cr trvels from point O to point A, which is 50km. wy in north-esterly direction, then the displcement of the cr from O is 50km.NE. The displcement of the cr is specified y the distnce trvelled (50km.) nd the direction of trvel (NE.) from O. Displcement is therefore vector, nd the mgnitude of the displcement (distnce), is sclr. On the digrm elow the displcement is represented y the directed line segment OA. The length of the line represents the mgnitude of the displcement nd is written OA. The rrowhed represents the direction of the displcement. N A OA =50km. O 45 0 E Exmple 2. A force of 50 Newton t n ngle of 20 0 to the horizontl downwrd, is pplied to wheelrrow. The digrm elow shows vector representing this force. A 50N 20 0 B AB = 50N Copyright 2006 RMIT University Lerning Skills Unit Pge 1 of 1

2 Geometric Vectors We will e considering vectors in three-dimensionl spce defined y three mutully perpendiculr directions. Definitions nd conventions. Vectors will e denoted y lower cse old letters such s,, c. Unit vectors i, j, k Vectors with mgnitude of one in the direction of the x-xis, y-xis nd z-xis will e denoted y i, j, nd k respectively. The nottion (,, c) will e used to denote the vector (i + j + ck) s well s the coordintes of point P (,, c). The context will determine which mening is correct. Exmple z k P (,4,5) i O j y x i = j = k =1 In the digrm ove the point P hs coordintes (,4,5). The vector OP is the vector i + 4j + 5k. This my lso e written (,4,5). Copyright 2006 RMIT University Lerning Skills Unit Pge 2 of 1

3 Directed Line Segment. The directed line segment, or geometric vector, PQ, from P( x, 1 y, 1 z1 ) to Q( x2, y2, z2 ) is found y sutrcting the co-ordintes of P (the initil point) from the co-ordintes of Q (the finl point). 2 1g 2 1g 2 1g c PQ = x x i + y y j + z z k Exmple. h z g g g PQ = 5 i j k PQ = 2i + 2j 2k g P,4,1 Q 5, 6, 1g y x The directed line segment PQ is represented y the vector 2i + 2j 2k, or (2, 2, 2). Any other directed line segment with the sme length nd sme directon s PQ is lso represented y 2i + 2j 2k or (2, 2, 2). The directed line segment QP hs the sme length s PQ ut is in the opposite direction. QP = PQ = (2i + 2j 2k) = 2i 2j + 2k or ( 2, 2, 2) Position Vector. The position vector of ny point is the directed line segment from the origin O (0,0,0) to the point nd is given y the co-ordintes of of the point. The position vector of P(, 4, 1) is i + 4j + k, or (, 4, 1). Exercise 1. Given the points A(, 0,4) B( 2, 4, ) nd C(1, 5,0), find: () AB () AC (c) CB (d) BC (e) CA (f) The position vectors of A, B nd C. Compre your nswers 1()nd 1(e), nd 1(c) nd 1(d). Wht do you notice? Copyright 2006 RMIT University Lerning Skills Unit Pge of 1

4 Length or Mgnitude of Vector. The length of vector = 1 i + 2 j + k is written nd is evluted y: 2 2 = g For exmple the length of the vector 2i + 2j - 2k equls = 12 = 2 1, 2, re often referred to s the components of vector Unit Vector A vector with mgnitude of one is clled unit vector. If is ny vector then unit vector prllel to is written ɵ ( ht ). The ht symolises unit vector. Vector cn then e written = ɵ therefore ɵ = Exmple A unit vector prllel to = (1, 2, ) is the vector ɵ = = 1, 2, g = g , 2, Adding nd Sutrcting Vectors. Vectors re dded or sutrcted y dding or sutrcting their corresponding components using the tringle rule y using the prllogrm rule. Exmple If = (-, 4, 2) nd = ( 1 2, ), find: (i) + ( ii). Adding or sutrcting components (i) + = (, 4, 2) + ( 1, 2, ) = ( 4, 2, 5) Similrly (ii) = (, 4, 2) ( 1, 2, ) = ( 2, 6, 1) Copyright 2006 RMIT University Lerning Skills Unit Pge 4 of 1

5 Tringle Rule (i) + + R P Q Plce the til of vector t the hed of vector (point Q). The directed line segment PR from the til of vector to the hed of vector is the vector +. (ii) To sutrct from, reverse the direction of to give then dd nd. = + ( ) Prllelogrm Rule (i) + S R + P nd re plced til-to-til ( point P) nd the prllelogrm (PQRS) completed. The digonl PR is the sum +. Q (ii) To find ( ), reverse the direction of to give then dd nd. Exercise 2. For vectors p (, 6, 5), q ( 4, 1, 0) nd r (1,, 5) find: () p + q () r + p (c) p q Copyright 2006 RMIT University Lerning Skills Unit Pge 5 of 1

6 Multipliction of vector y sclr. To multiply vector = 1i + 2j + k y sclr, m, multiply ech component of y m. m = m1i + m2j + mk The result is vector of length m If m > 0 the resultnt vector is in the sme direction s If m < 0 the resultnt vector is in the opposite direction from. Two vectors nd re sid to e prllel if nd only if = k where k is rel constnt. Exmple 1 = (i + j 2k) is multiplied y 7 7 = 7(i + j 2k) = 21i + 7j 14k. The mgnitude of is = = ( ) = 686 = 7 14 = = ( ) 2 Exmple 2 Find the vlue of m so tht the vector, ( 4, m, 8) is prllel to the vector, ( 6,, 12). For nd to e prllel = k Therefore ( 4, m, 8) = k ( 6,, 12) = ( 6k, k, 12k) equting i components 4 = 6k k = 2 equting j components m = k 2 m = m = 2 Exercise Find the following () (i + j 5k) () 8 (7i + 2j + 4k) (c) 4 ( j k) Copyright 2006 RMIT University Lerning Skills Unit Pge 6 of 1

7 Multipliction of vector y vector (1) Dot product or sclr product The dot product of two vectors (,, ) nd (,, ) is sclr, defined y = cosθ, where θ is the ngle etween nd nd = θ If is perpendiculr to then = 0 (cos(π/2)= 0). In prticulr i j = j k = k i = 0 If is prllel to then = (cos(0)=1) In prticulr i i = j j = k k = 1 Also ( 1i + 2j + k ) ( 1i + 2 j + k ) = ( i i + i j+ i k + j i + j j+ j k + k i + k j+ k k ) Thus cn e defined y = Exmple 1 Find the dot product of (2i + j + 4k) nd ( i 2j + k) (2i + j + 4k) ( i 2j + k) = 2( 1) + ( 2) + (4)(1) = = 4 (2i + j + 4k) ( i 2j + k) = 4 Copyright 2006 RMIT University Lerning Skills Unit Pge 7 of 1

8 Exmple 2 Find the sclr product of nd, s drwn, elow where = 14 nd = 6 = cosθ 0 0 = 14 6 cos 0 = = 7 0 = 7 Exercise 4 Find the dot product of the following vectors: () i nd 5j () 2i + k nd 7i + 2j + 4k (c) 5k nd j 2k (d) (2, 0, 4) nd (, 1, ) (e) (0, 5, 1) nd (4, 0, 0) (f) (g) (2) Cross product or vector product The cross product of two vectors nd is the vector, which is perpendiculr to oth nd nd is given y i j k = The mgnitude of is given y = sinθ where θ is the ngle etween nd. The direction of is tht in which your thum would point if the fingers of your right re curled from to. In prticulr i j = k, j k = i, k i = j i k = j k j = i j i = k Copyright 2006 RMIT University Lerning Skills Unit Pge 8 of 1

9 If is prllel to then = 0. (sin0 0 = 0) In prticulr i i = j j = k k = 0 If is perpendiculr to then = ( sin90 0 =1) = ( the cross product is not commuttive.) Exmple 1 Find if = 2i +j + k nd = 5j +k i j k = 2 1 = (9 5)i (6 0)j + (10 0)k 0 5 = 4i 6j + 10k Exmple 2 Find if = (2,1,1) nd = ( 2,4,0) i j k = = (0 4)i (0 + 2)j + (8 + 2)k = 4i 2j + 10k Exmple Find if = (2,1,1) nd = (8,4,4) Becuse = 4, is prllel to therefore = 0 Exercise 5 Find the cross product of the following vectors: () j k () i 4i (c) (2i + j k) (j + 2k) (d) j 5i (e) (i j + k) (2i + j k) Copyright 2006 RMIT University Lerning Skills Unit Pge 9 of 1

10 Projection of vectors. Consider the digrm elow: Q P θ R S Let PQ = nd PS =. Sclr projection The sclr projection of vector in the direction of vector is the length of the stright line PR or PR. PR = cosθ. Also cosθ = (ecuse = cosθ ) Therefore PR = ( ) = = ɵ (cncel, nd use = ˆ ) The sclr projection of vector in the direction of vector is given y = ɵ or cosθ Vector projection The vector projection of vector in the direction of vector is vector in the direction of with mgnitude equl to the length of the stright line PR or PR. Therefore the vector projection of in the direction of is the sclr projection multiplied y unit vector in the direction of. The vector projection of vector in the direction of vector is given y ( ɵ ) ˆ = ( ) 2 Copyright 2006 RMIT University Lerning Skills Unit Pge 10 of 1

11 Angle etween two vectors The ngle, θ etween two vectors cn e found from the definition of the dot product = cosθ therefore cosθ = θ cn lso e found from cosθ = ɵ Exmple Find () the sclr projection of vector = (2,, 1) in the direction of vector = (5, 2, 2). () the ngle etween nd. (c) the vector projection of in the direction of. () Sclr projection = = therefore ɵ (,, ) = ɵ = (2,, 1) ( 5, 2, 2) g = = 6 () Angle etween nd Therefore The sclr projection of in the direction of is The sclr projection of in the direction of is lso equl to cosθ, where θ is the ngle etween nd = cosθ. = = 14 = 14 cosθ 6 cos θ = = θ = 74 The ngle etween nd is Copyright 2006 RMIT University Lerning Skills Unit Pge 11 of 1

12 (c) Vector projection The vector projection in the direction of equls: (sclr projection in the direction of ) ˆ = 6 ( 5, 2, 2) = 6 (5, 2, 2) The vector projection of in the direction of is 6 (5, 2, 2) Exercise 6 For the following pirs of vectors find: (i) the sclr projection of on (ii) the ngle etween nd (iii) the vector projection of on () = (2,,1) = (5, 0, ) () = (0, 0, ) = (0, 0, 7) (c) = (5, 0, 0) = (0,, 0) (d) = (, 2, 1) = (2, 1, 2) Copyright 2006 RMIT University Lerning Skills Unit Pge 12 of 1

13 Answers 1.() ( 5, 4, 1) () ( 2, 5, 4) (c) (, 9, ) (d) (, 9, ) (e) (2, 5, 4) (f) OA = i + 4k, OB = 2i + 4j + k, OC = i 5j 2. () ( 1, 7, 5) () (4,, 10) (c) (7, 5, 5). () i + 9j 15k () 56i + 16j+ 2k (c) 4j+ 12k 4. () 0 () 26 (c) 10 (d) 6 (e) 0 (f) 10 2 (g) 0 5. () i () 0 (c) 9i 4j + 6k (d) 15k (e) i + 9j +7 k 1 (ii) ()(i) (iii) ( 5, 0, ) ()(i) (ii) 0 0 (iii) 0, 0, 7 g 7 (c)(i) 0 (ii) 90 0 (iii) 0 (d)(i) 2 (ii) (iii) 2 2, 1, 2 g Copyright 2006 RMIT University Lerning Skills Unit Pge 1 of 1

Analytically, vectors will be represented by lowercase bold-face Latin letters, e.g. a, r, q.

Analytically, vectors will be represented by lowercase bold-face Latin letters, e.g. a, r, q. 1.1 Vector Alger 1.1.1 Sclrs A physicl quntity which is completely descried y single rel numer is clled sclr. Physiclly, it is something which hs mgnitude, nd is completely descried y this mgnitude. Exmples