VECTOR ALGEBRA. Chapter Introduction Some Basic Concepts

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1 44 Chpter 0 VECTOR ALGEBRA In most sciences one genertion ters down wht nother hs built nd wht one hs estblished nother undoes In Mthemtics lone ech genertion builds new story to the old structure HERMAN HANKEL 0 Introduction In our dy to dy life, we come cross mny queries such s Wht is your height? How should footbll plyer hit the bll to give pss to nother plyer of his tem? Observe tht possible nswer to the first query my be 6 meters, quntity tht involves only one vlue (mgnitude) which is rel number Such quntities re clled sclrs However, n nswer to the second query is quntity (clled force) which involves musculr strength (mgnitude) nd direction (in which nother plyer is positioned) Such quntities re clled vectors In mthemtics, physics nd engineering, we frequently come cross with both types of WR Hmilton quntities, nmely, sclr quntities such s length, mss, ( ) time, distnce, speed, re, volume, temperture, work, money, voltge, density, resistnce etc nd vector quntities like displcement, velocity, ccelertion, force, weight, momentum, electric field intensity etc In this chpter, we will study some of the bsic concepts bout vectors, vrious opertions on vectors, nd their lgebric nd geometric properties These two type of properties, when considered together give full relistion to the concept of vectors, nd led to their vitl pplicbility in vrious res s mentioned bove 0 Some Bsic Concepts Let l be ny stright line in plne or three dimensionl spce This line cn be given two directions by mens of rrowheds A line with one of these directions prescribed is clled directed line (Fig 0 (i), (ii))

2 VECTOR ALGEBRA 45 Fig 0 Now observe tht if we restrict the line l to the line segment AB, then mgnitude is prescribed on the line l with one of the two directions, so tht we obtin directed line segment (Fig 0(iii)) Thus, directed line segment hs mgnitude s well s direction Definition A quntity tht hs mgnitude s well s direction is clled vector Notice tht directed line segment is vector (Fig 0(iii)), denoted s AB or simply s, nd red s vector AB or vector The point A from where the vector AB strts is clled its initil point, nd the point B where it ends is clled its terminl point The distnce between initil nd terminl points of vector is clled the mgnitude (or length) of the vector, denoted s AB, or, or The rrow indictes the direction of the vector Note Since the length is never negtive, the nottion < 0 hs no mening Position Vector From Clss XI, recll the three dimensionl right hnded rectngulr coordinte system (Fig 0(i)) Consider point P in spce, hving coordintes (x, y, z) with respect to the origin O(0, 0, 0) Then, the vector OP hving O nd P s its initil nd terminl points, respectively, is clled the position vector of the point P with respect to O Using distnce formul (from Clss XI), the mgnitude of OP (or r ) is given by OP = x + y + z In prctice, the position vectors of points A, B, C, etc, with respect to the origin O re denoted by, b, c, etc, respectively (Fig 0 (ii))

3 46 Fig 0 Direction Cosines Consider the position vector OP( or r ) of point P(x, y, z) s in Fig 03 The ngles α, β, γ mde by the vector r with the positive directions of x, y nd z-xes respectively, re clled its direction ngles The cosine vlues of these ngles, ie, cos α, cosβ nd cos γ re clled direction cosines of the vector r, nd usully denoted by l, m nd n, respectively Z C x z O r y P( x,y,z ) B Y P X A O 90 A Fig 03 X From Fig 03, one my note tht the tringle OAP is right ngled, nd in it, we x hve cos α= ( r stnds for r ) Similrly, from the right ngled tringles OBP nd r OCP, we my write cos β= y z nd cos r γ= r Thus, the coordintes of the point P my lso be expressed s (lr, mr,nr) The numbers lr, mr nd nr, proportionl to the direction cosines re clled s direction rtios of vector r, nd denoted s, b nd c, respectively

4 VECTOR ALGEBRA 47 Note One my note tht l + m + n = but + b + c, in generl 03 Types of Vectors Zero Vector A vector whose initil nd terminl points coincide, is clled zero vector (or null vector), nd denoted s 0 Zero vector cn not be ssigned definite direction s it hs zero mgnitude Or, lterntively otherwise, it my be regrded s hving ny direction The vectors AA, BB represent the zero vector, Unit Vector A vector whose mgnitude is unity (ie, unit) is clled unit vector The unit vector in the direction of given vector is denoted by â Coinitil Vectors Two or more vectors hving the sme initil point re clled coinitil vectors Colliner Vectors Two or more vectors re sid to be colliner if they re prllel to the sme line, irrespective of their mgnitudes nd directions Equl Vectors Two vectors nd b re sid to be equl, if they hve the sme mgnitude nd direction regrdless of the positions of their initil points, nd written s = b Negtive of Vector A vector whose mgnitude is the sme s tht of given vector (sy, AB ), but direction is opposite to tht of it, is clled negtive of the given vector For exmple, vector BA is negtive of the vector AB, nd written s BA = AB Remrk The vectors defined bove re such tht ny of them my be subject to its prllel displcement without chnging its mgnitude nd direction Such vectors re clled free vectors Throughout this chpter, we will be deling with free vectors only Exmple Represent grphiclly displcement of 40 km, 30 west of south Solution The vector OP represents the required displcement (Fig 04) Exmple Clssify the following mesures s sclrs nd vectors (i) 5 seconds (ii) 000 cm 3 Fig 04

5 48 Solution (iii) 0 Newton (iv) 30 km/hr (v) 0 g/cm 3 (vi) 0 m/s towrds north (i) Time-sclr (ii) Volume-sclr (iii) Force-vector (iv) Speed-sclr (v) Density-sclr (vi) Velocity-vector Exmple 3 In Fig 05, which of the vectors re: (i) Colliner (ii) Equl (iii) Coinitil Solution (i) Colliner vectors :, c nd d (ii) Equl vectors : nd c (iii) Coinitil vectors : b, c nd d EXERCISE 0 Fig 05 Represent grphiclly displcement of 40 km, 30 est of north Clssify the following mesures s sclrs nd vectors (i) 0 kg (ii) meters north-west (iii) 40 (iv) 40 wtt (v) 0 9 coulomb (vi) 0 m/s 3 Clssify the following s sclr nd vector quntities (i) time period (ii) distnce (iii) force (iv) velocity (v) work done 4 In Fig 06 ( squre), identify the following vectors (i) Coinitil (ii) Equl (iii) Colliner but not equl 5 Answer the following s true or flse (i) nd re colliner (ii) Two colliner vectors re lwys equl in mgnitude Fig 06 (iii) Two vectors hving sme mgnitude re colliner (iv) Two colliner vectors hving the sme mgnitude re equl

6 VECTOR ALGEBRA Addition of Vectors A vector AB simply mens the displcement from point A to the point B Now consider sitution tht girl moves from A to B nd then from B to C (Fig 07) The net displcement mde by the girl from point A to the point C, is given by the vector AC nd expressed s AC = AB + BC This is known s the tringle lw of vector ddition Fig 07 In generl, if we hve two vectors nd b (Fig 08 (i)), then to dd them, they re positioned so tht the initil point of one coincides with the terminl point of the other (Fig 08(ii)) b A + b (i) (ii) (iii) C Fig 08 For exmple, in Fig 08 (ii), we hve shifted vector b without chnging its mgnitude nd direction, so tht it s initil point coincides with the terminl point of Then, the vector + b, represented by the third side AC of the tringle ABC, gives us the sum (or resultnt) of the vectors nd b ie, in tringle ABC (Fig 08 (ii)), we hve AB + BC = AC Now gin, since AC = CA, from the bove eqution, we hve AB + BC + CA = AA = 0 This mens tht when the sides of tringle re tken in order, it leds to zero resultnt s the initil nd terminl points get coincided (Fig 08(iii)) C b B A b C b B b

7 430 Now, construct vector BC so tht its mgnitude is sme s the vector BC, but the direction opposite to tht of it (Fig 08 (iii)), ie, BC = BC Then, on pplying tringle lw from the Fig 08 (iii), we hve AC = AB + BC = AB + ( BC) = b The vector AC is sid to represent the difference of nd b Now, consider bot in river going from one bnk of the river to the other in direction perpendiculr to the flow of the river Then, it is cted upon by two velocity vectors one is the velocity imprted to the bot by its engine nd other one is the velocity of the flow of river wter Under the simultneous influence of these two velocities, the bot in ctul strts trvelling with different velocity To hve precise ide bout the effective speed nd direction (ie, the resultnt velocity) of the bot, we hve the following lw of vector ddition If we hve two vectors nd b represented by the two djcent sides of prllelogrm in mgnitude nd direction (Fig 09), then their sum + b is represented in mgnitude nd direction by the digonl of the prllelogrm through their common point This is known s Fig 09 the prllelogrm lw of vector ddition Note From Fig 09, using the tringle lw, one my note tht OA + AC = OC or OA + OB = OC (since AC = OB ) which is prllelogrm lw Thus, we my sy tht the two lws of vector ddition re equivlent to ech other Properties of vector ddition Property For ny two vectors + b nd b, = b + (Commuttive property)

8 Proof Consider the prllelogrm ABCD (Fig 00) Let AB = nd BC = b, then using the tringle lw, from tringle ABC, we hve AC = + b Now, since the opposite sides of prllelogrm re equl nd prllel, from Fig00, we hve, AD = BC = b nd DC = AB = Agin using tringle lw, from tringle ADC, we hve AC = AD + DC = b + Hence + b = b + Property For ny three vectors b, ndc VECTOR ALGEBRA 43 Fig 00 ( + b) + c = + ( b + c ) (Associtive property) Proof Let the vectors b, ndc be represented by PQ, QR nd RS, respectively, s shown in Fig 0(i) nd (ii) Fig 0 Then + b = PQ + QR = PR nd b + c = QR + RS = QS So ( + b) + c = PR + RS = PS

9 43 nd + ( b + c ) Hence ( + b) + c = + ( b + c ) Remrk The ssocitive property of vector ddition enbles us to write the sum of three vectors, b, c s + b + c without using brckets Note tht for ny vector, we hve + 0 = 0 + = Here, the zero vector 0 is clled the dditive identity for the vector ddition 05 Multipliction of Vector by Sclr Let be given vector nd λ sclr Then the product of the vector by the sclr λ, denoted s λ, is clled the multipliction of vector by the sclr λ Note tht, λ is lso vector, colliner to the vector The vector λ hs the direction sme (or opposite) to tht of vector ccording s the vlue of λ is positive (or negtive) Also, the mgnitude of vector λ is λ times the mgnitude of the vector, ie, λ = λ A geometric visulistion of multipliction of vector by sclr is given in Fig 0 Fig 0 When λ =, then λ =, which is vector hving mgnitude equl to the mgnitude of nd direction opposite to tht of the direction of The vector is clled the negtive (or dditive inverse) of vector nd we lwys hve + ( ) = ( ) + = 0 Also, if λ =, provided 0, ie λ = λ = = is not null vector, then

10 VECTOR ALGEBRA 433 So, λ represents the unit vector in the direction of We write it s â = Note For ny sclr k, k 0=0 05 Components of vector Let us tke the points A(, 0, 0), B(0,, 0) nd C(0, 0, ) on the x-xis, y-xis nd z-xis, respectively Then, clerly OA =, OB = nd OC = The vectors OA, OB nd OC, ech hving mgnitude, re clled unit vectors long the xes OX, OY nd OZ, respectively, nd denoted by iˆ, ˆj nd k ˆ, respectively (Fig 03) Fig 03 Now, consider the position vector OP of point P(x, y, z) s in Fig 04 Let P be the foot of the perpendiculr from P on the plne XOY We, thus, see tht P P is Fig 04 prllel to z-xis As iˆ, ˆj nd k ˆ re the unit vectors long the x, y nd z-xes, respectively, nd by the definition of the coordintes of P, we hve PP ˆ = OR= zk Similrly, QP ˆ = OS = yj nd OQ = xiˆ

11 434 Therefore, it follows tht OP = OQ + QP ˆ ˆ = xi + yj nd OP = OP ˆ ˆ ˆ + PP = xi + yj+ zk Hence, the position vector of P with reference to O is given by OP (or r ) = xiˆ+ yj ˆ+ zkˆ This form of ny vector is clled its component form Here, x, y nd z re clled s the sclr components of r, nd xiˆ, yj ˆ nd zk ˆ re clled the vector components of r long the respective xes Sometimes x, y nd z re lso termed s rectngulr components The length of ny vector r = xiˆ+ yj ˆ+ zkˆ, is redily determined by pplying the Pythgors theorem twice We note tht in the right ngle tringle OQP (Fig 04) OP = OQ + QP = x + y, nd in the right ngle tringle OP P, we hve OP = OP + PP = ( x + y ) + z Hence, the length of ny vector r = xiˆ+ yj ˆ+ zkˆ is given by r = xiˆ+ yj ˆ+ zkˆ = x + y + z If nd b re ny two vectors given in the component form i ˆ + ˆ+ j k 3ˆ nd bi ˆ + b ˆ j+ bk 3ˆ, respectively, then (i) the sum (or resultnt) of the vectors nd b is given by + b = ( + b ˆ ˆ ˆ ) i + ( + b) j+ ( 3 + b3) k (ii) the difference of the vector nd b is given by b = ( b ˆ ˆ ˆ ) i + ( b) j+ ( 3 b3) k (iii) the vectors nd b re equl if nd only if (iv) = b, = b nd 3 = b 3 the multipliction of vector by ny sclr λ is given by λ = ( λ ˆ ˆ ˆ ) i + ( λ ) j+ ( λ3) k

12 VECTOR ALGEBRA 435 The ddition of vectors nd the multipliction of vector by sclr together give the following distributive lws: Let nd b be ny two vectors, nd k nd m be ny sclrs Then (i) k + m = ( k + m) (ii) km ( ) = ( km ) (iii) k ( + b) = k+ kb Remrks (i) One my observe tht whtever be the vlue of λ, the vector λ is lwys colliner to the vector In fct, two vectors nd b re colliner if nd only if there exists nonzero sclr λ such tht b =λ If the vectors nd b re given in the component form, ie ˆ ˆ = i + j+ 3ˆ k nd b = bi ˆ ˆ + bj+ b3ˆ k, then the two vectors re colliner if nd only if bi+ b j+ bk = λ ( i ˆ ˆ + j+ k 3ˆ ) ˆ ˆ 3ˆ bi ˆ + b ˆ j+ bk 3ˆ = ( λ ˆ ˆ ˆ ) i + ( λ ) j+ ( λ3) k b =λ, b =λ, b3 =λ3 b = b b3 = =λ 3 (ii) If = i ˆ ˆ + j+ 3ˆ k, then,, 3 re lso clled direction rtios of (iii) In cse if it is given tht l, m, n re direction cosines of vector, then liˆ+ mj ˆ+ nkˆ = (cos α ) iˆ+ (cos β ) ˆj+ (cos γ ) kˆ is the unit vector in the direction of tht vector, where α, β nd γ re the ngles which the vector mkes with x, y nd z xes respectively Exmple 4 Find the vlues of x, y nd z so tht the vectors b = ˆ i + yj ˆ+ kˆ re equl = xiˆ+ ˆj+ zkˆ nd Solution Note tht two vectors re equl if nd only if their corresponding components re equl Thus, the given vectors nd b will be equl if nd only if x =, y =, z =

13 436 Exmple 5 Let = iˆ+ ˆj nd b = iˆ+ ˆj Is = b? Are the vectors nd b equl? Solution We hve = + = 5 nd b = + = 5 So, = b But, the two vectors re not equl since their corresponding components re distinct Exmple 6 Find unit vector in the direction of vector = iˆ+ 3ˆj+ kˆ Solution The unit vector in the direction of vector is given by ˆ = Now = = 4 Therefore ˆ = (iˆ+ 3 ˆj+ kˆ ) = iˆ+ 3 ˆj+ kˆ Exmple 7 Find vector in the direction of vector = iˆ ˆj tht hs mgnitude 7 units Solution The unit vector in the direction of the given vector is ˆ = = ( iˆ ˆj) = iˆ ˆj Therefore, the vector hving mgnitude equl to 7 nd in the direction of is 7 = 7 i j 5 5 = 7 ˆ 4 i 5 5 ˆj Exmple 8 Find the unit vector in the direction of the sum of the vectors, = iˆ+ ˆj 5kˆ nd b = iˆ+ ˆj+ 3kˆ Solution The sum of the given vectors is + b( = c,sy)=4iˆ+ 3ˆj kˆ nd c = ( ) = 9

14 Thus, the required unit vector is ˆ ˆ ˆ 4 ˆ 3 ˆ cˆ = c = (4i + 3 j k) = i + j kˆ c VECTOR ALGEBRA 437 Exmple 9 Write the direction rtio s of the vector = iˆ+ ˆj kˆ nd hence clculte its direction cosines Solution Note tht the direction rtio s, b, c of vector r = xiˆ+ yj ˆ+ zkˆ re just the respective components x, y nd z of the vector So, for the given vector, we hve =, b = nd c = Further, if l, m nd n re the direction cosines of the given vector, then b c l = =, m= =, n= = s r = 6 r 6 r 6 r 6 Thus, the direction cosines re,, Vector joining two points If P (x, y, z ) nd P (x, y, z ) re ny two points, then the vector joining P nd P is the vector PP (Fig 05) Joining the points P nd P with the origin O, nd pplying tringle lw, from the tringle OP P, we hve OP+ PP = OP Using the properties of vector ddition, the bove eqution becomes PP = OP OP ie PP = ( xiˆ+ y ˆ ˆ ˆ ˆ ˆ j+ zk) ( xi + yj+ zk) = ( x x ˆ ˆ ˆ ) i + ( y y) j+ ( z z) k The mgnitude of vector PP is given by Fig 05 = ( x x ) + ( y y ) + ( z z ) PP

15 438 Exmple 0 Find the vector joining the points P(, 3, 0) nd Q(,, 4) directed from P to Q Solution Since the vector is to be directed from P to Q, clerly P is the initil point nd Q is the terminl point So, the required vector joining P nd Q is the vector PQ, given by PQ = ( ) iˆ+ ( 3) ˆj+ ( 4 0) kˆ ie PQ = 3iˆ 5ˆj 4 kˆ 053 Section formul Let P nd Q be two points represented by the position vectors OP nd OQ, respectively, with respect to the origin O Then the line segment joining the points P nd Q my be divided by third point, sy R, in two wys internlly (Fig 06) nd externlly (Fig 07) Here, we intend to find the position vector OR for the point R with respect to the origin O We tke the two cses one by one Cse I When R divides PQ internlly (Fig 06) If R divides PQ such tht m RQ = n PR Fig 06, where m nd n re positive sclrs, we sy tht the point R divides PQ internlly in the rtio of m : n Now from tringles ORQ nd OPR, we hve RQ = OQ OR = b r nd PR = OR OP = r, Therefore, we hve m( b r) = n( r ) (Why?) or r mb + n = (on simplifiction) m+ n Hence, the position vector of the point R which divides P nd Q internlly in the rtio of m : n is given by mb + n OR = m+ n

16 VECTOR ALGEBRA 439 Cse II When R divides PQ externlly (Fig 07) We leve it to the reder s n exercise to verify tht the position vector of the point R which divides the line segment PQ externlly in the rtio PR m m : n ie = is given by QR n mb n OR = Fig 07 m n Remrk If R is the midpoint of PQ, then m = n And therefore, from Cse I, the midpoint R of PQ, will hve its position vector s + b OR = Exmple Consider two points P nd Q with position vectors OP = 3 b nd OQ = + b Find the position vector of point R which divides the line joining P nd Q in the rtio :, (i) internlly, nd (ii) externlly Solution (i) The position vector of the point R dividing the join of P nd Q internlly in the rtio : is ( OR b ) (3 b + + ) 5 = = + 3 (ii) The position vector of the point R dividing the join of P nd Q externlly in the rtio : is ( + b) (3 b) OR = = 4b Exmple Show tht the points A( iˆ ˆj+ kˆ),b( iˆ 3ˆj 5),C(3 kˆ iˆ 4j 4) kˆ re the vertices of right ngled tringle Solution We hve AB = BC = nd CA = ( ) iˆ+ ( 3+ ) ˆj+ ( 5 ) kˆ = iˆ ˆj 6kˆ (3 ) iˆ+ ( 4+ 3) ˆj+ ( 4+ 5) kˆ = iˆ ˆj+ kˆ ( 3) iˆ+ ( + 4) ˆj+ ( + 4) kˆ = iˆ+ 3ˆj+ 5kˆ

17 440 Further, note tht AB 4 = = BC + CA = Hence, the tringle is right ngled tringle EXERCISE 0 Compute the mgnitude of the following vectors: ˆ ˆ ˆ ˆ ˆ ˆ ˆ = i + j+ k; b = i 7j 3 k; c = i + j kˆ Write two different vectors hving sme mgnitude 3 Write two different vectors hving sme direction 4 Find the vlues of x nd y so tht the vectors iˆ+ 3 ˆj nd xiˆ+ yj ˆ re equl 5 Find the sclr nd vector components of the vector with initil point (, ) nd terminl point ( 5, 7) 6 Find the sum of the vectors = iˆ ˆj+ kˆ, b = iˆ+ 4ˆj+ 5kˆ nd c= iˆ 6 ˆj 7kˆ 7 Find the unit vector in the direction of the vector = iˆ+ ˆj+ kˆ 8 Find the unit vector in the direction of vector PQ, where P nd Q re the points (,, 3) nd (4, 5, 6), respectively 9 For given vectors, = iˆ ˆj+ kˆ nd b = iˆ+ ˆj kˆ, find the unit vector in the direction of the vector + b 0 Find vector in the direction of vector 5iˆ ˆj+ kˆ which hs mgnitude 8 units Show tht the vectors iˆ 3ˆj+ 4 kˆ nd 4iˆ+ 6ˆj 8kˆ re colliner Find the direction cosines of the vector iˆ+ ˆj+ 3kˆ 3 Find the direction cosines of the vector joining the points A (,, 3) nd B(,, ), directed from A to B 4 Show tht the vector iˆ+ ˆj+ kˆ is eqully inclined to the xes OX, OY nd OZ 5 Find the position vector of point R which divides the line joining two points P nd Q whose position vectors re iˆ+ ˆj kˆ nd iˆ+ ˆj+ kˆ respectively, in the rtio : (i) internlly (ii) externlly

18 VECTOR ALGEBRA 44 6 Find the position vector of the mid point of the vector joining the points P(, 3, 4) nd Q(4,, ) 7 Show tht the points A, B nd C with position vectors, = 3iˆ 4ˆj 4 kˆ, b = ˆ i ˆj+ kˆ nd c = iˆ 3ˆj 5kˆ, respectively form the vertices of right ngled tringle 8 In tringle ABC (Fig 08), which of the following is not true: (A) AB + BC + CA = 0 (B) AB + BC AC = 0 (C) AB + BC CA = 0 (D) Fig 08 AB CB+ CA = 0 9 If nd b re two colliner vectors, then which of the following re incorrect: (A) b =λ, for some sclr λ (B) =± b (C) the respective components of nd b re proportionl (D) both the vectors nd b hve sme direction, but different mgnitudes 06 Product of Two Vectors So fr we hve studied bout ddition nd subtrction of vectors An other lgebric opertion which we intend to discuss regrding vectors is their product We my recll tht product of two numbers is number, product of two mtrices is gin mtrix But in cse of functions, we my multiply them in two wys, nmely, multipliction of two functions pointwise nd composition of two functions Similrly, multipliction of two vectors is lso defined in two wys, nmely, sclr (or dot) product where the result is sclr, nd vector (or cross) product where the result is vector Bsed upon these two types of products for vectors, they hve found vrious pplictions in geometry, mechnics nd engineering In this section, we will discuss these two types of products 06 Sclr (or dot) product of two vectors Definition The sclr product of two nonzero vectors nd b, denoted by b, is

19 44 defined s b = b cos θ, where, θ is the ngle between nd b, 0 θ π (Fig 09) If either = 0orb = 0, then θ is not defined, nd in this cse, Fig 09 we define b = 0 Observtions b is rel number Let nd b be two nonzero vectors, then b = 0 if nd only if nd b re perpendiculr to ech other ie b = 0 b 3 If θ = 0, then b = b In prticulr, =, s θ in this cse is 0 4 If θ = π, then b = b In prticulr, ( ) =, s θ in this cse is π 5 In view of the Observtions nd 3, for mutully perpendiculr unit vectors iˆ, ˆj nd k ˆ, we hve iˆ iˆ= ˆj ˆj = k ˆ k ˆ =, iˆ ˆj = ˆj kˆ = k ˆ i ˆ = 0 6 The ngle between two nonzero vectors nd b is given by b b cos θ=, or θ = cos b b 7 The sclr product is commuttive ie b = b (Why?) Two importnt properties of sclr product Property (Distributivity of sclr product over ddition) Let, b ny three vectors, then ( b + c) = b + c nd c be

20 VECTOR ALGEBRA 443 Property Let nd b be ny two vectors, nd λ be ny sclr Then ( λ) b = ( λ) b = λ( b) = ( λb) If two vectors nd b re given in component form s i ˆ+ ˆj + k 3ˆ nd ˆ ˆ 3ˆ bi+ b j+ bk, then their sclr product is given s b ( iˆ+ ˆj + kˆ) ( biˆ+ b ˆj + bkˆ) Thus = 3 3 iˆ ( biˆ+ b ˆj + bkˆ) + ˆj ( biˆ+ b ˆj + b kˆ) + kˆ 3 ( bi ˆ ˆ+ b ˆj + bk 3 ) = 3 3 b( iˆ iˆ) + b ( iˆ ˆj) + b ( iˆ kˆ) + b( ˆj iˆ) + b ( ˆj ˆj) + b ( ˆj kˆ) = b 3 ( kˆ iˆ) + b ˆ ˆ ˆ ˆ 3 ( k j) + b 3 3( k k) (Using the bove Properties nd ) = b + b + 3 b 3 (Using Observtion 5) b = b + b+ b Projection of vector on line Suppose vector AB mkes n ngle θ with given directed line l (sy), in the nticlockwise direction (Fig 00) Then the projection of AB on l is vector p (sy) with mgnitude AB cosθ, nd the direction of p being the sme (or opposite) to tht of the line l, depending upon whether cos θ is positive or negtive The vector p A (i) B θ p C 0 0 (0 < θ < 90 ) l B θ C A p 0 0 (90 < θ < 80 ) (ii) l C p A θ l θ A p C l B 0 0 (80 < θ < 70 ) (iii) Fig 00 B 0 0 (70 < θ < 360 ) (iv)

21 444 is clled the projection vector, nd its mgnitude p is simply clled s the projection of the vector AB on the directed line l For exmple, in ech of the following figures (Fig 00(i) to (iv)), projection vector of AB long the line l is vector AC Observtions If ˆp is the unit vector long line l, then the projection of vector on the line l is given by pˆ Projection of vector on other vector b, is given by b b ˆ, or, or ( b) b b 3 If θ = 0, then the projection vector of AB will be AB itself nd if θ = π, then the projection vector of AB will be BA 4 If = π θ or 3π θ =, then the projection vector of AB will be zero vector Remrk If α, β nd γ re the direction ngles of vector = i ˆ ˆ + j+ 3ˆ k, then its direction cosines my be given s iˆ 3 cos α= =, cos, nd cos ˆ β= γ= i Also, note tht cos α, cos β nd cosγ re respectively the projections of long OX, OY nd OZ ie, the sclr components, nd 3 of the vector, re precisely the projections of long x-xis, y-xis nd z-xis, respectively Further, if is unit vector, then it my be expressed in terms of its direction cosines s = cosα iˆ+ cosβ ˆj+ cos γkˆ Exmple 3 Find the ngle between two vectors nd b with mgnitudes nd respectively nd when b= Solution Given b =, = nd b = We hve b π θ= cos = cos = b 3

22 VECTOR ALGEBRA 445 Exmple 4 Find ngle θ between the vectors = iˆ+ ˆj kˆ nd b= iˆ ˆj+ kˆ Solution The ngle θ between two vectors nd b is given by b cosθ = b Now b = ( iˆ+ ˆj kˆ) ( iˆ ˆj+ kˆ) = = Therefore, we hve cosθ = 3 hence the required ngle is θ = cos 3 Exmple 5 If = 5iˆ ˆj 3 kˆ nd b= iˆ+ 3ˆj 5kˆ, then show tht the vectors + b nd b re perpendiculr Solution We know tht two nonzero vectors re perpendiculr if their sclr product is zero Here + b = (5iˆ ˆj 3 kˆ) + ( iˆ+ 3ˆj 5 kˆ) = 6iˆ+ ˆj 8kˆ nd b = (5iˆ ˆj 3 kˆ) ( iˆ+ 3ˆj 5 kˆ) = 4iˆ 4ˆj+ kˆ So ( + b) ( b) = (6iˆ+ ˆj 8 kˆ) (4iˆ 4ˆj+ kˆ) = 4 8 6= 0 Hence + b nd b re perpendiculr vectors Exmple 6 Find the projection of the vector = iˆ+ 3ˆj+ kˆ on the vector b = iˆ+ ˆj+ kˆ Solution The projection of vector on the vector b is given by ( b ( ) 0 5 ) = = = 6 b () + () + () 6 3 Exmple 7 Find b, if two vectors nd b re such tht =, b = 3 nd b = 4 Solution We hve b = ( b) ( b) = b b + b b

23 446 = ( b) + b = () (4) + (3) Therefore b = 5 Exmple 8 If is unit vector nd ( x ) ( x + ) = 8, then find x Solution Since is unit vector, = Also, ( x ) ( x + ) =8 or x x+ x x =8 or x = 8 ie x = 9 Therefore x = 3 (s mgnitude of vector is non negtive) Exmple 9 For ny two vectors nd b, we lwys hve b b (Cuchy- Schwrtz inequlity) Solution The inequlity holds trivilly when either = 0 or b = 0 Actully, in such sitution we hve b = 0 = b So, let us ssume tht 0 b Then, we hve b = cos θ b Therefore b b Exmple 0 For ny two vectors nd b, we lwys hve + b + b (tringle inequlity) + b Solution The inequlity holds trivilly in cse either A B = 0orb = 0 (How?) So, let 0 b Then, + b Fig 0 = ( + b) = ( + b) ( + b) = + b + b + b b = + b + b (sclr product is commuttive) + b + b (since x x x R ) + b + b (from Exmple 9) = ( + b ) b C

24 VECTOR ALGEBRA 447 Hence b + + b Remrk If the equlity holds in tringle inequlity (in the bove Exmple 0), ie + b = + b, then AC = AB + BC showing tht the points A, B nd C re colliner Exmple Show tht the points A( iˆ+ 3ˆj+ 5 kˆ), B( iˆ+ ˆj+ 3 kˆ) nd C(7 iˆ kˆ ) re colliner Solution We hve AB = ( + ) iˆ+ ( 3) ˆj+ (3 5) kˆ= 3iˆ ˆj kˆ, BC = (7 ) iˆ+ (0 ) ˆj+ ( 3) k ˆ = 6iˆ ˆj 4k ˆ, AC = (7 + ) iˆ+ (0 3) ˆj+ ( 5) k ˆ = 9iˆ 3 ˆj 6 k ˆ AB = 4, BC = 4 nd AC = 3 4 Therefore AC = AB + BC Hence the points A, B nd C re colliner Note In Exmple, one my note tht lthough AB + BC + CA = 0 but the points A, B nd C do not form the vertices of tringle EXERCISE 03 Find the ngle between two vectors nd b with mgnitudes 3 nd, respectively hving b = 6 Find the ngle between the vectors ˆ iˆ ˆj+ 3 k nd 3iˆ ˆj+ kˆ 3 Find the projection of the vector iˆ ˆj on the vector iˆ+ ˆj 4 Find the projection of the vector iˆ+ 3ˆj+ 7kˆ on the vector 7iˆ ˆj+ 8kˆ 5 Show tht ech of the given three vectors is unit vector: ˆ ˆ (iˆ+ 3 ˆj+ 6 k), (3iˆ 6 ˆj+ k), (6iˆ+ ˆj 3 kˆ) Also, show tht they re mutully perpendiculr to ech other

25 448 6 Find nd b, if ( + b) ( b) = 8 nd = 8 b 7 Evlute the product (3 5 b) (+ 7 b) 8 Find the mgnitude of two vectors nd b, hving the sme mgnitude nd such tht the ngle between them is 60 o nd their sclr product is 9 Find x, if for unit vector, ( x ) ( x + ) = 0 If ˆ ˆ 3 ˆ, ˆ ˆ ˆ = i + j+ k b = i + j+ k nd c = 3iˆ+ ˆ j re such tht +λb is perpendiculr to c, then find the vlue of λ Show tht b+ b is perpendiculr to b b, for ny two nonzero vectors nd b If = 0 nd b = 0, then wht cn be concluded bout the vector b? 3 If bc,, re unit vectors such tht + b + c = 0, find the vlue of b + b c + c 4 If either vector = 0 or b = 0, then b = 0 But the converse need not be true Justify your nswer with n exmple 5 If the vertices A, B, C of tringle ABC re (,, 3), (, 0, 0), (0,, ), respectively, then find ABC [ ABC is the ngle between the vectors BA nd BC ] 6 Show tht the points A(,, 7), B(, 6, 3) nd C(3, 0, ) re colliner 7 Show tht the vectors iˆ ˆj+ kˆ, iˆ 3ˆj 5kˆ nd 3iˆ 4ˆj 4kˆ form the vertices of right ngled tringle 8 If is nonzero vector of mgnitude nd λ nonzero sclr, then λ is unit vector if (A) λ = (B) λ = (C) = λ (D) = / λ 063 Vector (or cross) product of two vectors In Section 0, we hve discussed on the three dimensionl right hnded rectngulr coordinte system In this system, when the positive x-xis is rotted counterclockwise

26 VECTOR ALGEBRA 449 into the positive y-xis, right hnded (stndrd) screw would dvnce in the direction of the positive z-xis (Fig 0(i)) In right hnded coordinte system, the thumb of the right hnd points in the direction of the positive z-xis when the fingers re curled in the direction wy from the positive x-xis towrd the positive y-xis (Fig 0(ii)) Fig 0 (i), (ii) Definition 3 The vector product of two nonzero vectors nd b, is denoted by b nd defined s b = b sinθ nˆ, where, θ is the ngle between nd b, 0 θ π nd ˆn is unit vector perpendiculr to both nd b, such tht b, nd nˆ form right hnded system (Fig 03) ie, the right hnded system rotted from to b moves in the Fig 03 direction of ˆn If either = 0orb = 0, then θ is not defined nd in this cse, we define b = 0 Observtions b is vector Let nd b be two nonzero vectors Then b = 0 if nd only if nd b re prllel (or colliner) to ech other, ie, b = 0 b

27 450 In prticulr, = 0 nd ( ) = 0, since in the first sitution, θ = 0 nd in the second one, θ = π, mking the vlue of sin θ to be 0 3 π If θ= then b = b 4 In view of the Observtions nd 3, for mutully perpendiculr unit vectors iˆ, ˆj nd k ˆ (Fig 04), we hve iˆ iˆ = ˆj ˆj = kˆ kˆ= 0 iˆ ˆj = k ˆ, ˆ j k ˆ = i ˆ, k ˆ i ˆ = ˆ j Fig 04 5 In terms of vector product, the ngle between two vectors nd b my be given s sin θ = b b 6 It is lwys true tht the vector product is not commuttive, s b = b Indeed, b = b sinθnˆ, where b, nd nˆ form right hnded system, ie, θ is trversed from to b, Fig 05 (i) While, = sin, where b, ndn form right hnded system ie θ is trversed from ˆ b to, Fig 05(ii) Fig 05 (i), (ii) Thus, if we ssume nd b to lie in the plne of the pper, then nˆ nd n ˆ both will be perpendiculr to the plne of the pper But, ˆn being directed bove the pper while ˆn directed below the pper ie nˆ ˆ = n

28 Hence b = b sinθnˆ b sin θnˆ 7 In view of the Observtions 4 nd 6, we hve = VECTOR ALGEBRA 45 = b ˆj iˆ= kˆ, kˆ ˆj= iˆ nd iˆ kˆ= ˆj 8 If nd b represent the djcent sides of tringle then its re is given s b By definition of the re of tringle, we hve from Fig 06, Are of tringle ABC = AB CD But AB = b (s given), nd CD = Fig 06 sinθ Thus, Are of tringle ABC = sin b θ = b 9 If nd b represent the djcent sides of prllelogrm, then its re is given by b From Fig 07, we hve Are of prllelogrm ABCD = AB DE But AB = b (s given), nd DE = sin θ Thus, Are of prllelogrm ABCD = b sinθ Fig = b 07 We now stte two importnt properties of vector product Property 3 (Distributivity of vector product over ddition): If, b nd c re ny three vectors nd λ be sclr, then (i) ( b + c) = b + c (ii) λ ( b ) = ( λ ) b = ( λb)

29 45 Let nd b ˆ ˆ 3ˆ be two vectors given in component form s i ˆ+ ˆj + k 3ˆ nd bi+ b j+ bk, respectively Then their cross product my be given by b iˆ ˆj kˆ = 3 b b b 3 Explntion We hve b ( iˆ+ ˆj + kˆ) ( biˆ+ b ˆj + bkˆ) = 3 3 b( iˆ iˆ) + b ( iˆ ˆj) + b ( iˆ kˆ ) + b( ˆj iˆ) = 3 b( ˆj ˆj) + b( ˆj kˆ ) b 3 ( kˆ iˆ) + b ˆ ˆ ˆ ˆ 3 ( k j) + b 3 3( k k) (by Property ) b ( iˆ ˆj) b ( kˆ iˆ) b ( iˆ ˆj) = 3 b( ˆj kˆ) + b( kˆ iˆ) b( ˆj kˆ) (s iˆ iˆ= ˆj ˆj = kˆ kˆ= 0 nd iˆ kˆ= kˆ iˆ, ˆj iˆ= iˆ ˆj nd kˆ ˆj= ˆj kˆ) bkˆ b ˆj bkˆ+ biˆ+ bj ˆ biˆ = (s iˆ ˆj = kˆ, ˆj kˆ= iˆ nd kˆ iˆ= ˆj) = ( b 3 b 3 ) i ˆ ( b ˆ ˆ 3 b 3 ) j+ ( b b ) k iˆ ˆj kˆ = 3 b b b3 Exmple Find b, if = iˆ+ ˆj+ 3 kˆ nd b = 3iˆ+ 5 ˆj kˆ Solution We hve b = iˆ ˆj kˆ = iˆ( 5) ( 4 9) ˆj+ (0 3) kˆ = 7iˆ+ 3 ˆj+ 7kˆ Hence b = ( 7) + (3) + (7) = 507

30 VECTOR ALGEBRA 453 Exmple 3 Find unit vector perpendiculr to ech of the vectors ( + b ) ( b ), where = iˆ+ ˆj+ kˆ, b = iˆ+ ˆj+ 3kˆ Solution We hve ˆ 3ˆ 4ˆ + b = i + j+ k nd b = ˆj kˆ A vector which is perpendiculr to both + b nd b is given by nd ( + b) ( b) = iˆ ˆj kˆ 3 4 ˆ 4ˆ ˆ = i + j k ( = c, sy) 0 Now c = = 4 = 6 Therefore, the required unit vector is c = i ˆ + ˆ j k ˆ c Note There re two perpendiculr directions to ny plne Thus, nother unit vector perpendiculr to + b nd b will be i ˆ ˆ j+ k ˆ But tht will be consequence of ( b) ( + b) Exmple 4 Find the re of tringle hving the points A(,, ), B(,, 3) nd C(, 3, ) s its vertices Solution We hve AB = ˆj + kˆ nd AC = iˆ+ ˆj The re of the given tringle is AB AC Now, iˆ ˆj kˆ AB AC = 0 = 4iˆ+ ˆj kˆ 0 Therefore AB AC = = Thus, the required re is

31 454 Exmple 5 Find the re of prllelogrm whose djcent sides re given by the vectors = 3iˆ+ ˆj+ 4kˆ nd b = iˆ ˆj+ kˆ Solution The re of prllelogrm with nd b s its djcent sides is given by b Now b Therefore b nd hence, the required re is 4 = iˆ ˆj kˆ 3 4 = 5iˆ+ ˆj 4kˆ = = 4 EXERCISE 04 Find b, if = iˆ 7ˆj+ 7kˆ nd b = 3iˆ ˆj+ kˆ Find unit vector perpendiculr to ech of the vector + b nd b, where = 3iˆ+ ˆj+ kˆ nd b = iˆ+ ˆj kˆ 3 If unit vector π mkes ngles with ˆ π i, with ˆj nd n cute ngle θ with 3 4 ˆk, then find θ nd hence, the components of 4 Show tht ( b) ( + b) = ( b ) 5 Find λ nd μ if (iˆ+ 6 ˆj+ 7 kˆ) ( iˆ+λ ˆj+μ kˆ) = 0 6 Given tht b = 0 nd b = 0 Wht cn you conclude bout the vectors nd b? 7 Let the vectors b,, c be given s i ˆ+ ˆ ˆ ˆj + k 3, bi ˆ+ b ˆj + bk 3, ˆ ˆ ci + cj+ c3ˆ k Then show tht ( b + c) = b + c 8 If either = 0 or b = 0, then b = 0 Is the converse true? Justify your nswer with n exmple 9 Find the re of the tringle with vertices A(,, ), B(, 3, 5) nd C(, 5, 5)

32 VECTOR ALGEBRA Find the re of the prllelogrm whose djcent sides re determined by the vectors = iˆ ˆj+ 3kˆ nd b = iˆ 7ˆj+ kˆ Let the vectors nd b be such tht = 3 nd b =, then b is 3 unit vector, if the ngle between nd b is (A) π/6 (B) π/4 (C) π/3 (D) π/ Are of rectngle hving vertices A, B, C nd D with position vectors ˆ iˆ+ ˆj+ 4 k, iˆ+ ˆj+ 4kˆ, ˆ ˆ ˆ i j+ 4k nd iˆ ˆj+ 4kˆ, respectively is (A) (B) (C) (D) 4 Miscellneous Exmples Exmple 6 Write ll the unit vectors in XY-plne Solution Let r = x i+ y j be unit vector in XY-plne (Fig 08) Then, from the figure, we hve x = cos θ nd y = sin θ (since r = ) So, we my write the vector r s r ( = OP ) = cosθ iˆ+ sinθ ˆj () Clerly, r = cos θ+ sin θ= Fig 08 Also, s θ vries from 0 to π, the point P (Fig 08) trces the circle x + y = counterclockwise, nd this covers ll possible directions So, () gives every unit vector in the XY-plne

33 456 Exmple 7 If iˆ+ ˆj+ kˆ, iˆ+ 5 ˆj, 3iˆ+ ˆj 3 kˆ nd iˆ 6 ˆj k ˆ re the position vectors of points A, B, C nd D respectively, then find the ngle between AB nd CD Deduce tht AB nd CD re colliner Solution Note tht if θ is the ngle between AB nd CD, then θ is lso the ngle between AB nd CD Now Therefore Similrly Thus AB = Position vector of B Position vector of A = (iˆ+ 5 ˆj) ( iˆ+ ˆj+ kˆ) = iˆ+ 4ˆj kˆ AB = () + (4) + ( ) = 3 CD = iˆ 8 ˆj+ kˆ nd CD = 6 AB CD cos θ = AB CD = ( ) + 4( 8) + ( )() = 36 = (3 )(6 ) 36 Since 0 θ π, it follows tht θ = π This shows tht AB nd CD re colliner Alterntively, AB = CD which implies tht AB nd CD re colliner vectors Exmple 8 Let b, nd c be three vectors such tht = 3, b = 4, c = 5 nd ech one of them being perpendiculr to the sum of the other two, find + b + c Solution Given ( b + c ) b ( c + ) = 0, c ( + b ) = 0 Now + b + c = ( + b + c) = ( + b + c) ( + b + c) = + ( b + c) + b b + b ( + c) + c( + b) + c c = + b + c = = 50 Therefore + b + c = 5

34 VECTOR ALGEBRA 457 Exmple 9 Three vectors, b nd c stisfy the condition + b + c = 0 Evlute the quntity μ= b + b c + c, if =, b = 4 nd c = Solution Since + b+ c = 0, we hve ( + b + c ) =0 or + b+ c =0 Therefore b+ c = = () Agin, b ( + b + c) =0 or b+ b c = b = 6 () Similrly c + b c = 4 (3) Adding (), () nd (3), we hve ( b+ b c+ c) = or μ =, ie, μ = Exmple 30 If with reference to the right hnded system of mutully perpendiculr unit vectors ˆ ˆ ˆ i, j nd k, α= 3 iˆ ˆj, β= iˆ+ ˆj 3kˆ, then express β in the form β=β +β,where β is prllel to α nd β is perpendiculr to α Solution Let β = λα, λ is sclr, ie, β ˆ ˆ = 3λi λj Now β =β β = ˆ ˆ ˆ ( 3 λ ) i + ( +λ) j 3k Now, since β is to be perpendiculr to α, we should hve α β = 0 ie, 3( 3 λ) ( +λ ) =0 or λ = Therefore β = 3 ˆ ˆ 3 i j nd β ˆ ˆ = i + ˆ j 3k

35 458 Miscellneous Exercise on Chpter 0 Write down unit vector in XY-plne, mking n ngle of 30 with the positive direction of x-xis Find the sclr components nd mgnitude of the vector joining the points P(x, y, z ) nd Q (x, y, z ) 3 A girl wlks 4 km towrds west, then she wlks 3 km in direction 30 est of north nd stops Determine the girl s displcement from her initil point of deprture 4 If = b + c, then is it true tht = b + c? Justify your nswer 5 Find the vlue of x for which x( iˆ+ ˆj+ kˆ ) is unit vector 6 Find vector of mgnitude 5 units, nd prllel to the resultnt of the vectors = iˆ+ 3 ˆj kˆ nd b = iˆ ˆj+ kˆ 7 If ˆ ˆ ˆ ˆ ˆ ˆ = i + j+ k, b = i j+ 3k nd c = iˆ ˆj+ kˆ, find unit vector prllel to the vector b + 3c 8 Show tht the points A (,, 8), B(5, 0, ) nd C(, 3, 7) re colliner, nd find the rtio in which B divides AC 9 Find the position vector of point R which divides the line joining two points P nd Q whose position vectors re ( + b) nd ( 3 b) externlly in the rtio : Also, show tht P is the mid point of the line segment RQ 0 The two djcent sides of prllelogrm re iˆ 4 ˆj+ 5 kˆ nd iˆ ˆj 3kˆ Find the unit vector prllel to its digonl Also, find its re Show tht the direction cosines of vector eqully inclined to the xes OX, OY nd OZ re,, Let ˆ ˆ ˆ ˆ ˆ ˆ = i + 4j+ k, b = 3i j+ 7 k nd c = iˆ ˆj+ 4kˆ Find vector d which is perpendiculr to both nd b, nd c d = 5 3 The sclr product of the vector iˆ+ ˆj+ kˆ with unit vector long the sum of vectors iˆ+ 4ˆj 5kˆ nd λ iˆ+ ˆj+ 3kˆ is equl to one Find the vlue of λ 4 If, b, c re mutully perpendiculr vectors of equl mgnitudes, show tht the vector + b + c is eqully inclined to, b nd c

36 VECTOR ALGEBRA Prove tht ( + b) ( + b) = + b, if nd only if, b re perpendiculr, given 0, b 0 Choose the correct nswer in Exercises 6 to 9 6 If θ is the ngle between two vectors nd b, then b 0 only when π π (A) 0 <θ< (B) 0 θ (C) 0 < θ < π (D) 0 θ π 7 Let nd b be two unit vectors nd θ is the ngle between them Then + b is unit vector if (A) π θ= (B) 4 π θ= (C) 3 π θ= (D) 8 The vlue of iˆ( ˆj kˆ) + ˆj ( iˆ kˆ) + kˆ ( iˆ ˆj) is (A) 0 (B) (C) (D) 3 9 If θ is the ngle between ny two vectors nd b, then b = b when θ is equl to (A) 0 (B) π 4 (C) π (D) π θ= π 3 Summry Position vector of point P(x, y, z) is given s OP( = r) = xiˆ+ yj ˆ+ zkˆ, nd its mgnitude by x + y + z The sclr components of vector re its direction rtios, nd represent its projections long the respective xes The mgnitude (r), direction rtios (, b, c) nd direction cosines (l, m, n) of ny vector re relted s: b c l =, m=, n= r r r The vector sum of the three sides of tringle tken in order is 0

37 460 The vector sum of two coinitil vectors is given by the digonl of the prllelogrm whose djcent sides re the given vectors The multipliction of given vector by sclr λ, chnges the mgnitude of the vector by the multiple λ, nd keeps the direction sme (or mkes it opposite) ccording s the vlue of λ is positive (or negtive) For given vector, the vector ˆ = gives the unit vector in the direction of The position vector of point R dividing line segment joining the points P nd Q whose position vectors re nd b respectively, in the rtio m : n n + mb (i) internlly, is given by m+ n mb n (ii) externlly, is given by m n The sclr product of two given vectors nd b hving ngle θ between them is defined s b = b cosθ Also, when b is given, the ngle θ between the vectors nd b my be determined by b cosθ = b If θ is the ngle between two vectors nd b, then their cross product is given s b = b sinθnˆ where ˆn is unit vector perpendiculr to the plne contining nd b Such tht bnform,, ˆ right hnded system of coordinte xes If we hve two vectors nd b, given in component form s i ˆ ˆ = + j+ k nd b = bi ˆ ˆ + bj+ b3ˆ k nd λ ny sclr, 3ˆ

38 VECTOR ALGEBRA 46 then + b nd b ( + b ) iˆ+ ( + b ) ˆj+ ( + b ) kˆ ; = 3 3 λ = ( λ ˆ ˆ ˆ ) i + ( λ ) j+ ( λ 3) k; b = b + b + b 3 3 ; iˆ ˆj kˆ b c b c = Historicl Note The word vector hs been derived from Ltin word vectus, which mens to crry The germinl ides of modern vector theory dte from round 800 when Cspr Wessel (745-88) nd Jen Robert Argnd (768-8) described tht how complex number + ib could be given geometric interprettion with the help of directed line segment in coordinte plne Willim Rowen Hmilton ( ) n Irish mthemticin ws the first to use the term vector for directed line segment in his book Lectures on Quternions (853) Hmilton s method of quternions (n ordered set of four rel numbers given s: + biˆ+ cj ˆ+ dkˆ, iˆ, ˆj, kˆ following certin lgebric rules) ws solution to the problem of multiplying vectors in three dimensionl spce Though, we must mention here tht in prctice, the ide of vector concept nd their ddition ws known much erlier ever since the time of Aristotle (384-3 BC), Greek philosopher, nd pupil of Plto ( BC) Tht time it ws supposed to be known tht the combined ction of two or more forces could be seen by dding them ccording to prllelogrm lw The correct lw for the composition of forces, tht forces dd vectorilly, hd been discovered in the cse of perpendiculr forces by Stevin-Simon (548-60) In 586 AD, he nlysed the principle of geometric ddition of forces in his tretise DeBeghinselen der Weeghconst ( Principles of the Art of Weighing ), which cused mjor brekthrough in the development of mechnics But it took nother 00 yers for the generl concept of vectors to form In the 880, Josih Willrd Gibbs ( ), n Americn physicist nd mthemticin, nd Oliver Heviside (850-95), n English engineer, creted wht we now know s vector nlysis, essentilly by seprting the rel (sclr)

39 46 prt of quternion from its imginry (vector) prt In 88 nd 884, Gibbs printed tretise entitled Element of Vector Anlysis This book gve systemtic nd concise ccount of vectors However, much of the credit for demonstrting the pplictions of vectors is due to the D Heviside nd PG Tit (83-90) who contributed significntly to this subject