not to be republishe NCERT VECTOR ALGEBRA Chapter Introduction 10.2 Some Basic Concepts

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1 44 MATHEMATICS Chpte 10 In most sciences one genetion tes down wht nothe hs built nd wht one hs estblished nothe undoes In Mthemtics lone ech genetion builds new stoy to the old stuctue HERMAN HANKEL 101 Intoduction In ou dy to dy life, we come coss mny queies such s Wht is you height? How should footbll plye hit the bll to give pss to nothe plye of his tem? Obseve tht possible nswe to the fist quey my be 16 metes, quntity tht involves only one vlue (mgnitude) which is el numbe Such quntities e clled scls Howeve, n nswe to the second quey is quntity (clled foce) which involves muscul stength (mgnitude) nd diection (in which nothe plye is positioned) Such quntities e clled vectos In mthemtics, physics nd engineeing, we fequently come coss with both types of WR Hmilton quntities, nmely, scl quntities such s length, mss, ( ) time, distnce, speed, e, volume, tempetue, wok, money, voltge, density, esistnce etc nd vecto quntities like displcement, velocity, cceletion, foce, weight, momentum, electic field intensity etc In this chpte, we will study some of the bsic concepts bout vectos, vious opetions on vectos, nd thei lgebic nd geometic popeties These two type of popeties, when consideed togethe give full elistion to the concept of vectos, nd led to thei vitl pplicbility in vious es s mentioned bove 10 Some Bsic Concepts VECTOR ALGEBRA not to be epublishe Let l be ny stight line in plne o thee dimensionl spce This line cn be given two diections by mens of owheds A line with one of these diections pescibed is clled diected line (Fig 101 (i), (ii))

2 VECTOR ALGEBRA 45 Fig 101 Now obseve tht if we estict the line l to the line segment AB, then mgnitude is pescibed on the line l with one of the two diections, so tht we obtin diected line segment (Fig 101(iii)) Thus, diected line segment hs mgnitude s well s diection Definition 1 A quntity tht hs mgnitude s well s diection is clled vecto uuu Notice tht diected line segment is vecto (Fig 101(iii)), denoted s AB o simply s uuu, nd ed s vecto AB o vecto uuu The point A fom whee the vecto AB stts is clled its initil point, nd the point B whee it ends is clled its teminl point The distnce between initil nd teminl points of vecto is clled the mgnitude (o length) of the vecto, denoted s uuu AB, o, o The ow indictes the diection of the vecto Note Since the length is neve negtive, the nottion < 0 hs no mening Position Vecto Fom Clss XI, ecll the thee dimensionl ight hnded ectngul coodinte system (Fig 10(i)) Conside point P in spce, hving coodintes (x, y, z) with uuu espect to the oigin O(0, 0, 0) Then, the vecto OP hving O nd P s its initil nd teminl points, espectively, is clled the position vecto of the point P with espect uuu to O Using distnce fomul (fom Clss XI), the mgnitude of OP (o ) is given by not to be epublishe uuu OP = x + y + z In pctice, the position vectos of points A, B, C, etc, with espect to the oigin O e denoted by, b, c, etc, espectively (Fig 10 (ii))

3 46 MATHEMATICS Fig 10 Diection Cosines uuu Conside the position vecto OP o of point P(x, y, z) s in Fig 103 The ngles α, β, γ mde by the vecto with the positive diections of x, y nd z-xes espectively, e clled its diection ngles The cosine vlues of these ngles, ie, cosα, cosβ nd cos γ e clled diection cosines of the vecto, nd usully denoted by l, m nd n, espectively Z X A x C z O P( x,y,z ) Fig 103 X Fom Fig 103, one my note tht the tingle OAP is ight ngled, nd in it, we x hve cos α= ( stnds fo ) Similly, fom the ight ngled tingles OBP nd OCP, we my wite cos β = y z nd cos γ= Thus, the coodintes of the point P my lso be expessed s (l, m,n) The numbes l, m nd n, popotionl to the diection cosines e clled s diection tios of vecto, nd denoted s, b nd c, espectively y O not to be epublishe B 90 A Y P

4 VECTOR ALGEBRA 47 Note One my note tht l + m + n = 1 but + b + c 1, in genel 103 Types of Vectos Zeo Vecto A vecto whose initil nd teminl points coincide, is clled zeo vecto (o null vecto), nd denoted s 0 Zeo vecto cn not be ssigned definite diection s it hs zeo mgnitude O, ltentively othewise, it my be egded s uuu uuu hving ny diection The vectos AA, BB epesent the zeo vecto, Unit Vecto A vecto whose mgnitude is unity (ie, 1 unit) is clled unit vecto The unit vecto in the diection of given vecto is denoted by â Coinitil Vectos Two o moe vectos hving the sme initil point e clled coinitil vectos Colline Vectos Two o moe vectos e sid to be colline if they e pllel to the sme line, iespective of thei mgnitudes nd diections Equl Vectos Two vectos nd b e sid to be equl, if they hve the sme mgnitude nd diection egdless of the positions of thei initil points, nd witten s = b Negtive of Vecto A vecto whose mgnitude is the sme s tht of given vecto uuu (sy, AB ), but diection is opposite to tht of it, is clled negtive of the given vecto Fo exmple, vecto BA uuu uuu uuu uuu is negtive of the vecto AB, nd witten s BA = AB Remk The vectos defined bove e such tht ny of them my be subject to its pllel displcement without chnging its mgnitude nd diection Such vectos e clled fee vectos Thoughout this chpte, we will be deling with fee vectos only Exmple 1 Repesent gphiclly displcement of 40 km, 30 west of south uuu Solution The vecto OP epesents the equied displcement (Fig 104) not to be epublishe Exmple Clssify the following mesues s scls nd vectos (i) 5 seconds (ii) 1000 cm 3 Fig 104

5 48 MATHEMATICS (iii) 10 Newton (iv) 30 km/h (v) 10 g/cm 3 (vi) 0 m/s towds noth Solution (i) Time-scl (ii) Volume-scl (iii) Foce-vecto (iv) Speed-scl (v) Density-scl (vi) Velocity-vecto Exmple 3 In Fig 105, which of the vectos e: (i) Colline (ii) Equl (iii) Coinitil Solution (i) Colline vectos :, c nd d (ii) Equl vectos : nd c (iii) Coinitil vectos : b, c nd d EXERCISE 101 Fig Repesent gphiclly displcement of 40 km, 30 est of noth Clssify the following mesues s scls nd vectos (i) 10 kg (ii) metes noth-west (iii) 40 (iv) 40 wtt (v) coulomb (vi) 0 m/s 3 Clssify the following s scl nd vecto quntities (i) time peiod (ii) distnce (iii) foce (iv) velocity (v) wok done 4 In Fig 106 ( sque), identify the following vectos (i) Coinitil (iii) Colline but not equl (ii) Equl 5 Answe the following s tue o flse (i) nd e colline not to be epublishe (ii) Two colline vectos e lwys equl in mgnitude (iii) Two vectos hving sme mgnitude e colline Fig 106 (iv) Two colline vectos hving the sme mgnitude e equl

6 VECTOR ALGEBRA Addition of Vectos uuu A vecto AB simply mens the displcement fom point A to the point B Now conside sitution tht gil moves fom A to B nd then fom B to C (Fig 107) The net displcement mde by the gil fom uuu point A to the point C, is given by the vecto AC nd expessed s uuu uuu uuu AC = AB + BC This is known s the tingle lw of vecto ddition In genel, if we hve two vectos nd b (Fig 108 (i)), then to dd them, they e positioned so tht the initil point of one coincides with the teminl point of the othe (Fig 108(ii)) b (i) A + b (ii) Fig 108 Fig 107 Fo exmple, in Fig 108 (ii), we hve shifted vecto b without chnging its mgnitude nd diection, so tht it s initil point coincides with the teminl point of Then, the vecto + b, epesented by the thid side AC of the tingle ABC, gives us the sum (o esultnt) of the vectos nd b ie, in tingle ABC (Fig 108 (ii)), we hve uuu uuu uuu AB + BC = AC uuu uuu Now gin, since AC = CA, fom the bove eqution, we hve uuu uuu uuu uuu AB + BC + CA = AA = 0 This mens tht when the sides of tingle e tken in ode, it leds to zeo esultnt s the initil nd teminl points get coincided (Fig 108(iii)) C b B A b not to be epublishe (iii) C b B b C

7 430 MATHEMATICS uuuu uuu Now, constuct vecto BC so tht its mgnitude is sme s the vecto BC, but the diection opposite to tht of it (Fig 108 (iii)), ie, uuuu uuu BC = BC Then, on pplying tingle lw fom the Fig 108 (iii), we hve uuuu uuu uuuu uuu uuu AC = AB + BC = AB + ( BC) = b uuuu The vecto AC is sid to epesent the diffeence of nd b Now, conside bot in ive going fom one bnk of the ive to the othe in diection pependicul to the flow of the ive Then, it is cted upon by two velocity vectos one is the velocity impted to the bot by its engine nd othe one is the velocity of the flow of ive wte Unde the simultneous influence of these two velocities, the bot in ctul stts tvelling with diffeent velocity To hve pecise ide bout the effective speed nd diection (ie, the esultnt velocity) of the bot, we hve the following lw of vecto ddition If we hve two vectos nd b epesented by the two djcent sides of pllelogm in mgnitude nd diection (Fig 109), then thei sum + b is epesented in mgnitude nd diection by the digonl of the pllelogm though thei common point This is known s Fig 109 the pllelogm lw of vecto ddition Note Fom Fig 109, using the tingle lw, one my note tht uuu uuu uuu OA + AC = OC uuu uuu uuu uuu uuu o OA + OB = OC (since AC = OB ) which is pllelogm lw Thus, we my sy tht the two lws of vecto ddition e equivlent to ech othe not to be epublishe Popeties of vecto ddition Popety 1 Fo ny two vectos nd b, + b = b + (Commuttive popety)

8 Poof Conside the pllelogm ABCD uuu uuu (Fig 1010) Let AB nd BC b, then using the tingle lw, fom tingle ABC, we hve uuu AC = + b Now, since the opposite sides of pllelogm e equl nd pllel, fom uuu uuu Fig1010, we hve, AD = BC = b nd uuu uuu DC = AB = Agin using tingle lw, fom tingle ADC, we hve uuuu uuu uuu AC = AD + DC = b + Hence + b = b + Popety Fo ny thee vectos b, ndc VECTOR ALGEBRA 431 ( + b) + c = ( + b + c ) (Associtive popety) uuu uuu uuu Poof Let the vectos b, ndc be epesented by PQ, QR nd RS, espectively, s shown in Fig 1011(i) nd (ii) Fig 1011 Then + b uuu uuu uuu = PQ + QR = PR nd uuu uuu uuu b + c = QR + RS = QS So uuu uuu uu ( + b) + c = PR + RS = PS Fig 1010 not to be epublishe

9 43 MATHEMATICS nd uuu uuu uu + ( b + c) = PQ + QS = PS Hence ( + b) + c = ( + b + c ) Remk The ssocitive popety of vecto ddition enbles us to wite the sum of thee vectos, b, c s + b + c without using bckets Note tht fo ny vecto, we hve + 0 = 0 + = Hee, the zeo vecto 0 is clled the dditive identity fo the vecto ddition 105 Multipliction of Vecto by Scl Let be given vecto nd λ scl Then the poduct of the vecto by the scl λ, denoted s λ, is clled the multipliction of vecto by the scl λ Note tht, λ is lso vecto, colline to the vecto The vecto λ hs the diection sme (o opposite) to tht of vecto ccoding s the vlue of λ is positive (o negtive) Also, the mgnitude of vecto λ is λ times the mgnitude of the vecto, ie, λ = λ A geometic visulistion of multipliction of vecto by scl is given in Fig Fig 101 When λ = 1, then λ =, which is vecto hving mgnitude equl to the mgnitude of nd diection opposite to tht of the diection of The vecto is clled the negtive (o dditive invese) of vecto nd we lwys hve + ( ) = ( ) + = 0 not to be epublishe 1 Also, if λ =, povided 0, ie λ = λ 1 = 1 1 is not null vecto, then

10 VECTOR ALGEBRA 433 So, λ epesents the unit vecto in the diection of We wite it s â = 1 Note Fo ny scl k, k0= Components of vecto Let us tke the points A(1, 0, 0), B(0, 1, 0) nd C(0, 0, 1) on the x-xis, y-xis nd z-xis, espectively Then, clely uuu uuu uuu OA = 1, OB = 1nd OC = 1 uuu uuu uuu The vectos OA, OB nd OC, ech hving mgnitude 1, e clled unit vectos long the xes OX, OY nd OZ, espectively, nd denoted by iˆ, ˆj nd k ˆ, espectively (Fig 1013) Fig 1013 uuu Now, conside the position vecto OP of point P(x, y, z) s in Fig 1014 Let P1 be the foot of the pependicul fom P on the plne XOY We, thus, see tht P 1 P is not to be epublishe Fig 1014 pllel to z-xis As iˆ, ˆj nd k ˆ e the unit vectos long the x, y nd z-xes, uuu uuu espectively, nd by the definition of the coodintes of P, we hve PP ˆ 1 = OR= zk uuu uuu uuu Similly, QP ˆ 1 = OS = yj nd OQ = xiˆ

11 434 MATHEMATICS Theefoe, it follows tht uuu uuu uuu OP1 = OQ + QP ˆ ˆ 1 = xi + yj nd uuu uuu uuuu OP = OP ˆ ˆ ˆ 1 + P1P = xi + yj+ zk Hence, the position vecto of P with efeence to O is given by uuu OP (o ) = xiˆ+ yj ˆ+ zkˆ This fom of ny vecto is clled its component fom Hee, x, y nd z e clled s the scl components of, nd xiˆ, yj ˆ nd zk ˆ e clled the vecto components of long the espective xes Sometimes x, y nd z e lso temed s ectngul components The length of ny vecto = xiˆ+ yj ˆ+ zkˆ, is edily detemined by pplying the Pythgos theoem twice We note tht in the ight ngle tingle OQP 1 (Fig 1014) uuuu uuuu uuu OP = OQ + QP = x + y, 1 nd in the ight ngle tingle OP 1 P, we hve uuu uuu uuu OP = OP 1 PP 1 ( x y ) z Hence, the length of ny vecto = xiˆ+ yj ˆ+ zkˆ is given by = xiˆ+ yj ˆ+ zkˆ = x + y + z If nd b e ny two vectos given in the component fom i ˆ 1 + ˆ+ j k 3ˆ nd ˆ ˆ 1 3ˆ bi + b j+ bk, espectively, then (i) the sum (o esultnt) of the vectos nd b is given by + b = ( 1+ b ˆ ˆ ˆ 1) i + ( + b) j+ ( 3 + b3) k (ii) the diffeence of the vecto nd b is given by b = ( 1 b ˆ ˆ ˆ 1) i + ( b) j+ ( 3 b3) k (iii) the vectos nd b e equl if nd only if not to be epublishe (iv) 1 = b 1, = b nd 3 = b 3 the multipliction of vecto by ny scl λ is given by λ = ( ˆ ˆ ˆ 1) i ( ) j ( 3) k 1

12 VECTOR ALGEBRA 435 The ddition of vectos nd the multipliction of vecto by scl togethe give the following distibutive lws: Let nd b be ny two vectos, nd k nd m be ny scls Then (i) k + m = ( k + m) (ii) km ( ) = ( km ) (iii) k ( b) k kb Remks (i) One my obseve tht whteve be the vlue of λ, the vecto λ is lwys colline to the vecto In fct, two vectos nd b e colline if nd only if thee exists nonzeo scl λ such tht b =λ If the vectos nd b e given in the component fom, ie ˆ ˆ = i 1 + j+ 3ˆ k nd b = bi ˆ ˆ 1 + bj+ b3ˆ k, then the two vectos e colline if nd only if bi + b j+ bk = λ ( i ˆ ˆ 1 + j+ k 3ˆ ) ˆ ˆ 1 3ˆ bi ˆ 1 + b ˆ j+ bk 3ˆ = ( λ ˆ ˆ ˆ 1) i + ( λ ) j+ ( λ3) k b1 1 =λ, b =λ, b3 =λ3 b1 = b b3 = =λ 1 3 (ii) If i ˆ ˆ 1 j 3ˆ k, then 1,, 3 e lso clled diection tios of (iii) In cse if it is given tht l, m, n e diection cosines of vecto, then liˆ+ mj ˆ+ nkˆ = (cos α ) iˆ+ (cos β ) ˆj+ (cos γ ) kˆ is the unit vecto in the diection of tht vecto, whee α, β nd γ e the ngles which the vecto mkes with x, y nd z xes espectively Exmple 4 Find the vlues of x, y nd z so tht the vectos b = ˆ i + yj ˆ+ kˆ e equl = xiˆ+ ˆj+ zkˆ nd not to be epublishe Solution Note tht two vectos e equl if nd only if thei coesponding components e equl Thus, the given vectos nd b will be equl if nd only if x =, y =, z = 1

13 436 MATHEMATICS Exmple 5 Let = iˆ+ ˆ j nd b = iˆ+ ˆj Is = b? Ae the vectos nd b equl? Solution We hve = 1 + = 5 nd b 1 5 So, = b But, the two vectos e not equl since thei coesponding components e distinct Exmple 6 Find unit vecto in the diection of vecto = iˆ+ 3ˆj+ kˆ Solution The unit vecto in the diection of vecto is given by Now = = 14 Theefoe 1 ˆ = (iˆ+ 3 ˆj+ kˆ ) = iˆ+ 3 ˆj+ 1 kˆ ˆ = Exmple 7 Find vecto in the diection of vecto = iˆ ˆj tht hs mgnitude 7 units Solution The unit vecto in the diection of the given vecto is 1 ˆ = = 1 ( iˆ ˆj) = 1 iˆ ˆj Theefoe, the vecto hving mgnitude equl to 7 nd in the diection of is 7 = 1 7 i j 5 5 = 7 ˆ 14 i ˆj 5 5 Exmple 8 Find the unit vecto in the diection of the sum of the vectos, = iˆ+ ˆj 5kˆ nd b = iˆ+ ˆj+ 3kˆ not to be epublishe Solution The sum of the given vectos is b( c,sy)=4iˆ 3ˆj kˆ nd c = ( ) = 9

14 Thus, the equied unit vecto is 1 1 ˆ ˆ ˆ 4 ˆ 3 ˆ cˆ = c = (4i + 3 j k) = i + j kˆ c VECTOR ALGEBRA 437 Exmple 9 Wite the diection tio s of the vecto = iˆ+ ˆj kˆ nd hence clculte its diection cosines Solution Note tht the diection tio s, b, c of vecto = xiˆ+ yj ˆ+ zkˆ e just the espective components x, y nd z of the vecto So, fo the given vecto, we hve = 1, b = 1 nd c = Futhe, if l, m nd n e the diection cosines of the given vecto, then 1 b 1 c l = =, m= =, n= = s = Thus, the diection cosines e,, Vecto joining two points If P 1 (x 1, y 1, z 1 ) nd P (x, y, z ) e ny two points, then the vecto joining P 1 nd P uuuu is the vecto PP 1 (Fig 1015) Joining the points P 1 nd P with the oigin O, nd pplying tingle lw, fom the tingle OP 1 P, we hve uuu uuuu uuuu OP1+ P1P = OP Using the popeties of vecto ddition, the bove eqution becomes uuuu uuuu uuu PP 1 = OP OP1 uuuu ie PP 1 = ( xiˆ+ y ˆ ˆ ˆ ˆ ˆ j+ zk) ( xi 1 + y1j+ z1k) = ( x x ˆ ˆ ˆ 1) i + ( y y1) j+ ( z z1) k uuuu The mgnitude of vecto PP 1 is given by uuuu PP = ( x x ) + ( y y ) + ( z z ) Fig 1015 not to be epublishe

15 438 MATHEMATICS Exmple 10 Find the vecto joining the points P(, 3, 0) nd Q( 1,, 4) diected fom P to Q Solution Since the vecto is to be diected fom P to Q, clely P is the initil point nd Q is the teminl point So, the equied vecto joining P nd Q is the vecto PQ uuu, given by ie PQ uuu = PQ uuu = ( 1 ) iˆ+ ( 3) ˆj+ ( 4 0) kˆ 3iˆ 5ˆj 4 kˆ 1053 Section fomul uuu uuu Let P nd Q be two points epesented by the position vectos OP nd OQ, espectively, with espect to the oigin O Then the line segment joining the points P nd Q my be divided by thid point, sy R, in two wys intenlly (Fig 1016) nd extenlly (Fig 1017) Hee, we intend to find uuu the position vecto OR fo the point R with espect to the oigin O We tke the two cses one by one Cse I When R divides PQ intenlly (Fig 1016) If R divides PQ uuu such tht m RQ uuu = n PR uuu Fig 1016, whee m nd n e positive scls, we sy tht the point R divides PQ uuu intenlly in the tio of m : n Now fom tingles ORQ nd OPR, we hve RQ uuu uuu uuu = OQ OR = b nd PR uuu uuu uuu = OR OP =, Theefoe, we hve m( b ) = n( ) (Why?) o mb + n = (on simplifiction) m+ n Hence, the position vecto of the point R which divides P nd Q intenlly in the tio of m : n is given by uuu mb + n OR = m+ n not to be epublishe

16 VECTOR ALGEBRA 439 Cse II When R divides PQ extenlly (Fig 1017) We leve it to the ede s n execise to veify tht the position vecto of the point R which divides the line segment PQ extenlly in the tio PR m m : n ie is given by QR n uuu mb n OR = Fig 1017 m n Remk If R is the midpoint of PQ, then m = n And theefoe, fom Cse I, the midpoint R of PQ uuu, will hve its position vecto s uuu + b OR = uuu Exmple 11 Conside two points P nd Q with position vectos OP = 3 b nd uuu OQ b Find the position vecto of point R which divides the line joining P nd Q in the tio :1, (i) intenlly, nd (ii) extenlly Solution (i) The position vecto of the point R dividing the join of P nd Q intenlly in the tio :1 is uuu ( OR b ) (3 + + ) 5 = = (ii) The position vecto of the point R dividing the join of P nd Q extenlly in the tio :1 is uuu ( + b) (3 b) OR = = 4b 1 Exmple 1 Show tht the points A( iˆ ˆj kˆ), B( iˆ 3ˆj 5 kˆ), C(3iˆ 4 j 4 k ˆ) e the vetices of ight ngled tingle not to be epublishe Solution We hve uuu AB = BC uuu = nd CA uuu = (1 ) iˆ+ ( 3 + 1) ˆj+ ( 5 1) kˆ iˆ ˆj 6kˆ (3 1) iˆ+ ( 4+ 3) ˆj+ ( 4+ 5) kˆ = iˆ ˆj+ kˆ ( 3) iˆ+ ( 1+ 4) ˆj+ (1 + 4) kˆ = iˆ+ 3ˆj+ 5kˆ

17 440 MATHEMATICS Futhe, note tht uuu AB uuu uuu 41 = = BC + CA = Hence, the tingle is ight ngled tingle EXERCISE 10 1 Compute the mgnitude of the following vectos: ˆ ˆ ˆ ˆ ˆ 1 ˆ 1 ˆ 1 = i + j + k; b = i 7j 3 k; c = i + j kˆ Wite two diffeent vectos hving sme mgnitude 3 Wite two diffeent vectos hving sme diection 4 Find the vlues of x nd y so tht the vectos iˆ+ 3 ˆj nd xiˆ+ yj ˆ e equl 5 Find the scl nd vecto components of the vecto with initil point (, 1) nd teminl point ( 5, 7) 6 Find the sum of the vectos = iˆ ˆj+ kˆ, b = iˆ+ 4ˆj+ 5kˆ nd c= iˆ 6 ˆj 7kˆ 7 Find the unit vecto in the diection of the vecto = iˆ+ ˆj+ kˆ 8 uuu Find the unit vecto in the diection of vecto PQ, whee P nd Q e the points (1,, 3) nd (4, 5, 6), espectively 9 Fo given vectos, = iˆ ˆj+ kˆ nd b = iˆ+ ˆj kˆ, find the unit vecto in the diection of the vecto + b 10 Find vecto in the diection of vecto 5iˆ ˆj+ kˆ which hs mgnitude 8 units 11 Show tht the vectos iˆ 3ˆj+ 4 kˆ nd 4iˆ+ 6ˆj 8kˆ e colline 1 Find the diection cosines of the vecto iˆ+ ˆj+ 3kˆ 13 Find the diection cosines of the vecto joining the points A (1,, 3) nd B( 1,, 1), diected fom A to B 14 Show tht the vecto iˆ+ ˆj+ kˆ is eqully inclined to the xes OX, OY nd OZ 15 Find the position vecto of point R which divides the line joining two points P nd Q whose position vectos e iˆ+ ˆj kˆ nd iˆ+ ˆj+ kˆ espectively, in the tio : 1 (i) intenlly (ii) extenlly not to be epublishe

18 VECTOR ALGEBRA Find the position vecto of the mid point of the vecto joining the points P(, 3, 4) nd Q(4, 1, ) 17 Show tht the points A, B nd C with position vectos, = 3iˆ 4ˆj 4 kˆ, b = ˆ i ˆj+ kˆ nd c = iˆ 3ˆj 5kˆ, espectively fom the vetices of ight ngled tingle 18 In tingle ABC (Fig 1018), which of the following is not tue: uuu uuuu uuu (A) AB + BC + CA = 0 uuu uuu uuu (B) AB + BC AC = 0 uuu uuu uuu (C) AB + BC CA = 0 uuu uuu uuu (D) Fig 1018 AB CB + CA = 0 19 If nd b e two colline vectos, then which of the following e incoect: (A) b = λ, fo some scl λ (B) = ± b (C) the espective components of nd b e not popotionl (D) both the vectos nd b hve sme diection, but diffeent mgnitudes 106 Poduct of Two Vectos So f we hve studied bout ddition nd subtction of vectos An othe lgebic opetion which we intend to discuss egding vectos is thei poduct We my ecll tht poduct of two numbes is numbe, poduct of two mtices is gin mtix But in cse of functions, we my multiply them in two wys, nmely, multipliction of two functions pointwise nd composition of two functions Similly, multipliction of two vectos is lso defined in two wys, nmely, scl (o dot) poduct whee the esult is scl, nd vecto (o coss) poduct whee the esult is vecto Bsed upon these two types of poducts fo vectos, they hve found vious pplictions in geomety, mechnics nd engineeing In this section, we will discuss these two types of poducts not to be epublishe 1061 Scl (o dot) poduct of two vectos Definition The scl poduct of two nonzeo vectos nd b, denoted by b, is

19 44 MATHEMATICS defined s b = b cos θ, whee, θ is the ngle between nd b, 0 (Fig 1019) If eithe = 0ob = 0, then θ is not defined, nd in this cse, we define b 0 Obsevtions 1 b is el numbe Let nd b be two nonzeo vectos, then b = 0 if nd only if nd b e pependicul to ech othe ie b = 0 b 3 If θ = 0, then b = b In pticul, =, s θ in this cse is 0 4 If θ = π, then b = b In pticul, ( ), s θ in this cse is π 5 In view of the Obsevtions nd 3, fo mutully pependicul unit vectos iˆ, ˆj nd k ˆ, we hve iˆ iˆ= ˆj ˆj = k ˆ k ˆ = 1, iˆ ˆj = ˆj kˆ = k ˆ i ˆ 0 6 The ngle between two nonzeo vectos nd b is given by Fig 1019 b 1 b cos, o θ = cos b b 7 The scl poduct is commuttive ie b = b (Why?) Two impotnt popeties of scl poduct Popety 1 (Distibutivity of scl poduct ove ddition) Let, b nd c be ny thee vectos, then ( b + c) = b + c not to be epublishe

20 A C (i) B θ p C 0 0 (0 < θ < 90 ) B 0 0 (180 < θ < 70 ) (iii) p A θ l l θ A VECTOR ALGEBRA 443 Popety Let nd b be ny two vectos, nd λ be ny scl Then ( λ) b = ( ) b ( b) ( b) If two vectos nd b e given in component fom s i 1 ˆ+ ˆj + k 3ˆ nd ˆ ˆ 1 3ˆ bi + b j+ bk, then thei scl poduct is given s b ( iˆ+ ˆj + kˆ) ( biˆ+ b ˆj + bkˆ) Thus = iˆ ( biˆ+ b ˆj + bkˆ) + ˆj ( biˆ+ b ˆj + bkˆ) + kˆ 3 ( bi ˆ 1 ˆ+ b ˆj + bk 3 ) = b( iˆ iˆ) + b ( iˆ ˆj ) + b ( iˆ kˆ) + b( ˆj iˆ) + b ( ˆj ˆj ) + b ( ˆj kˆ) = b 3 1( kˆ iˆ) + b ˆ ˆ ˆ ˆ 3 ( k j) + b 3 3( k k) (Using the bove Popeties 1 nd ) = 1 b 1 + b + 3 b 3 (Using Obsevtion 5) b b + b + b = Pojection of vecto on line uuu Suppose vecto AB mkes n ngle θ with given diected line l (sy), in the uuu nticlockwise diection (Fig 100) Then the pojection of AB on l is vecto p uuu (sy) with mgnitude AB cosθ, nd the diection of p being the sme (o opposite) to tht of the line l, depending upon whethe cosθ is positive o negtive The vecto p B (ii) C θ C A p 0 0 (90< θ <180) not to be epublishe Fig 100 p B 0 0 (70 < θ < 360 ) (iv) l l

21 444 MATHEMATICS is clled the pojection vecto, nd its mgnitude p is simply clled s the pojection uuu of the vecto AB on the diected line l Fo exmple, in ech of the following figues (Fig 100(i) to (iv)), pojection vecto uuu uuu of AB long the line l is vecto AC Obsevtions 1 If ˆp is the unit vecto long line l, then the pojection of vecto on the line l is given by pˆ Pojection of vecto on othe vecto b, is given by b 1 b ˆ, o, o ( b) b b uuu uuu 3 If θ = 0, then the pojection vecto of AB will be AB itself nd if θ = π, then the uuu uuu pojection vecto of AB will be BA π 3π uuu 4 If θ = o θ =, then the pojection vecto of AB will be zeo vecto Remk If α, β nd γ e the diection ngles of vecto = i ˆ ˆ 1 + j+ 3ˆ k, then its diection cosines my be given s iˆ 1 3 cos, cos, nd cos ˆ i Also, note tht cos α, cos β nd cosγ e espectively the pojections of long OX, OY nd OZ ie, the scl components 1, nd 3 of the vecto, e pecisely the pojections of long x-xis, y-xis nd z-xis, espectively Futhe, if is unit vecto, then it my be expessed in tems of its diection cosines s = cosα iˆ+ cosβ ˆj+ cos γkˆ Exmple 13 Find the ngle between two vectos nd b with mgnitudes 1 nd espectively nd when b= 1 Solution Given b 1, 1 nd b We hve not to be epublishe cos b b cos 3

22 VECTOR ALGEBRA 445 Exmple 14 Find ngle θ between the vectos = iˆ+ ˆj kˆ nd b= iˆ ˆj+ kˆ Solution The ngle θ between two vectos nd b is given by b cosθ = b Now b = ( iˆ+ ˆj kˆ) ( iˆ ˆj+ kˆ) = 1 1 1= 1 Theefoe, we hve cosθ = hence the equied ngle is θ = cos 3 Exmple 15 If = 5iˆ ˆj 3 kˆ nd b= iˆ+ 3 ˆj 5 kˆ, then show tht the vectos + b nd b e pependicul Solution We know tht two nonzeo vectos e pependicul if thei scl poduct is zeo Hee + b = (5iˆ ˆj 3 kˆ) + ( iˆ+ 3ˆj 5 kˆ) = 6iˆ+ ˆj 8 kˆ nd b = (5iˆ ˆj 3 kˆ) ( iˆ+ 3ˆj 5 kˆ) = 4iˆ 4ˆj+ kˆ So ( + b) ( b) = (6iˆ+ ˆj 8 kˆ) (4iˆ 4ˆj+ kˆ) = = 0 Hence + b nd b e pependicul vectos Exmple 16 Find the pojection of the vecto = iˆ+ 3ˆj+ kˆ on the vecto b = iˆ+ ˆj+ kˆ Solution The pojection of vecto on the vecto b is given by 1 ( b ( ) 10 5 ) = = = 6 b (1) + () + (1) 6 3 Exmple 17 Find b, if two vectos nd b e such tht, b 3 nd b = 4 not to be epublishe Solution We hve b = ( b) ( b) = b b + b b

23 446 MATHEMATICS = ( b) + b = () (4) + (3) Theefoe b = 5 Exmple 18 If is unit vecto nd ( x ) ( x+ ) = 8, then find x Solution Since is unit vecto, = 1 Also, ( x ) ( x + ) =8 o x x+ x x =8 o x 1 = 8 ie x = 9 Theefoe x = 3 (s mgnitude of vecto is non negtive) Exmple 19 Fo ny two vectos nd b, we lwys hve b b (Cuchy- Schwtz inequlity) Solution The inequlity holds tivilly when eithe = 0 o b = 0 Actully, in such sitution we hve b = 0 = b So, let us ssume tht 0 b Then, we hve b = cos θ 1 b Theefoe b b Exmple 0 Fo ny two vectos nd b, we lwys hve + b + b (tingle inequlity) + b A Solution The inequlity holds tivilly in cse eithe B = 0ob = 0 (How?) So, let 0 b Then, + b Fig 101 = ( + b) = ( + b) ( + b) = + b + b + b b = + b + b (scl poduct is commuttive) + b + b (since x x x R ) + b + b (fom Exmple 19) = ( b ) not to be epublishe b C

24 Hence b b VECTOR ALGEBRA 447 Remk If the equlity holds in tingle inequlity (in the bove Exmple 0), ie + b = + b, uuu uuu uuu then AC = AB + BC showing tht the points A, B nd C e colline Exmple 1 Show tht the points A( iˆ+ 3ˆj+ 5 kˆ), B( iˆ+ ˆj+ 3 kˆ) nd C(7 iˆ kˆ ) e colline Solution We hve Theefoe uuu AB = (1 ) iˆ ( 3) ˆj (3 5) kˆ 3iˆ ˆj k, ˆ BC (7 1) iˆ (0 ) ˆj ( 1 3) k ˆ 6iˆ ˆj 4k, ˆ AC (7 ) iˆ (0 3) ˆj ( 1 5) k ˆ 9iˆ 3 ˆj 6k ˆ uuu AB = uuu 14, BC uuu 14 nd AC 3 14 uuu AC uuu uuu = AB + BC Hence the points A, B nd C e colline uuu Note In Exmple 1, one my note tht lthough AB + uuu BC + uuu CA = 0 but the points A, B nd C do not fom the vetices of tingle EXERCISE Find the ngle between two vectos nd b with mgnitudes 3 nd, espectively hving b = 6 Find the ngle between the vectos ˆ iˆ ˆj+ 3 k nd 3iˆ ˆj+ kˆ 3 Find the pojection of the vecto iˆ ˆj on the vecto iˆ+ ˆj not to be epublishe 4 Find the pojection of the vecto iˆ+ 3ˆj+ 7kˆ on the vecto 7iˆ ˆj+ 8kˆ 5 Show tht ech of the given thee vectos is unit vecto: 1 ˆ 1 ˆ 1 (iˆ+ 3 ˆj+ 6 k), (3iˆ 6 ˆj+ k), (6iˆ+ ˆj 3 kˆ) Also, show tht they e mutully pependicul to ech othe

25 448 MATHEMATICS 6 Find nd b, if ( + b) ( b) = 8 nd = 8 b 7 Evlute the poduct (3 5 b) (+ 7 b) 8 Find the mgnitude of two vectos nd b, hving the sme mgnitude nd such tht the ngle between them is 60 o nd thei scl poduct is 1 9 Find x, if fo unit vecto, ( x ) ( x+ ) = 1 10 If ˆ ˆ 3 ˆ, ˆ ˆ ˆ = i + j+ k b = i + j+ k nd c = 3iˆ+ ˆ j e such tht +λb pependicul to c, then find the vlue of λ 11 Show tht b+ b is pependicul to b b, fo ny two nonzeo vectos nd b 1 If = 0 nd b = 0, then wht cn be concluded bout the vecto b? 13 If bc,, e unit vectos such tht + b + c = 0, find the vlue of b + b c + c 14 If eithe vecto = 0 o b = 0, then b = 0 But the convese need not be tue Justify you nswe with n exmple 15 If the vetices A, B, C of tingle ABC e (1,, 3), ( 1, 0, 0), (0, 1, ), espectively, then find ABC [ ABC is the ngle between the vectos BA uuu nd BC uuu ] 16 Show tht the points A(1,, 7), B(, 6, 3) nd C(3, 10, 1) e colline 17 Show tht the vectos iˆ ˆj+ kˆ, iˆ 3ˆj 5kˆ nd 3iˆ 4ˆj 4kˆ fom the vetices of ight ngled tingle 18 If is nonzeo vecto of mgnitude nd λ nonzeo scl, then λ is unit vecto if not to be epublishe (A) λ = 1 (B) λ = 1 (C) = λ (D) = 1/ λ 1063 Vecto (o coss) poduct of two vectos In Section 10, we hve discussed on the thee dimensionl ight hnded ectngul coodinte system In this system, when the positive x-xis is otted counteclockwise is

26 VECTOR ALGEBRA 449 into the positive y-xis, ight hnded (stndd) scew would dvnce in the diection of the positive z-xis (Fig 10(i)) In ight hnded coodinte system, the thumb of the ight hnd points in the diection of the positive z-xis when the finges e culed in the diection wy fom the positive x-xis towd the positive y-xis (Fig 10(ii)) Fig 10 (i), (ii) Definition 3 The vecto poduct of two nonzeo vectos nd b, is denoted by b nd defined s b = b sinθ nˆ, whee, θ is the ngle between nd b, 0 θ π nd ˆn is unit vecto pependicul to both nd b, such tht b, nd nˆ fom ight hnded system (Fig 103) ie, the ight hnded system otted fom to b moves in the Fig 103 diection of ˆn If eithe = 0ob = 0, then θ is not defined nd in this cse, we define b = 0 Obsevtions 1 b is vecto Let nd b be two nonzeo vectos Then b = 0 if nd only if e pllel (o colline) to ech othe, ie, b = 0 b not to be epublishe nd b

27 450 MATHEMATICS In pticul, = 0 nd ( ) = 0, since in the fist sitution, θ = 0 nd in the second one, θ = π, mking the vlue of sin θ to be 0 3 If then b b 4 In view of the Obsevtions nd 3, fo mutully pependicul unit vectos iˆ, ˆj nd k ˆ (Fig 104), 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 In tems of vecto poduct, the ngle between two vectos nd b my be given s sin θ = b b 6 It is lwys tue tht the vecto poduct is not commuttive, s b = b Indeed, b = b sinθnˆ, whee b, nd nˆ fom ight hnded system, ie, θ is tvesed fom to b, Fig 105 (i) While, = sin, whee 1 b, ndn fom ight hnded system ie θ is tvesed fom ˆ1 b to, Fig 105(ii) not to be epublishe Fig 105 (i), (ii) Thus, if we ssume nd b to lie in the plne of the ppe, then nˆ nd n ˆ1 both will be pependicul to the plne of the ppe But, ˆn being diected bove the ppe while ˆn 1 diected below the ppe ie nˆ ˆ 1 = n

28 Hence b = b sin nˆ b sin θnˆ 7 In view of the Obsevtions 4 nd 6, we hve = 1 VECTOR ALGEBRA 451 = b ˆj iˆ= kˆ, kˆ ˆj = iˆ nd iˆ kˆ= ˆj 8 If nd b epesent the djcent sides of tingle then its e is given s 1 b By definition of the e of tingle, we hve fom Fig 106, Ae of tingle ABC = 1 AB CD But AB = b (s given), nd CD = Fig 106 sinθ Thus, Ae of tingle ABC = 1 sin b θ = 1 b 9 If nd b epesent the djcent sides of pllelogm, then its e is given by b Fom Fig 107, we hve Ae of pllelogm ABCD = AB DE But AB = b (s given), nd DE = sin θ Thus, Fig Ae of pllelogm ABCD = b sinθ = b 107 We now stte two impotnt popeties of vecto poduct Popety 3 (Distibutivity of vecto poduct ove ddition): If, b nd c e ny thee vectos nd λ be scl, then (i) ( b + c) = b c (ii) λ ( b ) = ( λ ) b = ( λb) not to be epublishe

29 45 MATHEMATICS Let nd b be two vectos given in component fom s i 1 ˆ+ ˆj + k 3ˆ nd ˆ ˆ 1 3ˆ bi + b j+ bk, espectively Then thei coss poduct my be given by b iˆ ˆj kˆ = 1 3 b b b 1 3 Explntion We hve b = ( iˆ 1 + ˆ ˆ ˆ ˆ ˆ j+ k 3 ) ( bi 1 + bj+ bk 3 ) b( iˆ iˆ) + b ( iˆ ˆj ) + b ( iˆ kˆ ) + b( ˆj iˆ) = b( ˆj ˆj) + b( ˆj kˆ ) + 3 b( kˆ iˆ) + b( kˆ ˆj) + b( kˆ kˆ) (by Popety 1) b ( iˆ ˆj ) b ( kˆ iˆ) b( iˆ ˆj ) = 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 1 3 b 3 1) j+ ( b 1 b 1) k iˆ ˆj kˆ = 1 3 b1 b b3 Exmple Find b, if = iˆ+ ˆj+ 3 kˆ nd b = 3iˆ+ 5 ˆj kˆ Solution We hve b = iˆ ˆj kˆ not to be epublishe = iˆ( 15) ( 4 9) ˆj+ (10 3) kˆ = 17iˆ+ 13 ˆj+ 7kˆ Hence b = ( 17) + (13) + (7) = 507

30 VECTOR ALGEBRA 453 Exmple 3 Find unit vecto pependicul to ech of the vectos ( + b ) nd ( b ), whee = iˆ+ ˆj+ kˆ, b = iˆ+ ˆj+ 3kˆ Solution We hve ˆ 3ˆ 4ˆ + b = i + j+ k nd b = ˆj kˆ A vecto which is pependicul to both + b nd b is given by ( + b) ( b) = iˆ ˆj kˆ 3 4 ˆ 4ˆ ˆ = i + j k ( = c, sy) 0 1 Now c = = 4 = 6 Theefoe, the equied unit vecto is c 1 1 = i ˆ + ˆ j k ˆ c Note Thee e two pependicul diections to ny plne Thus, nothe unit vecto pependicul to + b nd b will be be consequence of ( b) ( + b) 1 1 i ˆ ˆ j+ k ˆ But tht will Exmple 4 Find the e of tingle hving the points A(1, 1, 1), B(1,, 3) nd C(, 3, 1) s its vetices uuu uuu Solution We hve AB = ˆj+ kˆ nd AC = iˆ+ ˆj The e of the given tingle is 1 uuu uuu AB AC Now, Theefoe Thus, the equied e is 1 1 iˆ ˆj kˆ uuu uuu AB AC = 0 1 = 4iˆ+ ˆj kˆ 1 0 not to be epublishe uuu uuu AB AC = = 1

31 454 MATHEMATICS Exmple 5 Find the e of pllelogm whose djcent sides e given by the vectos = 3iˆ+ ˆj+ 4kˆ nd b = iˆ ˆj+ kˆ Solution The e of pllelogm with nd b s its djcent sides is given by b Now b iˆ ˆj kˆ = = 5iˆ+ ˆj 4kˆ Theefoe b = = 4 nd hence, the equied e is 4 EXERCISE Find b, if = iˆ 7ˆj+ 7kˆ nd b = 3iˆ ˆj+ kˆ Find unit vecto pependicul to ech of the vecto + b nd b, whee = 3iˆ+ ˆj+ kˆ nd b = iˆ+ ˆj kˆ 3 If unit vecto 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 vectos nd b? 7 Let the vectos b,, c be given s i 1 ˆ+ ˆ ˆ ˆj + k 3, bi 1 ˆ+ b ˆj + bk 3, ˆ ˆ ci 1 + cj+ ck 3ˆ Then show tht ( b + c) = b + c 8 If eithe = 0 o b = 0, then b = 0 Is the convese tue? Justify you nswe with n exmple 9 Find the e of the tingle with vetices A(1, 1, ), B(, 3, 5) nd C(1, 5, 5) not to be epublishe

32 VECTOR ALGEBRA Find the e of the pllelogm whose djcent sides e detemined by the vectos = iˆ ˆj+ 3kˆ nd b = iˆ 7ˆj+ kˆ 11 Let the vectos nd b be such tht = 3 nd b =, then b is 3 unit vecto, if the ngle between nd b is (A) π/6 (B) π/4 (C) π/3 (D) π/ 1 Ae of ectngle hving vetices A, B, C nd D with position vectos 1 ˆ 1 iˆ+ ˆj+ 4 k, iˆ+ ˆj+ 4kˆ, ˆ 1 ˆ ˆ 1 i j+ 4k nd iˆ ˆj+ 4kˆ, espectively is 1 (A) (B) 1 (C) (D) 4 Miscellneous Exmples Exmple 6 Wite ll the unit vectos in XY-plne Solution Let x = i+ y j be unit vecto in XY-plne (Fig 108) Then, fom the figue, we hve x = cos θ nd y = sin θ (since = 1) So, we my wite the vecto s ( = OP uuu ) = cos iˆ sin ˆj (1) Clely, = cos θ+ sin θ= 1 not to be epublishe Fig 108 Also, s θ vies fom 0 to π, the point P (Fig 108) tces the cicle x + y = 1 counteclockwise, nd this coves ll possible diections So, (1) gives evey unit vecto in the XY-plne

33 456 MATHEMATICS Exmple 7 If iˆ ˆj kˆ, iˆ 5 ˆj, 3iˆ ˆj 3 kˆ nd iˆ 6 ˆj k ˆ e the position uuu vectos of points A, B, C nd D espectively, then find the ngle between AB nd CD uuu uuu uuu Deduce tht AB nd CD e colline Solution Note tht if θ is the ngle between AB nd CD, then θ is lso the ngle uuu uuu between AB nd CD uuu Now AB = Position vecto of B Position vecto of A = (iˆ+ 5 ˆj) ( iˆ+ ˆj+ kˆ) = iˆ+ 4 ˆj kˆ uuu Theefoe AB = (1) + (4) + ( 1) = 3 Similly CD uuu uuu = iˆ 8 ˆj+ kˆ nd CD = 6 uuu uuu AB CD Thus cos θ = uuu uuu AB CD = 1( ) + 4( 8) + ( 1)() = 36 = 1 (3 )(6 ) 36 uuu uuu Since 0 θ π, it follows tht θ = π This shows tht AB nd CD e colline uuu 1 uuu uuu uuu Altentively, AB CD which implies tht AB nd CD e colline vectos Exmple 8 Let b, nd c be thee vectos such tht 3, b 4, = = c = 5 nd ech one of them being pependicul to the sum of the othe two, find + b + c Solution Given ( b + c ) = 0, ( b c ) 0, + = c ( + b ) = 0 Now + b + c = ( + b + c) = ( + b + c) ( + b + c) = + ( b + c) + b b + b ( + c) + c( + b) + cc = + b + c = = 50 Theefoe + b + c = 50 = 5 not to be epublishe

34 VECTOR ALGEBRA 457 Exmple 9 Thee vectos, b nd c stisfy the condition + b + c = 0 Evlute the quntity μ= b + b c + c, if = 1, b = 4 nd c = Solution Since + b + c = 0, we hve ( b c ) =0 o + b+ c =0 Theefoe b+ c = = 1 (1) b + b + c Agin, ( ) =0 o b+ b c = b = 16 () Similly c + b c = 4 (3) Adding (1), () nd (3), we hve ( b+ b c+ c) = 1 o μ = 1, ie, μ = 1 Exmple 30 If with efeence to the ight hnded system of mutully pependicul unit vectos ˆ ˆ ˆ i, j nd k, α= 3 iˆ ˆj, β= iˆ+ ˆj 3kˆ, then expess β in the fom 1,whee is pllel to 1 nd is pependicul to α Solution Let 1, is scl, ie, β ˆ ˆ 1 = 3λi λj Now β =β β = ˆ ˆ ˆ 1 ( 3 λ ) i + (1 +λ) j 3k Now, since β is to be pependicul to α, we should hve α β = 0 ie, 3( 3 λ) (1 +λ ) =0 not to be epublishe o λ = 1 Theefoe β = 3 1 ˆ ˆ 1 3 i j nd β ˆ ˆ 1 = i + ˆ 3 j k

35 458 MATHEMATICS Miscellneous Execise on Chpte 10 1 Wite down unit vecto in XY-plne, mking n ngle of 30 with the positive diection of x-xis Find the scl components nd mgnitude of the vecto joining the points P(x 1, y 1, z 1 ) nd Q(x, y, z ) 3 A gil wlks 4 km towds west, then she wlks 3 km in diection 30 est of noth nd stops Detemine the gil s displcement fom he initil point of deptue 4 If = b + c = b + c? Justify you nswe 5 Find the vlue of x fo which x( iˆ+ ˆj+ kˆ ) is unit vecto 6 Find vecto of mgnitude 5 units, nd pllel to the esultnt of the vectos = 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 vecto pllel to the vecto b + 3c 8 Show tht the points A(1,, 8), B(5, 0, ) nd C(11, 3, 7) e colline, nd find the tio in which B divides AC 9 Find the position vecto of point R which divides the line joining two points P nd Q whose position vectos e ( + b ) nd ( 3 b ) extenlly in the tio 1 : Also, show tht P is the mid point of the line segment RQ 10 The two djcent sides of pllelogm e iˆ 4 ˆj+ 5 kˆ nd iˆ ˆj 3kˆ Find the unit vecto pllel to its digonl Also, find its e 11 Show tht the diection cosines of vecto eqully inclined to the xes OX, OY nd OZ e 1, 1, Let ˆ ˆ ˆ ˆ ˆ ˆ = i + 4j+ k, b = 3i j+ 7 k nd c = iˆ ˆj+ 4kˆ Find vecto d which is pependicul to both nd b, nd c d = The scl poduct of the vecto iˆ+ ˆj+ kˆ with unit vecto long the sum of vectos iˆ+ 4ˆj 5kˆ nd λ iˆ+ ˆj+ 3kˆ is equl to one Find the vlue of λ 14 If, b, c e mutully pependicul vectos of equl mgnitudes, show tht the vecto + b + c is eqully inclined to, b nd c not to be epublishe

36 VECTOR ALGEBRA Pove tht ( + b) ( + b) = + b, if nd only if, b e pependicul, given 0, b 0 Choose the coect nswe in Execises 16 to If θ is the ngle between two vectos nd b, then b 0 only when π π (A) 0 <θ< (B) 0 θ (C) 0 < θ < π (D) 0 θ π 17 Let nd b be two unit vectos nd θ is the ngle between them Then + b is unit vecto if (A) π θ= (B) 4 π θ = 3 (C) π θ = (D) π θ= 3 18 The vlue of iˆ( ˆj kˆ) ˆj ( iˆ kˆ) kˆ ( iˆ ˆj ) is (A) 0 (B) 1 (C) 1 (D) 3 19 If θ is the ngle between ny two vectos nd b, then b = b when θ is equl to (A) 0 (B) π 4 (C) π (D) π Summy uuu Position vecto of point P(x, y, z) is given s OP( = ) = xiˆ+ yj ˆ+ zkˆ, nd its mgnitude by x + y + z The scl components of vecto e its diection tios, nd epesent its pojections long the espective xes The mgnitude (), diection tios (, b, c) nd diection cosines (l, m, n) of ny vecto e elted s: not to be epublishe b c l =, m=, n= The vecto sum of the thee sides of tingle tken in ode is 0

37 460 MATHEMATICS The vecto sum of two coinitil vectos is given by the digonl of the pllelogm whose djcent sides e the given vectos The multipliction of given vecto by scl λ, chnges the mgnitude of the vecto by the multiple λ, nd keeps the diection sme (o mkes it opposite) ccoding s the vlue of λ is positive (o negtive) Fo given vecto, the vecto ˆ = gives the unit vecto in the diection of The position vecto of point R dividing line segment joining the points P nd Q whose position vectos e nd b espectively, in the tio m : n n + mb (i) intenlly, is given by m+ n mb n (ii) extenlly, is given by m n The scl poduct of two given vectos nd b hving ngle θ between them is defined s b = b cosθ Also, when b is given, the ngle θ between the vectos nd b my be detemined by b cosθ = b If θ is the ngle between two vectos nd b, then thei coss poduct is given s b = b sinθnˆ whee ˆn is unit vecto pependicul to the plne contining nd b Such tht bnfom,, ˆ ight hnded system of coodinte xes If we hve two vectos nd b, given in component fom s i ˆ ˆ = + j+ k nd b = bi ˆ ˆ 1 + bj+ b3ˆ k nd λ ny scl, 1 3ˆ not to be epublishe

38 then + b nd b ( + b ) iˆ+ ( + b ) ˆj+ ( + b ) kˆ ; = λ = ( λ ˆ ˆ ˆ 1) i + ( λ ) j+ ( λ 3) k; b b + b + b ; = iˆ ˆj kˆ b c = b c Histoicl Note VECTOR ALGEBRA 461 The wod vecto hs been deived fom Ltin wod vectus, which mens to cy The geminl ides of moden vecto theoy dte fom ound 1800 when Csp Wessel ( ) nd Jen Robet Agnd ( ) descibed tht how complex numbe + ib could be given geometic intepettion with the help of diected line segment in coodinte plne Willim Rowen Hmilton ( ) n Iish mthemticin ws the fist to use the tem vecto fo diected line segment in his book Lectues on Qutenions (1853) Hmilton s method of qutenions (n odeed set of fou el numbes given s: + biˆ+ cj ˆ+ dkˆ, iˆ, ˆj, kˆ following cetin lgebic ules) ws solution to the poblem of multiplying vectos in thee dimensionl spce Though, we must mention hee tht in pctice, the ide of vecto concept nd thei ddition ws known much elie eve since the time of Aistotle (384-3 BC), Geek philosophe, nd pupil of Plto ( BC) Tht time it ws supposed to be known tht the combined ction of two o moe foces could be seen by dding them ccoding to pllelogm lw The coect lw fo the composition of foces, tht foces dd vectoilly, hd been discoveed in the cse of pependicul foces by Stevin-Simon ( ) In 1586 AD, he nlysed the pinciple of geometic ddition of foces in his tetise DeBeghinselen de Weeghconst ( Pinciples of the At of Weighing ), which cused mjo bekthough in the development of mechnics But it took nothe 00 yes fo the genel concept of vectos to fom In the 1880, Josih Willd Gibbs ( ), n Ameicn physicist nd mthemticin, nd Olive Heviside ( ), n English enginee, ceted wht we now know s vecto nlysis, essentilly by septing the el (scl) not to be epublishe

39 46 MATHEMATICS pt of qutenion fom its imginy (vecto) pt In 1881 nd 1884, Gibbs pinted tetise entitled Element of Vecto Anlysis This book gve systemtic nd concise ccount of vectos Howeve, much of the cedit fo demonstting the pplictions of vectos is due to the D Heviside nd PG Tit ( ) who contibuted significntly to this subject not to be epublishe

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