, where P is called the tail and Q is called the nose of the vector.

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1 6 VECTORS 1 INTRODUCTION TO VECTOR ALGEBRA 11 Slrs nd Vetors Slr: A slr is quntity tht hs only mgnitude ut no diretion Slr quntity is expressed s single numer, followed y pproprite unit, eg length, re, mss, et In liner lger, rel numers re lled slrs Vetor: A vetor is quntity tht hs oth mgnitude nd diretion, eg displement, veloity, et 1 Representtion of Vetors () A vetor is represented digrmmtilly y direted line segment or n rrow A direted line segment hs oth mgnitude (length) nd diretion The length is denoted y V () If P nd Q re the given two points, then the vetor from P to Q is denoted y PQ, where P is lled the til nd Q is lled the nose of the vetor 13 Vetor Components In two-dimensionl oordinte system, ny vetor n e resolved into x-omponent nd y-omponent v = v,v x y Let us onsider the figure shown (djent) here In this figure, the omponents n e quikly red The vetor in the omponent form is v = 4,5 The reltion etween mgnitude of the vetor nd the omponents of the vetor n e lulted y using trigonometri rtios os v x Adjent side θ= = Hypotenuse v V V y v y Opposite side sin θ= = Hypotenuse v v = v os θ ; v = v sin θ x y V Figure 61 If vx y nd v re the known lengths of right tringle, then the length of the hypotenuse, V, is lulted y using the Pythgoren theorem x y v = v + v

2 6 Vetors TYPE OF VECTORS 1 Null Vetor/Zero Vetor A zero vetor or null vetor is vetor tht hs zero mgnitude, ie initil nd terminl points re oinident, so tht its diretion is in indeterminte form It is denoted y φ Unit Vetor A unit vetor is vetor of unit length A unit vetor is sometimes denoted y repling the rrow on vetor with ^ Unit vetors prllel to x-xis, y-xis nd z-xis re denoted y î, ĵ nd ˆk, respetively Unit vetor Û prllel to V V n e otined s Û = V Illustrtion 1: Find unit vetor of i J + 3k Sol: Here unit vetor of is given y ˆ = = i J + 3k If ˆ ˆ ˆ = i x + y j+ k z then it s mgnitude = x + y + = 14 1 â = = ( i J + 3k ) z Colliner or Prllel Vetors Two or more vetors re sid to e olliner, when they re long the sme lines or prllel lines irrespetive of their mgnitudes nd diretions 4 Like nd Unlike Vetors Vetors hving the sme diretion re lled like vetors Any two vetors prllel to one nother, hving unequl mgnitudes nd ting in opposite diretions re lled unlike vetors 5 Co-Initil Vetors All those vetors whose terminl points re sme, re lled o-terminl vetors 6 Co-Terminl Vetors Vetors tht hve the sme initil points re lled o-initil vetors Illustrtion : Whih re o-initil nd equl vetors in the given retngle digrm? Sol: By following ove mentioned onditions we n otin o-initil nd equl vetors A B d C Figure 6 D

3 Mthemtis 63 Here, nd re o-initil vetors, nd, nd d 7 Coplnr Vetors Vetors lie on the sme plne re lled oplnr 8 Negtive vetor re equl vetors A vetor tht points to diretion opposite to tht of the given vetor is lled negtive vetor 9 Reiprol of Vetor A vetor hving the sme diretion s tht of given vetor, ut mgnitude equl to the reiprol of the given vetor is known s the reiprol of nd is denoted y 1 10 Lolized nd Free vetors A vetor drwn prllel to given vetor through speified point unlike free vetor in spe is lled lolized vetor For exmple, the effet of fore ting on rigid ody depends not only on the mgnitude nd diretion ut lso on the line of tion of the fore A vetor tht depends only on its length nd diretion nd not on its position in the spe is lled free vetor, eg grvity In this hpter, we will del with free vetors, unless otherwise stted Thus free vetor n e determined in spe y hoosing n ritrry initil point llustrtion 3: Let = ˆi + j ˆ nd ˆ i ˆ ˆ = + j Is =? Are the vetors nd ˆ equl? Sol: Two vetors re equl if their modulus nd orresponding omponents oth re equl We hve = 1 + = 5 nd = + 1 So, = But, the two vetors re not equl, sine their orresponding omponents re distint Illustrtion 4: Find vetor of mgnitude 5 units whih is prllel to the vetor i ˆ ˆj Sol: As we know ˆ =, therefore required vetor will e 5 ˆ Let i ˆ ˆ = j Then, = + ( 1) = 5 Unit vetor prllel to ˆ 1 1 ˆ ˆ ˆ 1 = = = i j = i ˆj So, the required vetor is ˆ ˆ 5ˆ = 5 i j = 5i ˆ 5ˆj Position Vetor A vetor tht represents the position of point P in spe with respet to n ritrry referene origin O is lled position vetor (pv) It is lso known s lotion vetor or rdius vetor nd usully denoted s x, r or s; it orresponds to the displement from O to P r = OP

4 64 Vetors Illustrtion 5: Show tht, the three points A(-,3,5), B(1,,3) nd C(7,0,-1) re olliner Sol: By otining AB nd BC, we n onlude tht given points re olliner or not We hve AB = OB OA = i + j+ 3k i + 3 j+ 5k = 3 i j k BC = OC OB = 7 i + 0 j k i + j+ 3k = 6 i j 4k = 3 i j k Therefore, BC = AB This shows tht the vetors AB nd BC re prllel But, B is ommon point So, the given point A, B nd C re olliner 1 Equl Vetors Two vetors hving the sme orresponding omponents nd diretion nd represent the sme physil quntity re lled equl vetors Illustrtion 6: Find the vlues of x, y nd z, so tht the vetors = xi ˆ+ ˆj + zkˆ nd = i ˆ+ yj ˆ+ kˆ re equl Sol: Two vetors re equl, if their orresponding omponents re equl Note tht two vetors re equl, if their orresponding omponents re equl Thus, the given vetors nd will e equl, if nd only if x =, y =, z = 1 Illustrtion 7: Find the vetor joining the point P (, 3, 0) nd Q (-1, -, -4) direted from P to Q Sol: By sutrting the omponent of P from Q we will get PQ Sine the vetor is to e direted from P to Q Clerly, P is the initil point nd Q is the terminl point So, the required vetor joining P nd Q is the vetor PQ given y PQ = OQ OP = 1 ˆi + 3 ˆj Kˆ ie PQ = 3i ˆ ± 5ˆj 4kˆ Illustrtion 8: Show tht, the points A( i ˆ ˆj k ˆ),B( ˆi 3j ˆ 5k ˆ), C( 3i ˆ 4j ˆ 4kˆ) + re the verties of right-ngled tringle Sol: Here if AB = BC + CA then only the given points re the verties of right ngled tringle We hve AB = (1 ) ˆi + ( 3 + 1) ˆj + ( 5 1)kˆ = ˆi j ˆ 6kˆ BC = (3 1) ˆi + ( 4 + 3) ˆj + ( 4 + 5)kˆ = i ˆ ˆj + kˆ Moreover, AB = 41 = = BC + CA Hene, it is proved tht the points form right-ngled tringle nd CA = ( 3)i ˆ+ ( 1 + 4) ˆj + (1 + 4)kˆ = ˆi + 3j ˆ+ 5kˆ

5 Mthemtis 65 3 RESULTANT OF VECTORS When two or more vetors re dded, they yield the resultnt vetor If vetors A nd B re dded together, the result will e vetor R, ie R = A + B Sme tehnique n lso e pplied for multiple vetors 4 VECTOR ADDITION 41 Tringulr Lw of Addition It sttes tht if two vetors n e represented in mgnitude nd diretion y the two sides of tringle tken in the sme order, then their resultnt is represented y the third side of the tringle, tken in the opposite diretion of the sequene 4 Prllelogrm Lw of Addition It sttes tht if two vetors n e represented in mgnitude nd diretion y the two djent sides or prllelogrm, then their resultnt is represented y the digonl of the prllelogrm 43 Addition in Component Form Consider two vetors A nd B A= <,, 1 1 1> B= <,, > Then, A+ B= < 1 +,1 +,1 + > 44 Properties of Vetor Addition The properties of vetor ddition re listed s follows: () π / Commuttive () π /3 Assoitive () π /4 Null vetor is n dditive identity (d) Aˆ nd B ˆ Additive inverse (e) π (f) Aˆ Bˆ (g) π / 45 Vetor Sutrtion Sutrtion is tken s n inverse opertion of ddition If u nd v re two vetors, the differene u v of two vetors is defined to e the vetor dded to v to get u In order to otin u v, we put the tils of u nd v together, the direted segment from the nose of v to the nose of u is representtive of u v Illustrtion 9: If = i + j + 3k nd = i + 4 j 5k represent two djent sides of prllelogrm, find the unit vetors prllel to the digonls of the prllelogrm Sol: As mentioned ove, if two vetor quntities re represented y two djent sides or prllelogrm then the digonl of prllelogrm will e equl to the resultnt of these two vetors Let ABCD e prllelogrm suh tht, AB// ndbc//

6 66 Vetors Then, AB + BC = AC AC = + = 3i ˆ + 6ˆj kˆ nd AB + BD = AD BD = AD AB BD = = ˆi + ˆj 8kˆ Now, AC = 3i ˆ+ 6ˆj kˆ AC = = 7 And BD = ˆi + ˆj 8k ˆ BD = = 69 AC 1 Unit Vetor long AC = = ( 3i ˆ + 6 ˆ j k ˆ ) AC 7 BD 1 Unit vetor long BD = = ( i + J 8k ) BD 69 A D Figure 63 B C Illustrtion 10: ABCDE is pentgon Prove tht the resultnt of the fores AB, AE, BC, DC, ED nd AC is 3AC Sol: By using method of finding resultnt of vetor we n prove required result D Let R e the resultnt fore R = AB + AE + BC + DC + ED + AC R = ( AB + BC) + ( AE + ED + DC) + AC = AC + AC + AC = 3AC Hene proved E A B Figure 64 C Illustrtion 11: ABCD is prllelogrm If L nd M re the middle points of BC nd CD, respetively express 3 AL nd AM in terms of AB nd AD, lso show tht AL + AM = AC Sol: By using mid point formul nd method of finding resultnt of vetor we n prove given reltion Let nd e the position vetors of points B nd D, respetively e referred to A s the origin of referene Then AC = AD + DC = AD + AB DC = AB B L C = d + AB =, AD = d ie the position vetor of C referred to A is d+ M AL = pv of L, the midpoint of BC 1 1 AM = + d + = AD + AB A D AL + AM = + d + = + d = ( + d) = AC Figure 65 5 SCALAR MULTIPLE OF A VECTOR If is the given vetor, then k is vetor, whose mgnitude is k times the mgnitude of nd whose diretion is the sme or opposite s tht of ording to whether k is positive or negtive

7 Mthemtis 67 6 SECTION FORMULA () If nd re the position vetors of two points A nd B, then the position vetor of point whih divides A ( n + m) nd B in the rtio m:n is given y r = m+ n ( + ) () Position vetor of the midpoint of AB = MASTERJEE CONCEPTS If, nd re the position vetors of the verties of ny ABC Then the position vetor of entroid G will e The position vetor of inenter of tringle with position vetors of tringle ABC, re A ( ), B( ), C( + + ) is r = + + Anurg Srf (JEE 011, AIR 6) Illustrtion 1: If ABCD is qudrilterl nd E nd F re the mid points of AC nd BD, respetively, prove tht AB + AD + CB + CD = 4EF Sol: By using mid-point theorem we n prove given reltion Sine F is the midpoint of BD Applying the midpoint theorem in tringle ABD, we hve AB + AD = AF (i) D C Applying the midpoint theorem in tringle BCD, we hve CB + CD = CF (ii) Adding equtions (i) nd (ii), we otin AB + AD + CB + CD = ( AF + CF) Now pplying the midpoint theorem in tringle CFA, we hve AF + CF = EF AB + AD + CB + CD = AF + CF = 4EF Hene proved A E Figure 66 B

8 68 Vetors Illustrtion 13: If G is the entroid of the tringle ABC, show tht GA + GB + GC = 0 nd onversely GA + GB + GC = 0, then G is the entroid of tringle ABC (JEE ADVANCED) Sol: As G is the entroid of tringle ABC, hene G = + + Therefore A 3 y otining GA, GB nd GC we n prove this prolem Let the position vetor of the verties e, nd, respetively F E So, the position vetor of entroid, G, is + + G 3 GA + + = OA OG = = Similrly, GB = GC = B C D GA + GB + GC = ( ) = 0 3 Figure 67 Conversely if GA + GB + GC = 0 OA + OB + OC (OA OG) + ( OB OG) + ( OC OG) = 0 OA + OB + OC = 3OG OG = 3 Hene, G is the entroid of the points A, B nd C = x+ ˆ i x y ˆ j+ k ˆ Illustrtion 14: Find the vlues of x nd y, for whih the vetors re prllel = x 1 ˆ i + x + y ˆ j + k ˆ, Sol: Two vetors re prllel if rtio of there respetive omponents re equl nd x+ y x 1 re prllel if = = x=-5, y= 0 x 1 x + y 3 Illustrtion 15: If ABCD is prllelogrm nd E is the midpoint of AB, show y vetor method, tht DE trisets nd is triseted y AC Sol: By using setion formul, we n solve this prolem Let AB = nd BC = Then BC = AD = nd AC = AB + AD = + D A M K E Figure 68 Also, let K e point on AC, suh tht AK:AC = 1:3 1 AK AC AK 1 = = ( + ) (i) Let E e the midpoint of AB, suh tht AE = Let M e the point on DE suh tht DM: ME = :1 AD + AE + AM = = (ii) 1+ 3 ( + ) Compring equtions (i) nd (ii), we find tht AK = = AM, nd thus we onlude tht K nd M oinide, ie 3 DE trisets AC nd is triseted y AC Hene proved C B

9 Mthemtis 69 7 LINEAR COMBINATION OF VECTORS 71 Colliner nd Non-Colliner Vetors Let nd e non-zero vetors These vetors re sid to e olliner if there exists λ 0 suh tht α +λ + γ = 0 Given finite set of vetors,,, then the vetor r = x + y + z + is lled liner omintion of,,, for ny slr x, y, z R 7 Collinerity of Three Points Let three points with position vetors (non-zero),, nd e olliner Then there exists λ, γ oth not eing 0 suh tht α+λ +γ = 0 73 Coplnr Vetors Let nd, e non-zero, non-olliner vetors Then, ny vetor r oplnr with, s liner omintion of,, ie there exist some unique x, y R, suh tht n e uniquely expressed MASTERJEE CONCEPTS If,, re non-zero, non-oplnr vetors, then x + y + z = x' + y' + z' x = x',y = y',z = z' Let,, e non-zero, non-oplnr vetors in spe Then ny vetor r n e uniquely expressed s liner omintion of,, or there exists some unique x, y, z R, suh tht x + y + z = r Vihv Krishn (JEE 009, AIR ) 74 Liner Dependeny of Vetors A set of vetors { v 1, v, vp } hs only trivil solution The set { v 1,v,vp} ll 0, suh tht 1v1 + v + + pvp = 0 is sid to e linerly independent if the vetor eqution x1 v1 + x v + + xp vp = 0 is sid to e linerly dependent if there exists weights 1,, p, not MASTERJEE CONCEPTS Two non-zero, non-olliner vetors re linerly independent Any two olliner vetors re linerly dependent Any three non-oplnr vetors re linerly independent Any three oplnr vetors re linerly dependent Any four vetors in three -dimensionl spe re linerly dependent Nitish Jhwr (JEE 009, AIR 7)

10 610 Vetors Illustrtion 16: The position vetors of three points A = + 3, B = nd C = Prove tht the vetors AB nd AC re linerly dependent Sol: Here otin AB nd AC to hek its liner dependeny Let O e the point of referene, then, OA = + 3, OB = + 3 4, nd OC = AC = OC OA = = AB = OB OA = ( + 3 4) ( + 3) = AC =λab, where λ= 1 Hene AB nd AC re linerly dependent Illustrtion 17: Prove tht the vetors , nd re linerly dependent nd,,, eing linerly independent vetors Sol: We know tht if these vetors re linerly dependent, then we n express one of them s liner omintion of the other two Now, let us ssume tht the given vetors re oplnr, nd then we n write = ( ) + m( ), where nd m re slrs Compring the oeffiients of,nd on oth sides of the eqution 5 = 7 + 3m 6 = 8 + 0m 7 = 9 + 5m From equtions (i) nd (iii), we get 1 4 = 8 = = m, whih evidently stisfies eqution (ii) too Hene, the given vetors re linerly dependent (i) (ii) (iii) Illustrtion 18: Prove tht the four points + 3, + 3, 3+ 4 nd 6+ 6 re oplnr Sol: Let the given four points e P, Q, R nd S respetively These points re oplnr, if the vetors PQ, PR nd PS re oplnr These vetors re oplnr if one of them n e expressed s liner omintion of other two So, let PQ = xpr + yps 5+ 4 = x( + ) + y( 9+ 7) = ( x y) + ( x 9y) + ( x + 7y) x y = 1,x 9y = 5, x + 7y = Solving the first two of these three equtions, we get x =,y = On sustituting the vlues of x nd y in the third eqution, we find tht the third eqution is stisfied Hene, the given four points re oplnr

11 Mthemtis 611 Illustrtion 19: Show tht, the vetors + 3, + nd + 3 re non-oplnr vetors Sol: If vetors re oplnr then one of them n e expressed s liner omintion of other two otherwise they re non-oplnr Assume the given vetors re oplnr Then one of the given vetors is expressile in terms of the other two Let + 3 = x( + ) + y( + 3) for some slrs x nd y + 3 = (x + y) + (x + y) + ( x 3y) = x + y, 1 = x + y nd 3 = x 3y, Clerly, the first two equtions ontrdit eh other Hene, it is proved tht the given vetors re not oplnr 8 SCALAR OR DOT PRODUCT The slr produt of two vetors = (,, ) nd = (,, ) etween the two vetors The omponent form of the dot produt is s follows: = And in geometril form = osθ where θ is the ngle etween the two vetors nd 0 θ π From eqution (i), it n lso e written s osθ= =, whih n e used to find the ngle etween two vetors If nd re perpendiulr then θ= 90 osθ= 0 = 0 is written using dot s n opertor ( ) (i) (ii) MASTERJEE CONCEPTS > 0 Angle etween nd is ute < 0 Angle etween nd is otuse Shivm Agrwl (JEE 009, AIR 7) Geometril Interprettion of Dot Produt The slr produt is used to determine the projetion of r vetor long the given diretion is the omponent of vetor ON OB = in the diretion of vetor OA( = );ON = osθ Thus the projetion of long â = â ON = B M P P 1 O N Figure 69 A P

12 61 Vetors Projetion of long = ˆ OM = 81 Properties of Slr Produt The properties of slr produt re listed s follows: (), re vetors nd is numer () = () = (d) ( + ) = + (e) () = () (f) 0 = 0 (g) = os θ (h) = 0 = 0 or = 0 or Illustrtion 0: Find the ngle θ etween the vetors = ˆi + ˆj kˆ nd vetors = ˆi ˆj+ kˆ Sol: The ngle θ etween the two vetors nd is given y osθ= Now = ( ˆi + ˆj k ˆ) ( ˆi ˆj+ kˆ) = = Therefore, we hve osθ= Hene, the required ngle is θ= os 3 3 Illustrtion 1: Find the length of the projetion of vetor = i ˆ+ 3ˆj + kˆ on vetor ˆ = ˆ i + j ˆ + k ˆ Sol: The projetion of vetor on the vetor 1 is given y ( ) 1 ( ) 10 5 ( ) = = = Illustrtion : Let,, e the vetors of lengths 3, 4 nd 5, respetively Let e perpendiulr to (+ ), to nd to Then, find the length of the vetor Sol: By using property of slr produt of vetor we n solve this illustrtion Given = 3, = 4, = = ( + + )( + + ) = ( + ) + ( + ) + ( + ) = = 5 Illustrtion 3: Let ˆ ˆ ˆ ˆ ˆ = 4i + 5j k, = i 4j+ 5k nd = 3i ˆ+ ˆj kˆ Find vetor d, whih is perpendiulr to oth nd, nd stisfying d = 1 Sol: If two vetor re perpendiulr then their produt will e zero Let d = xi ˆ+ yj ˆ+ zk ˆ Sine d is perpendiulr to oth nd Therefore, d = 0 xi + yj + zk 4i + 5j k = 0 4x + 5y z = 0 d = 0 xi + yj + zk i 4 j + 5k = 0 x 4y + 5z = 0 d = 1 xi + yj + zk 3i + j k = 1 3x + y z = 1 ( ˆ ˆ ˆ) ( ˆ ˆ ) ( ˆ ˆ ˆ) ( ˆ ˆ ˆ) ( ˆ ˆ ˆ) ( ˆ ˆ ˆ) (i) (ii) (iii)

13 Mthemtis 613 Solving equtions (i), (ii) nd (iii), we get x = 7, y = z = 7 Hene, d = 7i ˆ 7 j 7kˆ Illustrtion 4: Three vetors, nd stisfy the ondition + + = 0 Evlute the quntity µ= + +, if = 1, = 4 nd = Sol: Simply using property of slr produt we n lulte the vlue of µ Sine + + = 0, + + = = 0 Therefore, + = = 1 we hve Similrly + = = 16, + = 4 1 On dding these equtions, we hve ( + + ) = 1 or µ= 1, ie, µ= Illustrtion 5: Prove, Cuhy Shwrz inequlity, (), nd hene show tht ( + + ) ( + + ) ( + + ) Sol: As we know os θ 1, solve it y multiplying oth side y We hve, os θ 1 os θ Let = ˆ ˆ ˆ ˆ ˆ ˆ 1i + j + 3k nd = 1i + j + 3k Then, = + +, = + + nd = ( ) ( )( ) (JEE ADVANCED) Illustrtion 6: If,, re three mutully perpendiulr vetors of equl mgnitude, prove tht + + is eqully inlined with vetors, nd (JEE ADVANCED) Sol: Here use formul of dot produt to solve the prolem Let = = = λ (sy) Sine,, re mutully perpendiulr vetors, We hve = = = 0 Now, + + = = + + = 3λ + + = 3λ Let + + mkes ngles θ 1, θ, θ 3 with, nd, respetively Then, ( + + ) + + λ osθ 1 = = = = = θ 1 = os Similrly, θ = os ndθ = os θ = θ = θ 3 3 Hene, + + is eqully inlined with, nd Illustrtion 7: Using vetors, prove tht os(a+b) = osa osb sina sinb (JEE ADVANCED) Sol: From figure, using vetor method we n esily prove tht os(a+b)=osa osb sina sinb

14 614 Vetors Let OX nd OY e the oordinte xes nd let ˆ i nd ˆ j e unit vetors long OX nd OY, respetively Let XOP = A nd XOQ = B Drw PL OX nd QM OX Therefore, the ngle etween OP nd OL is A & OQ nd OL is B Y In OLP, OL = OP osand LP = OP sina Therefore,OL = OP os A i nd LP = OP sina j Now, OL + LP = OP OP = OP ( osa) ˆi ( sina) ˆj In OMQ,OM = OQ osb nd MQ = OQ sinb Therefore, OM = ( OQ osb) ˆi, MQ = ( OQ sinb) ˆj OQ = OQ ( osβ ) ˆi + ( sinβ) ˆj ˆ ( ˆ) (i) (ii) X i O Y B M A Figure 610 j L P X From (i) nd (ii), weget OPOQ = OP ( os A) ˆi ( sina) ˆj OQ ( osb) ˆi + ( sinb) ˆj = OPOQ os A osb sinasinb But, OPOQ = OP OQ os(a + B) = OPOQ os(a + B) OP OQ os(a + B) = OPOQ[osAosB sinasinb] os(a + B) = osaosb sinasinb = log x ˆi 6ˆj + 3kˆ Illustrtion 8: Find the vlues of for whih the vetors ( ) mde n otuse ngle for ny x ( 0, ) log x ˆi j ˆ log x kˆ nd = + + (JEE ADVANCED) Sol: For otuse ngle osθ< 0, therefore y using formul osθ=, we n solve this prolem Let θ e the ngle etween the vetors nd Then, osθ= For θ to e n otuse ngle, we must hve os θ < 0,for ll x 0, < 0,for ll x ( 0, ) < 0, for ll x 0, < 0, for ll x 0, log x log x < 0, for llx 0, y + 6y 1 < 0, for ll y R, where y = log x [ x > 0 y = log x R] < 0 nd < 0 [ x + x + > 0 for ll x < 0 nd Disriminnt < 0] < 0 nd (3 + 4) < < 0 nd < < 0,0 3 3 Illustrtion 9: D is the midpoint of the side BC of tringle ABC, show tht AB + AC = ( AD + BD ) Sol: By using the formul of resultnt vetor we will get the required result Given D is midpoint of BC BD = DC We hve AB = AD + DB AB = ( AD + DB) AB = AD + DB + AD DB (i) B A C D Figure 611

15 Mthemtis 615 Also we hve AC = AD + DC AC = ( AD + DC) AC = AD + DC + AD DC (ii) Adding equtions (i) nd (ii), we get AB + AC = AD + BD + AD DB + DC = DA + DB 9 VECTOR OR CROSS PRODUCT Let nd e two vetors The vetor produt of these two vetors n e lulted s ( ) = sinθnˆ, where θ is the ngle etween the vetors nd, 0 θ π nd ˆn is the unit vetor t right ngles to oth nd, ie ˆn is vetor norml to the plne tht ontins nd, nd ˆn re three vetors whih form right-hnded set The onvention is tht we hoose the diretion speified y the right-hnd srew rule Imgine srew in your right hnd If you turn right-hnded srew from to, the srew dvnes long the unit vetor ˆn It is very importnt to relize tht the result of vetor produt is itself vetor n n Figure 61 Let us see how the order of multiplition mtters from the definition of the right-hnd srew rule: = The vetor given y points in the opposite diretion to So, ˆi ˆj kˆ We n define vetor produt in terms of mtrix nottion s = nd n terms of omponents s = 1,, 3, = 1,, 3 = 3 3,31 1 3, 1 1 x From the definition, the ngle n e lulted s sinθ= If nd re prllel then θ= 0 sinθ= 0 nd x = 0 91 Properties of Vetor Produt The properties of vetor produt re listed s follows:, nd re ll vetors in three dimensions () nd

16 616 Vetors () = sinθ () ˆi ˆj = k, ˆˆj kˆ = ˆˆ i,k ˆi = ˆj (d) = 0 = 0 or = 0 or (e) = (f) = = ( ) (g) ( + ) = + (h) ( ) = ( ) Geometril interprettion of = sinθ djent sides, denotes the re of prllelogrm, in whih nd re the two Vetor re of the plne figure Considering the oundries of losed, ounded surfe, whih hs een desried in speifi mnner nd tht do not ross, it is possile to ssoite direted line segment, suh tht C outwrd in the plne C Figure 613 not possile () () () is the numer of units of re enlosed y the plne figure The support of is perpendiulr to the re nd outside the surfe The sense of desription of the oundries nd the diretion of is in ordne with the RHS srew rule Vetor re of tringle If re the position vetors, then the vetor re of tringle is given y the formul 1 = ( ) If,, re the position vetors, then the vetor re of ABC is given y the formul O B() Figure 614 A() 1 = ( ) ( ) A() 1 = ( ) + ( ) + ( ) B() Figure 615 C( )

17 Mthemtis 617 MASTERJEE CONCEPTS (i) If three points with position vetors, nd re olliner, then + + = 0 (ii) Unit vetor perpendiulr to the plne of the + + ˆn = ± ABC, when,nd re the pv of its ngulr point is Nitish Jhwr (JEE 009, AIR 7) 10 ANGULAR BISECTOR As disussed erlier, the digonl of prllelogrm is not neessrily the isetor of the ngle formed y two djent sides However, the digonl of rhomus isets the ngle formed etween two djent sides Consider vetors AB = nd AD = forming prllelogrm ABCD s shown in the figure Consider the two unit vetors long the given vetors, forming rhomus AB C D C D D C A Figure 616 B B Now, AB = ndad = Therefore AC' = + Then, ny vetor long the internl isetor is λ + Similrly, ny vetor long the externl isetor is λ C Illustrtion 30: Find vetor of mgnitude 9, whih is perpendiulr to oth the vetors 4i ˆ ˆj + 3kˆ nd i ˆ+ ˆj kˆ Sol: By using property of vetor produt, we n solve this prolem Let = 4i ˆ ˆj + 3kˆ nd ˆ = i ˆ+ ˆj k ˆ Then, ˆ ˆ ˆ i j k = = 3 i 8+ 6 j+ 4 k = i + j+ k = = 3 1 ˆ ˆ ˆ ˆ ˆ ˆ 9 Required vetor = 9 = ( ˆi + ˆj + kˆ) = 3i ˆ+ 6ˆj + 6kˆ 3 Illustrtion 31: Find the re of prllelogrm, whose djent sides re given y the vetors = 3i ˆ+ ˆj + 4kˆ nd ˆ = ˆi ˆj + k ˆ Sol: The re of prllelogrm with nd s its djent sides is given y ˆi ˆj kˆ Now, = = 5i ˆ+ ˆj 4k ˆ Therefore, = = 4 ; Hene, the required re is 4

18 618 Vetors Illustrtion 3: Let,, e the unit vetors suh tht = = 0 = ± Prove tht nd the ngle etween nd is 6 π, Sol: Here = = 0, therefore is perpendiulr to the plne of nd nd it is prllel to We hve = = 0 nd is perpendiulr to the plne of nd is prllel to = λ for some slr λ π λ sin 1 = λ = = = 6 λ = λ=± = λ = ± Illustrtion 33: If,, re three non-zero vetors, suh tht = nd =, mutully t right ngles nd = 1 nd = prove tht,, re Sol: Use property of vetor or ross produt to prove this illustrtion We hve, = nd =, nd,, nd,, re mutully perpendiulr lines Agin = nd = = nd = π π sin = nd sin = [ nd ] = nd = = Putting = in = = 1 0 = 1 Putting = 1 in =, we otin = Illustrtion 34: Prove y vetor method, tht in ABC, = = sina sinb sinc Sol: As re of tringle ABC is equl to 1 AB AC 1 BC BA 1 = = CA CB, therefore y using ross produt method we n prove this prolem Let BC =, CA =, AB = Then The re of ABC = AB AC = BC BA = CA CB Dividing the ove expression y, we get sina sinb sinc = = sina = sinb = sinc

19 Mthemtis 619 Illustrtion 35: Given the vetors = pˆ + qˆ nd = pˆ + qˆ, where p nd q re unit vetors forming n ngle of 30 Find the re of the prllelogrm onstruted on these vetors Sol: Simply y pplying ross produt etween nd, we hve = ( pˆ + qˆ) ( pˆ + qˆ) = 3( pˆ qˆ) π 3 = 3 ( pˆ + qˆ) = 3 pˆ qˆ sin = 6 Illustrtion 36: Let OA =,OB = 10 + nd OC =, where O is the origin Let p denote the re of the qudrilterl OABC nd q denote the re of the prllelogrm with OA nd OC s djent sides Prove tht p = 6q Sol: We hve to otin the re of qudrilterl nd prllelogrm using ross produt method to get the required result We hve, p = re of the qudrilterl OABC = 1/ OB AC = 1/ OB ( OC OA) = 1/ ( 10 + ) ( ) = 1/ 10( ) 10( ) + ( ) ( ) = 1/10( ) ( ) = 6( ) nd q = re of the prllelogrm with djent sides OA nd OC = OA OC = From equtions (i) nd (ii), we get p = 6q (i) (ii) Illustrtion 37: Given tht D, E, F re the midpoints of the sides of tringle ABC, using the vetor method, prove 1 tht re of DEF = (re of ABC) 4 Sol: Tking A s the origin, let the position vetors of B nd C e nd respetively Then, the position vetor of D, E nd F re ( + ) 1, 1 nd 1 respetively Therefore first otin DE nd DF, nd fter tht y pplying formul of vetor re of tringle DEF we n otin the required result 1 1 A(origin) Now, DE = ( + ) = 1 1 nd DF = ( + ) = E( F ( 1 Vetor re of DEF = ( DE DE) = B () C( ) D = ( ) = ( AB AC ) = (vetor re of ABC) Figure Hene,re of DEF = (re of ABC) 4 ( (

20 60 Vetors Illustrtion 38: Given tht P, Q re the midpoints of the non-prllel sides BC nd AD of trpezium ABCD Show tht re of APD = CQB Sol: Use formul of vetor re of tringle to solve this prolem Let AB = nd AD = d Now DC is prllel to AB there exists slr t, suh tht DC = t AB = t AC = AD + DC = d + t D C AB + DC From geometry we know tht QP = QP = Now AP nd AQ re d + + t nd d, respetively 1 1 Now, APD = AP AD = ( + d + t) d = ( 1 + t)( d) d Also CQB = BC BQ = d t d t d 1 = ( d ) + = ( 1 + t) d = APD APD = CQB 11 TRIPLE PRODUCT OF VECTORS Two types of triple produts re listed elow: Vetor triple produt ( ) Slr triple produt ( ) 111 Slr Triple Produt The slr triple produt hs n interesting geometri interprettion: We know tht ( ) = sinθnˆ = (re of the prllelogrm defined y nd ) Thus, = (re of the prllelogrm) ˆn = (re of the prllelogrm) ˆn osφ h O A Q Figure 619 Figure 618 But osθ= h = height of the prllelepiped norml to the plne ontining nd ( φ is the ngle etween nd ˆn) So, ( ) = volume of the prllelepiped defined y, nd Thus, the following onlusions re rrived: () If ny two vetors re prllel, then ( ) = 0 (zero volume) () If the three vetors o-plnr, then ( ) = 0 (zero volume) () If ( ) = 0, then either (i) = 0, or (ii) = 0 or (iii) = 0 or (iv) two of the vetors re prllel or (v) the three vetors re o-plnr (d) ( ) = ( ) = ( ) = The sme volume (e) ( ) is lso known s ox produt, whih is represented s [ ] Also + d = d + d A C P B

21 Mthemtis 61 (f) If,, re non-oplnr, then [,, ]>0, for right-hnded system nd [,, ] < 0, for left hnded system (g) If O is the origin nd,, re the position vetors of A, B nd C, respetively, of the tetrhedron OABC, then the volume is given y the formul V = 1 6 Reiprol system of vetors () If,, nd ',',' re the two sets of non-oplnr vetors, suh tht ', = ' = ' = 1, ' = ' = 0, ' = ' = 0 nd ' = ' = 0, Then,, nd ',',' onstitute reiprol system of vetors () Reiprol system of vetors exists only in the se of dot produt () ',',' n e defined in terms of,, s ' = ;' = ;' = ( 0) Note: (i) ' + ' + ' = 0 ( ) + ( ) + ( ) = 0 (ii) ' = ' = ' = 1 (iii) ( + + )( ' + ' + ' ) = 3 1 (iv) If V then ' ' ' = = ' ' ' = 1 V Illustrtion 39: If,m,n three non-oplnr vetors, then prove tht m n ( ) = m m m (JEE ADVANCED) n n n Sol: Use slr triple produt method s mentioned ove to solve this prolem Let, ˆ ˆ ˆ ˆ ˆ ˆ = 1i + ˆ ˆ ˆ j+ 3k, m = m1i + mj + m3k, n = n1i + nj + n3k, nd ˆ ˆ ˆ = ˆ ˆ ˆ 1i + j + 3k, = 1i + j+ 3k 1 ˆ 3 i ˆj kˆ Now m n = m1 m m3 nd = ( ) = 1 3 n n n ˆ ˆ ˆ 1 ˆ ˆ ˆ i j k 3 i j k + + m n ( ) m m m mi ˆ mˆj mkˆ = = + + m m n n n ni ˆ+ nˆj+ nkˆ n n = 1i + j + 3k 1i + j + 3k = 1et Now, ( ˆ ˆ ˆ) ( ˆ ˆ ˆ)

22 6 Vetors m n ( ) = m m m = m m m, n n n n n n Hene proved Illustrtion 40: Find the volume of prllelepiped, whose sides re given y 3i ˆ+ 7j ˆ+ 5k, ˆ 5i ˆ+ 7j ˆ 3kˆ nd 7 ˆi 5ˆj 3kˆ Sol: We know tht, the volume of prllelepiped, whose three djent edges re,, is Let ˆ ˆ ˆ ˆ ˆ ˆ = 3i + 7 j + 5k, = 5 i+ 7 j 3k nd = 7 ˆi 5ˆj 3kˆ We know tht, the volume of prllelepiped, whose three djent edges re,, is Now, = = 3( 1 15) 7( ) + 5( 5 49) = = So, the required volume of the prllelepiped= = 64 = 64 ui units Illustrtion 41: Simplify Sol: Here y using slr triple produt we n simplify this = {( ) ( )}( ) [ydef] = ( + )( ) [y distlw] = ( + + ) ( ) = 0 = + + [y dist lw] = = = = 0 = (JEE ADVANCED) Illustrtion 4: Find the volume of the tetrhedron, whose four verties hve position vetors,, nd d, respetively (i) Sol: Here volume of tetrhedron is equl to 1 d d d 6 Let, four verties e A, B, C, D with pv,, nd d respetively DA = d,db = d,dc = d 1 1 = 6 = 6 ( d ) d d { d d d } { d d } Hene volume d d d d d d 1 1 = + = = 1 d 6

23 Mthemtis 63 Illustrtion 43: Let u nd v e unit vetors nd w is vetor, suh tht u v + u = w nd w u = v, then find the vlue of u v w (JEE ADVANCED) Sol: Here s given u v + u = w nd w u = v, solve it using slr triple produt Given, u v + u = w nd w u = v ( u v + u) u= w u ( u v) u + u x u= v (s,w u = v) ( uu ) v ( vu) u + u u = v (using uu = 1 nd u u = 0, sine unit vetor) v ( vu) u = v ( uv ) u = 0 uv = 0 (s; u 0) (i) u v w = u ( v w) = u ( v ( u v + u )) (given w = u v + u) = u ( v ( u v ) + v u) = u (( vv ) u ( vu) v + v u) = u ( u 0 + v u ) (s uv = 0 from (i)) = ( uu) u ( v u) = 1 0 = 1 u v w = 1 11 Vetor Triple Produt Definition: ( ) Hene ( ) = x + y ( ) = x + y 0 = x + y x y = = λ x = λ nd y = λ is vetor, whih is oplnr to nd nd perpendiulr to [Liner Comintion of nd ] Sustituting the vlues of x nd y we get, ( ) =λ λ This identity must hold true for ll vlues of,, Sustitute ˆi; ˆ = = jnd = kˆ ( ) =λ λ ˆi ˆj ˆi ˆˆ ji ˆi ˆˆ ii ˆj ˆj = λˆj λ = 1 = Note: Unit vetor oplnr with nd perpendiulr to is ± (i) (ii)

24 64 Vetors Illustrtion 44: Prove tht { ( d) } = ( d)( ) ( d) Sol: By using vetor triple produt s mention ove We hve, { ( d )} {( d ) = ( ) d } = {( d) d } [y distriutivelw] = d d Illustrtion 45: Let ˆ ˆ ˆ ˆ ˆ ˆ = i + j 3k, = i + j k, nd = i ˆ+ j ˆ kˆ Find the vlue (s) of, if ny, suh tht = 0 { } Sol: Here use vetor triple produt to otin the vlue of {( ) ( ) } ( ) = ( ) = { }, Given {( ) ( ) } ( ) = 0 = or = 0 = leds to three different equtions whih do not hve ommon solution 3 = 0 1 = = 0 = 3 1 Illustrtion 46: Solve for r, from the simultneous equtions r =,r = 0, provided is not perpendiulr to Sol: As given r =, solve this using vetor triple produt to get the result Given r = ( r ) = 0 ( r ) nd re olliner r = k r = + k (i) r = 0 ( + k ) = 0 ( ) k = putting in eq (i) we get r = = Illustrtion 47: If x + kx =, where k is slr nd, re ny two vetors, then determine x in terms of, nd k Sol: Here s given x + kx =, Apply ross produt of with oth side nd solve using vetor triple produt x + kx = (i) ( x ) + k( x) = ( ) x ( x ) + k ( x) = (ii) (i) ( x ) + k ( x ) = k x = (iii)

25 Mthemtis 65 Sustituting the vlues from equtions (i) nd (iii) in eqution (ii), we get, 1 x ( ) + k ( kx ) = k 1 1 ( + k ) x = ( ) + ( ) + k x = k + ( ) + k + k k 1 APPLICATION OF VECTORS IN 3D GEOMETRY () Diretion osines of r = i ˆ+ j ˆ+ kˆ re given y,, r r r () Inentre formul: The position vetor of the inentre of () Orthoentre formul: The position vetor of the orthoenter of tna + tnb + tnc ABC is tna + tnb + tnc + + ABC is + + (d) The vetor eqution of stright line pssing through fixed point with position vetor nd prllel to given vetor is given y r = +λ (e) The vetor eqution of line pssing through two points with position vetors nd is given y r = +λ α on the line r = +λ, where r is the position vetor of ny point on the give line Therefore, let the position vetor L e +λ P ( α ) ( α) PL = nd PL = α+λ = ( α) The length PL is the mgnitude of PL A B, nd the required length of perpendiulr r = + L = ( + ) (f) Perpendiulr distne of point from line: Let L e the foot of perpendiulr drwn P Figure 60 (g) Imge of point in stright line: If ( α) β= α Q β is the imge of P in r = +λ, then A L = ( + ) P B L = ( + ) QImge Figure 61

26 66 Vetors (h) Shortest distne etween two skew lines: Let l 1 nd I e two lines whose equtions re l 1 :r= 1 +λ 1 ndl :r= +λ, respetively ( 1 ) ( 1) 1 1 Then, shortest distne is given y PQ = = 1 1 Shortest distne etween two prllel lines: The shortest distne etween the two given prllel lines ( 1) r = 1 +λ nd r = +µ is given y d = If the lines r = 1 +λ 1 nd r = +µ interset, then the shortest distne etween them is zero Therefore, 1 1 = 0 (i) If the lines r = 1 +λ 1nd r = +λ re oplnr, then 1 1 = 1 nd the eqution of the plne ontining them is given y r 1 = 1 1 (j) The vetor eqution of plne through the point A ( ) nd perpendiulr to the vetor ( + ) = + r n = 0 is given y (k) Vetor The vetor eqution of plne norml to unit vetor n nd t distne d from the origin is given y rnˆ = d (l) The eqution of the plne pssing through point hving position vetor nd prllel to nd is given y r = +λ +µ r =, where λ nd µ re slrs (m) The vetor eqution of plne pssing through point,, is given y r =(1-s-t) +s +t r + + = or (n) The eqution of ny plne through the intersetion of plnes rn1 = d1 nd rn = d is r ( n1+λ n) = d1+λd, where λ is n ritrry onstnt (o) The perpendiulr distne of point hving position vetor n d from the plne rn = d is given y p = n n 1n (p) The ngle θ etween the plnes r 1 nˆ 1 = d 1 nd r nˆ = d is given y os θ = ± n n (q) The perpendiulr distne of point P( r ) from line pssing through nd prllel to is given y 1/ ( r ) P = ( r ) = ( r ) (r) The eqution of the plnes iseting the ngles etween the plnes r 1 n 1 = d 1 nd r n = d is r n n 1 ( ± ) = ± 1 d n d n 1 1 (s) The perpendiulr distne of point P( r ) from plne pssing through point nd prllel to points nd ( r ) ( ) is given y PM =

27 Mthemtis 67 (t) The perpendiulr distne of point P( r ) from plne pssing through the points, nd is given y ( r ) ( + + ) P = + + (u) Angle etween line nd the plne: If θ is the ngle etween line r ( ) sinθ= n n (v) The eqution of sphere with enter t C( ) nd rdius is r = (w) The plne rn = d touhes the sphere = +λ nd the plne rn = d, then If enter is the origin then r = n d r = R,if = R, ie the ondition of tngeny n (x) If nd re the position vetors of the extremities of dimeter of sphere, then its eqution is given y r r = 0 or r r + + = 0 or r + r = FORMULAE SHEET () OP = x î +y ĵ y () OP = x + y nd diretion is tnθ= x Vetor () Unit vetor Û = = Its modulus (d) Properties of vetor ddition: i + = = + + Assoitive ommuttive () ii + 0 = Null vetor is n dditive identity () ( ) + = + iii iv ( d) = ( d) + = 0 Additive inverse + d = + d () (d) 1 = (e) Setion formul: (i) If nd re the position vetors of two points A nd B, then the position vetor of point whih divides ( n + m) AB in the rtio m:n is given y r = ( m+ n) ( + ) (ii) Position vetor of mid-point of AB =

28 68 Vetors (f) Collinerity of three points: If,, nd re the position vetors (non-zero) of three points nd given they re olliner then there exists λγoth, not eing 0 suh tht +λ +γ (g) Coplnr vetors: Let, expressed uniquely s liner omintion of, (h) Produt of two vetors: (i) Slr Produt (dot produt) = + + If Note : osθ= e non-zero, non-olliner vetors Then, ny vetor r oplnr with, n e ie there exist some unique x, y R, suh tht x + y = r nd re perpendiulr if θ= 90 (ii) Properties of slr produt: i = + = + iii v If ˆ = ˆ = ˆ = i 1,0,0, j 0,1,0,k 0,0,1 then ˆˆ i j = ˆˆ jk = ki ˆˆ= 0 ii mn = mn = (mn ) iv + = + + (iii) Vetor (ross) Produt of two vetors: Let = (,, ),= (,, ) produt of is devoted y nd defined y =,,,, = =,, OR ˆi ˆj kˆ = = ˆi + ˆj + kˆ = sinθnˆ e two vetors then the ross Note: (i) θ eing ngle etween & (ii) If θ= 0, The = 0 ie = 0 nd & re prllel if =0

29 Mthemtis 69 (iv) Properties of ross produt i = 0 = 0 or = 0 or ii = + = + iii v is perpendiulr to oth nd iv ( n) = n( ) vi is Are of prllelogrm with sides nd (v) Slr Triple Produt: If ˆ ˆ ˆ ˆ ˆ ˆ = 1i + ˆ ˆ ˆ j+ 3k, = 1i + j+ 3k, = 1i + j+ 3k = = Then 1 3 = ( ) is lso represented s = = = If ny of the two vetors re prllel, then = 0 is the volume of the prllelepiped whose oterminous edges re formed y If re oplnr, = = re of tringle hving,, s position vetors of verties of tringle (vi) Vetor Triple Produt: ( ) = ( ) = Unit vetor oplnr with nd perpendiulr to is ±

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