03_L Stud Smrt H2 Mth Vectors Comple Numers_ns.pdf 1 11/10/2017 10:28:52 M -Level C M Y CM MY CY CMY K H 2 M t h e m t i c s Vectors nd Comple Numers Singpore si Pulishers Pte Ltd. LL RIGHTS RESERVED.
-Level Stud Smrt H2 Mthemtics VECTRS ND CMPLEX NUMBERS First Edition 2018 Pulished nd Distriuted : Singpore si Pulishers Pte Ltd 219 Henderson Rod #10-04 Henderson Industril Prk Singpore 159556 Tel : +65 6276 8280 F : +65 6276 8292 Emil: info.sg@spgrp.com Wesite: www.spgrp.com Fceook: Singpore-si-Pulishers LL RIGHTS RESERVED ll rights reserved. No prt of this puliction m e reproduced, stored in retrievl sstem, or trnsmitted in n form or n mens, electronic, mechnicl, photocoping, recording or otherwise, without the prior permission of the pulishers. ISBN-13 978-981-3210-99-8 ISBN-10 981-3210-99-0 Printed in Singpore SP Glol Prtners Network: ntigu & Brud ustrli Bhrin Bngldesh Brdos Bhutn Bostwn Brzil Brunei Cmodi Cnd Chile Chin Colomi Egpt Fiji Islnd Ghn Grend Gun Hong Kong Indi Indonesi Jmic Jpn Los Lenon Mcu Mlsi Mlwi Mldives Muritius Mnmr Nmii Nepl New Zelnd Nigeri Pkistn Ppu New Guine Philippines Polnd Portugl Qtr Rwnd Sudi ri Sechelles Singpore Solomon Islnds Sri Lnk Sint Luci Sint Vincent & the Grendines South fric South Kore Sri Tiwn Tnzni Thilnd Trinidd & Togo United r Emirtes United Kingdom United Sttes of meric Vietnm Zmi Zimwe For interntionl usiness enquiries, emil ig@spgrp.com www.spgrp.com
Prefce -Level Stud Smrt H2 Mthemtics Vectors nd Comple Numers is designed to provide students with structured prctice in H2 Mthemtics nd to epose students to vried question tpes to prepre them for mjor emintions. Wht cn ou epect to find in this ook? Generl Concepts These re included to help ou recp essentil concepts tht ou must know for ech topic. These summrise wht ou lern in school nd will sve ou precious time to get strted on the prctice questions. Ke Concepts These re quick reference of formule nd concepts to help ou work out specific question tpes quickl. Worked Emples These show detiled nd specific steps to solving prolem of prticulr question tpe. Lerning is lso reinforced through similr question tpes following the emples. Guided Eercise These provide ou with specific instructions to guide ou through solving prolem of prticulr question tpe. Prctice Questions with Quick nswer Ke These consolidte wht ou hve lernt in ech topic nd give comprehensive overview of the concepts covered. Quick nswer ke eside the questions lso mkes checking of nswers much esier for students. Pst -Level Emintion Questions nlsis This nlsis provides insight into the tpes of questions nd trends in emintions. Full Worked Solutions These re provided to help students etter understnd how ech prolem is eing solved nd lso serve s tool for self-stud nd ssessment. The Editoril Tem
Contents Chpter 1 Vectors I... 1 Generl Concepts Tpes of Questions Ke Concepts Worked Emples Guided Eercise Prctice Questions with Quick nswer Ke nlsis of Pst -Level Emintion Question Tpes Chpter 2 Vectors II... 55 Generl Concepts Tpes of Questions Ke Concepts Worked Emples Guided Eercise Prctice Questions with Quick nswer Ke nlsis of Pst -Level Emintion Question Tpes Chpter 3 Comple Numers 111 Generl Concepts Tpes of Questions Ke Concepts Worked Emples Guided Eercise Prctice Questions with Quick nswer Ke nlsis of Pst -Level Emintion Question Tpes Worked Solutions S1 S37 Complete our lerning with these resources! Grphs, functions nd series -Level 1 Grphs H 2 M t h e m tics Grphs, Functions nd Series Focus n Emintion Question Tpes 2 Trnsformtion of Grphs 1 Differentition -Level 1 Permuttion nd Comintion -Level 2 Mclurin Series H 2 M t h e m tics 3 Equtions nd Inequlities Clculus 4 Functions Focus n Emintion Question Tpes 5 Summtion Topicl Stud Notes Illustrted Emples Quick nswer Check Step-B-Step Solutions Sttistics CLCULUS Topicl Stud Notes Illustrted Emples Quick nswer Check Step-B-Step Solutions 3 Integrtion Techniques 4 D efinite Integrls 5 Differentil Equtions 6 rithmetic Progression 2 Proilit H 2 M t h e m tics Sttistics Focus n Emintion Question Tpes M SPeduction Like us on Fceook! Singpore-si-Pulishers 01 Pre&Cont_SS H2 MVCN.indd 2 7 Geometric Progression M SPeduction Enhnce our lerning eperience with our pp! Like us on Fceook! Singpore-si-Pulishers M SPeduction Enhnce our lerning eperience with our pp! 4 Norml Distriution 5 Smpling 6 Hpothesis Testing Topicl Stud Notes Illustrted Emples Quick nswer Check Step-B-Step Solutions 7 Regression nd Correltion Downlod M SPeduction pp Now! Enhnce our lerning eperience with our pp! 3 Binomil Distriution Like us on Fceook! Singpore-si-Pulishers 10/26/2017 4:26:43 PM
Chpter 1 Vectors I Bsic Properties of Vectors in Two nd Three Dimensions ddition nd sutrction of vectors, multipliction of vector sclr, nd their geometricl interprettions Use of nottions such s (, ( z, i + j, i + j + zk, B, Position vectors, displcement vectors nd direction vectors Unit vectors Distnce etween two points Concept of direction cosines Collinerit Use of the rtio theorem in geometricl pplictions Sclr nd Vector Products in Vectors Concepts of sclr product nd vector product of vectors nd their properties Clcultion of the mgnitude of vector nd the ngle etween two vectors Geometricl menings of n^ nd n^ where n^ is unit vector Eclude triple products c nd c Three-Dimensionl Vector Geometr Vector nd Crtesin equtions of lines Finding the foot of the perpendiculr nd distnce from point to line Finding the ngle etween two lines Reltionship etween two lines (coplnr or skew Eclude - finding the shortest distnce etween two skewed lines - finding n eqution for the common perpendiculr to two skew lines Generl Concepts Definition of Vector 1 vector is quntit hving oth mgnitude nd direction e.g. force, velocit. Nottion nd Representtion of Vectors B 2 vector is notted s B or nd is represented line segment where the length represents the mgnitude of the vector nd the direction of the line represents the direction of the vector. 3 vector cn e represented in ( column vector form e.g. B = ( 3 2 4 ( Crtesin form e.g. B = 3i + 2j + 4k z 4 2 B(3, 2, 4 3 -Level Stud Smrt H2 Mthemtics Vectors nd Comple Numers Chpter 1 Vectors I
Modulus of Vector 4 The modulus of vector B is the mgnitude of the vector nd is denoted B or B. 5 The mgnitude of column vector is clculted using Pthgors Theorem. N 2 = 2 + z 2 P(,, z P 2 = N 2 + 2 P = = 2 + z 2 + 2 _ 2 + 2 + z 2 Free Vector or Displcement Vector 6 free vector or displcement vector is vector which mkes no reference to n position. free vector such s B cn e represented n line segment of the sme length nd direction. It cn e locted nwhere (i.e. trnslted freel s it hs no prticulr position ssocited with it. z N z Position Vector 7 position vector is vector which defines the position of one point reltive to nother (usull the origin. vector P is clled the position vector of P reltive to the origin. Position vector P hs the origin s its initil point nd P s the end point with coordintes (,, z. z z i i i P(,, z 8 ln Crtesin form, P = i + j + zk where i is unit vector in the positive direction of the -is, j is unit vector in the positive direction of the -is nd k is unit vector is the positive direction of the z-is. Zero Vector (Null Vector 9 vector whose mgnitude is zero is clled zero vector. 10 zero vector is denoted 0. 11 zero vector hs no prticulr direction. Equl Vectors 12 Two vectors re equl when the hve the sme direction nd mgnitude. If B = C, then Vector conclusion B // CD Geometricl conclusion B // CD B = CD B = CD -Level Stud Smrt H2 Mthemtics Vectors nd Comple Numers Chpter 1 Vectors I
Equlit of Column Vectors p s q t 13 If B = ( r nd CD = ( u nd if B = CD, then ( Hence, p = s, q = t nd r = u. Negtive Vectors p s q r = ( t u. 14 Negtive vectors hve the sme mgnitude ut re in opposite directions. If B is negtive of B, then B = B nd B = B. Sclr Multiple of Vector 15 When vector is multiplied sclr k, the mgnitude of the vector chnges. The mgnitude of vector k is k times tht of (i.e. k = k. ( If k > 0, k is in the sme direction s. k k ( If k < 0, k is in the opposite direction to. Prllel Vectors 16 Prllel vectors cn e epressed s sclr multiple of ech other. If B = k CD where k is sclr, then Vector conclusion B // CD Geometricl conclusion B // CD B = k CD B = kcd B B Conversel, if B // CD, then B = k CD. Collinerit 17 If B = k BC where k is sclr, nd B is the common point, then, B nd C re colliner. Conversel if, B nd C re colliner, then B = k BC. k B C 1 Unit Vector 18 vector with mgnitude 1 is known s unit vector. The unit vector of vector is ^ = nd ^ = 1. ^ ddition of Vectors R 19 Tringulr Lw of Vector ddition PQ + QR = PR + = ( + P + Q -Level Stud Smrt H2 Mthemtics Vectors nd Comple Numers Chpter 1 Vectors I
20 Polgon Lw of Vector ddition + + c + d = e c 21 ddition of Column Vectors p s p s If = ( q r nd = ( t u, then + = ( q r + ( ( p + s t u = q + t r + u. Rtio Theorem 22 If P is the point which divides B internll in the rtio λ : μ, then P = μ + λ B μ + λ μ + λ = μ + λ Direction Rtio nd Direction Cosines 23 If P is vector such tht P is the point (,, c, then the direction rtio of the vector P is : : c. 24 If P mkes ngles of α, β nd γ with the, nd z es respectivel, then the direction cosines of P re cos α =, cos β = nd cos γ = c P P P where P = 2 + 2 + c 2. 25 cos 2 α + cos 2 β + cos 2 γ = ( P 2 + ( P 2 + ( c = 2 + 2 + c 2 ( P 2 _ = 2 + 2 + c 2 ( 2 + 2 + c 2 2 P 2 = 1 Thus, the sum of the squres of the direction cosines of n vector is unit. Sclr Product or Dot Product 26 The sclr product or dot product of two vectors nd is denoted nd is defined s cos θ where θ is the ngle etween nd. Thus, = cos θ where 0º θ 180º. Evluting the Sclr Product of Two Vectors 27 Form Two vectors Sclr product Crtesin = 1i + 2 j + 3 k nd = 1 i + 2 j + 3 k λ e = ( 1 i + 2 j + 3 k ( 1 i + 2 j + 3 k = 1 1 + 2 2 + 3 3 P μ d B Column vector ( 1 ( = 2 nd = 1 2 3 3 ( 1 ( 1 = 2 2 3 3 = 1 1 + 2 2 + 3 3 -Level Stud Smrt H2 Mthemtics Vectors nd Comple Numers Chpter 1 Vectors I
Results of Sclr Product 28 ngle θ Results of sclr product ( 0º Sclr Product of Like Prllel Vectors = cos 0 = = cos 0 = = 2 i i = j j = k k = 1 for the Crtesin unit vectors i, j nd k ( 180º Sclr Product of Unlike Prllel Vectors = cos ϖ = (c 90º Sclr Product of Perpendiculr Vectors = cos ϖ 2 = 0 i j = j k = k i = 0 for the Crtesin unit vectors i, j nd k (d (e cute 0º < θ < 90º tuse 90º < θ < 180º = cos θ > 0 since 0 < cos θ < 1 = cos θ < 0 since 1 < cos θ < 0 (f N.. Sclr Product of Zero Vector 0 = 0 cos θ = (0 cos θ = 0 Hence, = 0 if = 0 or = 0. Properties of Sclr Product 29 Sclr Product is Commuttive = cos θ nd = cos θ Hence, = 30 Sclr Product is Distriutive ( + c = + c 31 Sclr Multipliction of Sclr Product (m (n = mn( where m nd n -Level Stud Smrt H2 Mthemtics Vectors nd Comple Numers Chpter 1 Vectors I
pplictions f Sclr Product 32 ngle Between Two Vectors The ngle etween 2 vectors nd is unique. It is the ngle etween their directions when these directions oth converge or oth diverge from point. θ The cosine of the ngle θ etween two vectors is the sclr product of the respective unit vectors. cos θ = = ^ ^ 33 ngle Between Two Lines The ngle θ etween two lines is defined s the ngle etween their direction vectors or its supplementr ngle (i.e. 180º θ whichever is the cute ngle. cos θ = = ^ ^ θ = cos 1 = cos 1 ^ ^ 180º θ θ θ 34 Length of Projection nd Projection Vector θ θ N B B N ( The sclr length N is defined to e the length of the projection of onto. Length of projection of onto = N = cos θ = = ^ ^ is negtive when θ is otuse. However, since the length of projection must e positive, the length of projection of onto = ^ = ( The vector N is defined to e the projection vector of onto nd is given N = ( ^. ^ Vector Product or Cross Product 35 The vector product or cross product of two vectors nd is denoted nd is defined s ( sin θ ^n where θ is the ngle etween nd nd ^n is the unit vector perpendiculr to oth nd. Thus, = ( sin θ ^n where 0º θ 180º. θ -Level Stud Smrt H2 Mthemtics Vectors nd Comple Numers Chpter 1 Vectors I
36 The vector otined will e perpendiculr to oth vectors nd. i.e. ( nd ( Evluting the Vector Product of Two Vectors 37 If = 1 i + 2 j + 3 k nd = 1 i + 2 j + 3 k, then the vector product of the two vectors i j k = 1 2 3 1 2 3 = 2 3 2 3 i 1 3 1 3 j + 1 2 1 2 k = ( 2 3 3 2 i ( 1 3 3 1 j + ( 1 2 2 1 k R i j k i j k 1 2 3 1 2 3 1 2 3 1 2 3 = 2 3 i + 3 1 j + 1 2 k 3 2 i 1 3 j 2 1 k = ( 2 3 3 2 i + ( 3 1 1 3 j + ( 1 2 2 1 k Results of Vector Product 38 ngle θ Results of vector product ( 0º Vector Product of Like Prllel Vectors = ( sin 0º ^n = 0 i i = j j = k k = 0 for the Crtesin unit vectors i, j nd k ( 180º Vector Product of Unlike Prllel Vectors = ( sin 180º ^n = 0 i i = j j = k k = 0 for the Crtesin unit vectors i, j nd k (c 90º Vector Product of Perpendiculr Vectors = ( sin 90º ^n = ^n = i j = ( i j sin 90ºk = (1(1(1k = k for the Crtesin unit vectors i, j nd k i.e. i j = k, j k = i, k i = j j i = k, k j = i, i k = j (d N.. Vector Product of Zero Vector 0 = ( 0 sin θ ^n = ( (0 sin θ ^n = 0 Hence, = 0 if = 0 or = 0. -Level Stud Smrt H2 Mthemtics Vectors nd Comple Numers Chpter 1 Vectors I
Properties of Vector Product 39 Vector Product is Not Commuttive = ( 40 Vector Product is Distriutive ( + c = + c 41 Sclr Multipliction of Vector Product (m (n = (mn( where m nd n pplictions of Vector Product 42 Finding the Norml of Plne If nd re two vectors prllel to plne, then the norml to the plne n =. 43 Length of Sides of Right-ngled Tringle B = length of projection of onto = C cos θ = = ^ In generl, the length of projection B cn e found using sclr product. BC = length of projection of onto c = C sin θ = C ^ sin θ = C ^ = ^ In generl, the perpendiculr distnce BC from point C to line B cn e found using cross product. θ C c B -Level Stud Smrt H2 Mthemtics Vectors nd Comple Numers Chpter 1 Vectors I
Loction f Point 44 on 2-dimensionl plne in 3-dimensionl spce ( ( (c (d (e Representtion of Point on 2-dimensionl Plne P Defining Point on 2-dimesionl Plne (i lf nd re non-prllel free vectors nd is fied point on 2-dimensionl plne, then the position vector of n point P on the plne cn e epressed in terms of vectors nd. (ii The position vector of P, P = λ + μ where λ nd μ re sclrs. P is known s the liner comintion of vectors nd. Bse Vectors for 2-dimensionl Plne Vectors nd nd point form frme of reference for the position vector of n point on the 2-dimensionl plne. The vectors nd re known s se or sis vectors for 2-dimensionl plne. n vector tht is prllel to the plne cn e epressed in the form λ + μ. Position Vector of Point in Crtesin Form P(, j i Using the Crtesin frme of reference, the se vectors i nd j re unit vectors in the directions of the Crtesin es nd respectivel nd the fied point is the origin. The position vector of P(,, P = ( = i + j Points on the es ( If P is point on the -is, then P = ( = i (i.e. = 0 ( lf P is point on the -is, then P = ( = j (i.e. = 0 Representtion of Point in 3-dimensionl Spce R P Q c Defining Point in 3-dimensionl Spce (i lf, nd c re three non-prllel nd non-coplnr vectors nd is fied point in spce, then the position vector of n point P in spce cn e epressed in terms of vectors, nd c. (ii The position vector of P, P = λ + μ + ηc where λ, μ nd η re sclrs. P is known s the liner comintion of vectors, nd c. Bse Vectors for 3-dimensionl Spce Vectors, nd c nd point form frme of reference for the position vector of n point in the 3-dimensionl spce. The vectors, nd c re known s the se or sis vectors for 3-dimensionl spce. n vector in the 3-dimensionl spce cn e epressed in the form λ + μ + ηc. Position Vector of Point in Crtesin Form zi i P(,, z z Using the Crtesin frme of reference, the se vectors i, j nd k re unit vectors in the directions of the Crtesin es, nd z respectivel nd the fied point is the origin. The position vector of P(,, z, P = ( z = i + j + zk Points on the il Plnes ( If P is point on the plne, then P = ( = i + j (i.e. z = 0 0 ( If P is point on the z plne, then P = ( 0 z = i + zk (i.e. = 0 (c If P is point on the z plne, then P = ( 0 z = j + zk (i.e. = 0 i -Level Stud Smrt H2 Mthemtics Vectors nd Comple Numers Chpter 1 Vectors I
Eqution of Stright Line 45 prticulr line is uniquel locted on plne or in spce if ( it hs known direction nd psses through known fied point or ( it psses through two known fied points 46 Eqution of line on 2-dimensionl plne Eqution of line in 3-dimensionl spce If r is the position vector of n point P on the line, is the position vector of fied point on the line, λ is rel prmeter which gives the position vector of different point on the line nd m is the direction vector of the line, there re three ws of epressing the reltionship etween the coordintes of n point P on this line: ( Vector form r = + λm r = 1 i + 1 j + λ(i + j or r = ( 1 1 + λ ( ( Prmetric form = 1 + λ = 1 + λ (c Crtesin form _ 1 = _ 1 = λ When one or more of the direction vector components re zero, e.g. when = 0, the Crtesin eqution of the line is epressed s _ 1 = λ, = 1 m P ( Vector form r = + λm or r = 1 i + 1 j+z 1 k + λ(i + j + ck or r = ( 1 1 z 1 + λ ( c ( Prmetric form = 1 + λ, = 1 + λ, z = z 1 + λc (c Crtesin form z _ 1 = _ 1 = _ z z 1 c = λ When one or more of the direction vector components re zero, e.g. when = 0, the Crtesin eqution of the line is epressed s = 1, _ 1 = _ z z 1 c = λ. m P Note: The direction rtios of the line : : c pper in ll three forms. Vector Equtions of the es 47 The vector equtions of the es re s follows: on plne ( -is: r = λ ( 1 0, λ ( -is: r = λ ( 0 1, λ in spce ( -is: r = λ ( 1 0 0, λ ( -is: r = λ ( 0 1 0, λ (c z-is: r = λ ( 0 0 1, λ -Level Stud Smrt H2 Mthemtics Vectors nd Comple Numers 10 Chpter 1 Vectors I
Eqution of Line Prllel to il Plnes 48 Vector form Prmetric form Crtesin form - plne (z = 0 (0, 0, 1 z r = ( 1 1 z 1 + λ ( 0 = 1 + λ = 1 + λ z = z 1 _ 1 = _ 1, z = z 1 -z plne ( = 0 (1, 0, 0 z 1 r = ( 1 z 1 ( + λ 0 c = 1 = 1 + λ z = z 1 + λc _ = 1, 1 = z z 1 c -z plne ( = 0 (0, 1, 0 1 r = ( 1 z 1 ( + λ 0 c = 1 + λ = 1 z = z 1 + λc _ 1 = z z 1 c, = 1 z Pirs of Lines in 3-dimensionl Spce 49 The loction of two lines l 1 : r = 1 + λm 1 nd l 2 : r = 2 + μm 2 in spce m e such tht the lines re: Two lines Digrm Test ( Prllel nd intersect (i direction vectors m 1 nd m 2 of lines re prllel (i.e. m 1 = km 2 (ii point on l 1 lso lies on l 2 ( Prllel nd do not intersect (i direction vectors m 1 nd m 2 of lines re prllel (i.e. m 1 = km 2 (ii point on l 1 does not lie on l 2 (c Not prllel nd intersect (i direction vectors m 1 nd m 2 of lines re not prllel (i.e. m 1 km 2 (ii unique vlues of λ nd μ eist such tht 1 + λm 1 = 2 + μm 2 (d Not prllel nd do not intersect (i.e. skew lines (i direction vectors m 1 nd m 2 of lines re not prllel (i.e. m 1 km 2 (ii unique vlues of λ nd μ does not eist such tht 1 + λm 1 = 2 + μm 2 -Level Stud Smrt H2 Mthemtics Vectors nd Comple Numers 11 Chpter 1 Vectors I
Tpes of Questions Vector pertions Tpe (i Find constnt given two vectors with the unknown constnt nd reltion etween the moduli of the two vectors. (ii Find the position vector of point given the position vectors of two points nd the rtio in which the required point divides the line segment formed the two given points. (iii Find unit vector nd (or vector prllel to given vector. (iv Find the sum of two vectors given the two vectors. (v Find the sclr product ( ^ of two given vectors nd give its geometricl interprettion. (vi Find the cross product ( ^ of two given vectors nd give its geometricl interprettion. Vector Eqution of Line Tpe B Find eqution of line in Vector, Crtesin nd(or Prmetric forms given the (B1 position vector of point on the line nd direction vector of the line, (B2 position vectors of two points on the line, (B3 position vector of point on the line nd eqution of prllel or perpendiculr line. Reltionship Between Two Lines Tpe C Given equtions of two lines, (i show whether the two lines re prllel, perpendiculr, intersecting or skewed, (ii find the point of intersection if the two lines intersect, (iii find the ngle etween the two lines. Tpe D Find constnt, given two vectors with unknown constnt nd ngle etween the two vectors. Projection of Vector on Line Tpe E Find the length of the projection of vector onto (i line, (ii perpendiculr of the line, given the vector or coordintes of end points of the vector nd eqution of the line. Reltionship Between Point nd Line Tpe F Show tht point lies on line given the point nd the eqution of the line. Tpe G Given the eqution of line nd point which is not on the given line, (i find the foot of the perpendiculr from the point to the line, (ii find the reflection of the point in the line, (iii find the shortest distnce from the point to the line, (iv find the reflection of the line in nother line. -Level Stud Smrt H2 Mthemtics Vectors nd Comple Numers 12 Chpter 1 Vectors I
Tpe Ke Concepts 1 Modulus If B = ( z, then B = 2 Prllel Vectors _ 2 + 2 + z 2. If B = k CD, where k is sclr, then Vector conclusion Geometricl conclusion B // CD B // CD B = k CD B = kcd Conversel, if B // CD, then B = k CD. 3 Unit Vector The unit vector of vector is ^ = nd ^ = 1. 4 Tringulr Lw of Vector ddition PQ + QR = PR + = ( + + R P 5 Rtio Theorem If P is the point which divides B internll in the rtio λ : μ, then P = μ + λ B μ + λ = λ + μ λ + μ Q λ P μ B 6 Sclr Product If = 1 i + 2 j + 3 k nd = 1 i + 2 j + 3 k, then = 1 1 + 2 2 + 3 3 7 Vector Product or Cross Product If = 1 i + 2 j + 3 k nd = 1 i + 2 j + 3 k, then = i j 1 2 k 3 1 2 3 R or = 2 3 2 3 i 1 3 1 3 j + 1 2 1 2 k = ( 2 3 3 2 i ( 1 3 3 1 j + ( 1 2 2 1 k = ( i 1 j 2 k 3 i j 1 2 k 3 1 2 3 1 2 3 = 2 3 i + 3 1 j + 1 2 k 3 2 i 1 3 j 2 1 k = ( 2 3 3 2 i ( 3 1 1 3 j + ( 1 2 2 1 k -Level Stud Smrt H2 Mthemtics Vectors nd Comple Numers 13 Chpter 1 Vectors I
8 Length of Sides of Right-ngled Tringle B = length of projection of onto = C cos θ = = ^ BC = length of projection of onto c = C sin θ = C ^ sin θ θ C B c = C ^ = ^ Tpe Worked Emple Question Given Required Referred to the origin, the points nd B re such tht = mi 2j where m > 0 nd B = 4i + 6k. (i Given tht B = 2, find the vlue of m. (ii The point P on B is such tht P : B = 2 : 5. Find P. (iii If B =, find the unit vector ^. If C // B find C such tht C = 5. (iv The point Q is such tht BQ form prllelogrm. Find Q. (v Evlute ^ nd give its geometricl mening. (vi Evlute ^ nd give its geometricl mening. (i Vectors = mi 2j nd B = 4i + 6k with constnt m nd reltion etween their moduli B = 2 (ii Vectors = mi 2j nd B = 4i + 6k nd point P which divides B such tht P : PB = 2 : 3 (iii Vector B = 4i + 6k (iv Vectors = mi 2j nd B = 4i + 6k (v Vector = mi 2j nd unit vector ^ (vi Vector = mi 2j nd unit vector ^ (i Find constnt m. (ii Find P. (iii Find unit vector ^ nd prllel vector C such tht C = 5. (iv Find the sum of two vectors Q. (v Find sclr product ^ nd give its geometricl mening. (vi Find cross product ^ nd give its geometricl mening. -Level Stud Smrt H2 Mthemtics Vectors nd Comple Numers 14 Chpter 1 Vectors I
Steps 1 Find the moduli of the vector(s. (i B = ( 4 0 6 R B = 4i + 6k B = 4 2 + 6 2 = ( 2 m R = mi 2j = _ m 2 + ( 2 2 2 Form eqution sed on given condition. B = 2 4 2 + 6 2 = 2 m 2 + ( 2 2 3 Solve to find constnt. 52 = 4(m 2 + 4 m 2 + 4 = 13 m 2 = 9 m = 3 or 3 (N.. s m > 0 4 Find vector using rtio theorem. (ii 3i 2j 2 P 3 B 4i + 6k 2 ( 4 0 ( 6 + 3 2 3 P = 2 + 3 R 2(4i + 6k + (3(3i 2j P = 2 + 3 ( 8 0 ( 12 + 6 9 8i + 12k + 9i 6j = 5 = 5 = 1 5 ( 17 6 12 = 1 5 (17i 6j + 12k -Level Stud Smrt H2 Mthemtics Vectors nd Comple Numers 15 Chpter 1 Vectors I
Chpter 1 Tpe Guided Eercises Step 1 Find the moduli of the vector(s. (i = 4pi 12pj + 3k = (4p 2 + ( 12p 2 + 3 2 = 16p 2 + 144p 2 + 9 = 160p 2 + 9 2 Form eqution sed on given condition. 160p 2 + 9 = 7 3 Solve to find constnt. 160p 2 + 9 = 49 160p 2 = 40 p 2 = 40 160 p 2 = 1 4 p = 1 2 or p = 1 2 (N.. s p > 0 4 Find vector using rtio theorem. (ii _ B = 2 + P 3 2 3 B = 2 + P 1 B(3, 1, 2 P = 3 B 2 P = 3 ( 3 1 ( 2 2 6 2 3 P = ( 9 3 ( 6 4 P = ( 15 5 12 6 5 Find unit vector ^v = v v. (iii c = P P ( 15 5 = 5 2 + 15 2 ( 15 5 _ = 250 5 ( 1 3 _ = 5 10 = 1 10 ( 1 3 = 10 10 ( 1 3 (2, 6, 3 -Level Stud Smrt H2 Mthemtics Vectors nd Comple Numers P S Chpter 1 Step 6 Find prllel vector v = v ^v. Since DC // P, DC = 5c = 5 ( 10 10 ( 1 3 = 10 2 ( 1 3 7 Use tringle lw of vector ddition to find the sum of two vectors. P (iv P = P = ( 5 15 0 ( 2 6 3 = ( 3 21 3 ( 6 2 3 8 Evlute sclr product nd stte geometricl mening. (v c = ( 6 2 3 10 10 ( 1 3 = 10 10 ( 6 2 ( 3 1 3 = 10 10 [(2(1 + ( 6(3 + (3(0] = 10 10 2 18 = 10 10 (16 = 8 5 10 ( 5 15 0 c represents the length of the projection of onto P. 9 Evlute cross product nd stte geometricl mening. ( 6 2 ( 3 3 1 2 (vi ^ = 3 2 + 1 2 + 2 2 ( 6(2 (3(1 [(2(2 (3(3] (2(1 ( 6(3 = 14 = 1 14 ( 15 5 2 = 5 14 ( 3 1 4 = 5 ( 3 2 + 1 2 + 4 2 14 _ = 5 26 14 = 5 13 7 = 5 7 91