VECTORS. 2. Physical quantities are broadly divided in two categories viz (a) Vector Quantities & (b) Scalar quantities.

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1 J-Mthemtics. INTRDUCTIN : Vectors constitute one of the severl Mthemticl systems which cn e usefully employed to provide mthemticl hndling for certin types of prolems in Geometry, Mechnics nd other rnches of pplied Mthemtics. Vectors fcilitte mthemticl study of such physicl quntities s possess Direction in ddition to Mgnitude. Velocity of prticle, for exmple, is one such quntity.. Physicl quntities re rodly divided in two ctegories viz () Vector Quntities & () Sclr quntities. ( ) Vector quntities : ny quntity, such s velocity, momentum, or force, tht hs oth mgnitude nd direction nd for which vector ddition is defined nd meningful; is treted s vector quntities. Note : Quntities hving mgnitude nd direction ut not oeying the vector lw of ddition will not e treted s vectors. For exmple, the rottions of rigid ody through finite ngles hve oth mgnitude & direction ut do not stisfy the lw of vector ddition therefore not vector. ( ) Sclr quntities : quntity, such s mss, length, time, density or energy, tht hs size or mgnitude ut does not involve the concept of direction is clled sclr quntity.. MTHMTICL DSCRIPTIN F VCTR & SCL R : To understnd vectors mthemticlly we will first understnd directed line segment. Directed line segment : ny given portion of given stright line where the two end points re distinguished s Initil nd Terminl is clled Directed Line Segment. VCTRS ND6\\Dt\04\Kot\J-dvnced\SMP\Mths\Unit#0\NG\0-VCTR(THRY).p65 The directed line segment with initil point nd terminl point is denoted y the symol. The two end points of directed line segment re not interchngele nd the directed line segments. nd must e thought of s different. ( ) Vector : directed line segment is clled vector. very directed line segment hve three essentil chrcteristics. (i) (ii) Length : The length of will e denoted y the symol Clerly, we hve Support : The line of unlimited length of which directed line segment is prt is clled its line of support or simply the Support.

2 J-Mthemtics (iii) ( ) Sclr : Sense : The sense of is from to nd tht of from to so tht the sense of directed line segment is from its initil to the terminl point. ny rel numer is sclr. 4. CL SSIFICTIN F VCTRS : There is very importnt clssifiction of vector in which vectors re divided into two ctegories. ( ) Free vectors : If vector cn e trnslted nywhere in spce without chnging its mgnitude & direction, then such vector is clled free vector. In other words, the initil point of free vector cn e tken nywhere in spce keeping its mgnitude & direction sme. c c ( ) Loclised vectors : For vector of given mgnitude nd direction, if its initil point is fixed in spce, then such vector is clled loclised vector. xmples : Torque, Moment of Inerti etc. r F P r F Unless & until stted, vectors re treted s free vectors. 5. QULITY F TW VCTRS : Two vectors re sid to e equl if they hve ( ) the sme length, ( ) the sme or prllel supports nd ( c ) the sme sense. Note : Components of two equl vectors tken in ny ritrry direction re equl. i.e If ˆ ˆ ˆ ˆ ˆ ˆ i j k, i j k,, Illustrtion : Let ˆ ˆ ˆ r i j 5k, i ˆ ˆj k, ˆ ˆi ˆj kˆ find. Solution : i ˆ ˆj 5k ˆ (i ˆ ˆj k) ˆ (i ˆ ˆj k) ˆ ( i ˆ ˆj k) ˆ, where ˆ ˆ i, j & ˆk re the unit vectors tken long co-ordinte xes, then nd = ( )i ˆ ( ) ˆj ( )kˆ quting components of equl vectors =...(i) =...(ii) = 5...(iii) on solving (i), (ii) & (iii) we get,, c i ˆ ˆj kˆ. r c, then So 6 ns. ND6\\Dt\04\Kot\J-dvnced\SMP\Mths\Unit#0\NG\0-VCTR(THRY).p65

3 J-Mthemtics Do yourself - : (i) If = î + ĵ 7 ˆk nd = î + ĵ 7 ˆk re two equl vectors, then find +. (ii) If, re two vectors then which of the following sttements is/re correct - () = = () = = ± (C) = = (D) = = ± 6. LFT ND RIGHT - HNDD RINTTIN (CNFIGUR TINS) : z z x x y y Right - Hnded Triplet Left - Hnded Triplet For ech hnd tke the directions x, y nd z s shown in the figure. Thus we get two rectngulr coordinte systems. Cn they e mde congruent? They cnnot e, ecuse the two hnds hve different orienttions. Therefore these two systems re different. rectngulr coordinte system which cn e mde congruent with the system formed with the help of right hnd (or left hnd) is clled right hnded (or left hnded) rectngulr coordintes system. Thus we hve the following condition to identify these two systems using sense of rottion : ( ) If the rottion from x to y is in the nticlockwise direction nd z is directed upwrds (see right hnd), then the system is right hnded. ( ) If the rottion from x to y is clockwise nd z is directed upwrd (see left hnd) then the system is left hnded. Here fter we shll use the right-hnded rectngulr Crtesin coordinte system (or rtho-norml system). ND6\\Dt\04\Kot\J-dvnced\SMP\Mths\Unit#0\NG\0-VCTR(THRY).p65 7. LGR F VCTRS : It is possile to develop n lger of Vectors which proves useful in the study of Geometry, Mechnics nd other rnches of pplied Mthemtics. ( ) ddition of two vectors : The vectors hve mgnitude s well s direction, therefore their ddition is different thn ddition of rel numers. Let nd e two vectors in plne, which re represented y nd CD. Their ddition cn e performed in the following two wys : ( i ) Tringle lw of ddition of vectors : If two vectors cn e represented in mgnitude nd direction y the two sides of tringle, tken in order, then their sum will e represented y the third side in reverse order. Let e the fixed point in the plne of vectors. Drw line segment from, equl nd prllel to, which represents the vector. Now from, drw line segment F, equl nd prllel to CD, which represents the vector. Line segment F otined y joining nd F represents the sum of vectors nd. + C F D

4 J-Mthemtics i.e. F F or F This method of ddition of two vectors is clled Tringle lw of ddition of vectors. ( i i) Prllelogrm lw of ddition of vectors : If two vectors e represented in mgnitude nd direction y the two djcent sides of prllelogrm then their sum will e represented y the digonl through the co-initil point. Let nd e vectors drwn from point denoted y line segments Q R P nd Q. Now complete the prllelogrm PRQ. Then the vector represented y the digonl R will represent the sum of the vectors nd. i.e. P Q R or R This method of ddition of two vectors is clled Prllelogrm lw of ddition of vectors. ( i i i) Properties of vector ddition : () (commuttive) () ( ) c ( c) (ssocitivity) () 0 0 (dditive identity) (4) ( ) 0 ( ) (dditive inverse) + P ( ) Polygon lw of vector ddition (ddition of more thn two vectors): ddition of more thn two vectors is found to e y repetition of tringle lw. Suppose we hve to find the sum of five vectors,, c, d vectors e represented y line segment,, C, CD nd e. If these nd D respectively, then their sum will e denoted y. This is the vector e ++c+d+e D d + + c+ d ++c + C c represented y rest (lst) side of the polygon CD in reverse order. We cn lso mke it cler this wy : y tringle's lw c C C d D D e or or ( ) c C or ( c) d D or ( c d) e Here, we see tht is represented y the line segment joining the initil point of the first vector nd the finl point of the lst vector e. In order to find the sum of more tht two vectors y this method, polygon is formed. Therefore this method is known s the polygon lw of ddition. Note : If the initil point of the first vector nd the finl point of the lst vector re the sme, then the sum of the vectors will e null vector. 4 ND6\\Dt\04\Kot\J-dvnced\SMP\Mths\Unit#0\NG\0-VCTR(THRY).p65

5 J-Mthemtics ( c ) Sutrct ion of Vectors : rlier we hve descried the vector whose length is equl to vector ut direction is opposite. Sutrction of vector nd is defined s ddition of nd ( ). It is written s follows : ( ) Geometricl representtion : + C In the given digrm, nd re represented y nd. We extend the line in opposite direction upto C, where = C. The line segment C will represent the vector. y joining the points nd C, the vector represented y C is ( ). i.e. denotes the vector. Note : (i) ( ) 0 (ii) Hence sutrction of vectors does not oey the commuttive lw. (iii) ( c) ( ) c i.e. sutrction of vectors does not oey the ssocitive lw. ( d ) Multipliction of vector y sclrs : If is vector & m is sclr, then m( ) is vector prllel to whose modulus is m times tht of. This multipliction is clled SCLR MULTIPLICTIN. If & re vectors & m, n re sclrs, s, then : (i) m() ()m m (ii) m(n) n(m) (mn) (iii) (m n) m n (iv) m( ) m m ND6\\Dt\04\Kot\J-dvnced\SMP\Mths\Unit#0\NG\0-VCTR(THRY).p65 Illustrtion : Solution : Illustrtion : CD is prl lelogrm whose digonls meet t P. If is fixed poi nt, t hen + + C + D equls :- () P () P (C) P (D) 4 P Since, P isects oth the digonl C nd D, so + C = P nd + D = P + + C + D = 4 P ns. [D],, P, Q, R re five points in ny plne. If forces P, Q, R cts on point nd force P, Q, R cts on point then resultnt is :- () () (C) PQ (D) PR Solution : From figure P + P = Q + Q = R + R = So ( P + Q + R ) + ( P + Q + R ) = so required resultnt =. Illustrtion 4 : P Q R ns. [] Prove tht the line joining the middle points of two sides of tringle is prllel to the third side nd is of hlf its length. 5

6 J-Mthemtics Solution : Let the middle points of side nd C of C e D nd respectively. D nd C Now in C, y tringle lw of ddition D C C D C D C D C Hence, line D is prllel to third side C of tringle nd hlf of it. C Do yourself - : (i) If,, c e the vectors represented y the sides of tringle tken in order, then prove tht c = 0 (ii) If P Q Q R, then prove tht the points P, Q nd R re colliner. (iii) For ny two vectors nd prove tht () + + () + (c) + Note : In generl for ny non-zero vectors, lwys represent the three sides of tringle. & c one my note tht lthough c 0 8. CLLINR VCTRS : Two vectors re sid to e colliner if their supports re prllel disregrds to their direction. Colliner vectors re lso clled Prllel vectors. If they hve the sme direction they re nmed s like vectors otherwise unlike vectors. Note : (i) Symoliclly two non zero vectors & re colliner if nd only if, K 6 ut it will not c, where K R (ii) If ˆ ˆ i j ˆ k nd ˆ ˆ i j ˆ k re two colliner vectors then. (iii) If & re two non-zero, non-colliner vectors such tht x y 0 x = y = 0 9. C-INITIL VCTRS : Vectors hving sme initil point re clled Co-initil Vectors. Illustrtion 5 : If nd re non-colliner vectors, then find the vlue of x for which vectors : (x ) nd ( x) re colliner. Solution : Since the vectors nd re colliner. there exist sclr such tht = (x ) + = {( + x) } (x ( + x)) + ( + ) = 0 x ( + x) = 0 nd + = 0 x ( + x) = 0 nd = x + ( + x) = 0 4x = 0 x = 4 C D c d ns. ND6\\Dt\04\Kot\J-dvnced\SMP\Mths\Unit#0\NG\0-VCTR(THRY).p65

7 J-Mthemtics Illustrtion 6 : If ( î + ĵ ), (p î + 9 ĵ ) nd C ( î ĵ ) re colliner, then the vlue of p is :- Solution : () / () / (C) 7/ (D) 5/ = (p ) î + 6 ĵ, C = î 4 ĵ p 6 Now,, C re colliner C = p = 7/ 4 Illustrtion 7 : The vlue of when = î ĵ + ˆk nd = 8 î + ĵ + 4 ˆk re prllel is :- () 4 () 6 (C) (D) ns. [C] Solution : Since & re prllel 8 = = 4 = ns. [C] Do yourself - : ( i ) In the given figure, which vectors re (provided vertices,, C, D,, F re ll fixed) : () Prllel () qul (c) Coinitil (d) Prllel ut not equl. (ii) If i ˆ ˆj 4kˆ nd 8ˆi ˆj 6kˆ such tht =, then equls to (iii) If 5c nd 8 7 4c, then which sttement is/re true : () () c (C), nd c re colliner vectors. (D) F x y d z D c C 0. CPLNR VCTRS : given numer of vectors re clled coplnr if their supports re ll prllel to the sme plne. Note tht TW VCTRS R LWYS CPLNR. Note : Coplnr vectors my hve ny directions or mgnitude.. RPRSNTTIN F VCTR IN SPC IN TRMS F ND6\\Dt\04\Kot\J-dvnced\SMP\Mths\Unit#0\NG\0-VCTR(THRY).p65 RTHNRML TRID F UNIT VCTRS : Let P(x, y, z) e point in spce with reference to X, Y nd Z s the coordinte xes, then = x, = y nd C = z Let ˆi, ˆj, k ˆ e unit vectors long X, Y nd Z respectively, then xi, ˆ yj, ˆ C zkˆ P C ' C 'P = C If P r r xi ˆ yj ˆ zkˆ = C = xi ˆ yj ˆ zk ˆ r P x y z. PSITIN VCTR : [ C 'P C ] Let e fixed origin, then the position vector of point P is the vector P. If & re position vectors of two point nd, then = pv of pv of. ' C r P (x,y,z) C' ' 7

8 J-Mthemtics. ZR VCTR R NULL VCTR : vector of zero mgnitude i.e. which hs the sme initil & terminl point is clled zero vector. It is denoted y. It cn hve ny ritrry direction nd ny line s its line of support. 4. UNIT VCTR : vector of unit mgnitude in direction of vector is clled unit vector long nd is denoted y â symoliclly â (provided 0 ) 5. SCTIN FRMUL : If & re the position vectors of two points & then the p.v. of point C (r ) m : n is given y : which divides in the rtio ( ) Internl Division : m n C r m n Note : Position vector of mid point of = ( ) xternl division : m n C r m n ( ) m C( c ) n ( ) m ( ) ( ) n C( r ) Illustrtion 8 : Solution : Prove tht the medins of tringle re concurrent. Let C e tringle nd position vectors of three vertices, nd C with respect to the origin e, nd c respectively.,, C c gin, let D e the middle point of the side C, c so the position vector of point D is D [ Position vector of the middle point of ny line =/ (Sum of position vectors of end point of line)] Now tke point G, which divides the medin D in the rtio :...D Position vector of point G is G.. c 8 c G C D Similrly, the position vector of the middle points of the other two medins, which divide the medins in the rtio : will comes out to the sme c, which is the position vector of G. Hence, the medins of the tringles meet in G i.e. re concurrent. ND6\\Dt\04\Kot\J-dvnced\SMP\Mths\Unit#0\NG\0-VCTR(THRY).p65

9 J-Mthemtics Illustrtion 9 : If the middle points of sides C, C & of tringle C re respectively D,, F then position vector of centroid of tringle DF, when position vector of,,c re respectively i + j, j + k, k + i is - () (i + j + k) () (i + j + k) (C) (i + j + k) (D) (i + j + k) Solution : The position vector of points D,, F re respectively i j So, position vector of centroid of DF = + k, i + k j i j k j i k k i j = nd i k + j [i + j + k]. ns. [D] Do yourself - 4 : (i) Find the position vectors of the points which divide the join of the points nd internlly nd externlly in the rtio :, (ii) CD is prllelogrm nd P is the point of intersection of its digonls. If is the origin of reference, show tht C D 4P (iii) Find the unit vector in the direction of î 6 ĵ + ˆk. 6. VCTR QUTIN F LIN : () R(r) R(r) () () ND6\\Dt\04\Kot\J-dvnced\SMP\Mths\Unit#0\NG\0-VCTR(THRY).p65 r=+µ( ) Prmetric vector eqution of line pssing through two points () where t is prmeter. If the line psses through the point () is r = + t. Note : (i) qutions of the isectors of the ngles etween the lines r r = + p(c ˆ ˆ). (ii) (iii) In plne, two lines re either intersecting or prllel. Two non prllel nor intersecting lines re clled skew lines. r=+ & () is given y, r = + t( ) & is prllel to the vector then its eqution & r c is, r = + t( ˆ + c) ˆ Illustrtion 0 : In tringle C, D nd re points on C nd C respectively, such tht D = DC nd = C. Let P e the point of intersection of D nd. Find P/P using vector methods. (J- 99) & 9

10 J-Mthemtics Solution : Let the position vectors of nd e nd respectively. qutions of D nd re r t( / )... (i) r s( / 4 )... (ii) C(0) If they intersect t P we must hve identicl vlues of r. Compring the coefficients of nd in (i) nd (ii), we get t = s 4, t = s 9 8 solving we get t =, s. Putting for t or s in (i) or (ii), we get the point P s. k. Let P divide in the rtio k :, then P is 4. k () P D k () Illustrtion Compring nd, we get k = 8(k + ) nd = (k + ) k = 8 nd this stisfies the nd reltion lso. Hence the required rtio is 8 :. ns. : Find whether the given lines re coplnr or not r ˆi ˆj 0k ˆ (i ˆ ˆj 8k) ˆ r 4i ˆ ˆj k ˆ (i ˆ 4ˆj 7k) ˆ Solution : L : r ( )i ˆ ( ) ˆj (8 0)kˆ L : r (4 )i ˆ (4 ) ˆj (7 )kˆ The given lines re not prllel. For coplnrity, the lines must intersect. ( )i ˆ ( ) ˆj (8 0)k ˆ (4 )i ˆ (4 ) ˆj (7 )kˆ 4...(i) 4...(ii) (iii) Solving (i) & (ii), =, µ = nd =, µ = stisfies eqution (iii) Given lines re intersecting & hence coplnr. ns. 7. TST F CLLINRITY F THR PINTS : ( ) points C will e colliner if = C, ( ) Three points,, C with position vectors,, c respectively re colliner, if & only if there exist sclrs x, y, z not ll zero simultneously such tht ; x y zc 0, where x + y + z = 0 ( c ) Collinerly cn lso e checked y first finding the eqution of line through two points nd stisfying the third point. Illustrtion : Prove tht the points with position vectors ˆ ˆ i j k, ˆ i ˆ ˆ j 4kˆ colliner. 0 & c + c 7ˆj 0kˆ re ND6\\Dt\04\Kot\J-dvnced\SMP\Mths\Unit#0\NG\0-VCTR(THRY).p65

11 Solution : If we find, three sclrs, m & n such tht m nc 0 where m n 0 then points re colliner. (i ˆ ˆj k) ˆ m(i ˆ ˆj 4k) ˆ n( 7ˆj 0k) ˆ 0 ( m)i ˆ ( m 7n) ˆj ( 4m 0n)kˆ 0 + m = 0, + m 7n = 0, 4m + 0n = 0 Solving, we get =, m =, n = since + m + n = 0 Hence, the points re colliner. liter : i ˆ ˆj 4kˆ ˆi ˆj kˆ ˆi 5ˆj 7kˆ C c 7ˆj 0kˆ ˆi ˆj 4kˆ i ˆ 0ˆj 4kˆ ˆi 5ˆj 7kˆ C Hence, & c re colliner. J-Mthemtics Do yourself - 5 : (i) The position vectors of the points P, Q, R re î + ĵ + ˆk, î + ĵ +5 ˆk nd 7 î ˆk respectively. ely. Prove tht P, Q nd R re colliner. 8. SCLR PRDUCT F TW VCTRS (DT PRDUCT) : Definition : Let nd e two non zero vectors inclined t n ngle. Then the sclr product of with is denoted y. nd is defined s. = cos ; 0. Geometricl Interprettion of Sclr product : ND6\\Dt\04\Kot\J-dvnced\SMP\Mths\Unit#0\NG\0-VCTR(THRY).p65 Now =,. = cos = ( cos ) gin,. = cos = = (mgnitude of ) (Projection of on ) = ( cos ) = (Mgnitude of ) (Projection of on ) ( ). cos ( 0 ) Note tht if is cute then. 0 ( ) (i). & if is otuse then. 0 (ii).. (commuttive) M N

12 J-Mthemtics ( c ). ( c ).. c (distriutive) ( d ). 0 ; (, 0 ) (e) ˆi. ˆi ˆj. ˆj k ˆ. kˆ ; ˆi. ˆj ˆj. kˆ k ˆ. ˆi 0 ( f ) Projection of on.. (Provided 0 ) Note : Projection of on (i) The vector component of. long nd perpendiculr. to [y tringle lw of vector ddition] (ii) The ngle etween & is given y. cos 0 (iii) If ˆ ˆ i j ˆ k & ˆ ˆ ˆ i j k, then., (iv) Mximum vlue of. (v) Minimum vlues of. (vi) ny vector cn e written s, ˆˆ (. i)i (. ˆˆ j) j (.k)k ˆ ˆ ( g ) Vector equt ion of ngle isector : vector in the direction of the isector of the ngle etween the two vectors & is. Hence isector of the ngle etween the two vectors & is (ˆ ) ˆ, where R. isector of the exterior ngle etween & is ( ˆ ) ˆ, R ' C' Illustrtion :,, c, d re the position vectors of four coplnr points,, C nd D respectively. Solution : ( d) If ( d).( c) 0 ( d).(c ), then for the C, D is :- () incentre () orthocentre (C) circumcentre (D) centroid. ( c) Similrly ( d).(c ) = 0 ( d) ( c) = 0 D C D C D is the orthocentre of C. ns. [] C ND6\\Dt\04\Kot\J-dvnced\SMP\Mths\Unit#0\NG\0-VCTR(THRY).p65

13 J-Mthemtics Illustrtion 4 : The vector c, directed long the internl isector of the ngle etween the vector 7 î 4 ĵ 4 ˆk nd î ĵ + ˆk with c = 5 6 is - () 5 ( î 7 ĵ + ˆk ) () 5 (5 î + 5 ĵ + ˆk ) (C) 5 ( î + 7 ĵ + ˆk ) (D) none of these Solution : Let = 7 î 4 ĵ 4 ˆk nd = î ĵ + ˆk internl isector divides the C in the rtio of = 9, C = : C, C C c D D = 9( i ˆ ˆj k) ˆ (7i ˆ 4ˆj 4k) ˆ ˆ 9 = ˆ i 7ˆj k 4 c = ± D 5 6 = ± 5 ( î 7 ĵ + k) D ns.[] Illustrtion 5 : If modulii of vectors,, c re, 4 nd 5 respectively nd nd + c, nd c +, c nd + re perpendiculr to ech other, then modulus of + + c is - () 5 () 5 (C) 50 (D) 0 Solution : ( + c ). +. c = 0 Similrly ( c + ). c +. = 0 nd c ( + ) c. + c. = c + c. = 0 Now + + c = + + c + (. +. c + c. ) = = 50 ND6\\Dt\04\Kot\J-dvnced\SMP\Mths\Unit#0\NG\0-VCTR(THRY).p c = 5 ns. [] Illustrtion 6 : If p th, q th, r th terms of G.P. re the positive numers,, c then ngle etween the vectors Solution : log ˆi log ˆj log c kˆ nd (q r)i ˆ (r p) ˆj (p q)kˆ is :- () () Let x 0 e first term nd x the common rtio of the G.P. (C) sin c = x 0 x p, = x 0 x q, c = x 0 x r log = log x 0 + (p ) log x; If log = log x 0 + (q ) log x; log c = log x 0 + (r ) log x log i log j log c k nd ( q r) i ( r p) j ( p q) k (D) none of these. (log ) ( q r) (log x 0 (p ) log x)(q r) 0, ns. Illustrtion 7 : Find the distnce of the point ( î + ĵ + ˆk ) from the line which is pssing through (4 î + ĵ + ˆk ) nd which is prllel to the vector C i ˆ ˆj 6kˆ. (Roorkee 99)

14 J-Mthemtics Solution : = 0 (,,) (i ˆ ˆj 6k) ˆ M.i ˆ ( i ˆ k). ˆ 7 = = 0 M = M (4i+j+k) M C So, M = = 0 ns. Illustrtion 8 : Prove tht the medins to the se of n isosceles tringle is perpendiculr to the se. Solution : The tringle eing isosceles, we hve C(c) = C c Now P lso C c... (i) where P is mid-point of C. c P.C.(c ) (c ) () P (0) (C ) 0 {y (i)} Medin P is perpendiculr to se C. Do yourself - 6 : (i) Find the ngle etween two vectors & with mgnitude nd respectively nd such tht. =. (ii) Find the vlue of ( + ). ( ) if = î + ĵ + ˆk, = î + ĵ ˆk. (iii) The sclr product of the vector î + ĵ + ˆk with unit vector long the sum of the vectors î + 4 ĵ 5 ˆk nd î + j + ˆk is equl to, find. (iv) Find the projection of the vector 4i ˆ ˆj kˆ on the vector 4 i ˆ 6ˆj kˆ. lso find component of long nd perpendiculr to. ( v ) Find the unit vectors long the ngle isectors etween the vectors ˆi ˆj kˆ nd i ˆ 6ˆj kˆ. 9. LINR CMINTINS : Given finite set of vectors,, c,... then the vector r x y zc... is clled liner comintion of,, c,... for ny x, y, z... R. FUND MNTL THRM IN PLN : Let, e non zero, non colliner vectors. then ny vector r coplnr with, cn e expressed uniquely s liner comintion of, i.e. there exist some unique x, y R such tht x y r. Illustrtion 9 : Find vector c in the plne of i ˆ ˆj kˆ nd nd c.( i ˆ ˆj k) ˆ Solution : ny vector in the plne of & cn e written s x y let c x y [y fundmentl theorem in plne] ˆi ˆj kˆ such tht c is perpendiculr to ND6\\Dt\04\Kot\J-dvnced\SMP\Mths\Unit#0\NG\0-VCTR(THRY).p65

15 J-Mthemtics Now, given tht c. 0 (x y). 0 x. y 0 x( + ) + y() = 0 x + y = 0...(i) lso (x y).( i ˆ ˆj k) ˆ x.( i ˆ ˆj k) ˆ y.( i ˆ ˆj k) ˆ x( ) + y( + ) = y 4 y x 8 ˆ ˆ Hence the required vector c (i ˆ ˆj k) ( ˆi ˆj k) 8 4 = [ 6ˆi ˆj kˆ i ˆ ˆj k] ˆ [ 4ˆi 5ˆj k] ˆ ns. 8 8 Do yourself - 7 : (i) Find vector r in the plne of p ˆi ˆj r.q nd q ˆj kˆ such tht r is perpendiculr to p nd 0. VCTR PRDUCT F TW VCTRS (CRSS PRDUCT) : ( ) If & re two vectors & is the ngle etween them, then sin nˆ, where ˆn is the unit vector perpendiculr to oth C = & such tht, & n forms right hnded screw system. Sign convention : C = ND6\\Dt\04\Kot\J-dvnced\SMP\Mths\Unit#0\NG\0-VCTR(THRY).p65 Right hnded screw system :, nd ˆn form right hnded system it mens tht if we rotte vector towrds the direction of. through the ngle, then ˆn dvnces in the sme direction s right hnded screw would, if turned in the sme wy. ( ) Lgrnges Identity : For ny two vectors &.. ; ( ) (. ).. ( c ) Formultion of vector product in terms of sclr product : The vector product is the vector c, such tht (i) c (.) (ii) c. 0; c. 0 nd (iii),, c form right hnded system ( d ) (i) 0 & re prllel (colliner) ( 0, 0 ) (ii) (not commuttive) (iii) ( m ) ( m ) m ( ) where m is sclr. i.e. K, where K is sclr 5

16 J-Mthemtics (iv) ( c) ( ) ( c ) (distriutive over ddition) (v) ˆi ˆi ˆj ˆj kˆ kˆ 0 (vi) ˆi ˆj k ˆ, ˆj kˆ ˆi, kˆ ˆi ˆj j i (e) If ˆ ˆ ˆ i j k ( f ) Geometriclly re represented y & k ˆi ˆj kˆ & ˆ ˆ ˆ i j k, then = re of the prllelogrm whose two djcent sides. ( g ) (i) Unit vector perpendiculr to the plne of & (ii) vector of mgnitude r & perpendiculr to the plne of & (iii) If is the ngle etween &, then is ˆn r ( ) is s in (h ) Vector re : (i) If, & c re the pv s of points, & C then the vector re of tringle C c c. (ii) The points, & C re colliner if c c 0 (iii) re of ny qudrilterl whose digonl vectors re d & d is given y d d. Illustrtion 0 : Find the vectors of mgnitude 5 which re perpendiculr to the vectors ˆi ˆj kˆ. Solution : Unit vectors perpendiculr to & ˆi ˆj kˆ = 5ˆi 5ˆj 5kˆ Unit vectors ( 5i ˆ 5ˆj 5k) ˆ 5 6 i ˆ ˆj kˆ 5 Hence the required vectors re (i ˆ ˆj k) ˆ ns. Illustrtion : If,, c re three non zero vectors such tht c nd c, prove tht,,c re mutully t right ngles nd nd c. nd ND6\\Dt\04\Kot\J-dvnced\SMP\Mths\Unit#0\NG\0-VCTR(THRY).p65

17 Solution : c nd c c, c nd, c, c nd c,,c re mutully perpendiculr vectors. gin, c nd c c nd c sin c nd c sin c nd c c c putting in c c nd c J-Mthemtics Illustrtion : Show tht the re of the tringle formed y joining the extremities of n olique side of trpezium to the midpoint of opposite side is hlf tht of the trpezium. Solution : Let CD e the trpezium nd e the midpoint of C. Let e the initil point nd let e the position vector of nd d tht of D. Since DC is prllel to, t is vector long DC, so ND6\\Dt\04\Kot\J-dvnced\SMP\Mths\Unit#0\NG\0-VCTR(THRY).p65 tht the position vector of c is d t. d t d ( t ) the position vector of is re of D = d ( t) d = ( t) d 4 re of the trpezium Illustrtion : Let & e two non-colliner unit vectors. If (0) D (d) = re (CD) + re (C). = (d t) (d t) d t d d ( t) d = D u. v, then v is - () u () u u. (C) u u. Solution : u.. (.)(.) = cos (where is the ngle etween nd ) = cos = sin v sin u u.u. (.) (.) = (.) = sin v u lso u. 0 Hence, v u u u. & C () [J 99] u u. (D) ns. (, C) 7

18 J-Mthemtics Do yourself - 8 : (i) If = c d nd c = d, then show tht ( d ) is prllel to ( c ) when d nd c. (ii) Find, if = î + ˆk nd = î + ĵ + ˆk. (iii) For ny two vectors u & v, prove tht [J 98] () (u. v) u v u v () ( u )( v ) ( u.v) u v (u v). SHRTST DISTNC TWN TW LINS : If two lines in spce intersect t point, then oviously the shortest distnce etween them is zero. Lines which do not intersect & re lso not prllel re clled skew lines. In other words the lines which re not coplnr re skew lines. For Skew lines the direction of the shortest distnce vector would e perpendiculr to oth the lines. The mgnitude of the shortest distnce vector would e equl to tht of the projection of long the direction of the line of shortest distnce, LM is prllel to p q i.e. LM = Projection of on LM =. (p q) ( ).(p q) Projection of on p q p q p q ( ) The two lines directed long p & q will intersect only if shortest distnce = 0 i.e. ( ).(p q) 0 i.e. ( ) lies in the plne contining p & q p q 0 ( ) If two lines re given y r K & r K i.e. they ( d ) re prllel then, d Illustrtion 4 : Find the shortest distnce etween the lines r (4i ˆ ˆj) (i ˆ ˆj k) ˆ nd r (i ˆ ˆj k) µ(i ˆ 4ˆj 5k) ˆ Solution : We known, the shortest distnce etween the lines r nd r ( ).( ) d = Compring the given eqution with the equtions r we hve ˆ ˆ ˆ ˆ ˆ 4 i j, i j k Now, ˆ ˆ ˆ i 0 j k, 8 L M p q nd r ˆi ˆj kˆ nd ˆ ˆ ˆ i 4 j 5k nd = ˆi ˆj kˆ i ˆ ˆj 0kˆ 4 5 p q is given y respectively, ( ˆ ˆ ˆ ˆ ˆ ˆ ).( ) ( i 0 j k).(i j 0k) = 6 nd ( ).( ) 6 6 Shortest distnce d =. ns. 5 5 ND6\\Dt\04\Kot\J-dvnced\SMP\Mths\Unit#0\NG\0-VCTR(THRY).p65

19 J-Mthemtics Do yourself - 9 : (i) Find the shortest distnce etween the lines : r (i ˆ ˆj k) ˆ (i ˆ ˆj 4k) ˆ r (i ˆ 4ˆj 5k) ˆ (i ˆ 4ˆj 5k) ˆ. &. SCLR TRIPL PRDUCT / X PRDUCT / MIXD PRDUCT : (i) The sclr triple product of three vectors, & c is defined s :. c c s in cos ccos c where is the ngle etween & & is the ngle etween & c. It is lso defined s [ c], spelled s ox product. ( i i) Sclr triple product geometriclly represents the volume of the prllelopiped whose three coterminous edges re represented y, & c i.e. V [ c ] (iii) In sclr triple product the position of dot & cross cn e interchnged i.e.. ( c ) ( ). c R [ c ] [ c ] [ c ] (iv). ( c). ( c ) i.e. [ c] = [ c ] If ˆ ˆ i j ˆ k ; ˆ ˆ ˆ i j k & ˆ ˆ ˆ c ci c j ck then [ c] = In generl, if l m n l m n ; ; & c c l cm cn c c c ND6\\Dt\04\Kot\J-dvnced\SMP\Mths\Unit#0\NG\0-VCTR(THRY).p65 then [ c] [ l m n ] c c c ( v ) If,, c re coplnr [ c] ; where l, m & n re non coplnr vectors. = 0,, c re linerly dependent. (vi ) Sclr product of three vectors, two of which re equl or prllel is 0 i.e. [ c ] 0 Note : If,, c re non- coplnr then [ c] 0 for right hnded system & [ c] 0 left hnded system. (vii) [ i j k ] (viii) [ K c] K [ c ] (ix) [( ) c d ] [ c d ] [ c d ] (viii) The Volume of the tetrhedron C with s origin & the pv s of, nd C eing, & c given y V [ c ] c 6 The position vector of the centroid of tetrhedron if the pv s of its ngulr vertices re,, c & d re given y ( c d) 4 Note tht this is lso the point of concurrency of the lines joining the vertices to the centroids of the opposite fces nd is lso clled the centre of the tetrhedron. In cse the tetrhedron is regulr it is equidistnt from the vertices nd the four fces of the tetrhedron. (ix) [ c c ] 0 & [ c c ] [ c] for re C 9

20 J-Mthemtics Illustrtion 5 : For ny three vectors,, c prove tht [ c c ] [ c] Solution : We hve [ c c ] = {( ) ( c)}.(c ) { c c}.(c ) { 0} = { c c}.(c ) = ( ).c ( c).c ( c).c ( ). ( c). ( c). = [ c] [ c ] { [ c c] 0,[ c c] 0,[ ] 0,[ c ] 0}. ns. Illustrtion 6 : If, = [ c] [ c] [ c] re non-zero nd non-colliner vectors then show [ ˆˆ i]i [ ˆˆ j]j [ k]k ˆ ˆ Solution : Let xi ˆ yj ˆ zkˆ ( ). ˆi (xi ˆ yj ˆ zk). ˆ ˆi ( ).i ˆ = x lso ( ). ˆ j = y & ( ).k ˆ = z [ ˆˆ i]i [ ˆˆ j]j [ k]k ˆ ˆ Do yourself - 0 : (i) If,, c re three non coplnr mutully perpendiculr unit vectors then find [,, c ]. (ii) If r e vector perpendiculr to + + c, where [ c ] = z nd r = ( c ) + m( c ) + n( ), then find l + m + n. (iii) Find the volume of the prllelepiped whose coterminous edges re represented y i ˆ ˆj 4k, ˆ ˆi ˆj kˆ nd c i ˆ ˆj kˆ (iv) xmine whether the vectors i ˆ ˆj k, ˆ ˆi ˆj kˆ nd c i ˆ ˆj 4kˆ form left hnded or right hnded system. ( v ) For three vectors u, v, w which of the following expressions is not equl to ny of the remining three? [J 98] () u.(v w) () (v w).u (C) v.(u w) (D) (u v).w ns.. VCTR TRIPL PRDUCT : Let, & c e ny three vectors, then the expression ( c) 0 is vector & is clled vector triple product. Geometricl interprettion of ( c) c Consider the expression ( c) which itself is vector, since it is ( ) c cross product of two vectors & ( c). Now ( c) is vector perpendiculr to the plne contining & ( c) ut ( c) is vector perpendiculr to the plne & c, therefore ( c) is vector lies in the plne of & c nd perpendiculr to. Hence we cn express ( c) & c i.e. ( c) = x yc where x & y re sclrs. () ( c ) (. c ) (. )c () ( ) c (. c ) (. c) (c) ( ) c ( c ) in terms of ND6\\Dt\04\Kot\J-dvnced\SMP\Mths\Unit#0\NG\0-VCTR(THRY).p65

21 Illustrtion 7 : Prove tht c c c Solution : We hve, c c = c. c d c. c = = d.c d.c. c c 0. c =.c. c. c J-Mthemtics (where, d ( ) ) ( [] 0 ) c. c c c c c c Illustrtion 8 : Show tht ( c).( d) (c ).( d) ( ).(c d) 0 Solution : Let c u, c v, c d w L.H.S = u.( d) v.( d) ( ).w = (u ).d (v ).d.( w) [(c) ].d [(c ) ].d.[ (c d)] [(.)c (c.)].d [(c.) (.)c].d.[(.d)c (.c)d] {(.)(c.d)} {(c.) (.d)} {(c.)(.d)} {(.)(c.d)} {(.c) (.d)} {(.d) (.c)} 0 = R.H.S. Do yourself - : (i) If = î 4 ĵ + 7 ˆk, = î + 5 ĵ 9 ˆk nd c = î + ĵ + ˆk then find [ c ] nd lso ( c ). ND6\\Dt\04\Kot\J-dvnced\SMP\Mths\Unit#0\NG\0-VCTR(THRY).p65 4. LINR INDPNDNC ND DPNDNC F VCTRS : ( ) If x, x,...x n re n non zero vectors, & k, k,...k n re n sclrs & if the liner comintion k x k x...k n x n 0 k 0, k 0...k n 0 then we sy tht vectors x, x,...x n re linerly independent vectors. ( ) If x, x,...x n re not linerly independent then they re sid to e linerly dependent vectors. i.e. if k x k x...k n x n = 0 & if there exists t lest one k r 0 then x, x,...x n re sid to e linerly dependent. FUNDMNTL THRM IN SPC : Let,, c e non-zero, non-coplnr vectors in spce. Then ny vector r, cn e uniquely expressed s liner comintion of,, c i.e. There exist some unique x, y, z R such tht r x y zc. Note : (i) If i ˆ ˆj 5kˆ then is expressed s liner comintion of vectors ˆi, ˆj, k ˆ. lso,, ˆi, ˆj, kˆ form linerly dependent set of vectors. In generl, every set of four vectors is linerly dependent system. (ii) If,, c re three non-zero, non-coplnr vectors then x y zc 0 x = y = z = 0 (iii) ˆi, ˆj, k ˆ re linerly independent set of vectors. For K ˆ i K ˆ ˆ j K k 0 K 0 K K (iv) Two vectors & re linerly dependent is prllel to i.e. 0 liner dependence of &. Conversely if 0 then & re linerly independent. (v) If three vectors,, c re linerly dependent, then they re coplnr i.e. [ c] = 0 conversely, if [ c] 0, then the vectors re linerly independent.

22 J-Mthemtics Illustrtion 9 : Show tht points with position vectors c, c, 4 7 7c re colliner. It is given tht vectors,, c re non-coplnr. Solution : The three points re colliner, if we cn find, nd, such tht ( c) ( c) (4 7 7c) 0 with + + = 0 equting the coefficients, nd c seprtely to zero, we get + 4 = 0, + 7 = 0 nd + 7 = 0 on solving we get, =, =, = So tht + + = 0 Hence the given vectors re colliner. Illustrtion 0 : If ˆi ˆj k, ˆ 4ˆi ˆj 4kˆ nd c ˆi ˆj kˆ Solution : then - re linerly dependent vectors nd c = () =, = () =, = ± (C) =, = ± (D) = ±, = If,, c re linerly dependent vectors, then c should e liner comintion of nd. Let c p q i.e. ˆi ˆj kˆ p(i ˆ ˆj k) ˆ q(4i ˆ ˆj 4k) ˆ quting coefficients of from first nd third, =. ˆi, ˆj,k ˆ we get = p + 4q, = p + q, = p + 4q Now c = + + = + + = {Using = } = ± Hence, = ±, =. ns. (D) 5. CPLNRITY F FUR PINTS : Four points,, C, D with position vectors,, c, d respectively re coplnr if nd only if there exist sclrs x, y, z, w not ll zero simultneously such tht x y zc wd 0 where, x + y + z + w = 0 6. RCIPRCL SYSTM F VCTRS : If,, c & ', ', c ' re two sets of non coplnr vectors such tht. '. ' c. c ', then the two systems re clled Reciprocl System of vectors. c c Note : ' ; ' ; c ' [ c] [ c] [ c] 7. PRPRTIS F RCIPRCL SYSTM F VCTRS : ( ). '.c '.c ' c. ' c. ' 0 ( ) The sclr triple product [ c] formed y three non-coplnr vectors,, c is the reciprocl of the sclr triple product formed from reciprocl system. ND6\\Dt\04\Kot\J-dvnced\SMP\Mths\Unit#0\NG\0-VCTR(THRY).p65

23 Illustrtion : Find set of vectors reciprocl to the vectors, Solution : Let the given vectors e denoted y, [ c ] ( ).c ( ).( ) ( ) nd. nd c where c nd let the reciprocl system of vectors e ' 'nd c ' c ( ) c ( ) ' ; ' ; c ' [ c] [ c] [ c] Illustrtion : If ', ',c ' re required reciprocl system of vectors for, nd c c ', ',c ' then shown tht ; ' ' c c ' 0 [ c] [ c] [ c] where,,c re non-coplnr vectors. J-Mthemtics. ns. Solution : ( c) Here ' [ c] (.c) (.)c ' [ c] (.)c (.c) Similrly ' & (c.) (c.) c c ' [ c] [ c] ' ' c c ' (.c) (.)c (.)c (.c) (c.) (c.) = [.. etc.] [ c] = 0. ns. ND6\\Dt\04\Kot\J-dvnced\SMP\Mths\Unit#0\NG\0-VCTR(THRY).p65 Do yourself - : c (i) If p, [ c ] c q, [c ] r [ c ], then find the vlue of ( + ). p + ( + c ). q + ( c + ). r. (ii) Find the set of vectors reciprocl to the set i ˆ ˆj k, ˆ ˆi ˆj kˆ nd ˆ i ˆj kˆ. Hence prove tht. '. ' c.c '. (iii) If, nd c re non zero, non coplnr vectors determine whether the vectors r c, r 5 c nd r 4 5 c re linerly independent or dependent. 8. PPLICTIN F VCTRS : ( ) Work done ginst constnt force F over displcement s is defined s W F.s ( ) The tngentil velocity V of ody moving in circle is given y V w r where r is the pv of the point P. w r p L v

24 J-Mthemtics ( c ) The moment of F out is defined s M r F where r is the pv of P wrt. The direction of M is long to the norml to the plne PN such tht r, F & M form r' r P F right hnded system. ( d ) Moment of couple = (r r ) F where r & r re pv s of the point of the ppliction of the forces F & F. Q N Do yourself - : (i) Force F 4ˆi ˆj kˆ cting on prticle displces it from the point 5ˆi ˆj 4kˆ. Find the work done y the force F. ˆi ˆj kˆ to the point Miscellneous Illustrtions : Illustrtion : Forces of mgnitudes 5, 4, units ct on prticle in the directions i ˆ ˆj kˆ, ˆ i ˆj kˆ nd ˆi ˆj kˆ respectively, nd the prticle gets displced from the point whose position vector is 6ˆi ˆj kˆ, to the point whose position vector is 9 ˆ i 7ˆj 5kˆ. Find the work done. 5 Solution : If the forces re F,F,F then F ˆ ˆ ˆ 4 (i j k) ; F ˆ ˆ ˆ (i j k) ˆ nd hence the sum force F F ˆ ˆ F F (8 i j 7k) Displcement vector = 9i ˆ 7ˆj 5k ˆ (6i ˆ ˆj k) ˆ = i ˆ 5ˆj kˆ ˆ nd F ˆ ˆ ( i j k) Work done = (8i ˆ ˆj 7k) ˆ. (i ˆ 5ˆj k) ˆ = (4 5 4) = 4 units. ns. Illustrtion 4 : Let u nd v e unit vectors. If w is vector such tht w + ( w u ) = v, then prove tht ( u v ). w nd tht the equlity holds if nd only if u is perpendiculr to v. Solution : w (w u) v...(i) w u v w (w u) v w v.w v.w w (u w )...(ii) lso tking dot product of (i) with v, we get w.v (w u).v v.v v.(w u) w.v v.v v Now ; w (u w ) v.(w u) ( w (u w) ) 4...(iii) (using (ii) nd (iii)) ( 0 cos ) = ( w + w sin )...(iv) ND6\\Dt\04\Kot\J-dvnced\SMP\Mths\Unit#0\NG\0-VCTR(THRY).p65

25 J-Mthemtics s we know ;0 w cos w w cos w w cos...(v) from (iv) nd (v) v.(w u) qulity holds only when cos = 0 = i.e., u w u.w 0 w (w u) v u w u.(w u) u.v (tking dot with u ) = u v u v = 0 u v Illustrtion 5 : point (x, y ) with sciss x = nd point (x, y ) with ordinte y = re given in rectngulr crtesin system of co-ordintes XY on the prt of the curve y = x x + which lies in the first qudrnt. Find the sclr product of nd. Solution : Since (x, y ) nd (x, y ) lies on y = x x +. y = x x + y = () + (s ; x = ) y = so the co-ordintes of (, ) lso, y = x x + = x x + x = 4, x (s lie in st qudrnt) co-ordintes of (4, ). Hence, ˆi ˆj nd 4i ˆˆj. = 4 + = 6. ND6\\Dt\04\Kot\J-dvnced\SMP\Mths\Unit#0\NG\0-VCTR(THRY).p65 Illustrtion 6 : If is rel constnt nd,, C re vrile ngles nd then find the lest vlue of tn + tn + tn C Solution : The given reltion cn e re-written s ; ( 4ˆi j ˆ 4k) ˆ. (tn i ˆ tn j ˆ tn Ck) ˆ = 6 ( 4) ( 4). tn tn tn C. cos = 6 (s,. = cos ). tn tn tn C cos 6 tn + tn + tn C = sec...(i) lso, sec (s, sec )...(ii) from (i) nd (ii), tn + tn + tn C lest vlue of tn + tn + tn C =. 4 tn + tn + 4 tn C = 6 5

26 J-Mthemtics Illustrtion 7 : Prove tht the right isectors of the sides of tringle re concurrent. Solution : Let the right isectors of sides C nd C meet t nd tking s origin, let the position vectors of, nd C e tken s,, c respectively. Hence the mid-points D,, F re c c,, () c D C,.(c ) 0 F i.e. = c c () gin since D C(c) C,.( c) 0 or = c = = c... (i) Now we hve to prove tht F is lso to which will e true if.( ) 0 i.e. = which is true y (i) Illustrtion 8 :,, C nd D re four points such tht m(i ˆ 6ˆj k), ˆ C (i ˆ ˆj) nd CD n( 6ˆi 5ˆj k) ˆ. Find the conditions on the sclrs m nd n so tht CD intersects t some point. lso find the re of the tringle C. (Roorkee 995) Solution : m(i ˆ 6ˆj k), ˆ C (i ˆ ˆj) CD n( 6ˆi 5ˆj k) ˆ If nd CD intersect t, then p, C qcd D where oth p nd q re positive quntities less thn Now we know tht C C 0 p C q CD 0 {y (i)} or pm( î 6 ĵ + ˆk ) + ( î ĵ ) + q.n( 6 î + 5 ĵ ˆk ) = 0 Since î, ĵ, ˆk re non-coplnr, the ove reltion implies tht if x î + y ĵ + z ˆk = 0, then x = 0, y = 0 nd z = 0 mp + 6qn = 0, 6pm + 5qn = 0 pm qn = 0 Solving these for pm nd qn, we get pm =, qn = 0, 0 m n or p =, q m n 6 m, n gin re of C = C = qcd p Put pm, qn =.. i ˆ 6ˆj 6k ˆ ˆi ˆj kˆ = pqnm C ND6\\Dt\04\Kot\J-dvnced\SMP\Mths\Unit#0\NG\0-VCTR(THRY).p65

27 J-Mthemtics Illustrtion 9 :,, c re three non-coplnr unit vectors such tht ngle etween ny two is If + c = + m + n c, then determine, m, n in terms of. (J-997) Solution : = = c =, [c] 0. =. c = c. = cos... (i) Multiply oth sides of given reltion sclrly y, nd c, we get 0 + [ c ] =. + (m + n) cos... (ii) 0 = m + (n + ) cos... (iii) [ c ] + 0 = ( + m)cos + n... (iv) ND6\\Dt\04\Kot\J-dvnced\SMP\Mths\Unit#0\NG\0-VCTR(THRY).p65 dding, we get [ c ] = ( + m + n) + ( + m + n) cos or [ c ] = ( + m + n) ( + cos)... (v) [ c] From (ii), (m + n) = cos [ c] cos Putting in (v), we get [ c] cos or cos [ c] = ( cos ) cos cos [ c] n {s ove} ( cos )( cos ) [ c] cos nd m = (n + ) cos = ( cos )( cos ) Thus the vlues of, m, n depend on [ c ] Hence we now find the vlue of sclr [ c ] in terms of. Now [ c ] = = ( + cos) cos...c cos cos...c = cos cos c. c. c.c cos cos cos cos cos [ c ] = ( + cos)( cos) [ c] cos cos Putting in the vlue of, m, n we hve (pply C + C + C ) (pply R R nd R R ) cos n,m ns. ( cos ) ( cos ) 7

28 J-Mthemtics Illustrtion 40 : Vectors x, y nd z ech of mgnitude, mke ngles of 60 with ech other. If x (y z), y (z x) nd x y c, then find x, y nd z in terms of, nd c. (Roorkee 997) Solution : x.y cos 60 y.z z. x... (i) lso x = y = z = gin (x.z)y (x.y)z y z y z, z x 8 {y (i)}...(ii) Now c (y z) (x y) y (x y) z (x y) = [(y. y)x (y. x)y] [(z. y)x (z. x)y] = (x y) (x y) or c = x Similrly, c = y Now z = y or z = + x z = ( c ) Illustrtion 4 : If x y =, y z = of, or + ( c), x. =, x. y {y (ii)} {y (i)} ns. = nd y. z =, then find x, y nd z in terms nd (Roorkee 998) Solution : x y =... (i) y z = lso x.... (ii) =, x. y =, y. z =... (iii) We hve to mke use of the reltions given ove. From (i) x. (x y) = x. x. = 0 [x x y] Similrly y. = 0, y. = 0, z. = 0 Multiplying (i) vectorilly y, (x y) = or or 0 y = ( ) y using reltions is (iii) nd (iv). gin multiplying (i) vectorilly y y, (x y) y = y y y = y x x [y y] y = 0 (. y)x (. x)y = ( ) y or (x. y)y (y. y)x = y {y (iii)}... (iv)... (v) ND6\\Dt\04\Kot\J-dvnced\SMP\Mths\Unit#0\NG\0-VCTR(THRY).p65

29 where y {y (v)} Hence x is known in terms of, nd. J-Mthemtics gin multiplying (ii) vectorilly y y, we get (y z) y = y or z [ y y] y or y z (y. z)y = y or y z = y + y {y (iii)} where y is given y (v)... (vi) Results (v) nd (vi) give the vlues of x, y nd z in terms of, nd. ns. NSWRS FR D YURSLF ND6\\Dt\04\Kot\J-dvnced\SMP\Mths\Unit#0\NG\0-VCTR(THRY).p65 : (i) 7 (ii) : (i) (),d ;, x, z ; c, y (ii) 4 : (i) 6 : (i) ( v ) 4 5 6, 5 (), x ;, d (iii),,c (iii) ˆ 6 i ˆ j k ˆ ; c, y (c), y, z (ii) 5 (iii) (iv), (i ˆ 6ˆj k) ˆ nd ˆ i ˆ j k ˆ & 6 ˆ i 4 ˆj 0 kˆ 7. (i) r ˆ i ˆj k ˆ 8 : (ii) ˆ i ˆj kˆ (iii) î + 6 ĵ + 6 ˆk 9 : (i) 6 0 : ( i ) ± (ii) 0 (iii) 7 (iv) Right hnded system ( v ) C : (i) 6, 9i ˆ 0ˆj kˆ : (i) (ii) i ˆ kˆ 8 ˆ i ˆj 7kˆ,, : (i) 0 units 7i ˆ ˆj 5kˆ (d), z; x, z 90ˆi 0ˆj 45kˆ 49 (iii) linerly dependent. 9