FREE Download Study Package from website: &

Size: px

Start display at page:

Download "FREE Download Study Package from website: &"

Rodger McBride
6 years ago
Views:

1 .. Linea Combinations: (a) (b) (c) (d) Given a finite set of vectos a b c,,,... then the vecto xa + yb + zc +... is called a linea combination of a, b, c,... fo any x, y, z... R. We have the following esults: If a, b ae non zeo, non collinea vectos then xa + yb x' a + y' b x x' ; y y' FundamentalTheoem: Let a, b be non zeo, non collinea vectos. Then any vecto coplana with a, b can be expessed uniquely as a linea combination of a, b i.e. Thee exist some uniquly x, y R such that xa + yb. If a, b, c ae non zeo, non coplana vectos then: xa + yb + zc x' a + y' b + z' c x x', y y', z z' Fundamental Theoem In Space: Let a, b, c be non zeo, non coplana vectos in space. Then any vecto, can be uniquly expessed as a linea combination of a, b, c i.e. Thee exist some unique x,y R such that xa + yb + zc. (e) If x, x,... x n ae n non zeo v ectos, & k, k,...k n ae n scalas & if the linea combination kx + kx +... k nxn 0 k 0, k 0... k n 0 then we say that vectos x, x,... x n ae LINEARLY INDEPENDENT VECTORS. (f) If x, x,... x n ae not LINEARLY INDEPENDENT then they ae said to be LINEARLY DEPENDENT vectos. i.e. if k x k x k n x n 0 & if thee exists at least one k 0 then x, x,... x n ae said to be L INEARLY D EPENDENT. Note : If k 0; k x + k x + k x k x knx n 0 k x k x + k x + + k x + k + x k nx n k x k x + k x + + k + + k k k k x k n k x n x c x + c x c x + c x c x n n i.e. x is expessed as a linea combination of vectos. x, x,... x, x+,... xn Hence x with x, x,... x, x+... xn foms a linealy dependent set of vectos. Note : If a 3 î + ĵ + 5 kˆ then a is expessed as a LINEAR COMBINATION of vectos î, ĵ, kˆ Also, a, î, ĵ, kˆ fom a linealy dependent set of vectos. In geneal, evey set of fou vectos is a linealy dependent system. î, ĵ, kˆ ae Linealy Independent set of vectos. Fo K î + K ĵ + K 3 kˆ 0 K K K 3 0 Two vectos a & b ae linealy dependent a is paallel to b i.e. a x b 0 linea dependence of a & b. Convesely if a x b 0 then a & b ae linealy independent. If thee vectos a, b, c ae linealy dependent, then they ae coplana i.e. [ a, b, c ] 0, convesely, if [ a, b, c] 0, then the vectos ae linealy independent. Solved Example: Given A that the points a b + 3 c, a + 3 b 4 c, 7 b + 0 c, A, B, C have position vecto pove that vectos AB and AC ae linealy dependent. Solution. Let A, B, C be the given points and O be the point of efeence then OA a b + 3 c, OB a + 3 b 4 c and OC 7 b + 0 c Now AB p.v. of B p.v. of A OB OA ( a + 5 b 7 c ) AB AC λ AB whee λ. Hence AB and AC ae linealy dependent Solved Example: Pove that the vectos 5 a + 6 b + 7 c, 7 a 8 b + 9 c and 3 a + 0 b + 5 c ae linealy dependent a, b, c being linealy independent vectos. Solution. We know that if these vectos ae linealy dependent, then we can expess one of them as a linea combination of the othe two. Teko Classes, Maths : Suhag R. Kaiya (S. R. K. Si), Bhopal Phone : , page 36 of 77

2 Now let us assume that the given vecto ae coplana, then we can wite 5 a + 6 b + 7 c l( 7 a 8 b + 9 c ) + m (3 a + 0 b + 5 c ) whee l, m ae scalas Compaing the coefficients of a, b and c on both sides of the equation 5 7l (i) 6 8l + 0 m...(ii) 7 9l + 5m...(iii) Fom (i) and (iii) we get 4 8l l m which evidently satisfies (ii) equation too. Hence the given vectos ae linealy dependent. Self Pactice Poblems :. Does thee exist scalas u, v, w such that ue + ve + we3 i whee e k, e j + k, e 3 j + k? Ans. No. Conside a base a, b, c and a vecto a + 3b c. Compute the co-odinates of this vecto elatively to the base p, q, whee p a 3b, q a b + c, 3a + b + c. Ans. (0, 7/5, /5) 3. If a and b ae non-collinea vectos and A (x + 4y) a + (x + y + ) b and B (y x + ) a + (x 3y ) b, find x and y such that 3A B. Ans. x, y 4. If vectos a, b, c be linealy independent, then show that :(i) a b + 3c, a + 3b 4c, b + c ae linealy dependent (ii) a 3b + c, a 4b c, 3a + b c ae linealy independent. 5. Given that î ĵ, to the fist vecto î ĵ ae two vectos. Find a unit vecto coplana with these vectos and pependicula î ĵ. Find also the unit vecto which is pependicula to the plane of the two given vectos. Do you thus obtain an othonomal tiad? Ans. ( î + ĵ) ; k; Yes 6. If with efeence to a ight handed system of mutually pependicula unit vectos î, ĵ, kˆ α 3 i j, β i + j 3k expess β in the fom β β + β whee β is paallel to α & β is pependicula to α. 3 3 Ans. β i j, β i + j 3k 7. Pove that a vecto in space can be expessed linealy in tems of thee non-coplana, non-null [ b c] a + [ c a] b + [ a b] c vectos a, b, c in the fom [a b c] Note: Test Of Collineaity: Thee points A,B,C with position vectos a, b, c espectively ae collinea, if & only if thee exist scalas x, y, z not all zeo simultaneously such that; xa + yb + zc 0, whee x + y + z 0. Note: Test Of Coplanaity: Fou points A, B, C, D with position vectos a, b, c, d espectively ae coplana if and only if thee exist scalas x, y, z, w not all zeo simultaneously such that xa + yb + zc + wd 0 whee, x + y + z + w 0. Solved Example Show that the vectos a b + 3c, a + b c and a + b 3c ae non-coplana vectos. Solution. Let, the given vectos be coplana. Then one of the given vectos is expessible in tems of the othe two. Let a b + 3c x ( a + b c) + y ( a + b 3c), fo some scalas x and y. a b + 3c (x + y) a (x + y) b + ( x 3y) c x + y, x + y and 3 x 3y. Solving, fist and thid of these equations, we get x 9 and y 7. Clealy, these values do not satisfy the thid equation. Hence, the given vectos ae not coplana. Solved Example: Pove that fou points a + 3b c, a b + 3c, 3a + 4b c and a 6b + 6c ae coplana. Solution. Let the given fou points be P, Q, R and S espectively. These points ae coplana if the vectos PQ, PR and PS ae coplana. These vectos ae coplana iff one of them can be expessed as a linea combination of othe two. So, let PQ x PR + y PS a 5b + 4c a + b c + y ( a 9b 7c) a 5b + 4c (x y) a + (x 9y) b + ( x + 7y) c x y, x 9y 5, x + 7y 4 [Equating coeff. of a, b, c on both sides] x ( ) Teko Classes, Maths : Suhag R. Kaiya (S. R. K. Si), Bhopal Phone : , page 37 of 77 Solving the fist of these thee equations, we get x, y.

3 These values also satisfy the thid equation. Hence, the given fou points ae coplana. Self Pactice Poblems :. If, a, b, c, d ae any fou vectos in 3-dimensional space with the same initial point and such that 3a b + c d 0, show that the teminal A, B, C, D of these vectos ae coplana. Find the point at which AC and BD meet. Find the atio in which P divides AC and BD.. Show that the vecto a b + c, b c a and a 3b 4c ae non-coplana, whee a, b, c, ae any noncoplana vectos. 3. Find the value of λ fo which the fou points with position vectos ĵ kˆ, 4î + 5ĵ + λkˆ. 3 î + 9ĵ + 4kˆ and 4 î + 4ĵ + 4kˆ ae coplana. Ans. λ. Application Of Vectos:(a) Wok done against a constant foce F ove a displacement s (c) is defined as W F. s (b) The tangential velocity V of a body moving in a cicle is given by V w x whee is the pv of the point P. The moment of F about O is defined as M x F whee is the pv of P wt O. The diection of M is along the nomal to the plane OPN such that, F & M fom a ight handed system. (d) Moment of the couple ( ) x F whee F & F. & ae pv s of the point of the application of the foces Solved Example: Foces of magnitudes 5 and 3 units acting in the diections 6 î + ĵ + 3kˆ and 3 î + ĵ + 6kˆ espectively act on a paticle which is displaced fom the point (,, ) to (4, 3, ). Find the wok done by the foces. Solution. Let F be the esultant foce and d be the displacement vecto. Then, F (6î + ĵ + 3kˆ) (3î + ĵ + 6kˆ) ( 39î + 4ĵ + 33kˆ) and, d ( 4î + 3ĵ + kˆ ) ( î + ĵ kˆ ) î + ĵ + kˆ Total wok done F. d 7 Self Pactice Poblems :. ( 39î + 4ĵ + 33kˆ). ( î + ĵ + kˆ ) 48 ( ) units. 7 7 A point descibes a cicle unifomly in the î, ĵ plane taking seconds to complete one evolution. If its initial position vecto elative to the cente is î, and the otation is fom î to ĵ, find the position vecto at the end of 7 seconds. Also find the velocity vecto. Ans. / ( ĵ 3 î), p/ ( î 3 ĵ). The foce epesented by 3 î + kˆ is acting though the point 5î + 4ĵ 3kˆ. Find its moment about the point î + 3ĵ + kˆ. Ans. î 0 ĵ 3kˆ 3. Find the moment of the comple fomed by the foces 5 î + kˆ and 5î kˆ acting at the points (9,, ) and (3,, ) espectively Ans. î ĵ 5kˆ Miscellaneous Solved Examples Solved Example: Show that the points A, B, C with position vectos î ĵ + kˆ, î 3ĵ 5kˆ and 3î 4ĵ 4kˆ espectively, ae the vetices of a ight angled tiangle. Also find the emaining angles of the tiangle. Solution. We have, AB Position vecto of B Position vecto of A BC ( î 3ĵ 5kˆ ) ( î ĵ + kˆ ) î ĵ 6kˆ Position vecto of C Position vecto of B Teko Classes, Maths : Suhag R. Kaiya (S. R. K. Si), Bhopal Phone : , page 38 of 77

4 ( 3î 4ĵ 4kˆ ) ( î 3ĵ 5kˆ ) î ĵ + kˆ and, CA Position vecto of A Position vecto of C ( î ĵ + kˆ ) ( 3î 4ĵ 4kˆ ) î + 3ĵ + 5kˆ Since AB + BC + CA ( î ĵ 6kˆ ) + ( î ĵ + kˆ ) + ( î + 3ĵ + 5kˆ ) 0 So, A, B and C ae the vetices of a tiangle. Now, BC. CA ( î ĵ + kˆ ). ( î + 3ĵ + 5kˆ ) BC CA BCA π Since a is the angle between the vectos AB and AC. Theefoe cos A AB. AC AB AC ( ) ( î ĵ 6kˆ) + ( ) 4 + ( 6). (î Hence, ABC is a ight angled tiangle. 3ĵ 5kˆ) + ( 3) 4 + ( 5) A cos BA (î + ĵ + 6kˆ) (î ĵ + kˆ) cos B cos B B cos BA BC ( ) + () Solved Example: If a, b, c ae thee mutually pependicula vectos of equal magnitude, pove that a + b + c is equally inclined with vectos a, b and c. Solution.: Let a b c λ (say). Since a, b, c ae mutually pependicula vectos, theefoe a. b b. c c. a 0...(i) Now, a + b + c a. a + b. b + c. c + a. b + b. c + c. a a + b + c [Using (i) ] 3λ [ a b c λ] a + b + c 3 λ...(ii) Suppose a + b + c makes angles θ, θ, θ 3 with a, b and c espectively. Then, a. (a + b + c) a. a + a. b + a. c cosθ a a + b + c a a + b + c a a λ [Using (ii)] a a + b + c a + b + c 3λ 3 θ cos 3 Similaly, θ cos and θ 3 3 cos θ 3 θ θ 3. Hence, a + b + c is equally inclineded with a, b and c Solved Example: Pove using vectos : If two medians of a tiangle ae equal, then it is isosceles. Solution. : Let ABC be a tiangle and let BE and CF be two equal medians. Taking A as the oigin, let the position vectos of B and C be b and c espectively. Then, P.V. of E c and, P.V. of F b BE (c b) CF (b c) Now, BE CF BE CF BE CF (c b) (b c) c b b c c b b c 4 4 (c b). (c b ) (b c ). (b c) 4 Teko Classes, Maths : Suhag R. Kaiya (S. R. K. Si), Bhopal Phone : , page 39 of 77

5 c. c 4b. c + 4b. b b. b 4b. c + 4c. c c 4b. c + 4 b b 4b. c + 4 c 3 b 3 c b c AB AC Hence, tiangle ABC is an isosceles tiangle. Solved Example: Using vectos : Pove that cos (A + B) cos A cos B sin A sin B Solution. Let OX and OY be the coodinate axes and let î and ĵ be unit vectos along OX and OY espectively. Let XOP A and XOQ B. Dawn PL OX and QM OX. Clealy angle between OP and OQ is A + B In OLP, OL OP cos A and LP OP sin A. Theefoe OL (OP cos A) î and LP Now. OL + LP OP OP OP [(cos A ) î (sin A) ĵ ] In OMQ, OM OQ cos B and MQ OQ sin B. Theefoe, OM (OQ cos B) î, MQ (OQ sin B) ĵ Now, OM + MQ OQ Fom (i) and (ii), we get OP. OQ OP [(cos A) î (sin A) ĵ ]. OQ [(cos B) î + (sin B) ĵ ] OP. OQ [cos A cos B sin A sin B] But, OP. OQ OP OQ cos (A + B) OP. OQ cos (A + B) OP. OQ cos (A + B) OP. OQ [cos A cos B sin A sin B] cos (A + B) cos A cos B sin A sin B Solved Example: Pove that in any tiangle ABC (i) c a + b ab cos C (ii) c bcosa + acosb. Solution. (i) In ABC, AB + BC + CA 0 o, BC + CA AB Squaing both sides ( BC ) + ( CA ) + ( BC ). CA + ( AB )...(i) a + b + ( BC. CA ) c c a + b ab cos (π C) c a + b ab cosc (ii) ( BC + CA ). AB AB. AB (OP sin A) A) ( ĵ) BC. AB + CA. AB c ac cosb bc cos A c acosb + bcosa c. Solved Ex.: If D, E, F ae the mid-points of the sides of a tiangle ABC, pove by vecto method that aea of DEF 4 (aea of ABC) Solution. Taking A as the oigin, let the position vectos of B and C be b and c espectively. Then, the position vectos of D, E and F ae (b + c), c and b espectively. b Now, DE c (b + c) c and DF b ( ( b + c) b c Vecto aea of DEF ( DE DF) (b c) (AB AC) (vecto aea of ABC) Hence, aea of DEF 4 aea of ABC. Solved Example: P, Q ae the mid-points of the non-paallel sides BC and AD of a tapezium ABCD. Show that APD CQB. Solution. Let AB b and AD d Now DC is paallel to AB thee exists a acala t sush that DC t DB t b AC AD + DC d + t b Teko Classes, Maths : Suhag R. Kaiya (S. R. K. Si), Bhopal Phone : , page 40 of 77

6 The position vectos of P and Q ae (b + d + t b) and d espectively. FREE Download Study Package fom website: & Now APD AP AD (b + d + t b) d ( + t) ( b d) Also CQB CQ CB d (d + tb [b (d + t b)] d t b [ d + ( t) b] ( t) d b + t b d ( t t) b d + ( t) b d + APD Hence the esult. Solved Example: Let u and v ae unit vectos and w is a vecto such that u v + u w and w u v then find the value of [ u v w]. Solution. Given u v + u w and w u v ( u v + u) u w u ( u v ) u u u + v (as, w u v ) ( u. u ) v ( v. u ) u + u u v (using u. u and u u 0, since unit vecto) v (v. u) u v ( u. v) u 0 u. v 0 (as; u 0)...(i) u. (v w) u. (v (u v + u)) (given w u v + u) u. (v (u v) + v u) u. ((v. v) u (v. u) v + v u) u. ( v u 0 + v u) (as; u. v 0 fom (i)) v (u. u) u. (v u) v u 0 (as, [ u v u] 0) (as; u v ) [ u v w] Sol. Ex.: In any tiangle, show that the pependicula bisectos of the sides ae concuent. Solution. Let ABC be the tiangle and D, E and F ae espectively middle points of sides BC, CA and AB. Let the pependicula of D and E meet at O join OF. We ae equied to pove that OF is to AB. Let the position vectos of A, B, C with O as oigin of efeence be a, b and c espectively. OD ( b + c ), OE ( c + a ) and OF ( a + b ) Also BC c b, CA a c and AB b a Since OD BC, ( b + c ). ( c b ) 0 b c Similaly ( c + a ). ( a + c ) 0 a c fom (i) and (ii) we have a b 0...(i)...(ii) ( a + b ). ( b + a ) 0 ( b + a ). ( b a ) 0 Solved Example: A, B, C, D ae fou points in space. using vecto methods, pove that AC + BD + AD + BC AB + CD what is the implication of the sign of equaility. Solution.: Let the position vecto of A, B, C, D be a, b, c and d espectively then AC + BD + AD + BC ( c a). ( c a ) + ( d b ). ( d b ) + ( d a ). ( d a ) + ( c b). ( c b ) c + a a. c + d + b d. b + d + a a. d + c + b b. c a + b a. b + c + d c. d + a + b + c + d + a. b + c. d a. c b. d a. d b. c ( a b ). ( a b ) + ( c d ). ( c d ) + ( a b c d ) AB + CD AB + CD + ( a + b c d). ( a + b c d) AB + CD AC + BD + AD + BC AB + CD fo the sign of equality to hold, a + b c d 0 a c d b AC and BD ae collinea the fou points A, B, C, D ae collinea Teko Classes, Maths : Suhag R. Kaiya (S. R. K. Si), Bhopal Phone : , page 4 of 77

e.g: If A = i 2 j + k then find A. A = Ax 2 + Ay 2 + Az 2 = ( 2) = 6

e.g: If A = i 2 j + k then find A. A = Ax 2 + Ay 2 + Az 2 = ( 2) = 6 MOTION IN A PLANE 1. Scala Quantities Physical quantities that have only magnitude and no diection ae called scala quantities o scalas. e.g. Mass, time, speed etc. 2. Vecto Quantities Physical quantities