.. 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 3 3 +... + 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 : 0 903 903 7779, 0 98930 5888. page 36 of 77
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 + 3...(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 : 0 903 903 7779, 0 98930 5888. page 37 of 77 Solving the fist of these thee equations, we get x, y.
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) 5 + 3 ( 39î + 4ĵ + 33kˆ) 36 + 4 + 9 9 + 4 + 36 7 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 (78 + 4 + 66) 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 : 0 903 903 7779, 0 98930 5888. page 38 of 77
( 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 ) 3 + 5 0 BC CA BCA π Since a is the angle between the vectos AB and AC. Theefoe cos A AB. AC AB AC + 4 + 36 + 6 + 30 ( ) + 9 + 5 ( 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) + 6 6 6 cos B cos B B cos BA BC + + 6 + ( ) + () 4 6 4 4 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 : 0 903 903 7779, 0 98930 5888. page 39 of 77
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) 8 4 4 (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 : 0 903 903 7779, 0 98930 5888. page 40 of 77
The position vectos of P and Q ae (b + d + t b) and d espectively. FREE Download Study Package fom website: www.tekoclasses.com & www.mathsbysuhag.com 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 : 0 903 903 7779, 0 98930 5888. page 4 of 77