# CLASS XI CHAPTER 3. Theorem 1 (sine formula) In any triangle, sides are proportional to the sines of the opposite angles. That is, in a triangle ABC

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1 CLASS XI Anneue I CHAPTER.6. Poofs and Simple Applications of sine and cosine fomulae Let ABC be a tiangle. By angle A we mean te angle between te sides AB and AC wic lies between 0 and 80. Te angles B and C ae similaly defined. Te sides AB, BC and CA opposite to te vetices C, A and B will be denoted by c, a and b, espectively (see Fig..5). c A b Teoem (sine fomula) In any tiangle, sides ae popotional to te sines of te opposite angles. Tat is, in a tiangle ABC B sin A sin B sin C a b c Poof Let ABC be eite of te tiangles as sown in Fig..6 (i) and (ii). a Fig..5 C B B c a c a A b C D A b D C (i) (ii) Fig..6 Te altitude is dawn fom te vete B to meet te side AC in point D [in (i) AC is poduced to meet te altitude in D]. Fom te igt angled tiangle ABD in Fig..6(i), we ave sin A, i.e., c sin A () c and sin (80 C) asinc a Fom () and (), we get c sin A a sin C, i.e., sin A sin C a c () ()

2 Similaly, we can pove tat sin A sin B a b Fom () and (4), we get sin A sin B sin C a b c Fo tiangle ABC in Fig..6 (ii), equations () and (4) follow similaly. Teoem (Cosine fomulae) Let A, B and C be angles of a tiangle and a, b and c be lengts of sides opposite to angles A, B and C, espectively, ten a b + c bccosa b c + a cacosb c a + b abcosc Poof Let ABC be tiangle as given in Fig..7 (i) and (ii) (4) B B c a c a A b C D A D b C (i) Fig..7 Refeing to Fig..7 (ii), we ave BC BD + DC BD + (AC AD) BD + AD + AC AC.AD + o a b + c bc cosa Similaly, we can obtain AB AC AC ABcos A b c + a ca and cosb c a + b abcosc Same equations can be obtained fo Fig..7 (i), wee C is obtuse. A convenient fom of te cosine fomulae, wen angles ae to be found ae as follows: (ii) b + c a cos A bc c + a b cos B ac a + b c cos C ab

3 Eample 5 In tiangle ABC, pove tat B C b c A tan cot b + c C A c a B tan cot c + a A B a b C tan cot a + b Poof By sine fomula, we ave Teefoe, a b c ksay ( ). sin A sin B sin C b c k(sin B sin C) b+ c k(sin B + sin C) B+ C B C cos sin B+ C B C sin cos (B+C) (B C) cot tan π A B C cot tan B C tan A cot B C b c A Teefoe, tan cot b + c Similaly, we can pove ote esults. Tese esults ae well known as Napie s Analogies. Eample 6 In any tiangle ABC, pove tat a sin (B C) + b sin (C A) + C sin (A B) 0 Solution Conside a sin (B C) a [sinb cosc cosb sinc] () sin A sin B sin C Now k(say) a b c Teefoe, sin A ak, sin B bk, sin C ck Substituting te values of sinb and sinc in () and using cosine fomula, we get a + b c c + a b asin(b C) a bk ck ab ac

4 k k( b ( a + b c c a + b ) c ) Similaly, b sin (C A) k (c a ) and csin (A B) k (a b ) Hence L.H.S k (b c + c a + a b ) 0 R.H.S. Eample 7 Te angle of elevation of te top point P of te vetical towe PQ of eigt fom a point A is 45 and fom a point B, te angle of elevation is 60, wee B is a point at a distance d fom te point A measued along te line AB wic makes an angle 0 wit AQ. Pove tat d ( ) Poof Fom te Fig..8, we ave PAQ 45, BAQ 0, PBH 60 P 5 45 A d 0 60 B H Q Fig..8 Clealy APQ 45, BPH 0, giving APB 5 Again PAB 5 ABP 50 Fom tiangle APQ, we ave AP + (Wy?) o AP Applying sine fomula in Δ ABP, we get AB AP d sin5 sin50 sin5 sin50 i.e., d sin5 sin 0 ( ) (wy?) Eample 8 A lamp post is situated at te middle point M of te side AC of a tiangula plot ABC wit BC 7m, CA 8m and AB 9 m. Lamp post subtends an angle 5 at te point B. Detemine te eigt of te lamp post. Solution Fom te Fig..9, we ave AB 9 c, BC 7 a and AC 8 b. 4

5 Fig..9 M is te mid point of te side AC at wic lamp post MP of eigt (say) is located. Again, it is given tat lamp post subtends an angle θ (say) at B wic is 5. Applying cosine fomula in ΔABC, we ave a + b c cos C () ab Similaly using cosine fomula in ΔBMC, we get BM BC + CM BC CM cos C. Hee CM CA4, since M is te mid point of AC. Teefoe, using (), we get BM o BM 7 Tus, fom ΔBMP igt angled at M, we ave PM tan θ BM 7 o tan(5 ) 7 (wy?) o 7( )m. EXERCISE.5 In any tiangle ABC, if a 8, b 4, c 0, find 4. cosa, cosb, cosc (Ans. 5, 5, 0). sina, sinb, sinc (Ans. 5, 4 5, ) Fo any tiangle ABC, pove tat. A B cos a+ b c C sin 4. A B sin a b c C cos 5

6 5. B C b c A sin cos 6. a (b cos C c cos B) b a c 7. a (cos C cos B) (b c) cos A 8. sin(b C) sin(b + C) b c a 9. B C B C ( b + + c)cos acos 0. a cos A + b cos B + c cos C a sin Bsin C.. cos A cos B cos C a + b + c + +. (b c ) cota + (c a ) cotb + (a b ) cotc 0 a b c abc b c c a a b sina+ sinb+ sinc 0 a b c 4. A tee stands vetically on a ill side wic makes an angle of 5 wit te oizontal. Fom a point on te gound 5m down te ill fom te base of te tee, te angle of elevation of te top of te tee is 60. Find te eigt of te tee. (Ans. 5 m) 5. Two sips leave a pot at te same time. One goes 4 km pe ou in te diection N45 E and ote tavels km pe ou in te diection S75 E. Find te distance between te sips at te end of ous. (Ans km (appo.)) 6. Two tees, A and B ae on te same side of a ive. Fom a point C in te ive te distance of te tees A and B is 50m and 00m, espectively. If te angle C is 45, find te distance between te tees (use.44). (Ans. 5.5 m) 5.7. Squae-oot of a Comple Numbe CHAPTER 5 We ave discussed solving of quadatic equations involving comple oots on page of tetbook. Hee we eplain te paticula pocedue fo finding squae oot of a comple numbe epessed in te standad fom. We illustate te same by an eample. Eample Find te squae oot of 7 4i Solution Let + iy 7 4i Ten ( + iy) 7 4i o y + yi 7 4i Equating eal and imaginay pats, we ave y 7 () y 4 We know te identity ( ) ( ) + y y + ( y) Tus, + y 5 () 6

7 Fom () and (), 9 and y 6 o ± and y ±4 Since te poduct y is negative, we ave, y 4 o,, y 4 Tus, te squae oots of 7 4i ae 4i and + 4i EXERCISE 5.4 Find te squae oots of te following:. 5 8i ( Ans. 4i, + 4i). 8 6i (Ans. i, + i). i (Ans. + ± μ i ) 4. i (Ans. ± m i ) 5. i (Ans. i + ± ± ) 6. + i (Ans. ± ± i ) 9.7. Infinite G.P. and its Sum CHAPTER 9 G. P. of te fom a, a, a, a,... is called infinite G. P. Now, to find te fomula fo finding sum to infinity of a G. P., we begin wit an eample. Let us conside te G. P., 4,,,... 9 Hee a,. We ave n n Sn n Let us study te beaviou of as n becomes lage and lage: n n We obseve tat as n becomes lage and lage, becomes close and close to zeo. Matemati- cally, we say tat as n becomes sufficiently lage, becomes sufficiently small. In ote wods as n n 7

8 n n, 0. Consequently, we find tat te sum of infinitely many tems is given by S. Now, fo a geometic pogession, a, a, a,..., if numeical value of common atio is less tan, ten n n a( ) a a Sn ( ) In tis case as n, n 0 since <. Teefoe S n a Symbolically sum to infinity is denoted by S o S. Tus, we ave S a. Fo eamples (i) (ii) EXERCISE 9.4 Find te sum to infinity in eac of te following Geometic Pogession..,,,... (Ans..5). 6,.,.4,... (Ans. 7.5) 9. 5, 0 80,, (Ans. 5 ) 4.,,, (Ans. ) 5 5. Pove tat Let + a + a +... and y + b + b +..., wee a < and b <. Pove tat y + ab + a b y CHAPTER Equation of family of lines passing toug te point of intesection of two lines Let te two intesecting lines l and l be given by A + B y + C 0 () and A+ By + C 0 () Fom te equations () and (), we can fom an equation ( ) A+ By+ C+ k A+ By + C 0 () 8

9 wee k is an abitay constant called paamete. Fo any value of k, te equation () is of fist degee in and y. Hence it epesents a family of lines. A paticula membe of tis family can be obtained fo some value of k. Tis value of k may be obtained fom ote conditions. Eample 0 Find te equation of line paallel to te y-ais and dawn toug te point of intesection of 7y and + y 7 0 Soluion Te equation of any line toug te point of intesection of te given lines is of te fom 7 y k( + y 7) 0 () i. e., ( + k) + ( k 7) y + 5 7k 0 If tis line is paallel to y-ais, ten te coefficient of y sould be zeo, i.e., k 7 0 wic gives k 7. Substituting tis value of k in te equation (), we get 44 0, i.e., 0, wic is te equied equation. EXERCISE 0.4. Find te equation of te line toug te intesection of lines + 4y 7 and y + 0 and wose slope is 5. (Ans. 5 7y ). Find te equation of te line toug te intesection of lines + y 0 and 4 y and wic is paallel to 5 + 4y 0 0 (Ans. 5 + y 7 0). Find te equation of te line toug te intesection of te lines + y 4 0 and 5y 7 tat as its -intecept equal to 4. (Ans y ) 4. Find te equation of te line toug te intesection of 5 y and + y 0 and pependicula to te line 5 y 0. (Ans y 78 0.) 0.7. Sifting of oigin An equation coesponding to a set of points wit efeence to a system of coodinate aes may be simplified by taking te set of points in some ote suitable coodinate system suc tat all geometic popeties emain uncanged. One suc tansfomation is tat in wic te new aes ae tansfomed paallel to te oiginal aes and oigin is sifted to a new point. A tansfomation of tis kind is called a tanslation of aes. Te coodinates of eac point of te plane ae canged unde a tanslation of aes. By knowing te elationsip between te old coodinates and te new coodinates of points, we can study te analytical poblem in tems of new system of coodinate aes. Fig. 0. To see ow te coodinates of a point of te plane canged unde a tanslation of aes, let us take a point P (, y) efeed to te aes OX and OY. Let O X and O Y be new aes paallel to OX and OY espectively, wee O is te new oigin. Let (, k) be te coodinates of O efeed to te old aes, i.e., OL and LO k. Also, OM and MP y (see Fig.0.) 9 Y O L Y' 0' M' k P{(, y) ( ', y')} M X' X

10 Let O M and M P y be espectively, te abscissa and odinates of a point P efeed to te new aes O X and O Y. Fom Fig.0., it is easily seen tat OM OL + LM, i.e., + and MP MM + M P, i.e., y k + y Hence, +, y y + k Tese fomulae give te elations between te old and new coodinates. Eample Find te new coodinates of point (, 4) if te oigin is sifted to (, ) by a tanslation. Solution Te coodinates of te new oigin ae, k, and te oiginal coodinates ae given to be, y 4. Te tansfomation elation between te old coodinates (, y) and te new coodinates (, y ) ae given by + i.e., and y y + k i.e., y y k Substituting te values, we ave and y 4 6 Hence, te coodinates of te point (, 4) in te new system ae (, 6). Eample Find te tansfomed equation of te staigt line y + 5 0, wen te oigin is sifted to te point (, ) afte tanslation of aes. Solution Let coodinates of a point P canges fom (, y) to (, y ) in new coodinate aes wose oigin as te coodinates, k. Teefoe, we can wite te tansfomation fomulae as + and y y. Substituting, tese values in te given equation of te staigt line, we get ( + ) (y ) o y Teefoe, te equation of te staigt line in new system is y EXERCISE 0.5. Find te new coodinates of te points in eac of te following cases if te oigin is sifted to te point (, ) by a tanslation of aes. (i) (, ) (Ans (4, )) (ii) (0, ) (Ans. (, )) (iii) (5, 0) (Ans. (8, ) ) (iv) (, ) (Ans. (, 0)) (v) (, 5) (Ans. (6, )). Find wat te following equations become wen te oigin is sifted to te point (, ) (i) + y y y + 0 (Ans. y + y + 6y+ 0) (ii) y y + y 0 (iii) y y + 0 (Ans. y y 0 ) (Ans. y 0 ) CHAPTER.5. Limits involving eponential and logaitmic functions Befoe discussing evaluation of limits of te epessions involving eponential and logaitmic functions, we intoduce tese two functions stating tei domain, ange and also sketc tei gaps ougly. 0

11 Leonad Eule (707AD 78AD), te geat Swiss matematician intoduced te numbe e wose value lies between and. Tis numbe is useful in defining eponential function and is defined as f () e, R. Its domain is R, ange is te set of positive eal numbes. Te gap of eponential function, i.e., y e is as given in Fig... Y O gap of y e X Fig.. Similaly, te logaitmic function epessed as log e : R + Ris given by log e y, if and only if e y. Its domain is R + wic is te set of all positive eal numbes and ange is R. Te gap of logaitmic function y log e is sown in Fig... Y O X gap of y log e Fig.. e In ode to pove te esult lim, we make use of an inequality involving te epession 0 e wic uns as follows: + e + (e ) olds fo all in [, ] ~ {0}.

12 e Teoem 6 Pove tat lim 0 Poof Using above inequality, we get Also + e lim lim + (e ), [, ] ~ {0} and lim[ + ( e ) ] + ( e )lim + ( e )0 0 0 Teefoe, by Sandwic teoem, we get e lim 0 loge( + ) Teoem 7 Pove tat lim 0 Poof Let log (+ ) e log e( + ) y + e y y e e y o. y y y. Ten y e lim lim y (since 0 gives y 0) y 0 y 0 y e lim y as lim 0 y 0 y log e ( + ) lim 0 e Eample 5 Compute lim 0 Solution We ave e e lim lim 0 0 y e lim, weey y 0 y.

13 e Eample 6 Compute lim 0 sin e sin e sin Solution We ave lim lim 0 0 e sin lim lim loge Eample 7 Evaluate lim Solution Put +, ten as 0. Teefoe, loge log e( + ) log e( + ) lim lim since lim 0 0 EXERCISE. Evaluate te following limits, if eist 4 + e e e. lim (Ans. 4). lim sin e e e. lim (Ans. e 5 ) 4. lim e lim e loge( + ) lim 0 (Ans. e ) 6. (Ans. ) 8. ( e ) lim 0 cos log ( + ) lim 0 sin (Ans. e ) (Ans. ) (Ans. ) (Ans. )

14 Anneue II CLASS XII CHAPTER 5 Teoem 5 (To be inseted on page 7 unde te eading teoem 5) (i) Deivative of Eponential Function f() e. If f() e, ten f '() lim Δ 0 e lim Δ 0 e f ( +Δ) f( ) Δ e Δ +Δ Δ e lim Δ 0 Δ e e [sincelim ] 0 d Tus, ( e ) e. d (ii) Deivative of logaitmic function f() log e. If f() log e, ten f '() log e( +Δ) loge lim Δ Δ 0 Δ loge + lim Δ Δ 0 Δ loge + lim Δ Δ 0 [since log e( + ) lim ] 0 Tus, d log e d 4

15 7.6.. ( p + q) a + b + c d. We coose constants A and B suc tat p + q CHAPTER 7 d a b c d A ( + + ) + B A(a + b) + B Compaing te coefficients of and te constant tems on bot sides, we get aa p and Ab + B q Solving tese equations, te values of A and B ae obtained. Tus, te integal educes to A ( ) B a + b a + b + c d+ a + b + c d wee I AI + BI Put a + b + c t, ten (a + b)d dt ( a + b) a + b + c d So I Similaly, I ( ) C a + b + c + a + b + c d is found, using te integal fomula discussed in [7.6., Page 8 of te tetbook]. Tus ( ) p+ q a + b+ cd is finally woked out. Eample 5 Find + d Solution Following te pocedue as indicated above, we wite d A + + B d ( ) A ( ) + B Equating te coefficients of and constant tems on bot sides, We get A and A + B 0 Solving tese equations, we get A and B. Tus te integal educes to 5

16 d d + d ( ) Conside I Put + t, ten ( )d dt I + I () ( ) + d Tus I ( ) + d t dt t + C ( ) + + C, wee C is some constant. Fute, conside I 5 + d d 4 Put t. Ten d dt Teefoe, I 5 t dt 5 5 t t sin t + + C ( ) 5 5 ( ) + sin C 5 5 ( ) sin C, wee C Putting values of I and I in (), we get is some constant. + d wee C ( + ) + ( ) C 5 sin, C + C is anote abitay constant. 6

17 Inset te following eecises at te end of EXERCISE 7.7 as follows:. +. ( + ) + 4. ( + ) 4 Answes.. 4. (+ ) + ( + ) + log C 8 6 ( + ) log C ( + ) 4 ( 4 ) + sin + + C 7 CHAPTER Scala tiple poduct Let ab, and c be any tee vectos. Te scala poduct of a and ( b c), i.e., a ( b c ) is called te scala tiple poduct of ab, and c in tis ode and is denoted by [ ab,, c ] (o [ abc ]). We tus ave [ ab,, c ] a ( b c) Obsevations. Since ( b c) is a vecto, a ( b c ) is a scala quantity, i.e. [ ab,, c ] is a scala quantity.. Geometically, te magnitude of te scala tiple poduct is te volume of a paallelopiped fomed by adjacent sides given by te tee vectos ab, and c (Fig. 0.8). Indeed, te aea of te paallelogam foming te base of te paallelopiped is b c. Te eigt is te pojection of a along te nomal to te plane containing b and c wic is te magnitude of te component of a in te diection of b c a.( b c) i.e.,. So te equied volume of te paallelopiped ( b c) a.( b c) is b c a.( b c ), ( b c) Fig If ˆ ˆ ˆ ˆ ˆ ˆ a ai ˆ + a ˆ ˆ j+ ak, b bi + bj+ bk and c ci + cj+ ck, ten 7

18 iˆ ˆj kˆ b b b b c c c c (b c b c ) î + (b c b c ) ĵ + (b c b c ) ˆk and so a.( b c) a( bc bc) + a( bc bc) + a ( bc bc) a a a b b b c c c 4. If ab, and c be any tee vectos, ten [ ab,, c ] [ bca,, ] [ cab,, ] (cyclic pemutation of tee vectos does not cange te value of te scala tiple poduct). Let ˆ ˆ ˆ ˆ ˆ ˆ a ai ˆ + a ˆ ˆ j+ ak, b bi + bj+ bk and c ci + cjk. Ten, just by obsevation above, we ave a a a [ abc,, ] b b b c c c a (b c b c ) + a (b c b c ) + a (b c b c ) b (a c a c ) + b (a c a c ) + b (a c a c ) b b b c c c a a a [ bca,, ] Similaly, te eade may veify tat [ ab,, c ] [ cab,, ] Hence [ ab,, c ] [ bca,, ] [ cab,, ] 5. In scala tiple poduct a.( b c), te dot and coss can be intecanged. Indeed, a.( b c) [ abc,, ] [ bca,, ] [ cab,, ] c.( a b ) ( a b ). c 6. [ abc,, ] [ a, c, b ]. Indeed 8

19 [ abc,, ] a.( b c ) a.( c b ) ( a.( c b )) a, c, b 7. [ aab,, ] 0. Indeed [ aab,, ] [ aba,,,] [ baa,, ] b.( a a) b.0 0. (as a a 0) Note: Te esult in 7 above is tue iespective of te position of two equal vectos Coplanaity of tee vectos Teoem Tee vectos a,b and c ae coplana if and only if a ( b c) 0. Poof : Suppose fist tat te vectos a,b and c ae coplana. If b and c ae paallel vectos, ten, b c 0 and so a ( b c) 0. If b and c ae not paallel ten, since a,b and c ae coplana, b c is pependicula to a. So a ( b c) 0. Convesely, suppose tat a ( b c) 0. If a and b c ae bot non-zeo, ten we conclude tat a and b c ae pependicula vectos. But b c is pependicula to bot b and c. Teefoe a andb and c must lie in te plane, i.e. tey ae coplana. If a 0, ten a is coplana wit any two vectos, in paticula wit b and c. If ( b c) 0, ten b and c ae paallel vectos and so, a, b and c ae coplana since any two vectos always lie in a plane detemined by tem and a vecto wic is paallel to any one of it also lies in tat plane. Note: Coplanaity of fou points can be discussed using coplanaity of tee vectos. Indeed, te fou uuuuuu uuu points A, B, C and D ae coplana if te vectos AB, AC and AD ae coplana. Eample 6: Find.( ), ˆ ˆ ˆ, ˆ a b c if a i + j+ k b i + j+ k and c iˆ+ ˆj+ kˆ. Solution : We ave a.( b c) 0. Eample 7: Sow tat te vectos ˆ ˆ ˆ ˆ ˆ a i j+ k, b i + j 4k and c iˆ ˆj+ 5kˆae coplana. Solution : We ave a.( b c)

20 Hence, in view of Teoem, ab, and c ae coplana vectos. Eample 8: Find λ if te vectos ˆ ˆ ˆ ˆ ˆ ˆ a i + j+ k, b i j k and cλ iˆ+ 7ˆj+ kˆ ae coplana. Solution : Since ab, and c ae coplana vectos, we ave abc,, 0, i.e., 0. λ 7 ( + 7) (6 + λ) + ( 4 + λ) 0 λ 0. Eample 9: Sow tat te fou points A, B, C and D wit position vectos 4iˆ + 5 ˆj+ kˆ, ( ˆj+ kˆ),iˆ+ 9ˆj+ 4kˆand 4( iˆ+ ˆj+ kˆ), espectively ae coplana. uuuuuu uuu Solution : We know tat te fou points A, B, C and D ae coplana if te tee vectos AB, AC and AD ae coplana, i.e., if uuuuuuuuu AB, AC, AD 0 uuu Now AB ( ˆj+ kˆ) (4iˆ + 5 ˆj+ kˆ) 4iˆ 6 ˆj kˆ) uuu AC (iˆ+ 9 ˆj+ 4 kˆ) (4iˆ + 5 ˆj+ kˆ) iˆ+ 4 ˆj+ kˆ uuu and AD 4( iˆ+ ˆj+ kˆ) (4iˆ + 5 ˆj+ kˆ) 8iˆ ˆj+ kˆ Tus 4 6 uuu uuu uuu AB, AC, AD Hence A, B, C and D ae coplana. Eample 0 : Pove tat a+ b, b c, c a + + a, b, c. Solution : We ave a b, b c, c a ( a+ b).(( b+ c) ( c+ a)) ( a+ b).( b c+ b a+ c c+ c a) ( a+ b).( b c+ b a+ c a) (as c c 0 ) a.( b c) + a.( b a) + a.( c a) + b.( b c) + b.( b a) + b.( c a) abc,, aba,, [ aca,, ] bbc,, + bba,, + bca,, 0

21 abc,, (Wy?) Eample : Pove tat abc,, d + abc,, + [ abd,, ] Solution We ave abc,, d + a.( b ( c+ d)) a.( b c+ b d) a.( b c) + a.( b d) a, b, c + a, b, d EXERCISE 0.5. Find abc if a iˆ ˆj + kˆ, b iˆ ˆj + kˆand c i+ j kˆ (Answe : 4). Sow tat te vectos ˆ ˆ ˆ ˆ ˆ ˆ a i j+ k, b i + j 4kand c iˆ ˆj+ 5kˆ ae coplana.. Find λ if te vectos iˆ ˆj+ kˆ,iˆ+ ˆj+ kˆand iˆ+λˆj kˆae coplana. (Answe : λ 5) 4. Let ˆ ˆ ˆ ˆ a i + j+ k, b iand c c ˆ ˆ ˆ i+ cj+ ck Ten (a) If c and c, find c wic makes ab, and c coplana (Answe : c ) (b) If c and c, sow tat no value of c can makes ab, and c coplana. 5. Sow tat te fou points wit position vectos 4iˆ+ 8ˆj+ kˆ,iˆ+ 4ˆj+ 6 kˆ,iˆ+ 5ˆj+ 4kˆ and 5iˆ+ 8ˆj+ 5kˆae coplana. 6. Find suc tat te fou points A (,, ) B (4,, 5), C (4,, ) and D (6, 5, ) ae coplana. (Answe 5) 7. Sow tat te vectos ab, and c coplana if a b, + b+ c and c+ a ae coplana.

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