SUPPLEMENTARY MATERIAL CHAPTER 7 A (2 ) B. a x + bx + c dx

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1 SUPPLEMENTARY MATERIAL CHAPTER 7 ( px + q) a x + bx + c dx. We choose constants A and B such that d px + q A ( ax + bx + c) + B dx A(ax + b) + B Compaing the coefficients of x and the constant tems on both sides, we get aa p and Ab + B q Solving these equations, the values of A and B ae obtained. T hus, the integal educes to SUPPLEMENTARY MATERIAL A ( ) B ax + b ax + bx + c dx + ax + bx + c dx AI 1 + BI whee I 1 ( ) Put ax + bx + c t, then (ax + b)dx dt So I 1 Similaly, I a x + b a x + b x + c d x not to be epublished 3 ( ax + bx + c ) + C 1 3 a x + bx + c dx

2 614 MATHEMATICS is found, using the integal fomulae discussed in [7.6., Page 38 of the textbook]. Thus ( px + q ) ax + bx + c dx is finally woked out. Example 5 Find x 1 + x x d x Solution Following the pocedue as indicated above, we wite d A B dx A (1 x) + B x ( x x ) Equating the coefficients of x and constant tems on both sides, We get A 1 and A + B Solving these equations, we get A and B. T hus the integal educes to x 1+ x x dx Conside I (1 ) x x x dx x x dx 1 1 I + I (1) 1 (1 ) 1+ Put 1 + x x t, then (1 x)dx dt Thus I 1 (1 x ) 1+ x x dx ( ) 3 x x x x x dx not to be epublished 1 3 t d t t + C C1, whee C 3 1 is some constant. 1

3 Futhe, conside I Put 1 x t. Then dx dt Theefoe, I whee C is some constant. 1 SUPPLEMENTARY MATERIAL x x dx x dx 4 5 Putting values of I 1 and I in (1), we get x 1+ x x dx whe e C t dt t t t + s i n + ( x 1) x 1 ( x ) + si n + C x 1 ( 1) 1 sin x + x x + + C 1 1 (1 + x x ) + ( x 1) 1 + x x 3 8 C C 3 5 x C 1 s in C, not to be epublished 1 is anothe abitay constant.,

4 616 MATHEMATICS Inset the following execises at the end of EXERCISE 7.7 as follows: 1. x x + x 13. ( x + 1) x ( x + 3) 3 4x x Answe s x + x + x 1 ( 1) 1 1 ( x + x) + lo g x + + x + x + C x 3 3 ( x + 3 ) + x lo g x + x + + C x + ( x + ) 3 4 x x (3 4 x x ) + sin + + C 3 7 CHAPTER Scala Tiple Poduct Let a, b and c be any thee vectos. The scala poduct of a and ( b c ), i.e., a ( b c ) is called the scala tiple poduct of a, b and c in this ode and is denoted by [ a, b, c ] (o [ a b c ]). We thus have [ a, b, c ] a ( b c ) Obsevations 1. Since ( b c ) is a vecto, a ( b c ) is a scala quantity, i.e. [ a, b, c ] is a scala not to be epublished quantity.. Geometically, the magnitude of the scala tiple poduct is the volume of a paallelopiped fomed by adjacent sides given by the thee O b c a θ c A B b C Fig. 10.8

5 SUPPLEMENTARY MATERIAL 617 vectos a, b and c (Fig. 10.8). Indeed, the aea of the paallelogam foming the base of the paallelopiped is b c. The height is the pojection of a along the nomal to the plane containing b and c which is the magnitude of the component of a a.( b c ) in the diection of b c i.e.,. So the equied ( b c) volume of the paallelopiped is a.( b c) b c a.( b c ), ( b c ) 3. I f a a 1i + a j+ a3k, b b1 i + b j + b3k and c c1i + c j + c3k, then i j k b c b1 b b3 c c c 1 3 (b b 3 c ) i + (b 3 b 1 ) j + (b 1 c b ) k and so a. ( b c ) a ( b c b c ) + a ( b c b c ) + a ( b c b c ) a a a 1 3 b b b c1 c c3 4. If a, b and c be any thee vectos, then [ a, b, c ] [ b, c, a ] [ c, a, b ] not to be epublished (cyclic pemutation of thee vectos does not change the value of the scala tiple poduct). a a i + a j + a k, b b i + b j + b k and c c i + c j k. Let

6 618 MATHEMATICS Then, just by obsevation above, we have a1 a a3 [ a, b, c ] b b b 1 3 c c c 1 3 a 1 (b b 3 c ) + a (b 3 b 1 ) + a 3 (b 1 c b ) b 1 (a 3 c a ) + b (a 1 a 3 ) + b 3 (a a 1 c ) b b b 1 3 c c c 1 3 a a a 1 3 [ b, c, a ] Similaly, the eade may veify that [ a, b, c ] [ c, a, b ] Hence [ a, b, c ] [ b, c, a ] [ c, a, b ] 5. In scala tiple poduct a.( b c), the dot and coss can be intechanged. Indeed, a.( b c ) [ a, b, c ] [ b, c, a ] [ c, a, b ] c.( a b ) ( a b ). c 6. [ a, b, c ] [ a, c, b ]. Indeed [ a, b, c ] a.( b c ) a.( c b ) ( a.( c b )) a, c, b not to be epublished

7 7. [ a, a, b ] 0. Indeed [ a, a, b ] [ a, b, a, ] [ b, a, a] b.( a a) b.0 0. Note: SUPPLEMENTARY MATERIAL 619 (as a a 0) The esult in 7 above is tue iespective of the position of two equal vectos Coplanaity of Thee Vectos Theoem 1 Thee vectos a, b and c ae coplana if and only if a ( b c) 0 Poof Suppose fist that the vectos a, b and c ae coplana. If b and c ae paallel vectos, then, b c 0 and so a ( b c) 0. If b and c ae not paallel then, since a,b and c ae coplana, b c is pependicula to a. So a ( b c) 0. Convesely, suppose that a ( b c) 0. If a and b c ae both non-zeo, then we conclude that a and b c ae pependicula vectos. But b c is pependicula to both b and c. Theefoe, a andb and c must lie in the plane, i.e. they ae coplana. If a 0, then a is coplana with any two vectos, in paticula with b and c. If ( b c) 0, then 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 them and a vecto which is paallel to any one of it also lies in that plane. Note: Coplanaity of fou points can be discussed using coplanaity of thee vectos. uuu uuu uuu Indeed, the fou points A, B, C and D ae coplana if the vectos AB,AC and AD not to be epublished ae coplana..

8 60 MATHEMATICS Example 6 Find.( ), 3, a b c if a i + j + k b i + j + k and c 3i + j + k. 1 3 Solution We have a.( b c ) Example 7 Show that the vectos a i j + 3 k, b i + 3 j 4 k and c i 3 j + 5k ae coplana. 1 3 Solution We have a.( b c ) Hence, in view of T heoem 1, a, b and c ae coplana vectos. Example 8 Find l if the vectos a i + 3 j + k, b i j k and c λ i + 7 j + 3k ae coplana. Solution Since a, b and c ae coplana vectos, we have a, b, c 0, i.e., λ ( 3 + 7) 3 (6 + l) + 1 ( 14 + l) 0 l 0. Example 9 Show that the fou points A, B, C and D with position vectos 4i + 5 j + k, ( j + k ), 3i + 9 j + 4k and 4( i + j + k ), espectively ae coplana. not to be epublished Solution We know that the fou points A, B, C and D ae coplana if the thee vectos uuu uuu uuu AB,AC and AD ae coplana, i.e., if uuu uuu uuu AB, AC, AD 0

9 uuu No w AB ( j + k ) (4i + 5 j + k ) 4i 6 j k ) uuu AC (3i + 9 j + 4 k ) (4i + 5 j + k ) i + 4 j + 3k uuu and AD 4( i + j + k ) (4i + 5 j + k ) 8i j + 3k Thus 4 6 uuu uuu uuu AB,AC, AD SUPPLEMENTARY MATERIAL 61 Hence A, B, C and D ae coplana. Example 30 Pove that a + b, b c, c a + + a, b, c. Solution We have 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) a, b, c a, b, a [ a, c, a] b, b, c + b, b, a + b, c, a a, b, c (Why?) Example 31 Pove that a, b, c d + a, b, c + [ a, b, d ] Solution We have a, b, c d + a.( b ( c + d )) a.( b c + b d ) a.( b c ) + a.( b d) a, b, c + a, b, d. not to be epublished

10 6 MATHEMATICS Execise Find a b c if 3 a i j + k, b i 3j + k andc 3i + j k (Ans. 4). Show that the vectos a i j + 3 k, b i + 3j 4k and c i 3 j + 5k ae coplana. 3. Find λ if the vectos i j + k,3i + j + k and i + λj 3k ae coplana. (Ans. λ 15) 4. Let a i + j+ k, b i and c c 1i + c j+ c3k Then (a) If 1 and c, find which makes a, b and c coplana (Ans. c3 ) (b) If c 1 and 1, show that no value of can make a, b and c coplana. 5. Show that the fou points with position vectos 4i + 8 j + 1 k, i + 4 j + 6 k,3i + 5 j + 4k and 5i + 8 j + 5k ae coplana. 6. Find x such that the fou points A (3,, 1) B (4, x, 5), C (4,, ) and D (6, 5, 1) ae coplana. (Ans. x 5) 7. Show that the vectos a, b and c coplana if a + b, b + c and c + a ae coplana. not to be epublished