![614 MATHEMATICS
is found, using the integral formulae discussed in [7.6.2, Page 328 of the textbook].
Thus
2
( )
px q ax bx c dx
+ + +
∫ is finally worked out.
Example 25 Find 2
1
x x x dx
+ −
∫
Solution Following the procedure as indicated above, we write
x = ( )
2
A 1 B
d
x x
dx
+ − +
= A (1 – 2x) + B
Equating the coefficients of x and constant terms on both sides,
We get – 2A = 1 and A + B = 0
Solving these equations, we get A =
1
2
− and
1
B .
2
= Thus the integral
reduces to
2
1
x x x dx
+ −
∫ =
2 2
1 1
(1 2 ) 1 1
2 2
x x x dx x x dx
− − + − + + −
∫ ∫
= 1 2
1 1
I I
2 2
− + (1)
Consider I1
=
2
(1 2 ) 1
x x x dx
− + −
∫
Put 1 + x – x2
= t, then (1 – 2x)dx = dt
Thus I1
= 2
(1 2 ) 1
x x x dx
− + −
∫ =
1 3
2 2
1
2
C
3
t dt t
= +
∫
= ( )
3
2 2
1
2
1 C
3
x x
+ − + , where C1
is some constant.
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![616 MATHEMATICS
Fig. 10.28
Insert the following exercises at the end of EXERCISE 7.7 as follows:
12. 2
x x x
+ 13. 2
( 1) 2 3
x x
+ + 14. 2
( 3) 3 4
x x x
+ − −
Answers
12.
3 2
2 2
2
1 (2 1) 1 1
( ) log | | C
3 8 16 2
x x x
x x x x x
+ +
+ − + + + + +
13.
3
2 2 2
2
1 3 2 3
(2 3) 2 3 lo g C
6 2 4 2
x
x x x x
+ + + + + + +
14.
3 2
2 1
2
1 7 2 ( 2) 3 4
(3 4 ) sin C
3 2 2
7
x x x x
x x − + + − −
− − − + + +
CHAPTER 10
10.7 Scalar Triple Product
Let be any three vectors. The scalar product of , i.e.,
is called the scalar triple product of in this order and is denoted by [ ]
(or [ ]). We thus have
[ ] =
Observations
1. Since ( )
b c
×
is a vector, ( )
a b c
⋅ ×
is a
scalar quantity, i.e. [ ] is a scalar
quantity.
2. Geometrically, the magnitude of the scalar
tripleproductisthe volume ofa parallelopiped
formed by adjacent sides given by the three
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![SUPPLEMENTARY MATERIAL 617
vectors (Fig. 10.28). Indeed, the area of the parallelogram forming
the base of the parallelopiped is . The height is the projection of along
the normal to the plane containing and which is the magnitude of the
component of in the direction of i.e., . So the required
volume of the parallelopiped is ,
3. If then
= 1 2 3
1 2 3
ˆ
ˆ ˆ
i j k
b b b
c c c
= (b2
c3
– b3
c2
) ˆ
i + (b3
c1
– b1
c3
) ĵ + (b1
c2
– b2
c1
) k̂
and so
1 2 3
1 2 3
1 2 3
a a a
b b b
c c c
=
4. If be any three vectors, then
[ ] = [ ] = [ ]
(cyclic permutation of three vectors does not change the value of the scalar
triple product).
Let
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![618 MATHEMATICS
Then, just by observation above, we have
[ ]
1 2 3
1 2 3
1 2 3
a a a
b b b
c c c
=
= a1
(b2
c3
– b3
c2
) + a2
(b3
c1
– b1
c3
) + a3
(b1
c2
– b2
c1
)
= b1
(a3
c2
– a2
c3
) + b2
(a1
c3
– a3
c1
) + b3
(a2
c1
– a1
c2
)
1 2 3
1 2 3
1 2 3
b b b
c c c
a a a
=
= [ ]
Similarly, the reader may verify that
= [ ] = [ ]
Hence [ ] = [ ] = [ ]
5. In scalar triple product , the dot and cross can be interchanged.
Indeed,
= [ ] = [ ] = [ ] =
6. = [ ] = – [ ]. Indeed
= [ ] =
=
=
=
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![620 MATHEMATICS
Example 26 Find .
Solution We have
Example 27 Show that the vectors
are coplanar.
Solution We have
r
c) .
=
−
− −
−
=
1 2 3
2 3 4
1 3 5
0
Hence, in view of Theorem 1, are coplanar vectors.
Example 28 Find λ if the vectors
are coplanar.
Solution Since are coplanar vectors, we have , i.e.,
1 3 1
2 1 1
7 3
0
− − =
λ
.
⇒ 1 (– 3 + 7) – 3 (6 + λ) + 1 ( 14 + λ) = 0
⇒ λ = 0.
Example 29 Show that the four points A, B, C and D with position vectors
ˆ ˆ ˆ ˆ
ˆ ˆ ˆ ˆ ˆ ˆ ˆ
4 5 , ( ),3 9 4 and 4(– )
i j k j k i j k i j k
+ + − + + + + + , respectively are coplanar.
Solution We know that the four points A, B, C and D are coplanar if the three vectors
are coplanar, i.e., if
[ ] = 0
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![SUPPLEMENTARY MATERIAL 621
Now ˆ ˆ ˆ
ˆ ˆ ˆ ˆ ˆ
– ( ) – (4 5 ) – 4 6 2 )
j k i j k i j k
= + + + = − −
ˆ ˆ ˆ
ˆ ˆ ˆ ˆ ˆ ˆ
(3 9 4 ) –(4 5 ) – 4 3
i j k i j k i j k
= + + + + = + +
and ˆ ˆ ˆ
ˆ ˆ ˆ ˆ ˆ ˆ
4( ) – (4 5 ) – 8 3
i j k i j k i j k
= − + + + + = − +
Thus [ ]
r
u
.
=
− − −
−
− −
=
4 6 2
1 4 3
8 1 3
0
Hence A, B, C and D are coplanar.
Example 30 Prove that
Solution We have
(as )
(Why ?)
Example 31 Prove that
Solution We have
.
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lemh2sm.pdf
- 1. SUPPLEMENTARY MATERIAL 613 CHAPTER 7 7.6.3 2 ( ) . px q ax bx c dx + + + ∫ We choose constants A and B such that px + q = 2 A ( ) B d ax bx c dx + + + = A(2ax + b) + B Comparing the coefficients of x and the constant terms on both sides, we get 2aA = p and Ab + B = q Solving these equations, the values of A and B are obtained. Thus, the integral reduces to 2 2 A (2 ) B ax b ax bx c dx ax bx c dx + + + + + + ∫ ∫ = AI1 + BI2 where I1 = 2 (2 ) ax b ax bx c dx + + + ∫ Put ax2 + bx + c = t, then (2ax + b)dx = dt So I1 = 3 2 2 1 2 ( ) C 3 ax bx c + + + Similarly, I2 = 2 ax bx c dx + + ∫ SUPPLEMENTARYMATERIAL 2022-23
- 2. 614 MATHEMATICS is found, using the integral formulae discussed in [7.6.2, Page 328 of the textbook]. Thus 2 ( ) px q ax bx c dx + + + ∫ is finally worked out. Example 25 Find 2 1 x x x dx + − ∫ Solution Following the procedure as indicated above, we write x = ( ) 2 A 1 B d x x dx + − + = A (1 – 2x) + B Equating the coefficients of x and constant terms on both sides, We get – 2A = 1 and A + B = 0 Solving these equations, we get A = 1 2 − and 1 B . 2 = Thus the integral reduces to 2 1 x x x dx + − ∫ = 2 2 1 1 (1 2 ) 1 1 2 2 x x x dx x x dx − − + − + + − ∫ ∫ = 1 2 1 1 I I 2 2 − + (1) Consider I1 = 2 (1 2 ) 1 x x x dx − + − ∫ Put 1 + x – x2 = t, then (1 – 2x)dx = dt Thus I1 = 2 (1 2 ) 1 x x x dx − + − ∫ = 1 3 2 2 1 2 C 3 t dt t = + ∫ = ( ) 3 2 2 1 2 1 C 3 x x + − + , where C1 is some constant. 2022-23
- 3. SUPPLEMENTARY MATERIAL 615 Further, consider I2 = 2 2 5 1 1 4 2 x x dx x dx + − = − − ∫ ∫ Put 1 . 2 − = x t Then dx = dt Therefore, I2 = 2 2 5 2 t d t − ∫ = 2 1 2 1 5 1 5 2 sin C 2 4 2 4 5 t t t − − + ⋅ + = ( ) 2 1 2 2 1 1 5 1 5 2 1 ( ) sin 2 2 4 2 8 5 x x x C − − − − − + + = 2 1 2 1 5 2 1 (2 1) 1 sin 4 8 5 x x x x C − − − + − + + , where C2 is some constant. Putting values of I1 and I2 in (1), we get 2 1 x x x dx + − ∫ = 3 2 2 2 1 1 (1 ) (2 1) 1 3 8 x x x x x − + − + − + − 1 5 2 1 sin , 16 5 x C − − + + where C = 1 2 C C 2 + − is another arbitrary constant. 2022-23
- 4. 616 MATHEMATICS Fig. 10.28 Insert the following exercises at the end of EXERCISE 7.7 as follows: 12. 2 x x x + 13. 2 ( 1) 2 3 x x + + 14. 2 ( 3) 3 4 x x x + − − Answers 12. 3 2 2 2 2 1 (2 1) 1 1 ( ) log | | C 3 8 16 2 x x x x x x x x + + + − + + + + + 13. 3 2 2 2 2 1 3 2 3 (2 3) 2 3 lo g C 6 2 4 2 x x x x x + + + + + + + 14. 3 2 2 1 2 1 7 2 ( 2) 3 4 (3 4 ) sin C 3 2 2 7 x x x x x x − + + − − − − − + + + CHAPTER 10 10.7 Scalar Triple Product Let be any three vectors. The scalar product of , i.e., is called the scalar triple product of in this order and is denoted by [ ] (or [ ]). We thus have [ ] = Observations 1. Since ( ) b c × is a vector, ( ) a b c ⋅ × is a scalar quantity, i.e. [ ] is a scalar quantity. 2. Geometrically, the magnitude of the scalar tripleproductisthe volume ofa parallelopiped formed by adjacent sides given by the three 2022-23
- 5. SUPPLEMENTARY MATERIAL 617 vectors (Fig. 10.28). Indeed, the area of the parallelogram forming the base of the parallelopiped is . The height is the projection of along the normal to the plane containing and which is the magnitude of the component of in the direction of i.e., . So the required volume of the parallelopiped is , 3. If then = 1 2 3 1 2 3 ˆ ˆ ˆ i j k b b b c c c = (b2 c3 – b3 c2 ) ˆ i + (b3 c1 – b1 c3 ) ĵ + (b1 c2 – b2 c1 ) k̂ and so 1 2 3 1 2 3 1 2 3 a a a b b b c c c = 4. If be any three vectors, then [ ] = [ ] = [ ] (cyclic permutation of three vectors does not change the value of the scalar triple product). Let 2022-23
- 6. 618 MATHEMATICS Then, just by observation above, we have [ ] 1 2 3 1 2 3 1 2 3 a a a b b b c c c = = a1 (b2 c3 – b3 c2 ) + a2 (b3 c1 – b1 c3 ) + a3 (b1 c2 – b2 c1 ) = b1 (a3 c2 – a2 c3 ) + b2 (a1 c3 – a3 c1 ) + b3 (a2 c1 – a1 c2 ) 1 2 3 1 2 3 1 2 3 b b b c c c a a a = = [ ] Similarly, the reader may verify that = [ ] = [ ] Hence [ ] = [ ] = [ ] 5. In scalar triple product , the dot and cross can be interchanged. Indeed, = [ ] = [ ] = [ ] = 6. = [ ] = – [ ]. Indeed = [ ] = = = = 2022-23
- 7. SUPPLEMENTARY MATERIAL 619 7. = 0 Indeed = Note: The result in 7 above is true irrespective of the position of two equal vectors. 10.7.1 Coplanarity of Three Vectors Theorem 1 Three vectors are coplanar if and only if . Proof Suppose first that the vectors are coplanar. If are parallel vectors, then, and so . If are not parallel then, since are coplanar, is perpendicular to . So . Conversely, suppose that . If and are both non-zero, then we conclude that and are perpendicular vectors. But is perpendicular to both . Therefore, and and must lie in the plane, i.e. they are coplanar. If = 0, then is coplanar with any two vectors, in particular with . If ( ) = 0, then are parallel vectors and so, , and are coplanar since any two vectors always lie in a plane determined by them and a vector which is parallel to any one of it also lies in that plane. Note: Coplanarity of four points can be discussed using coplanarity of three vectors. Indeed, the four points A, B, C and D are coplanar if the vectors are coplanar. 2022-23
- 8. 620 MATHEMATICS Example 26 Find . Solution We have Example 27 Show that the vectors are coplanar. Solution We have r c) . = − − − − = 1 2 3 2 3 4 1 3 5 0 Hence, in view of Theorem 1, are coplanar vectors. Example 28 Find λ if the vectors are coplanar. Solution Since are coplanar vectors, we have , i.e., 1 3 1 2 1 1 7 3 0 − − = λ . ⇒ 1 (– 3 + 7) – 3 (6 + λ) + 1 ( 14 + λ) = 0 ⇒ λ = 0. Example 29 Show that the four points A, B, C and D with position vectors ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ 4 5 , ( ),3 9 4 and 4(– ) i j k j k i j k i j k + + − + + + + + , respectively are coplanar. Solution We know that the four points A, B, C and D are coplanar if the three vectors are coplanar, i.e., if [ ] = 0 2022-23
- 9. SUPPLEMENTARY MATERIAL 621 Now ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ – ( ) – (4 5 ) – 4 6 2 ) j k i j k i j k = + + + = − − ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ (3 9 4 ) –(4 5 ) – 4 3 i j k i j k i j k = + + + + = + + and ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ 4( ) – (4 5 ) – 8 3 i j k i j k i j k = − + + + + = − + Thus [ ] r u . = − − − − − − = 4 6 2 1 4 3 8 1 3 0 Hence A, B, C and D are coplanar. Example 30 Prove that Solution We have (as ) (Why ?) Example 31 Prove that Solution We have . 2022-23
- 10. 622 MATHEMATICS Exercise 10.5 1. Find (Ans. 24) 2. Show that the vectors are coplanar. 3. Find λ if the vectors ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ,3 2 and 3 i j k i j k i j k − + + + + λ − are coplanar. (Ans. λ = 15) 4. Let Then (a) If c1 = 1 and c2 = 2, find c3 which makes coplanar (Ans. c3 = 2) (b) If c2 = –1 and c3 = 1, show that no value of c1 can make coplanar. 5. Show that the four points with position vectors ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ 4 8 12 ,2 4 6 ,3 5 4 and 5 8 5 i j k i j k i j k i j k + + + + + + + + are coplanar. 6. Find x such that the four points A (3, 2, 1) B (4, x, 5), C (4, 2, –2) and D (6, 5, –1) are coplanar. (Ans. x = 5) 7. Show that the vectors coplanar if are coplanar. 2022-23