Comprehensive CBSE Question Bank in Mathematics-XII (Term-II)

CBSE II Question Bank in Mathematics CLASS 12 Features Long Answer-I Type Questions Long Answer-II Type Questions Strictly Based on the Latest CBSE Term-wise Syllabus Case Study Based MCQs Chapter-wise Important Results and Formulae Very Short/Short Answer Type Questions



Comprehensive CBSE Question Bank in Mathematics Term–II (For Class XII) (According to the Latest CBSE Examination Pattern) By Parmanand Gupta A.P. Prabhakaran B.Sc. (Hons.). M.Sc. (Delhi) Formerly Principal M.Phil (KU), Pre. Ph.D. (IIT Delhi) Gokulum Public School, Vatakara Associate Professor of Mathematics Former Head of Department of Mathematics Kerala Indira Gandhi National College, Ladwa Kurukshetra University Haryana Arun Singh B.Tech., U.P.   laxmi Publications (P) Ltd (An iso 9001:2015 company) bengaluru • chennai • guwahati • hyderabad • jalandhar Kochi • kolkata • lucknow • mumbai • ranchi new delhi

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Contents Chapters Page No. Unit III: Calculus 1 84 1. Integrals 107 2. Applications of Integrals 3. Differential Equations 171 245 Unit IV: Vectors and Three Dimensional Geometry 311 4. Vector Algebra 5. Three Dimensional Geometry Unit VI: Probability 6. Probability



Syllabus Mathematics-XII (2021-22) Subject Code-041 Course Structure (Term-II) One Paper Max Marks: 40 No. Units Marks III. Calculus 18 IV. Vectors and Three-Dimensional Geometry 14 VI. Probability 8 40 Total 10 Internal Assessment 50 Total Unit III: Calculus 1. Integration Integration as inverse process of differentiation. Integration of a variety of functions by substitution, by partial fractions and by parts, Evaluation of simple integrals of the following types and problems based on them. ∫ ∫ ∫ ∫ ∫dx x2 ± a2 , dx , dx , ax2 dx + , dx x2 ± a2 a2 − x2 + bx c ax2 + bx + c ∫ ∫ ∫ ∫px ax2 + +q c dx, px + q dx, a2 ± x2 dx, x2 − a2 dx bx + ax2 + bx + c Fundamental Theorem of Calculus (without proof). Basic Properties of definite integrals and evaluation of definite integrals. 2. Applications of the Integrals Applications in finding the area under simple curves, especially lines, parabolas; area of circles/ ellipses (in standard form only) (the region should be clearly identifiable). 3. Differential Equations Definition, order and degree, general and particular solutions of a differential equation. Solution of differential equations by method of separation of variables, solutions of homogeneous

differential equations of first order and first degree of the type: dy = f(y/x). Solutions of linear dx differential equation of the type:           dy + py = q, where p and q are functions of x or constants. dx Unit IV: Vectors and Three-Dimensional Geometry 1. Vector Algebra Vectors and scalars, magnitude and direction of a vector. Direction cosines and direction ratios of a vector. Types of vectors (equal, unit, zero, parallel and collinear vectors), position vector of a point, negative of a vector, components of a vector, addition of vectors, multiplication of a vector by a scalar, position vector of a point dividing a line segment in a given ratio. Definition, Geometrical Interpretation, properties and application of scalar (dot) product of vectors, vector (cross) product of vectors. 2. Three-dimensional Geometry Direction cosines and direction ratios of a line joining two points. Cartesian equation and vector equation of a line, coplanar and skew lines, shortest distance between two lines. Cartesian and vector equation of a plane. Distance of a point from a plane. Unit VI: Probability 1. Probability Conditional probability, mutliplication theorem on probability, independent events, total probability, Bayes’ theorem, Random variable and its probability distribution. INTERNAL ASSESSMENT 10 MARKS Periodic Test 5 Marks Mathematics Activities: Activity file record +Term end assessment of one activity & Viva 5 Marks

Integrals 1 UNIT III: CALCULUS CHAPTER 1: Integrals Syllabus: Integration as inverse process of differentiation, Integration of a variety of functions by substitution, by partial fractions and by parts, Evaluation of simple integrals of the following types and problems based on them. dx dx dx dx , , ax2 + bx + c , ax2 + bx + c x2 ± a2 a2 − x2     dx x2 ± a2 ,    px +q px + q ax2 + bx + c dx, ax2 + bx + c dx, a2 ± x2 dx, x2 − a2 dx Fundamental Theorem of Calculus (without proof). Basic Properties of definite integrals and evaluation of definite integrals. A. INDEFINITE INTEGRALS IMPORTANT RESULTS AND FORMULAE 1. Integration is the reverse process of differentiation. In the differential calculus, we are given a function and we have to find the derivative or differential coefficient of this function. But in the integral calculus, we have to find a function whose differential coefficient is given. Thus integration is the process which is the reverse process of differentiation. d f (x) f (x) dx = g(x) + C . i.e., dx ( g( x) ) = ⇔ 2. Derivative of a function is unique but a function can have many integrals. 3. f (x) dx = g(x) + C , where C is a constant is known as indefinite integral. C is called as constant of integration, dx is known as the element of integration, and f(x) is known as integrand. bb  4. f (x) dx is known as definite integral and is given by f (x) dx = g(b) − g(a) . aa 5. Algebra of integration: ( )(i)d dx f (x) dx = f (x) i.e., the differentiation of an integral is the integrand.  (ii) kf (x) dx = k f (x) dx , where k is a constant. (iii) dx = x + C 1

2 Mathematics—XII   (iv) [ f (x) ± g(x)]dx = f (x)dx ± g(x)dx i.e., the integral of a sum or difference of a finite number of functions is equal to the sum or difference of their integrals. (v) kdx = kx + C . 6. Standard results on integration: xn+1 I. (i) xndx = n+1 + C, n ≠ −1 (ii) 1 dx = log | x | + C x ax (iii) exdx = ex + C (iv) ax dx = log a + C, a > 0, a ≠ 1. II. (i) sin x dx = − cos x + C (ii) cos x dx = sin x + C (iii) tan x dx = log|sec x|+ C = − log|cos x|+ C (iv) cot x dx = log|sin x|+ C (v) sec x dx = log|sec x + tan x|+ C = log tan  π + x  + C  4 2  (vi) cosec x dx = log | cosec x − cot x | + C = log tan x + C 2 (vii) sec2 x dx = tan x + C (viii) cosec2x dx = − cot x + C (ix) sec x tan x dx = sec x + C (x) cosec x cot x dx = −cosec x + C III. (i) 1 dx = sin−1 x + C or –cos–1 x + C 1 − x2 (ii) 1 1 dx = tan−1 x + C or – cot–1 x + C + x2 (iii) 1 dx = sec–1x or – cosec–1 x + C x(x2 − 1) IV. Extension of the standard results: (i) (ax + b)ndx = (ax + b)n+1 + C (ii) sin(ax + b) dx = − cos(ax + b) + C (n + 1)a a (iii) e ax dx = eax +C (iv) dx = log | ax + b| and so on. a (ax + a b) V. Special Integrals:  (a) (i) x2 1 a2 dx = 1 log x−a +C (ii) a2 1 x2 dx = 1 log a+x +C − 2a x+a − 2a a−x (iii) a2 1 x2 dx = 1 tan−1 x +C + a a (b) (i) 1 x2 dx = sin−1  x + C a2 −  a 

Integrals 3 (ii) 1 dx = log|x + x2 − a2 |+ C x2 − a2 (iii) 1 dx = log|x + x2 + a2 |+ C x2 + a2 (c) (i) a2 − x2 dx = x a2 − x2 + a2 sin−1  x + C 2 2  a  (ii) x2 − a2 dx = x x2 − a2 − a2 log x + x2 − a2 + C 22 (iii) x2 + a2 dx = x x2 + a2 + a2 log x + x2 + a2 + C 22 7. Methods of integration: (a) Integration by the decomposition of the integrand: In this method, the integrand is decomposed into number of parts so that each part can be integrated easily using standard results. Example 1. x3 + 5x2 + 4x + 1 dx . Evaluate x2 x3 + 5x2 + 4x + 1  x 4 1  dx x2  x x2   Solution. Here, dx = + 5 + + = x2 + 5x + 4 log|x|+ x−1 +C 2 −1 = x2 + 5x + 4 log | x |− 1 + C. 2 x Example 2. Evaluate  x+ 1 2 dx .  x  Solution. Here,  x+ 1 2 dx =  x + 1 + 2 dx = x2 + log| x| + 2x + C .  x   x 2 Example 3. Evaluate dx x − 2. x+1− Solution. Here,  dx = ( x+1− x+1+ x−2 (x − 2) dx x+1− x − 2)( x + 1 + x−2 x+1 + x−2 dx 1 x+1+ x − 2)dx (x + 1) − (x − 2) 3  = = ( 3 (x 3 (x 3 3 1  (x + 1)2  + (x − 2)2  + C. =  + − 2)2  = 2 + 1)2  3   3 3 9  2 2

4 Mathematics—XII Example 4. Evaluate 1 + sin 2x dx . Solution.  Here, 1 + sin 2x dx = sin2 x + cos2 x + 2 sin x cos x dx = (sin x + cos x)2 dx = (sin x + cos x)dx = − cos x + sin x + C. Example 5. Evaluate: tan−1(sec x + tan x)dx. Solution. Here, tan−1(sec x + tan x)dx −1  1 + sin x  dx tan−1  1 − cos  π + x   dx  cos x   2 x    = tan =   π   2 +  sin   2 sin2  π + x    4 2  = tan −1   dx  π x  π x    4 2   4 2    2 sin + cos +  = tan−1 tan  π + x  dx    4 2    = π + x dx = π x + x2 + C.  4 2  4 4 (b) Integration by substitution: The method of evaluating an integral by reducing it to a standard form by a proper substitution is called integration by substitution. This method is mainly used in the following types of integrals: (i) f (x)[ f (x)]n dx (ii) f ′(x) dx . f (x) In this method we will make a suitable substitution for f(x) and change dx accord- ingly. Then the given integrals will be reduced to any one of the standard integrals which can be easily integrated. Evaluate: 1 − tan x dx Example 6. 1 + tan x . I= 1− tan x dx cos x − sin x dx Solution.  Let 1+ tan x = cos x + sin x Now put cos x + sin x = t ⇒ (–sin x + cos x)dx = dt I= dt log(t) x x C . ⇒ t = = log | cos + sin | + Evaluate sinx dx. Example 7. sin(x − a) Solution. Let I = sin x a) dx . Put (x – a) = t and so dx = dt sin(x −

Integrals 5 I= sin(a + t) dt sin a cos t+ cos a sin t dt sin t sin t  ⇒ =  = sin a cot t dt + cos a dt = sin a log|sin t|+ t cos a = sin a. log |sin(x – a)| + (x – a) cos a + C. Example 8. Evaluate sin 2x dx . Solution. a2sin2x + b2cos2x I= sin 2x dx Let a2 sin2 x + b2 cos2 x . Put a2 sin2 x + b2 cos2 x = t ⇒ (a2 – b2) sin 2x dx = dt I 1 dt 1 t (a2 − b2 ) t − ⇒ = = (a2 b2 ) log | | = (a2 1 b2 ) log | a2 sin 2 x + b2 cos2 x|+ C. − Example 9. Evaluate sec x.log(sec x + tan x)dx . Solution. Let I = sec x. log (sec x + tan x) dx Put log (sec x + tan x) = t ⇒ sec x dx = dt ⇒ I = t dt = t2 = 1 [log(sec x + tan x)]2 + C . 22 Example 10. Evaluate sin3x cos4x dx. Solution. Let I = sin3 x cos4 x dx. Put cos x = t ⇒ – sin x dx = dt  ⇒ I = sin2 x.t4(−dt) = − (1 − t2 )t4dt = − (t4 − t6 ) dt = − t5 + t7 = − cos5 x + cos7 x + C . 57 5 7 8. Evaluation of special integrals (a) Evaluation of Integrals of the type ax2 1 + c dx + bx Reduce this integral in any one of the standard forms of the type  x2 1 dx = 1 log x−a +C − a2 2a x+a  a2 1 dx = 1 log a+x +C − x2 2a a−x  a2 1 dx = 1 tan −1  x  + C . + x2 a  a 

6 Mathematics—XII Example 11. Evaluate 3x2 + 1 − 10 dx. 13x Solution. Let  I = 1 dx = 1 1 dx 3x2 + 13x − 10 3 13 x − 10 x2 + 33 = 1 1 dx = 1 × 1 log x + 13 − 17 3 3 × 17 66  x + 13  2 −  17  2 2  6   6  6 x + 13 + 17 66 = 1 log 6x − 4 = 1 log 3x − 2 +C. 17 6x + 30 17 3x + 15 Example 12. Evaluate 4x2 1 + 3 dx . – 4x Solultion. Let I = 1 dx 4x2 − 4x + 3  = 1 1 dx = 1 1 1 2 dx 4 −x 4 2  x2 + 3  x − 1 2 +  2   4 1× 1 tan−1  x −1  1 tan−1  2x − 1  C 4 1  2  22  2  =   = + . 2  1  2 Example 13. Evaluate: e2x ex + 5 dx. + 6ex Solution. Let I = e2x ex + 5 dx . Put ex = t so that exdx = dt. + 6ex I= 1 dt 1 dt  ∴ t2 + 6t + 5 = (t + 3)2 − 22 (It is of the form x2 1 a2 dx ) − ⇒ I = 1 log t+3−2 = 1 log ex + 1 +C. 2×2 t+3+2 4 ex + 5 (b) Evaluation of integrals of the type 1 dx. ax2 + bx + c Reduce this integral in any one of the standard forms of the type  1 dx = sin−1  x  + C a2 − x2  a   1 dx = log x + x2 − a2 + C x2 − a2  1 dx = log x + x2 + a2 + C x2 + a2

Integrals 7 Example 14. Evaluate (x − 1 − 2) dx. 1)(x 1 dx = 1 dx x2 − 3x + 2 Solution.  Let I = 2 2    x − 3 −  1 2 2 ⇒ I= log  x − 3 +  x − 3 2 −  1 2 +C  2   2   2  = log  x − 3  + x2 − 3x + 2 + C .  2  Example 15. Evaluate: 1 dx. 9 + 8x − x2 Solution. Let I = 1 dx 9 + 8x − x2  = 1 dx = 1 dx −(x2 − 8x − 9) −{(x − 4)2 − 52 } = 1 dx = sin−1  x − 4  + C . − (x − 4)4  5  52 Example 16. Evaluate x2 1 + 2 dx . − 4x Solution. Let  I = 1 dx = 1 dx x2 − 4x + 2 (x − 2)2 − ( 2)2 = log (x − 2) + x2 − 4x + 2 + C . Example 17. Evaluate cos x dx . Solution. Let sin2x − 2sin x − 3 I = cos x dx . sin2 x − 2 sin x − 3 Put sin x = t so that cos x dx = dt. 1 dt 1 dt − 2t − 3 (t − 1)2 − 22  ∴ I= t2 = = log (t − 1) + t2 − 2t − 3 = log (sin x − 1) + sin2 x − 2 sin x − 3 + C . (c) Evaluation of the integrals of the type  px+q c dx or px + q dx ax2 +bx + ax2 + bx + c In this type we have to find two values λ and µ such that px + q = λ ddx (ax2 + bx + c) + μ  Then the above integral reduces to any of the standard results which we can integrate.

8 Mathematics—XII Example 18. Evaluate x2 4x + 1 2 dx. + 3x + Solution. We have to find two values of λ and µ such that 4x + 1 = λ d (x2 + 3x + 2) + μ dx = λ(2x + 3) + µ Solving, we get λ = 2, µ = – 5. I= 4x + 1 dx 2(2x + 3) − 5 dx + 3x + x2 + 3x + 2  ∴ x2 2 = 2x + 3 dx 5 dx I1 I2 + 3x + 3x  = 2 x2 2 − x2 + + 2 = + Now, I1= 2 2x + 3 2 dx = 2 log|x2 + 3x + 2|+ C1. x2 + 3x +  I2 = x2 + 5 + 2 dx = 5 3 1 1 2 dx 3x 2 2 2   x +  −  = 5. 1 log x+ 3−1 + C2 = 5 log x+1 + C2 2. 1 x+ 22 x+2 3+1 2 22 ∴ I = 2 log | x2 + 3x + 2| − 5 log x+1 +C. x+2 Example 19. Evaluate 2x + 3 dx. x2 + 4x + 1 Solution. Let I = 2x + 3 dx . We have to find two values of λ and µ such that x2 + 4x + 1 2x + 3 = λ d (x 2 + 4x + 1) + µ = λ(2x + 4) + μ dx Solving, we get λ = 1, µ = – 1.  ∴ I = 2x + 3 dx = (2x + 4) − 1 dx x2 + 4x + 1 x2 + 4x + 1 2x + 4 dx − 1 dx I1 I2 x2 + 4x + 1 + 4x  = x2 = − . + 1 2x + 4 dx = dt t x2 + 4x + 1 . x2 + 4x + 1 t Now,  I1 = = 2 =2 1 dx = 1 dx x2 + 4x + 1 (x + 2)2 − ( 3)2 Also,  I2 = = log x + 2 + x2 + 4x + 1 + C ∴ I = 2 x2 + 4x + 1 − log x + 2 + x2 + 4x + 1 + C .

Integrals 9  (d) Evaluation of the integrals of the form 1 dx , 1 dx , a sin2 x + b cos2 x a + b cos2 x   a 1 dx , 1 dx , (a x 1 x)2 dx . + b sin2 x a + b sin2 x + c cos2 x sin + b cos To evaluate these types of integrals, we have to do the following: (i) Divide both numerator and denominator by cos2 x (ii) Put tanx = t so that sec2x dx = dt. This substitution reduces the integral of the form  at2 1 + c dx which can be integrated. + bt Example 20. Evaluate a2 sin2 x 1 b2cos2x dx . + Solution. Dividing both numerator and denominator by cos2 x, we get I = a2 sec2 x b2 dx. Put tan x = t so that sec2 x dx = dt. tan2 x + 1 dt 1 1 dt a2t2 + a2   ∴ I= b2 =  b 2 a  t2 + = 1 ×  1  tan −1   t   a2  b    b   a  a  = 1 tan−1  at  = 1 tan−1  a tan x  + C . ab  b  ab  b  Example 21. Evaluate 3 + 1 2x dx . sin Solution.  Let I= 3 + 1 2x dx = 3(sin2 x + cos2 1 + 2 sin x cos x dx sin x) Dividing both numerator and denominator by cos2 x, we get I = 3 tan2 sec2 x x + 3 dx . x+2 tan Put tan x = t so that sec2 x dx = dt  ∴ I= 1 dt = 1 1 dt 3 t2 + 2t + 3 3 t2 + 2 t + 1 3 1 1 dt 1. 1 tan−1  t + 1  3  3  = 2 =   2  3 2 2  22   t + 1  +  2 2    3  3  3    3 = 1 tan−1  3t +1  = 1 tan−1  3 tan x+ 1  + C . 2  2 2  22  2 2  2

10 Mathematics—XII  (e) Evaluation of integrals of the form a x 1 b x dx , a + 1 x dx , sin + cos b sin + a 1 x dx , a sin x + 1 cos x + c dx . b cos b To evaluate these types of integrals, we do the following: 2 tan x 1 − tan2 x (i) Put sin x = 2 cos x = 2 1 + tan2 x , x 2 1+ tan2 2 (ii) Replace 1 + tan2 x = sec2 x 22 (iii) Put tan x = t so that sec2 x dx = 2dt . 22 Now the integral will be reduced to any one of the standard forms. Example 22. Evaluate 1 + sin 1 + cos x dx. x I= 1 dx . Solution. Let 1 + sin x + cos x 2 tan x 1 − tan2 x sin x 2 cos x 2 Put = tan2 x , = x 1 + 1 + tan2 22 I= 1 1 − tan2 x dx ⇒ 2 tan x 1+ 2 + 2 tan2 x x 1+ 1+ tan2 22 1 + tan2 x 2 dx = tan2 x x+ tan2 x 1 + + 2 tan 1 − 22 2 sec2 x 2 dx [ 1 + tan2θ = sec2 θ] = tan x . 2 + 2 2 Now put tan x = t so that sec2 x dx = 2dt . 22 2dt dt log | t x +C 2 + 2t t+1 2  ⇒ I= = = + 1| = log tan +1 . 9. Integration by parts.    If u and v are two functions of x, du vdx uvdx = u vdx − dx  dx .

Integrals 11 i.e., The integral of the product of two functions = First function × integral of the second – Integral of (derivative of the first × integral of the second) Here the choice of the first function is important. We can use the order ILATE, where I = Inverse trigonometric functions L = Logarithmic function A = Algebraic function T = Trigonometric functions E = Exponential function If the integrand contains only one function, we take that function as the first func- tion and 1 as the second function. Example 23. Evaluate x sec2x dx . Solution. Let I = x sec2 x dx Integrating by parts by taking, u = x, v = sec2 x, we get   {  }I = x sec2 x dx − 1. sec2 x dx dx = x tan x − 1. tan x dx = x tan x + log|cos x|+ C. Example 24. Evaluate sin−1x dx . Solution. Let I = sin−1 x dx Take sin–1x = t, then x = sin t ⇒ dx = cos t dt, then I = t cos t dt . Now integrating by parts, we get I = t sin t − 1. sin t dt = t sin t + cos t = x sin−1 x + 1 − sin2 t + C = x sin−1 x + 1 − x2 + C. Example 25. Evaluate sin–1x dx . 3 (1 − x2 )2 Solution. Take sin–1 x = t ⇒ x = sin t ⇒ dx = cos t dt Then,  I = t t sec2 t dt sin 2 t)3/2 ⋅ cos t dt = (1 −

12 Mathematics—XII Now integrating by parts, I = t tan t − 1. tan tdt = t tan t + log|cos t| = t. sin t + log 1 − sin2 t 1 − sin2 t = (sin−1 x). x + log 1 − x2 + C 1 − x2 = x sin−1 x + 1 log |1 − x2 |+ C. 1 − x2 2 10. Evaluation of integrals of the form ex { f (x) + f ′(x)}dx For evaluating this integral, we use the standard result ex { f (x) + f ′(x)}dx = ex f (x) + C. Example 26. Evaluate ex (sin x + cos x)dx. Solution. We know that if f(x) = sin x, f ′(x) = cos x. The given integral is of the form  ex { f (x) + f ′(x)}dx = ex f (x) + C = ex sin x + C . Example 27. Evaluate ex  1 − 1  dx.  x x2  Solution. This is of the form ex{ f (x) + f ′(x)}dx where f (x) = 1 , f ′(x) = −1 . x x2 ∴ ex  1 − 1  dx = ex + C.  x x2  x Example 28. Evaluate ex  2 + sin 2x  dx.  1 + cos 2x    Solution.   ex2+sin 2x  dx = ex  2 + 2 sin x cos x  dx = ex (sec2 x + tan x)dx  1 + cos 2x   2 cos2 x  We know that if f(x) = tan x, f ′ (x) = sec2x ∴ ex  2 + sin 2x  dx = extan x + C.  1 + cos 2x  11. Integration of Rational Algebraic Functions by partial fractions Partial fractions: If a function can be written as the sum of two or more fractions, those fractions are known as partial fractions. Fundamental rules for resolving into partial fractions. (i) The integrand must be a proper fraction i.e., the degree of the numerator must be less than the degree of the denominator. If not, divide and convert the integrand into a proper fraction.

Integrals 13 (ii) If the rational fraction is of the form (x px + q b) , the partial fraction is of the form − a)(x − px + q A B (x − a)(x − b) = x − a + x−b. (iii) If the rational fraction is of the form px + q , the partial fraction is of the form (x − a)2 px + q A B (x − a)2 = (x − a)2 + x− a . (iv) If the rational fraction is of the form px2 + qx + r (x − a)(x − b)(x − c) , the partial fraction is of the form px2 + qx + r = A + B + x C c . (x − a)(x − b)(x − c) x−a x−b − (v) If the rational fraction is of the form (x px2 + qx + r c) , the partial fraction is of the − a)(x2 + bx + form (x px2 + qx + r c) = A + Bx + C c , where x2 + bx + c cannot be factorised − a)(x2 + bx + x−a x2 + bx + further. Example 29. Evaluate (x – 2x – 1 – 3) dx . 1)(x + 2)(x Solution. Let 2x − 1 = A + B + C (x − 1)(x + 2)(x − 3) (x − 1) (x + 2) (x − 3) ⇒ 2x – 1 = A(x + 2)(x – 3) + B(x – 1)(x – 3) + C(x – 1) (x + 2) Now, x = – 2 ⇒ B = − 1 , x = 3  C = 1 , x = 1  A = − 1 32 6 2x − 1 dx 1 1 1 1 1 1 1)(x + 2)(x 6 − 3 + 2 −    ∴(x − − 3) = − (x dx − (x dx + (x 3) dx 1) 2) = − 1 log| x − 1| − 1 log | x + 2| + 1 log| x − 3| + C . 63 2 Example 30. Evaluate (x2 + 2x + 2) dx. 1)(x2 Putting x2 = t, 2x dx = dt, we get I = 1 dt. Solution. (t + 1)(t + 2) Let 1 = A + B (t + 1)(t + 2) (t + 1) (t + 2) ⇒ A(t + 2) + B(t + 1) = 1 ⇒ A = 1, B = –1 1  1 1  dt + 1)(t  + +   ∴   Ι= (t + dt = (t 1) − (t 2) 2) = log|t + 1|− log|t + 2| = log |x2 + 1| – log|x2 + 2| + C.

14 Mathematics—XII Example 31. Evaluate (x − 3x + 1 2) dx . 2)2(x + Solution. Let 3x + 1 2) = A + (x B + C (x − 2)2(x + (x − 2) − 2)2 (x + 2) ⇒ A(x – 2) (x + 2) + B(x + 2) + C(x – 2)2 = 3x + 1 Eliminating, we get x = 2 ⇒ B = 7 , x = −2  C = − 5 4 16 Now comparing co-efficients of x2 on both sides, we get A + C = 0 ⇒ A = 5 16 I= 5 1 dx 7 1 dx 5 1 dx 16 − 4 − 2)2 16 +   ∴ x 2 + (x − x 2 = 5 log | x − 2| − 7 2) − 5 log | x + 2|+ C. 16 4(x − 16 12. Evaluation of the integrals of the form or which can be reduced to the form   a2 − x2dx or x2 − a2dx or x2 + a2dx . For these type of integrals we use the following standard results: (i) a2 − x2 dx = x a2 − x2 + a2 sin−1  x  + C 2 2  a  (ii) x2 − a2 dx = x x2 − a2 − a2 log x + x2 − a2 + C 22 (iii) x2 + a2 dx = x x2 + a2 + a2 log x + x2 + a2 + C. 22 Example 32. Evaluate x2 + 2x + 5 dx . Solution. We have x2 + 2x + 5 dx  = (x + 1)2 + 22 dx (It is of the form x2 + a2 dx ) = 1 (x + 1) (x + 1)2 + 22 + 1 × 22 log x + 1 + (x + 1)2 + 22 +C 22 = 1 (x + 1) x2 + 2x + 5 + 2 log x +1+ x2 + 2x + 5 + C. 2 Example 33. Evaluate (x − 3)(5 − x) dx. Solution. We have (x − 3)(5 − x) dx   = −x2 + 8x − 15 dx = −(x2 − 8x + 15)dx = −{(x − 4)2 − 12} dx  = 12 − (x − 4)2 dx (It is of the form a2 − x2 dx )

Integrals 15 = 1 (x − 4) 12 − (x − 4)2 + 1 × 12 sin−1  x − 4  + C 2 2  1  = 1 (x − 4) (x − 3)(5 − x) + 1 sin−1(x − 4) + C. 22 x2 + 1 dx , x2 − 1 dx , 1 dx + λx2 + + λx2 + λx2   13. Integrals of the form x4 x4 1 x4 1 + + 1 where λ ∈ R. To evaluate these types of integrals we follow the following steps: (i) Divide both numerator and denominator by x2. (ii) Express the denominator in the form  x + 1 2 ± k2 or  x − 1 2 ± k2  x   x  (iii) Put x + 1 = t if we have 1 − 1 in the numerator and x − 1 = t if we have 1 + 1 x x2 x x2 in the numerator. (iv) This substitution reduces the integral to any one of the forms 1 dx x2 + a2 or x2 1 a2 dx which can be integrated. − Example 34. Evaluate the integral x4 x2 − 1 1 dx + x2 + I= x2 −1 dx. Solution. Let x4 + x2 + 1 Divide both numerator and denominator of the integrand by x2. Then, 1 − 1 1 − 1 x2 + x2 dx = dx.  I = 12 x2 + 1 + 1  x x  x2  − 1 Put x + 1 = t so that  1 − 1  dx = dt. x  x2  I= 1 dt. ⇒ t2 − 12 (This is of the form x2 1 a2 dx = 1 log x−a + C) − 2a x+a I = 1 log t−1 + C = 1 log x + 1 −1 +C 2⋅1 t+1 2 x ⇒ 1 x + x + 1 = 1 log x2 − x + 1 + C. 2 x2 + x + 1

16 Mathematics—XII PRACTICE EXERCISE 1.1 Very Short/Short Answer Type Questions 1. Find sin 6 x dx 2. Find sin2x − cos2x dx . cos8 x sin2x cos2x 3. Find the antiderivative of  3 x+ 1 4. Evaluate:  cos−1 (sin x) dx .  x  5. Evaluate:  ex (sin x − cos x) dx . 6.  dx Evaluate: sin2 x cos2x . 7. Write the value of  x + cos 6x dx . 8. Write the value of sec 2 x dx 3x2 + sin 6x cosec 2 x 9. Given  ex (tan x + 1) sec x dx = ex f (x) + C . Write f(x) satisfying the given function. 10. Evaluate:  (1 − x) x dx . 11. If  x− 1  ex dx = f (x) ex + C , find the value of f(x).  x2  Evaluate the following integrals: 12.  1 + 2 2x dx. 13.  (ax + b)3 dx . cos 14.  (log x)2 dx. 15.  sec x (sec x + tan x) dx . x 16.  dx 17.  2 − 3 sin x dx. . x2 + 16 cos2 x 18.  (1 + log x)2 dx. 19. e2x − e−2x dx x e2x + e−2x 20.  cos x dx. 21. tan2 x dx. x 22. 1 cos x x dx. 23. x2 − 1 dx. + sin2 x2 + 1 24. 2x cos x2 dx. 25.  x− 1 2 dx.  x  26. cos 3x cos x dx. ( )tan−1 x 2 27. 1 + x2 dx. 28. cos 2x x dx. 29. 1 ex dx. sin2 x cos2 + e2x  dx 30. x cos2 (1 + log x).

Integrals 17 Long Answer-I Type Questions Evaluate the following integrals: 1. x4 x2 − 2 dx 2. (x + 3) ex dx + x2 (x + 5)3 3.  log x dx 4. x2 + x + 1 dx (x + 1)2 (x2 + 1) (x + 2) 5. x3 dx 6.  sin x − x cos x dx (x − 1) (x2 + 1) x (x + sin x) 7.  dx 8. x2 − 3x + 1 dx sin x + sin 2x 1 − x2 9. x3 dx 10. x cos−1 x dx x4 + 3x2 + 2 1 − x2 11.  5x − 2 dx 12.  x+2 dx 1 + 2x + 3x2 x2 + 5x + 6 13. sin6 x − cos6 x dx 14. 1 − sin x . e− x dx sin2 x cos2 x 1 + cos x 2 15.  8 dx 16.  3x + 1 dx (x + 2) (x2 + 4) (x + 1)2 (x + 3) 17.  (3 sin x − 2) cos x dx 18. e2x 1− sin 2x  dx 5 − cos2 x − 4 sin x  cos   1 − 2x  19. 3x + 5 dx 20.  x dx + 1) (x − 1) x3 − x2 − x + 1 (x2 21.  cos 2x − cos 2α dx 22.  x+2 dx cos x − cos α x2 + 2x + 3 23.  dx 24.  sin (x − a) dx x (x5 + 3) sin (x + a) 25. (x2 x2 + 1 25) dx 26.   1 + sin x  ex dx + 4) (x2 +  1 + cos x    27. x2 dx 28.  sin x sin 2x sin 3x dx (x sin x + cos x)2 29. x2 tan−1 x dx 30. 2x dx (x2 + 1) (x2 + 3) 31. e2x sin x dx 32. x tan−1 x dx.  dx 34. x3 cos x4dx. 33. 32 − 2x2 .

18 Mathematics—XII 35. sec2(2 tan−1 x) dx. 36. x tan−1 x2 dx. (1 + x2) (1 + x4) 37. sec x cosec x dx. 38. x sin3(x2) cos(x2) dx. log(cot x) 39. sec4 x tan x dx. 40. (1 dx ) ⋅ + ex 41. sin x dx. Long Answer-II Type Questions Evaluate the following integrals: 1. (x2 + x2 + 4) dx. 2. (x x2 + x + 1 dx. 1) (x2 + 1)2 (x + 2) ( )3.  cot x + tan x dx 4. sin 4 x + sin 2 1 x + cos4 x dx. x cos2 5. x sin−1 x dx. 6.  (x 6x + 7 − 4) dx. 1 − x2 − 5) (x 7. (x + 1)(x + log x)2 dx. 8. ecos2 x sin 2x dx. 2x 9. (ex + e−x ) dx. 10. (ex dx ) ⋅ (ex − e−x ) + e−x 11. ex(tan x + log sec x) dx. e x (sin x cos x − 1) dx. sin2 x 12. 13. ex(x2 + 1) dx. 14. ex(x + 1) dx. (x + 1)2 cos 2(xe x ) 1 dx. ( )e x sin e x 15. + 16. x dx. sin 4 x cos4 x 17. 5 + dx 2 x ⋅ 18. 1 + dx x ⋅ 3 sin 2 cos2 19. 3 + dx x ⋅ 20. 5 dx + 2 ⋅ 4 sin cos x 21. tan−1  1 2x 2  dx. 22.  x) + 1  dx.  −x  log(log (log x)2   

58 Mathematics—XII ∫ ∫π − x) π x) I= = 0 0 4(π − x) sin (π dx 4(π − x) sin x dx ...(ii) 1 + cos2 (π − 1 + cos2 x Adding (i) and (ii), we get π 4x sin x π 4(π − x)sin x dx 0 1 + cos2 x 1 + cos2 x ∫ ∫2I= dx + 0 ∫π sin x x dx . π sin x x dx . 1 + cos2 1 + cos2 = 4π I = 2π 0 0 ∫⇒ Put cos x = t ⇒ sin x dx = – dt. When x = 0, t = 1 and when x = π, t = – 1. −1 1 ( − dt) 1 1 dt 1 + t2 1 + t2 I = 2π = 2π 1 −1 ∫ ∫⇒ = 2π ⎡⎣tan−1 t⎦⎤1−1 = 2π ⎣⎡tan−1(1) − tan−1 (−1)⎦⎤ = 2π ⎡ π − ⎛ − π⎞ ⎤ = 2π × π = π2 . ⎢⎣ 4 ⎜⎝ 4 ⎠⎟ ⎥⎦ 2 π x tan x π x sin2 sec x cosec 0 ∫ ∫11.I= x dx = x dx ...(i) ...(ii) 0 ππ ∫ ∫⇒ I = (π − x) sin2 (π − x) dx = (π − x) sin2x dx 00 Adding (i) and (ii), we get π sin 2 π π π ⎡⎣⎢x sin 2x ⎤π π2 π2 ∫ ∫2I 2 2 2 ⎦⎥0 2 4 = π xdx = (1 − cos 2x) dx = − = ⇒ I = . 0 0 π 4 sin 2θ dθ sin4 θ + cos4 θ Let I = 0 ∫12. π ∫⇒ 4 2 sin θ cos θ dθ . sin4 θ + cos4 θ I= 0 Dividing both numerator and denominator by cos4 θ, we get ππ 4 4 2 tan θ sec2 0 tan2 θ 2 + I= ∫ ∫ ( )0 2 tan θ sec2 θ dθ = θ dθ . tan4 θ + 1 1 Put tan2 θ = t ⇒ 2 tan θ sec2 θdθ = dt.

Integrals 59 When θ = 0, t = 0 and θ = π , t = 1 . 4 1⇒ t2 1 1 dt =  tan −1 t10 =  tan −1 1 − tan−1 0 = π −0 = π . + 4 4 I= 0 2π dx ...(i) 1 + esin x ...(ii) Let I = 0 13. aa Applying  f (x) dx =  f (a − x) dx , we get 00  2π 2π dx 2π esin x dx 1 + e− sin x esin x + 1 I= = = 0 0 0 dx 1 + esin(2π − x) Adding (i) and (ii), we get  2π 2π esin x dx esin x + 1 2I = + 0 0 dx 1 + esin x 2π (1 + esin x ) dx 2π 0 esin x + 1  = = 1 dx = [x]02π = 2π ⇒ I = π. 0 4  ( )14. Let I = x + x − 2 + x − 4 dx 0 42 44    = x dx + x − 2 dx + x − 2 dx + x − 4 dx . 00 20 42 44    = x dx + − (x − 2) dx + (x − 2) dx + − (x − 4) dx . 00 20 = x2 4 + 2x − x2 2 +  x2 − 2 x  4 + 4x − x2 4 .      2  2   2 0 2 0  2 0 ⇒ I = 8 + (4 – 2) + (8 – 8) – (2 – 4) + (16 – 8) = 8 + 2 + 0 + 2 + 8 = 20. 15. If x3 – x = 0, then x(x + 1)(x – 1) = 0 ⇒ x = – 1, 0, 1. Clearly, in the interval [–1, 0], x3 – x ≥ 0; in [0, 1], x3 – x ≤ 0 and in [1, 2], x3 – x ≥ 0. 20 12    ∴ x3 − x dx = x3 − x dx + x3 − x dx + x3 − x dx . −1 −1 01 01 2   = (x3 − x) dx + − (x3 − x) dx + (x3 − x) dx −1 0 1

60 Mathematics—XII = ⎡ x4 − x2 ⎤0 + ⎡ x2 − x4 ⎤1 + ⎡ x4 − x2 ⎤2 ⎢ 4 ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 2 ⎦ −1 ⎣ 2 4 ⎦0 ⎣ 4 2 ⎦1 = 0 − ⎛ 1 − 1⎞ + ⎛ 1 − 1⎞ − 0 + (4 − 2) − ⎛ 1 − 1⎞ . ⎝⎜ 4 2 ⎟⎠ ⎜⎝ 2 4 ⎟⎠ ⎜⎝ 4 2 ⎟⎠ = 1 + 1 + 2 + 1 = 11 4 4 4 4 π π x 2 sin x cos x ππ 2 2 2 2 x + sin x + x dx 1 2 x sec2 x dx 2 tan x dx Let I = 1 + cos x = 2 2 0 2 0 2 0 0 ∫ ∫ ∫ ∫16. dx = + . 2 cos2 Integrating by parts the first integral by taking u = x and v = sec2 x , we get 2 1 ⎡ x tan x π 1 π tan x dx π tan x dx . ∫ ∫I = 2 ⎢⎣ ⋅2 2 − 2 2 1· 2 2 2 2 ⎤2 ⎥⎦0 0 + 0 π ππ π ∫ ∫⇒ I = ⎡ x tan x⎤2 − 2 tan x dx 2 tan x dx = ⎣⎢⎡x tan x⎤2 = π tan π −0= π . ⎢⎣ 2 ⎦⎥0 0 2 2 2 ⎥⎦0 2 4 2 + 0 1 log ⎛ 1 1⎞⎟⎠ dx 1 ⎛ 1 − x⎞ ⎝⎜ x 0 ⎝⎜ x ⎠⎟ Let I = 0 ∫ ∫17. − = log dx ...(i) 1 ⎛ 1 − (1 − x) ⎞ 1 log ⎛ x ⎞ dx ∫ ∫⇒ log ⎜ 1− x ⎟ dx ⎜ − ⎟ ...(ii) I= ⎝ ⎠ = ⎝ 1 x ⎠ 0 0 ⎛a a ⎞ ∫ ∫⎜∵ f (x) dx = f (a − x) dx ⎟ ⎜⎝ 0 0 ⎠⎟ Adding (i) and (ii), we get 1 log ⎛ 1 − x ⎞ dx 1 ⎛ x ⎞ ∫ ∫2I = 0 ⎝⎜ x ⎟⎠ + 0 log ⎜ − ⎟ dx = log 1 = 0 ⇒ I = 0. ⎝ 1 x ⎠ ∫18.Let π 1 + x x dx ...(i) sin I= 0 Applying the property, aa ∫ ∫f (x) dx = f (a − x) dx , we get 00 ∫ ∫π 1 + π−x x) dx = π 1 π −x x dx ...(ii) sin(π − 0 + sin I= 0

Integrals 61 Adding (i) and (ii) , we get    πx + π− x π π π 1 π 1 − sin x 1 + sin x sin sin 1 − sin2 x 2I = dx = 1 + x dx =π 1+ x dx =π dx 0 0 0 0  π π =π = π (sec2 x − sec x tan x) dx 1 − sin x dx = π tan x − sec x0π cos2 x 00 2I = π [(tan π – sec π) – (tan 0 – sec 0)] = π [0 – (– 1) – (0 – 1)] = 2π π x sin ∴ 1 + x dx = π . 0 1 19. Let I = x2 (1 − x)n dx . 0 aa  Applying the property, f (x) dx = f (a − x) dx , we get 00 11  I = (1 − x)2 [1 − (1 − x)]n dx = (1 − 2x + x2 ) xn dx 00 1 (xn − 2xn + 1 + xn + 2 ) dx =  xn + 1 1  xn + 2 1  xn + 3 1   − 2   +   =  n + 1 0  n + 2   n + 3 0 0 0 = n 1 1 − n 2 2 + n 1 3 = (n + 1) (n 2 2) (n + 3) . + + + + sin 5x π 5 20. I= 1 sin x −  4 = 3⋅ 2  0 52 21. I = log x + x2 +4  4 = log 4+ 20  3 3+ 13 22. I = log (x + 2) + x2 + 4x + 3 2 = log 4+ 15 . 1 3+ 8 Put e2x = t. I = 1 e2 dt 1 [tan−1 t]1e2 = 1  tan −1(e 2 ) − π 23. 2 1 1+ t2 = 2 2  4  . 24. Put sec x = t, sec x tan x = dt. I= 2 dt = [tan−1 t]12 = tan−1(2) − π . 1 1+ t2 4 25. I= 1  tan −1(2x)0k =π  tan−1 (2k) = π  2k = 1  k = 1 . 4 16 42

Vector Algebra 171 UNIT IV: VECTORS AND THREE DIMENSIONAL GEOMETRY CHAPTER 4: Vector Algebra Syllabus: Vectors and scalars, magnitude and direction of a vector. Direction cosines and direction ratios of a vector. Types of vectors (equal, unit, zero, parallel and collinear vectors), position vector of a point, negative of a vector, components of a vector, addition of vectors, multiplication of a vector by a scalar, position vector of a point dividing a line segment in a given ratio. Definition, Geometrical Interpretation, properties and application of scalar (dot) product of vectors, vector (cross) product of vectors. IMPORTANT RESULTS AND FORMULAE 1. Scalars: A Physical quantity which has only magnitude and no direction is a scalar quantity or simply a scalar. Examples: Mass, length, time, area, volume, speed, density, work, temperature, power etc. are examples of scalar. 2. Vectors: A physical quantity which has magnitude as well as direction is a vector quantity or a vector. Examples: Velocity, acceleration, force, displacement etc. Three characteristics are needed to specify a vector. (i) A non-negative quantity, which is the magnitude of a vector. (ii) Unit in terms of which the magnitude is measured. (iii) Direction and the line of action. 3. Directed Line Segment: A line segment is a directed line segment if its initial point and terminal points are specified. l l is a line. A, B are two points of the line. Then A⎯→B is l directed line segment with initial point A and terminal point B. B Directed Line Segment has three characteristics B AB (i) Length: The length is the distance between the initial BA point and the terminal point. A A (ii) Support: The support of a directed line segment is the line containing the line segment. (iii) Sense: Sense of A⎯→B is from A to B. 4. Representation of a vector as a directed line segment A vector can be represented by a directed line segment. The length of the directed line segment is the magnitude or modulus of the vector. The direction of the vector is the sense of the directed line segment. 171

172 Mathematics—XII The vector initial point A and the terminal point B is denoted by A⎯→B and the modulus is denoted by ⎯A→B or AB. 5. Types of vectors (i) Zero (null) vector: A vector whose initial and terminal points are coincides (same) is a zero vector. It is denoted by A⎯→A or ⎯B→B or ⎯→ 0 . The modulus of the zero vector is 0 and the direction is not specified. The vectors other than zero vector are known as proper vectors. (ii) Unit vector: A vector whose magnitude is 1 is a unit vector. The unit vector in the ⎯→ direction of a⎯→ is denoted by a . Also, a = a a⎯→ . (iii) Like vectors: Two vectors are said to be like vectors if they have the same or parallel line of support and having the same direction. (iv) Unlike vectors: Two vectors are said to be unlike vectors if they have the same or parallel line of support and having opposite directions. (v) Free vectors: Vectors whose initial point and terminal point are not specified is a free vector. Such vectors are denoted by ⎯a→ . (vi) Equal vectors: Two vectors are equal, if they have (i) same or parallel lines of support (ii) same direction (iii) equal magnitude. ⎯→ ⎯→ ⎯→ ⎯→ Note: a = b  a = b . But the converse is not true, i.e., if the magnitude of two vectors are equal, the vectors need not be equal. (vii) Co-initial vectors: Vectors having the same initial point are called co-initial vectors. (viii) Collinear vectors: Vectors having the same or parallel supports are called collinear vectors. To prove two vectors are collinear, we have to show that one vector is a scalar multiple of the other. (ix) Coplanar vectors: Two or more vectors are said to be coplanar if they lie in the same plane or their supports are parallel to the same plane. (x) Negative of a vector: Negative of a vector is defined as the vector having the same magnitude but opposite in direction. If ⎯a→ = ⎯A→B , then −⎯→a can be represented by the vector ⎯B→A or any directed line segment equal in magnitude of ⎯A→B but in opposite direction of ⎯A→B . (xi) Position vector of a point: If P is any point in a plane and O is any other point in the same plane, then O⎯→P is called the Position Vector of P with respect to O. The point O is called the point of reference or origin. In other words the Position Vector of a point is the vector attached to the point by the virtue of its position with respect to some other point.

Vector Algebra 173 6. Addition of vectors B (i) Triangle law of vector addition If two vectors are represented in magnitude and direction by a+b b the two sides of a triangle in the same order, then their sum is represented by the third side taken in the reverse order. O⎯→A + ⎯A→B = O⎯→B . O a A (ii) Parallelogram law of vector addition C B If two vectors are represented in magnitude and direction by the two adjacent sides of a parallelogram, then their sum is represented by the diagonal of the parallelogram which is co-initial with the given vectors, O⎯→A + O⎯→C = O⎯→B . OA 7. Properties of Addition of Vectors (i) Commutative: For any two vectors ⎯a→ and ⎯→ a⎯→ + b⎯→ = b⎯→ + ⎯a→ . b, (ii) Associative: For any three vectors a⎯→ , b⎯→ and ⎯c→ , ⎯→ ⎯→ ⎯→ ⎯→ ⎯→ ⎯→ (a + b) + c= a + (b + c) . (iii) Additive identity: If →a is any vector, there exists → such that → → → → → 0 a +0 =0 +a = a. ⎯→ Here, 0 is called the additive identity for vector addition. (iv) Additive inverse: If →a is any vector, then there exists −⎯→a such that →a + (−⎯→a) = ⎯→ →→ Here, – a⎯→is called the additive inverse of →a . (−a) + a =0. 8. Difference of Vector : ⎯a→ − b⎯→ is defined as ⎯→ ⎯→ a + ( −b ) . 9. Multiplication of a Vector by a Scalar If a⎯→ is a vector and k is a scalar, then k a⎯→ is a vector whose magnitude is k times of the vector ⎯a→ and the direction is (i) Same as that of ⎯a→if k > 0 (ii) Opposite as that of a⎯→ if k < 0. 10. Properties of Multiplication of a Vector by Scalar If m is a scalar and a⎯→is any vector, then (i) m→a = →a m (ii) m(n→a ) = (m→a )n = (mn)→a (iii) → = m →a + n →a (iv) m(→a + →b ) = m →a + m →b (m + n) a (v) → = −m →a m(− a ) 11. To prove two vectors →a and →b are collinear, we have to show that one vector is a scalar multiple of the other. This condition gives the idea of proving →a || →b , or proving three points A, B and C collinear.

174 Mathematics—XII (i) To prove →a ||→b , we have to show →a = k →b , where k is a scalar. (ii) To prove the points A, B and C are collinear, we have to show that A⎯→B = k B⎯C→. 12. Expression of a vector in terms of the position vector of the initial point and the terminal point Any vector = P⋅V of the Terminal point – P⋅V of the initial point. ∴ A⎯→B = P⋅V of B – P⋅V of A. Example 1. If the position vectors of A and B are respectively 2iˆ + 10ˆj + 3kˆ and 3iˆ + 5ˆj − 6kˆ , find A⎯→B . Sol. We have A⎯→B = P⋅V of B – P⋅V of A = (3iˆ + 5ˆj − 6kˆ) − (2iˆ + 10ˆj + 3kˆ) = iˆ − 5ˆj − 9kˆ . 13. Section Formula (i) Internal Division: The position Vector →r of the point which divides the line segment joining the points with position vectors →a and →b in the ratio m : n internally is →r = m →b + n →a . m+ n (ii) External Division: The position Vector →r of the point which divides the line segment joining the points with position vectors →a and →b in the ratio m : n externally is: →→ →r mb− n a = m−n . (iii) Midpoint Formula: The position vector of the midpoint of the line segment joining two points with position vectors →a and →b is →a + →b . 2 We can prove many of the geometrical or trigonometric results using vectors. Example 2. If →a , →b , →c are the three vectors represented by the sides of triangle, taken in order, prove that: →a + →b + →c = 0 . A Proof. Let B⎯C→ = → ⎯→ → and ⎯→ → a, CA = b AB = c . ( )Then, →a + →b + →c = B⎯C→ + C⎯A→ + A⎯→B c b = B⎯A→ + A⎯→B = B⎯B→ = → . B a C 0

Vector Algebra 175 Note: From the above example, we can deduce a method for proving a triangle (or a polygon). To prove a triangle (or a polygon), we have to show that the sum of the vectors taken in order along → the sides is 0 . Example 3. If ⎯PO→ + O⎯Q→ = Q⎯O→ + O⎯→R , prove that the points P, Q, R are collinear. Proof. Given P⎯O→ + O⎯Q→ = Q⎯O→ + O⎯→R ⇒ P⎯Q→ = Q⎯→R ( )⇒ P⎯Q→ = 1 Q⎯→R ⇒ P⎯Q→ , Q⎯→R are collinear (since one vector is scalar multiple of other). Since there is a common point Q, the points P, Q, R are collinear. Example 4. In the figure, A, B, P, Q and R are 5 points. Show that the sum of the vectors ⎯A→P , A⎯Q→ , ⎯A→R , ⎯PB→ , Q⎯→B , and ⎯R→B is 3 ⎯A→B . R Proof. We have, by triangle law of addition ⎯A→P + ⎯PB→ = ⎯A→B Q ...(1) ⎯→ ⎯→ = ⎯A→B ...(2) A B AQ + QB A⎯→R + ⎯R→B = ⎯A→B ...(3) Adding (1), (2) and (3), ⎯A→P + A⎯Q→ + A⎯→R + ⎯PB→ + Q⎯→B + ⎯R→B = 3 ⎯A→B . P 14. Component of a Vector: If →r = →a + →b , then →a and →b are known as the components of →r . 15. Components of a Vector in Two Dimensions Let P(x, y) be any point and →r be the position vector of P with respect to origin O and i , j are the unit vectors along x-axis and y-axis respectively. Then, → xi + yj Y (i) r = (ii) xi and yj are known as component vectors P (x, y) of →r along x-axis and y-axis respectively. (iii) x and y are known as the scalar components j r of →r along x-axis and y-axis respectively. (iv) If →r = xi + yj , then the magnitude O i X r = →r = x2 + y2 . L

176 Mathematics—XII 16. Components of a Vector in Three Dimensions Let P(x, y, z) be any point in the three dimensional co-ordinate system and →r be the position vector of P. Let i , j , k be the unit vectors along x-axis, y-axis and z-axis respectively. Let α, β and γ be the angles the vector →r makes with x-axis, y-axis and z-axis respectively. Then, (i) →r = xi + yj + zk (ii) The vector components of →r along Z the axes are xi, yj , and zk respectively. k r P (x, y, z) The scalar components of →r are x, j Y y and z. Sometimes, x, y and z are γ β termed as rectangular components O of →r . α (iii) If →r = xi + yj + zk , then the i X magnitude →r = x2 + y2 + z2 . If P(x, y, z) is any point in the space, then position vector of P(x, y, z) is given by →r = xi + yj + zk . 17. Vector when co-ordinates of the end points are given If P(x1, y1, z1); Q(x2, y2, z2) are any two points, then P⎯Q→ = (x2 − x1)i + (y2 − y1)j + (z2 − z1)k . ⎯PQ→ = (x2 − x1)2 + (y2 − y1)2 + (z2 − z1)2 . 18. Direction Cosines and Direction Ratios of a vector Z If α, β and γ are the angles, →r makes with x-axis, y-axis and z-axis respectively, then l = cos α, m = cos β and n = cos γ are called direction cosines. Any three numbers which are proportional to the k r P (x, y, z) direction cosines are called the direction ratios. j Y γ β If →r = ai + bj + ck , then directions ratios are a, b, c and O is usually denoted by < a, b, c >, i.e., the direction ratios of a vector are the components of the vector. α i Note: The directon ratios of a vector are not unique. X 19. Properties of Direction Cosines and Direction Ratios If →r = ai + bj + ck is a vector, then (i) l2 + m2 + n2 = 1 (ii) x = l →r , y = m →r , z = n →r

Three Dimensional Geometry 245 CHAPTER 5: Three Dimensional Geometry Syllabus: Direction cosines and direction ratios of a line joining two points. Cartesian equation and vector equation of a line, coplanar and skew lines, shortest distance between two lines. Cartesian and vector equation of a plane. Distance of a point from a plane. IMPORTANT RESULTS AND FORMULAE 1. A point in the space is represented by an ordered triplet (x, y, z). 2. Distance formula: The distance between two points A(x1, y1, z1) and B(x2, y2, z2) is AB = (x1 − x2 )2 + (y1 − y2 )2 + (z1 − z2 )2 . 3. Section formula: The point of internal division: The coordinate of a point P, which divides the join of two given points A(x1, y1, z1) and B(x2, y2, z2) in the ratio m : n internally is ⎛ mx2 + nx1 , my2 + ny1 , mz2 + nz1 ⎞ . ⎜⎝ m + n m + n m + n ⎟⎠ The point of external division: The co-ordinate of a point Q, which divides the join of two given points A(x1, y1, z1) and B(x2, y2, z2) in the ratio m : n externally is ⎛ mx2 − nx1 , my2 − ny1 , mz2 − nz1 ⎞ . ⎝⎜ m − n m − n m − n ⎟⎠ Midpoint formula: The coordinate of the midpoint P of the line segment joining A(x1, y1, z1) and B(x2, y2, z2) is ⎛ x1 + x2 , y1 + y2 , z1 + z2 ⎞ . ⎝⎜ 2 2 2 ⎠⎟ 4. Direction cosines of a line: The direction cosines of any line is defined as the direction cosines of any vector whose support is the given line. It follows from the definition that if A and B are two points of a line, then the direction cosines of the line are the direction cosines of A⎯→B or B⎯A→ . i.e., if α, β, γ are the angles at which the line makes with x-axis, y-axis and z-axis respectively, then cos α, cos β, cos γ or – cos α, – cos β, – cos γ are the direction cosines of the line. Therefore if l, m, n are the direction cosines of a line, then – l, – m, – n are also the direction ratios of the line. Also, l2 + m2 + n2 = 1, i.e., cos2α + cos2 β + cos2γ = 1. The direction cosines of a line passing through the points A(x1, y1, z1) and B(x2, y2, z2) is x2 − x1 = y2 − y1 = z2 − z1 . AB AB AB 5. Direction ratios of a line: The direction ratios of a line are any three numbers which are proportional to the direction cosines of the line. 245

246 Mathematics—XII (i) If a, b, c are the direction ratios of a line with direction cosines l, m, n then, l = m = n. a b c (ii) If a, b, c are the direction ratios of a line, then its direction cosines are ± a ,± b ,± c. a2 + b2 + c2 a2 + b2 + c2 a2 + b2 + c2 (iii) Direction ratios of a line passing through the points A(x1, y1, z1) and B(x2, y2, z2) are x2 – x1, y2 – y1, z2 – z1. 6. Projection of a line segment on a given line: The projection of a line segment AB, where A(x1, y1, z1) and B(x2, y2, z2) respectively, on the line whose direction cosines are l, m, n is l(x2 – x1) + m(y2 – y1) + n(z2 – z1). 7. The equation of a line passing through a point and parallel to a given vector (i) The vector equation of a line passing through a point whose position vector is →a and parallel to a given vector (or in the direction) of →b is →r = →a + λ →b . (ii) Cartesian equation of a line passing through the point A(x1, y1, z1) and having the x − x1 y − y1 z − z1 direction ratios a, b, c is a = b = c . (This form of the equation of the line is also known as symmetric form of the equation of the line). Note: If the equation of the line is x − x1 = y − y1 = z − z1 = λ , then any point on this line is of a b c the form (x1 + aλ, y1 + bλ, z1 + cλ). 8. The equation of a line passing through two points (i) The vector equation of a line passing through two points with position vectors →a and →b is →r = →a + λ(→b − →a ) . (ii) The Cartesian equations of a line passing through (x1, y1, z1) and (x2, y2, z2) is x − x1 y − y1 z − z1 x2 − x1 = y2 − y1 = z2 − z1 . Example 1. Find the equation of a line which passes through the point with position vector 2i − j + 4k and is in the direction of i + j − 2k . Also, deduce it to the Cartesian form. Solution. We have the equation of the line passing through →a and parallel to →b is →r = →a + λ →b . Hence, the required vector equation is (2i − j + 4k) + λ(i + j − 2k) ...(1) Cartesian form : Let →r = xi + y j + zk . Then the equation (1) becomes →r = xi + y j + zk = (2i − j + 4k) + λ(i + j − 2k) . ⇒ x = 2 + λ; y = –1 + λ; z = 4 – 2λ. ⇒ x−2 = y+1 = z−4 is the required cartesian equation. 1 1 −2

Three Dimensional Geometry 247 Example 2. Find the vector equation of the line through A(3, 4, –7) and B(1, –1, 6). Find Solution. also, its Cartesian equation. We have the vector equation of the line passing through two points with position vectors →a and →b is →r = a + λ(→b − →a ) . Here, → , →b = i− j + 6k . a = 3i + 4j − 7k Hence, the vector equation of the line is →r = (3i + 4j − 7k) + λ[(i − j + 6k) − (3i + 4 j − 7k)] . = (3iˆ + 4 ˆj − 7kˆ) + λ(−2iˆ − 5ˆj + 13kˆ) is the vector equation. The above equation can be written in the Cartesian form as x−3 = y−4 = z+7 ⋅ −2 −5 13 9. The shortest distance between two skew-lines (Vector form) The distance between two skew-lines →r = →a 1 + λ →b 1 and →r = →a 2 + μ →b 2 is given by S.D. = ( )(a→2 − a→1) ⋅ b→1× b→2 b→1× b→2 Notes 1. If the S.D. = 0, then the lines are intersecting. 2 . If two lines intersect, then they are coplanar. Hence, two lines →r = →a 1 + λ →b 1 and →r = →a 2 + μ →b 2 are coplanar if ⎛ → − → ⎞ · ⎛ → → ⎞ = 0. ⎜ ⎟ ⎜ ⎟ a2 a1 b1× b2 ⎝ ⎠⎝ ⎠ 10. The shortest distance between two skew-lines (Cartesian form) The shortest distance between the lines x − x1 = y − y1 = z − z1 and x − x2 = y − y2 = z − z2 is given by a1 b1 c1 a2 b2 c2 x2 − x1 y2 − y1 z2 − z1 a1 b1 c1 a2 b2 c2 S.D. = . (b1c2 − b2c1)2 + (c1a2 − c2a1)2 + (a1b2 − a2b1)2 11. The condition for two lines to intersect or coplanar The condition for two lines x − x1 = y − y1 = z − z1 and x − x2 = y − y2 = z − z2 to a1 b1 c1 a2 b2 c2 x2 − x1 y2 − y1 z2 − z1 intersect or coplanar is a1 b1 c1 = 0. b2 c2 a2

248 Mathematics—XII 12. The distance between two parallel lines The distance between two parallel lines →r = a1 + λb and → is given by r = a2 + μb d= (a→2 − a→1) × →b . →b Example 3. Find the shortest distance between the lines →r = (4iˆ − ˆj) + λ(iˆ + 2ˆj − 3kˆ) and →r = (iˆ − ˆj + 2kˆ) + μ(2iˆ + 4ˆj − 5kˆ) . ⎛ a→2 − a→1 ⎞ . ⎛ b→1× b→2 ⎞ ⎜⎝ ⎠⎟ ⎜⎝ ⎟⎠ Solution. We have, S.D. = Now, b→1× b→2 Example 4. and a→2 − a→1 = (iˆ − ˆj + 2kˆ) − (4iˆ − ˆj) = −3iˆ + 0 ˆj + 2kˆ Solution. b→1× b→2 = iˆ ˆj kˆ 1 2 −3 = 2iˆ − ˆj + 0kˆ . 2 4 −5 ∴ (a→2 − a→1) . (b→1× b→2 ) = – 6 + 0 + 0 = – 6 and b→1× b→2 = 4 + 1 = 5 Hence, S.D. = −6 = 6 . 5 5 Find the shortest distance between the lines →r = (iˆ + 2ˆj + 3kˆ) + λ(2iˆ + 3ˆj + 4kˆ) and →r = (2iˆ + 4ˆj + 5kˆ) + μ(4iˆ + 6ˆj + 8kˆ). [AI CBSE 2015] These equations can be written as →r = (iˆ + 2ˆj + 3kˆ) + λ(2iˆ + 3ˆj + 4kˆ) and → = (2iˆ + 4 ˆj + 5kˆ) + 2μ(2iˆ + 3 ˆj + 4kˆ) . These two lines are parallel. r ∴ S.D. = (a→2 − a→1) × →b . | →b | Here a→1 = iˆ + 2ˆj + 3kˆ , a→2 = 2iˆ + 4 ˆj + 5kˆ . and →b = 2i + 3j + 4k. iˆ ˆj kˆ Now, (a→2− a→1) × →b = 1 2 2 = 2iˆ − 0ˆj − kˆ . 234 ( )⇒ 5 and →b = 4 + 9 + 16 = 29 ∴ S.D = 5 →→ → = 29 . a2 − a1 × b 13. The general equation of a plane in the vector form is →→ = d where →n is the vector r⋅ n normal (perpendicular) to the plane. 14. The general equation of a plane in the Cartesian form is ax + by + cz + d = 0 where a, b, c are the d.r’s of the normal to the plane.

Three Dimensional Geometry 249 15. The vector equation of the plane passing through a point whose position vector is →a is (→r − →a ) . →n = 0 where →n is the normal (perpendicular) to the plane. 16. The equation of the plane passing through naoprmoianltt(oxt1h, ey1p, lza1n) eis. a(x – x1) + b(y – y1) + c(z – z1) = 0, where a, b, c are the d.r’s of the 17. The equation of the plane passing through three points (x1, y1, z1); (x2, y2, z2) and x − x1 y − y1 z − z1 (x3, y3, z3) is x2 − x1 y2 − y1 z2 − z1 = 0. x3 − x1 y3 − y1 z3 − z1 18. Intercept form of the equation of the plane The intercept form of the equation of the plane which cuts intercepts a, b, c on x-axis, y-axis and z-axis respectively is x + y + z = 1. a b c 19. The equation of the plane in the normal form The vector equation of the plane in the normal form is →r ⋅ nˆ = p where nˆ is the unit vector normal to the plane and p is the distance of the plane from origin. The Cartesian equation of the plane in the normal form is lx + my + nz + p where l, m, n are the direction cosines of the normal to the plane and p is the distance of the plane from origin. Example 5. Find the equation of the plane through the points A(2, 2, –1), B(3, 4, 2) and Solution. C(7, 0, 6). Example 6. We have the equation of the plane passing through the three points (x1, y1, z1), Solution. x − x1 y − y1 z − z1 (x2, y2, z2) and (x3, y3, z3) is x2 − x1 y2 − y1 z2 − z1 = 0. y3 − y1 z3 − z1 x3 − x1 x−2 y−2 z+1 Hence, the required equation of the plane is 1 2 3 = 0 . 5 −2 7 ⇒ 20(x – 2) + 8(y – 2) –12(z + 1) = 0. ⇒ 5x + 2y – 3z – 17 = 0 is the required equation of the plane. A plane meets the co-ordinate axes in A, B, C such that the centroid of the triangle is (p, q, r). Show that the equation of the plane is x + y + z = 3. p q r Let the equation of the plane be x + y + z = 1. a b c Then, the co-ordinates of A, B, C are A(a, 0, 0), B(0, b, 0) and C(0, 0, c). So, the centroid of ΔABC is ⎛ a , b , c ⎞ . ⎝⎜ 3 3 3 ⎠⎟ But the given centroid is (p, q, r) ⇒ a = 3p, b = 3q, c = 3r. x y z ∴ The required equation of the plane is 3p + 3q + 3r = 1 ⇒ x + y + z = 3 . p q r

250 Mathematics—XII 20. Equation of a plane parallel to a given plane Since the parallel planes have the same normal, the equations of the parallel planes differ by a constant. Vector form: The equation of any plane parallel to the plane →r . →n = d is →r . →n = k where k is to be determined by some other condition. Cartesian form: The equation of any plane parallel to the plane ax + by + cz + d = 0 is ax + by + cz + k = 0 where k is to be determined by some other condition. Example 7. Find the equation of the plane parallel to the plane 2x – y + 3z + 7 = 0 and passing through the point (1, 4, – 2). Solution. The equation of any plane parallel to the given plane is 2x – y + 3z + k = 0. It passes through (1, 4, –2) ⇒ 2 – 4 – 6 + k = 0 ⇒ k = 8. Hence, the required equation of the plane is 2x – y + 3z + 8 = 0. 21. The equation of the plane passing through the intersection of two planes Vector form: The equation of the plane passing through the intersection of the planes →r . n→1 = d1 and →r . n→2 = d2 is given by →r .(n→1 + λ n→2 ) = d1 + λd2 . Cartesian form: The equation of the plane passing through the intersection of the planes a1x + b1y +c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 is (a1x + b1y +c1z + d1) + λ(a2x + b2y + c2z + d2) = 0 where λ is a constant. Example 8. Find the equation of the containing the plane line of intersection of the planes x + y + z – 6 = 0 and 2x + 3y + 4z + 5 = 0 and passing through the point (1, 1, 1). Solution. The equation of the plane passing through the line of intersection of the given planes is (x + y + z – 6) + λ(2x + 3y + 4z + 5) = 0. ...(1) Given, the required plane passes through (1, 1, 1) is (1 + 1 + 1 – 6) + λ(2 + 3 + 4 + 5) = 0. ⇒ – 3 + 14λ = 0 ⇒ λ= 3 . 14 Substituting λ = 3 in (1), we get the equation of the plane as 14 (x + y + z – 6) + 3 (2x + 3y + 4z + 5) = 0. 14 ⇒ 20x + 23y + 26z – 69 = 0. 22. Distance of a point from a plane Vector form: The distance of the point having position vector →a from the plane →r . →n = d is given by →a . →n− d . →n Cartesian form: The distance of the point (x1, y1, z1) from the plane ax + by + cz + d = 0 is given by ax + by + cz + d . a2 + b2 + c2

Three Dimensional Geometry 251 Example 9. Find the distance of the point 2iˆ + ˆj − kˆ from the plane →r .(iˆ − 2ˆj + 4kˆ) = 9 . Solution. The distance of the point having position vector →a from the plane →r . →n = d is given by →→ . a .n− d → n Here, →a = 2iˆ + ˆj − kˆ , → = iˆ − 2ˆj + 4kˆ . n ∴ The distance of the given point from the plane (2iˆ + ˆj − kˆ).(iˆ − 2ˆj + 4kˆ) − 9 = 1 + 4 + 16 = | 2 − 2 −4 − 9| = 13 units. 21 21 Example 10. Find the distance of the point (2, 1, 0) from the plane 2x + y + 2z + 5 = 0. Solution. The required distance = |2 × 2 +1+ 2×0 + 5| = 10 . 4 +1+ 4 3 23. The distance between two parallel planes The distance between two parallel planes ax + by + cz + d1 = 0 and ax + by + cz + d2 = 0 is |d1 − d2 | . a2 + b2 + c2 Example 11. Find the distance between the parallel planes x + y – z + 4 = 0 and x + y – z + 5 = 0. Solution. The distance between the parallel planes = |4 − 5| = 1 . 1+1+1 3 24. Method of finding the intersection of a line and a plane. The intersection of a line and a plane is a point. The following are the steps. (i) Find the co-ordinates of any point of the line in terms of λ. (ii) Let this point be the point of intersection. (iii) This point is also a point of the plane. It must satisfy the equation of the plane. Hence, we will get the value of λ and the point of intersection of the line and the plane. Example 12. Find the distance between the point with position vector −iˆ − 5ˆj − 10kˆ and the point of the intersection of the line x−2 = y+1 = z−2 with the plane 3 4 12 x – y + z = 5. Solution. Let P(– 1, – 5, – 10) be the given point and Q be the point of intersection of the line with the plane.

252 Mathematics—XII Let x−2 = y + 1 = z−2 = λ . x–23 =y+14 =z–212 P(–1, –5, –10) 3 4 12 Then any point on the line is (3λ + 2, 4λ – 1, 12λ + 2). It lies on the plane Q (2, –1, 2) x–y+z–5=0 x – y + z – 5 = 0. ⇒ 3λ + 2 – 4λ + 1 + 12λ + 2 = 5. ⇒ 11λ = 0 ⇒ λ = 0. Hence Q(2, – 1, 2). ∴ The required distance = PQ = (2 + 1)2 + (−1 + 5)2 + (2 + 10)2 = 9 + 16 + 144 = 13 . 25. The condition of coplanarity of two lines and equation of the plane containing them. Two lines x − x1 = y − y1 = z − z1 and x − x2 = y − y2 = z − z2 are coplanar if a1 b1 c1 a2 b2 c2 x2 − x1 y2 − y1 z2 − z1 a1 b1 c1 = 0 and the equation of the plane containing them is a2 b2 c2 x − x1 y − y1 z − z1 a1 b1 c1 = 0 . a2 b2 c2 Example 13. Prove that the lines x+1 = y+ 3 = z +5 and x−2 = y−4 = z−6 are 3 5 7 1 4 7 coplanar. Also, find the plane containing these lines. Solution. We know that lines x − x1 = y − y1 = z − z1 and x − x2 = y − y2 = z − z2 a1 b1 c1 a2 b2 c2 x2 − x1 y2 − y1 z2 − z1 are coplanar if a1 b1 c1 = 0 . b2 c2 a2 Now, x2 − x1 y2 − y1 z2 − z1 3 7 11 a1 b1 c1 = 3 5 7 = 21 − 98 + 77 = 0 . a2 b2 c2 1 4 7 So, the given lines are coplanar. The equation of the plane containing these lines is given by x − x1 y − y1 z − z1 ⇒ x+1 y+3 z+5 a1 b1 c1 = 0 3 5 7 = 0. a2 b2 c2 147

Three Dimensional Geometry 253 ⇒ (x + 1)7 – (y + 3)14 + (z + 5)7 = 0. ⇒ (x + 1) – (y + 3)2 + (z + 5) = 0. ⇒ x – 2y + z = 0. 26. Image of a point in a plane P(x1, y1 , z1) Let P and Q be two points and π be a plane such that (i) PQ is perpendicular to the plane π. (ii) The midpoint of PQ lies in the plane. ax+by+cz+d = 0 Then either of the points is the image of the other. Steps for obtaining the image of a point in a plane. R (i) Write the equation for the perpendicular PQ. π Find any point of the line PQ in terms of λ. Let this point be foot of the perpendicular R. (ii) This point is also a point of the plane. Therefore, Q we get the value of λ and obtain the co-ordinates (x1 + aλ, y1 + bλ, z1+cλ) of R. (iii) R is the midpoint on PQ and hence using midpoint formula, we will get Q. (iv) This point is the image of P. Example 14. Find the image of the point (3, – 2, 1) in the plane 3x – y + 4z = 2. Solution. Let P(3, – 2, 1) be the given point and R(a, b, c) be its image. Let Q be the foot of the perpendicular from P to the plane, then Q is the midpoint of PR. Now PQ is the normal to the plane. The d.r’s of PQ are 3, − 1, 4 . Therefore, the equation of PQ is x − 3 = y+2 = z − 1 = λ. 3 −1 4 Let the co-ordinates of Q be (3λ + 3, – λ – 2, 4λ + 1). This is a point of the plane 3x – y + 4z = 2. ⇒ 3(3λ + 3) – (– λ – 2) + 4(4λ + 1) – 2 = 0 ⇒ 26λ + 13 = 0. ⇒ λ = − 1 . 2 Therefore, the point Q is ⎛ 3 , − 3 , −1 ⎠⎟⎞ . P(3,–2, 1) ⎝⎜ 2 2 ⇒ ⎛3+ a , −2 + b , 1 + c⎞ = ⎛ 3 , − 3 , −1 ⎞ . ⎝⎜ 2 2 2 ⎟⎠ ⎝⎜ 2 2 ⎟⎠ Q ⇒ a = 0, b = –1, c = –3. ⇒ R (0, –1, –3) is the image of P (3, –2, 1). R (a, b, c)

254 Mathematics—XII A. STRAIGHT LINE IN SPACE Example 1. Can a line have direction angles 45°, 60°, 135°? Solution. Example 2. Let l = cos 45° = 1, m = cos 60° = 1 and n = cos 135° = – 1 . Solution. 2 2 2 Example 3. ∴ l2 + m2 + n2 = ⎛ 1 ⎞2 + ⎛ 1 ⎞2 + ⎛ − 1 ⎞2 = 1 + 1 + 1 = 5 ≠1 Solution. ⎝⎜ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠ ⎝⎜ 2 ⎠⎟ 2 4 2 4 ∴ A line cannot have direction angles 45°, 60°, 135°. Find the direction cosines of a line which is equally inclined with the axes. How many such lines are there? (NCERT) Let the required line make angles α, β, γ with the positive direction of the axes. Since the line is equally included to the axes, we have β = α or π – α and γ = α or π – α. ∴ cos β = cos α or – cos α and cos γ = cos α or – cos α ∴ The d.c.’s of the line are cos α, ±cos α, ±cos α. Also, (cos α)2 + (±cos α)2 + (±cos α)2 = 1 (Using l2 + m2 + n2 = 1) ⇒ 3 cos2 α = 1 ⇒ cos α = ± 1 3 ∴ The d.c.’s of the line are ± 1 , ± 1 , ± 1 . 3 3 3 ∴ Possible d.c.’s are: 1 , 1 , 1 ; – 1 , 1 , 1 ; 1 , – 1 , 1 , and 1 , 1 , – 1 3 3 3 3 3 3 3 3 3 3 3 3 because d.c.’s 1 , 1 , 1 and – 1 , – 1 , – 1 ; 333 333 – 1, 1, 1 and 1 ,– 1 ,– 1 ; 333 333 1 , – 1, 1 and – 1 , 1 , – 1 ; 3 3 3 3 3 3 1 , 1 , – 1 and – 1 , – 1 , 1 3 3 3 3 33 represent the same lines. ∴ There are exactly four lines passing through a point and equally inclined to the axes. A line passes through the points (6, –7, –1) and (2, –3, 1). Find direction ratios and the direction cosines of the line so directed that its angle with the x-axis is acute. The line passes through (6, –7, –1) and (2, –3, 1). d.r.’s of this line are 2 – 6, –3 + 7, 1 + 1, i.e., –4, 4, 2 or –2, 2, 1. Let d.c.’s of this line be –2λ, 2λ, λ. ∴ (–2λ)2 + (2λ)2 + λ2 = 1

Three Dimensional Geometry 255 ⇒ 4λ2 + 4λ2 + λ2 = 1 ⇒ 9λ2 = 1 ⇒ λ2 = 1 ⇒ λ = ± 1 9 3 Since α is acute, cos α is +ve. ⇒ –2λ is +ve ⇒ λ is –ve ∴ λ = – 1 3 ∴ The d.c.’s of the line are – 2 ⎛ − 1 ⎞ , 2 ⎛ − 1 ⎞ , – 1 , i.e., 2 ,– 2 ,– 1 . ⎜⎝ 3 ⎟⎠ ⎜⎝ 3 ⎟⎠ 3 3 3 3 Example 4. If A(1, 2, –2) and B(3, – 4, 5) are two points in space, find the d.c.’s of the lines Solution. OA, OB and AB, O being the origin. Given points are A(1, 2, –2), B(3, – 4, 5) and O(0, 0, 0). Direction cosines of OA d.r.’s of OA are 1 – 0, 2 – 0, –2 – 0, i.e., 1, 2, – 2. OA = (1)2 + (2)2 + (−2)2 = 1 + 4 + 4 = 3 A(1, 2, –2) ∴ The d.c.’s of OA are 1/3, 2/3, –2/3. Direction cosines of OB d.r.’s of OB are 3 – 0, –4 – 0, 5 – 0, i.e., 3, –4, 5. OB = (3)2 + (− 4)2 + (5)2 = 9 + 16 + 25 = 50 = 5 2 O B (0, 0, 0) (3, –4, 5) ∴ The d.c.’s of OB are 3/ 5 2 , –4/ 5 2 , 5/ 5 2 , i.e., 3/ 5 2 , –4/ 5 2 , 1 / 2 . Direction cosines of AB d.r.’s of AB are 3 – 1, –4 – 2, 5 – (–2), i.e., 2, –6, 7. AB = (2)2 + (− 6)2 + (7)2 = 4 + 36 + 49 = 89 ∴ The d.c.’s of AB are 2 / 89 , – 6 / 89 , 7 / 89 . Example 5. (i) Show that the point (2, 3, 7) lies on the line: x − 2 = y − 3 = z−7 . Solution. 4 2 3 (ii) Show that the point (1, 1, 3) lies on the line: →r = iˆ + ˆj + 3 kˆ + λ(2 iˆ – ˆj – kˆ ). (i) Given line is x −2 = y− 3 = z −7 . ...(1) 4 2 3 (2, 3, 7) lies on the line (1) if 2−2 = 3 −3 = 7−7 or if 0 = 0 = 0, which 4 2 3 is true. ∴ (2, 3, 7) lies on (1). (ii) Given line is →r = iˆ + ˆj + 3kˆ + λ(2iˆ − ˆj − kˆ) . ...(1) (1, 1, 3) lies on (1) if its positive vector iˆ + ˆj + 3kˆ satisfies (1) or if iˆ + ˆj + 3kˆ = iˆ + ˆj + 3kˆ + λ(2iˆ − ˆj − kˆ)

256 Mathematics—XII or if iˆ + ˆj + 3kˆ = (1 + 2λ)iˆ + (1 − λ)ˆj + (3 − λ)kˆ or if 1 = 1 + 2λ, 1 = 1 – λ, 3 = 3 – λ or if λ = 0, λ = 0, λ = 0, which is true for λ = 0. ∴ (1, 1, 3) lies on (1). Example 6. If the equations of a line AB are 3 – x = y+2 = z + 2 , find the direction cosines Solution. 3 –2 6 of a line parallel to AB. (CBSE 2012 C) Given line is 3 – x = y+2 = z + 2 i.e., x–3 = y+2 = z + 2 3 –2 6 –3 –2 6 ∴ –3, –2, 6 are d.r.’s of AB. (–3)2 + (–2)2 + (6)2 = 9 + 4 + 36 = 7 Example 7. ∴ The d.c.’s of AB are –3/7, –2/7, 6/7. Solution. The d.c’s of any line parallel to AB are – 3/7, – 2/7, 6/7. The cartesian equations of a line are 3x + 1 = 6y – 2 = 1 – z. Find a point on the line and its d.r.’s and write it in vector form. We have 3x + 1 = 6y – 2 = 1 – z. ⇒ 3 ⎛ x + 1⎞ =6 ⎛ y – 1⎞ = – (z – 1) (Note this step) ⎜⎝ 3 ⎟⎠ ⎝⎜ 3 ⎟⎠ x − ⎛ − 1 ⎞ y − 1 z−1 x − ⎛ − 1⎞ y − 1 z−1 ⎝⎜ 3 ⎟⎠ 3 −1 ⎜⎝ 3 ⎟⎠ 3 −6 ⇒ = = ⇒ = = 1/3 1/6 2 1 (Dividing by 6) ∴ Given line passes through the point (–1/3, 1/3, 1) and has d.r.’s 2, 1, –6. Let → = P.V. of ⎛ − 1 , 1 , 1⎞⎠⎟ = − 1 iˆ + 1 ˆj + kˆ and → = 2iˆ + ˆj − 6kˆ . ⎝⎜ 3 3 3 3 a b ∴ The vector equation of the given line is → = → + → . r a λb ⇒ → = − 1 iˆ + 1 jˆ + kˆ + λ(2iˆ + jˆ − 6kˆ ) . 3 3 r Example 8. A line passes through the point with position vector 2iˆ − ˆj + 4kˆ and is in the Solution. direction of iˆ + 2ˆj − kˆ . Find the equations of the line in the vector and in cartesian forms. (NCERT) Let → = 2iˆ − ˆj + 4kˆ and → = iˆ + 2 ˆj – kˆ . a b Let →r be the position vector of a general point (x, y, z) on the line. ∴ The vector equation of the line is →r = →a + λ→b .

Three Dimensional Geometry 257 ⇒ →r = 2iˆ − jˆ + 4kˆ + λ(iˆ + 2 jˆ − kˆ) . ...(1) This is the vector equation of the given straight line. (1) ⇒ xiˆ + yˆj + zkˆ = (2 + λ)iˆ + (–1 + 2λ)ˆj + (4 − λ)kˆ ⇒ x = 2 + λ, y = –1 + 2λ, z = 4 – λ ⇒ x − 2 = λ, y + 1 = λ, z−4 =λ ⇒ x − 2 = y + 1 = z−4 . 1 2 −1 1 2 −1 These are the cartesian equations of the given line. Example 9. The cartesian equations of a line are x − 5 = y + 4 = z − 6 . Find the vector Solution. 3 7 2 equation of the line. (NCERT) We have x − 5 = y + 4 = z − 6 = λ (say) 3 7 2 ∴ x = 3λ + 5, y = 7λ – 4, z = 2λ + 6 Let → be the position vector of a general point (x, y, z) on the line. r ∴ →r = xiˆ + yˆj + zkˆ = (3λ + 5)iˆ + (7λ − 4)ˆj + (2λ + 6)kˆ ⇒ →r = 5iˆ − 4 jˆ + 6kˆ + λ(3iˆ + 7 jˆ + 2kˆ) . This is the vector equation of the given line. Example 10. The cartesian equations of a line are 6x – 2 = 3y + 1 = 2z – 2. Find the direction cosines of the line. Write down the cartesian and vector equations of a line passing through (2, –1, –1) which is parallel to the given line. (CBSE 2013 C) Solution. We have 6x – 2 = 3y + 1 = 2z – 2. ...(1) ⇒ 6 ⎛ x – 1⎞ = 3 ⎛ y + 1⎞ = 2(z – 1) ⎝⎜ 3 ⎠⎟ ⎝⎜ 3 ⎠⎟ x – 1 y + 1 z–1 x – 1 y + 1 z – 1 3 3 1/2 3 2 3 3 ⇒ 1/6 = 1/3 = ⇒ = = 1 ∴ d.r.’s of the given line (1) are 1, 2, 3. Now, 12 + 22 + 32 = 14 ∴ The d.c.’s of (1) are 1 , 2 , 3 . 14 14 14 The required line passes through (2, –1, –1) and is parallel to (1) and hence 1, 2, 3 are also its d.r.’s. Cartesian equations of the required line are x– 2 = y – (–1) = z – (–1) i.e., x–2 = y+1 = z+1 . 1 2 3 1 2 3 Let →a = P.V. of (2, –1, –1) = 2iˆ – ˆj – kˆ and → = iˆ + 2 ˆj + 3kˆ . b ∴ The vector equation of the required line is →→ → i.e., → = 2iˆ − jˆ – kˆ + λ(iˆ + 2 jˆ + 3kˆ ) . r = a + λb r

258 Mathematics—XII Example 11. Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the yz-plane. Solution. Given points are (5, 1, 6) and (3, 4, 1). ∴ d.r.’s of the line are 3 – 5, 4 – 1, 1 –6 or –2, 3, –5 or 2, –3, 5. The point (5, 1, 6) is on the line. ∴ Using x − x1 = y − y1 = z − z1 , the equations of the line are b1 b2 b3 x − 5 = y−1 = z − 6 . ...(1) 2 −3 5 Let this line crosses the yz-plane at the point (0, a, b). ∴ (1) ⇒ 0 − 5 = a−1 = b − 6 ...(2) 2 −3 5 (2) ⇒ a−1 = − 5 ⇒ 2a – 2 = 15 ⇒ a = 17/2 −3 2 (2) ⇒ b − 6 = − 5 ⇒ 2b – 12 = –25 ⇒ b = –13/2 5 2 ∴ The required point is (0, 17/2, –13/2). Example 12. The points A(4, 5, 10), B(2, 3, 4) and C(1, 2, –1) are three vertices of a parallelogram ABCD. Find vectorial and cartesian equations of the side AB and the coordinates of D. Solution. Let the coordinates of D be (x′, y′, z′). The position vectors of points A, B, C, D are → = 4iˆ + 5 ˆj + 10kˆ = →a (say) D (x¢, y¢, z¢) C (1, 2, –1) OA → = 2iˆ + 3 ˆj + 4kˆ = →b (say) OB → = iˆ + 2 ˆj − kˆ = →c (say) OC → = x′iˆ + y′ ˆj + z′ kˆ = →d (say) A (4, 5, 10) B (2, 3, 4) OD Equation of AB. The vector equation of line AB is →r = →a + λ(→b − →a ) , where →r is the position vector of a general point (x, y, z) on AB. ∴ →r = 4iˆ + 5 ˆj + 10kˆ + λ [(2iˆ + 3 ˆj + 4kˆ) − (4iˆ + 5 ˆj + 10kˆ)] ⇒ → = 4 iˆ + 5 jˆ + 10kˆ + λ(− 2iˆ − 2 jˆ − 6kˆ ) . r The equation of AB implies xi + yj + zk = (4 − 2λ)i + (5 − 2λ) j + (10 − 6λ)k. ⇒ x = 4 – 2λ, y = 5 – 2λ, z = 10 – 6λ ⇒ x−4 = λ, y−5 = λ, z − 10 = λ ⇒ x− 4 = y − 5 = z − 10 . −2 −2 −6 1 1 3