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### Vector & 3D complex & Misc(12th).pdf

1. MATHEMATICS
2. Question bank on Vector, 3D and Complex Number & Misc. Subjective Problem Select the correct alternative : (Only one is correct) Q.12/vec If  a = 11 1 ,  b = 23 ,   a b  = 30 , then   a b  is : (A) 10 (B*) 20 (C) 30 (D) 40 Q.25/vec ThepositionvectorofapointPmovinginspaceisgivenby k̂ ) t sin 5 ( ĵ ) t cos 4 ( î ) t cos 3 ( R OP      . The time 't' when the point P crosses the plane 4x – 3y + 2z = 5 is (A) 2  sec (B*) 6  sec (C) 3  sec (D) 4  sec [Hint: put x = 3 cos t ; y = 4 cos t ; z = 5 sin t in the equation of the plane, we get 12 cos t – 12 cos t + 10 sin t = 5 sin t = 2 1  t = 6  sec ] Q.36/vec Indicatethe correct order sequence in respect of the following : I. The lines 3 4 x   = 1 6 y   = 1 6 y   and 1 1 x   = 2 2 y   = 2 3 z  are orthogonal. II. The planes 3x – 2y – 4z = 3 and the plane x – y – z = 3 are orthogonal. III. The function f (x) = ln(e–2 + ex) is monotonic increasing  x R. IV. If g is the inverse of the function, f (x) = ln(e–2 + ex) then g(x) = ln(ex – e–2). (A) FTFF (B)TFTT (C*) FFTT (D)FTTT [Sol. I. L1 is | | to 1 V k̂ ĵ î 3      L2 is | | to 2 V k̂ 2 ĵ 2 î      2 1 V · V   = 3 + 2 – 2 = 3  Lis not perpendicular to L2  False II. 3·1 – (2) (–1) – (4)(–1) = 3 + 2 + 4  0  planes are not perpendicular  False III. f (x) = ln(e–2 + ex) f ' (x) = x 2 x e e e · 1   > 0  f is increasing  x  R  True IV. y = ln(e–2 + ex) e–2 + ex = ey ex = ey – e–2  f–1(y) = ln(ey – e–2) g (x) = ln(ex – e–2)  True] Q.42/complex If 2 1 z z is purelyimaginarythen 2 1 2 1 z z z z   is equal to : (A*) 1 (B) 2 (C) 3 (D) 0 [Hint: E = 1 ) z z ( ) z z ( 1 2 1 2 1   = 1 x x 1   i i = 1 ]
3. Q.57/vec Ina regular tetrahedron, thecentres of the fourfaces are the verticesof asmaller tetrahedron.The ratio of the volume ofthe smaller tetrahedron to that of the larger is n m ,where m and n arerelativelyprime positive integers. The value of (m + n)is (A) 3 (B) 4 (C) 27 (D*) 28 [Hint: Vl = ] c b a [ 6 1    ; Vs = ] c b a [ 27 1 · 6 1    Hence 27 1 V Vs  l = n m or 27 n = 1 m = k  m and n arerelativelyprime  k = 1, (m + n) = 28 further hintfor Vs =       3 c · 3 b · 3 a 6 1    = ] c b a [ 27 1 · 6 1    ] Q.69/vec If    a b c , , arenon-coplanarunitvectorssuchthat      a x bxc b c ( ) ( )   1 2 thentheanglebetween b & a   is (A*) 3 /4 (B) /4 (C) /2 (D)  Q.712/vec Thesine ofangleformedbythelateral faceADC and planeofthebaseABC ofthe tetrahedronABCD where A  (3, –2, 1); B  (3, 1, 5); C  (4, 0, 3) and D  (1, 0, 0) is (A) 2 29 (B*) 5 29 (C) 3 3 29 (D) 2 29 Q.86/complex Let z be a complex number, then the region represented by the inequalityz + 2< z + 4 is given by : (A*) Re (z) >  3 (B) Im(z) < 3 (C) Re (z) <  3 & Im (z) >  3 (D) Re (z) <  4 & Im (z) >  4 Q.914/vec The volume of the parallelpiped whose edges are represented by the vectors  a i j k    2 3 4   ,  b i j k    3 2    ,  c i j k       2 is : (A*) 7 (B) 5 (C) 4 (D) none Q.1015/vec Let w , v , u    be thevectors such that 0 w v u       , if 5 w & 4 v , 3 u       then the value of u . w w . v v . u         is : (A) 47 (B*) 25 (C) 0 (D) 25 Q.1116/vec Let  a = j ˆ î  ,  b= k̂ j ˆ  ,  c= î k̂  . If  d is a unit vector such that      a d b c d . [ , , ]   0 then  d (A*) ) k̂ 2 j ˆ î ( 6 1    (B) ) k̂ j ˆ î ( 3 1    (C) ) k̂ j ˆ î ( 3 1    (D) ± k̂ Q.128/complex If z be a complex number for which z z   1 2 , then the greatest value of z is :
4. (A*) 2 1  (B) 3 1  (C) 2 2 1  (D) none [Hint: z z z z z z      1 1 1 | | r r r r     1 2 1 Nowconsiderallinequalities ] Q.1322/vec If the nonzero vectors   a b & are perpendicular to each other, then the solution of the equation,    r a b   is : (A*)         r xa a a a b    1 . (B)         r xb b b a b    1 . (C)    r xa b   (D)    r xb a   where x is anyscalar. [Hint: b a 2 b y a x r          take cross with a  a ) b a ( z a b y a r              ] a ) a · b ( b ) a · a [( z ) a b ( y b              ] b ) a · a ( z ) a b ( y b          since b a b     arenon coplanar  0 y & 1 ) a · a ( z     z = 2 a 1   ) b a ( a 1 a x r 2         ] Q.1423/vec The lines k 4 z 1 3 y 1 2 x       and 1 5 z 2 4 y k 1 x      are coplanar if (A) k = 1 or – 1 (B*) k = 0 or – 3 (C) k = 3 or – 3 (D) k = 0 or – 1 Q.1524/vec Which one of the followingstatement is INCORRECT ? (A) If   n a . = 0 ,   n b . = 0 &   n c . = 0 for some non zero vector  n , then     a b c = 0 (B*) there exist a vector having direction angles  = 30º &  = 45º (C) locus of point for which x = 3 & y = 4 is a line parallel to the z - axis whose distance from the z-axis is 5 (D)InaregulartetrahedronOABCwhere'O'istheorigin,thevector OA OB OC      isperpendicular to the planeABC. [Explanation: (A)   n is perpendicular to   a b , as well as  c     a b c , , must be in the same plane       a b c = 0
5. (B) If one direction angle is  then the remaining twocannot be less than 90  (D) verify that     a a b c          3 .      a b c   = 0 where    a b c   ] Q.1612/complex Given that the equation, z2 +(p+ iq) z + r+ is = 0 has a real root where p, q, r, s  R. Then which oneis correct (A) pqr = r2 + p2s (B) prs = q2 + r2p (C) qrs = p2 + s2q (D*) pqs = s2 + q2r [Hint: Let z =  be the real root 2 + (p + iq) + r + is = 0 (2 + p + r) + i(q + s) = 0 + 0 i  q + s = 0 ..........(1) and 2 + p + r = 0 ..........(2) From (1)  = – s q . Put in (2)to get the result ] Q.1727/vec The distance of the point (3, 4, 5) from x-axis is (A) 3 (B) 5 (C) 34 (D*) 41 [Hint: distance from x-axis of x, y, z = 2 1 2 1 z y  ] Q.1829/vec Given nonzero vectors C and B , A    , thenwhich one of thefollowing is false? (A)Avector orthogonal to C and B A     is ± C ) B A (      (B)Avector orthogonal to B A    and B A    is ± B A    (C)Volume of the parallelopiped determined by C and B , A    is | C · B A |     (D*) Vector projection of B onto A   is | B | B · A    [Hint: It should be B B ) B · A ( 2     ] Q.1930/vec Given three vectors    a b c , , such that theyare nonzero, noncoplanar vectors, then which of the followingarenoncoplanar. (A*)       a b b c c a    , , (B)       a b b c c a    , , (C)       a b b c c a    , , (D)       a b b c c a    , , [Hint: Verify 3 2 1 v v v      in ordertoquicklyanswer ] Q.2016/complex The sum i + 2i2 + 3i3 + ........ + 2002i2002, where i = 1  is equal to (A) – 999 + 1002i (B) – 1002 + 999i (C) – 1001 + 1000i (D*) – 1002 + 1001i [Sol. S = i + 2i2 + 3i3 + ......... + 200i2002 iS = i2 + 2i3 + .........+ 2001i2002 + 2002i2003 – – – – – ——————————————————— S(1 – i) = i + i2 + i3 + ......... + i2002 – 2002i2003 = i i i   1 ) 1 ( 2002 + 2002 i = i i  1 2 + 2002 i = i(1 + i) + 2002i
6. S = i i    1 2003 1 = 2 ) 1 ( ) 2003 1 ( i i    = – 1 – i + 2003i – 2003 = 2 2002 2004 i   = – 1002 + 1001i ] Q.2131/vec Locus of the point P, for which OP  represents a vector with direction cosine cos  = 1 2 ( 'O' is the origin) is : (A)Acircle parallel to yz plane withcentre on the xaxis (B*)aconeconcentricwithpositive xaxis havingvertex at theoriginand theslant heightequal to the magnitudeofthevector (C)a rayemanatingfrom the origin and making an angleof60ºwith xaxis (D) a disc parallel to yz plane withcentre on xaxis & radius equal to| | OP  sin 60º [Hint: ] Q.2238/vec A line with direction ratios (2, 1, 2) intersects the lines ) k̂ j ˆ î ( j ˆ r        and ) k̂ ĵ î 2 ( î r        atAand B, then l (AB) is equal to (A*) 3 (B) 3 (C) 2 2 (D) 2 [Hint: L1 : 1 1 x  = 1 1 y  = 1 0 z  =  ; L2 : 2 1 x  = 1 0 y  = 1 0 z  =  Hence any point on L1 and L2 can be (, – 1, ) and (2 – 1, , )  2 1 2     = 1 1     = 1    solving  = 1 and  = 3 A = (3, 2, 3) and B = (1, 1, 1) ] Q.2347/vec The vertices of a triangle areA(1, 1, 2), B(4, 3, 1) & C(2, 3, 5).The vector representing the internal bisector of theangleAis : (A)    i j k   2 (B) 2 2    i j k   (C) 2 2    i j k   (D*) 2 2    i j k   Q.2428/complex Lowest degree of a polynomial with rational coefficients if one of its root is, i  2 is (A) 2 (B*) 4 (C) 6 (D) 8 [Sol. Let x = i  2   2 2 x  = – 1  x2 + 2 – x 2 2 = – 1  x2 + 3 = x 2 2  x4 + 9 + 6x2 = 8x2  x4 – 2x2 + 9 = 0 ] Q.2555/vec Aplane vector has components 3 & 4 w.r.t. therectangularcartesian system.This system is rotated
7. through an angle  4 in anticlockwisesense.Then w.r.t. thenew system the vectorhascomponents : (A) 4, 3 (B*) 7 2 1 2 , (C) 1 2 7 2 , (D) none [Hint: cos  = 3 5 sin  = 4 5 now w.r.t. new system XY Xcomponent is5 cos ( 45º) Ycomponentis5 sin ( 45º) ] Q.2656/vec Let  a = xi + 12j  k ;  b = 2i + 2xj + k &  c = i + k . If the ordered set      b c a is left handed , then: (A) x  (2, ) (B) x  (,  3) (C*) x  ( 3, 2) (D) x  { 3, 2} [Sol. For a right hand set ] c b a [    > 0 and for a left handed system ] c b a [    < 0 ] Q.2773/vec If k̂ j ˆ î cos    , k̂ j ˆ cos î    & k̂ cos j ˆ î    (      2n) are coplanar then the value of            2 ec cos 2 ec cos 2 ec cos 2 2 2 equal to (A) 1 (B*)  (C) 3 (D) none of these [Hint: cos cos cos    1 1 1 1 1 1 = 0  cos cos cos cos cos          1 1 0 0 1 1 1 1 = 0  (2)   2 2 2 2 0 0 2 2 2 2 1 1 2 2 2 2 sin sin sin sin cos      = 0 + 2 sin2  2         2 2 2 2 2 2 sin cos sin    + 2 sin2  2 2 sin2  2 sin2  2 sin sin sin 2 2 2 2 1 2 2 2                  + sin2  2 sin2  2 = 0 multiplyby cosec2  2 cosec2  2 cosec2  2 cosec2  2  2 + cosec2  2 + cosec2  2 = 0 Alternatively: Expandnumber2 (cos  1) [ cos  (cos  1)  (1  cos ) ] + (1  cos ) (1  cos ) = 0 or (1  cos ) (1  cos ) cos  + (1  cos ) (1  cos ) + (1  cos ) (1  cos ) = 0 Now proceed ] Q.2834/complex Thestraight line (1+ 2i)z + (2i– 1) z =10i on thecomplex plane, has intercepton the imaginary
8. axis equal to (A*) 5 (B) 2 5 (C) – 2 5 (D) – 5 [Hint: put z = iy (1 + 2i) iy – (2i – 1) i y = 10 i y + 0 y = 10  y = 5 ] Q.2975/vec Theperpendiculardistanceofanangularpoint of acubeofedge 'a'from thediagonal whichdoes not pass thatangularpoint, is (A) a 3 (B) a 2 (C) a 2 3 (D*) a 3 2 [Sol. Consider aunit cube equation ofAB is ) k̂ j ˆ î ( k̂ r       p.v. of N  , , (–1 – ) k̂ ) 1 ( ĵ î ON         now   AB . ON = 0  +  + l +  = 0   = – 1/3 Hence k̂ 3 2 ĵ 3 1 î 3 1 ON      ; 3 2 9 4 9 1 9 1 | ON |      ] Q.3088/vec Which one of the following does not hold for the vector V  =      a x b x a ? (A) perpendicular to  a (B*) perpendicular to  b (C) coplanar with  a &  b (D) perpendicular to  a x  b . Q.3153/complex Let z1, z2 & z3 be thecomplex numbers representing thevertices ofa triangleABC respectively. If P is a point representing the complex number z0 satisfying ; a(z1 z0) + b(z2 z0) + c(z3 z0) = 0, then w.r.t. the triangleABC, the point Pis its : (A) centroid (B) orthocentre (C) circumcentre (D*) incentre [Hint: az1 + bz2 + cz3 = z0(a + b + c)  z0 = az bz cz a b c 1 2 3      z0 is the incentre] Q.3296/vec Given the positionvectors of the vertices ofatriangleABC, ) a ( A  ; ) b ( B ; ) c ( C .Avector r  is parallel to the altitude drawn from the vertexA, making an obtuse angle with the positiveY-axis. If | r |  = 34 2 ; k̂ 3 j ˆ î 2 a     ; k̂ 4 j ˆ 2 î b     ; k̂ 2 j ˆ î 3 c     , then r  is (A*) k̂ 6 j ˆ 8 î 6    (B) k̂ 6 j ˆ 8 î 6   (C) k̂ 6 j ˆ 8 î 6    (D) k̂ 6 j ˆ 8 î 6   [Sol. | r |  = 2 34
9. Equation ofline BC: k̂ 4 j ˆ 2 î   + ) ( BC k 2 j 3 i 2         p.v. of N is 2 +1 , 2 – 3 , 2 – 4 vector  AN = (2 – 1) î + (3– 3) j ˆ + (2–1)k̂ now  AN .  BC = 0 2(2 – 1)–3(3– 3) +2 (2–1) (4 + 9 + 4) = 2 + 9 + 2 = 13 = 13/17  AN = 17 k̂ 9 j ˆ 12 î 9   ; 17 34 3 17 306 AN    ) AN ( P r     | AN | . | P | | r |    hence 17 34 3 | P | 34 2  | P | = 3 34  P = 3 34 or 3 34  r  = + 3 34           17 k̂ 9 ĵ 12 î 9 = + 2 ( k̂ 3 j ˆ 4 î 3   )  angle with yaxis is –ve +ve sign to be rejected k̂ 6 j ˆ 8 î 6 r       (A) ] Q.3372/complex The complex number, z = 5 + i has an argument which is nearlyequal to : (A) /32 (B*) /16 (C) /12 (D) /8 [Hint: z = 5 + i 5 + i = 26 (cos + i sin) +24 + 10i = 26(cos2 + i sin2) +476 + 480i = 676(cos4 + i sin4)  676 4 476 676 4 480 sin cos        and  tan 4= 476 480 ~ 1 4 =  4 =  16 ] Q.3497/vec If k̂ j ˆ î a     , k̂ j ˆ î b     , k̂ j ˆ 2 î c     , then the value of c . c b . c a . c c . b b . b a . b c . a b . a a . a                   equal to (A) 2 (B) 4 (C*) 16 (D) 64 [Hint:  2 c b a    = 2 1 2 1 1 1 1 1 1 1   = 16 ] Q.3596/complex If the equation x2 + ax + b = 0 where a, b  R has a non real root whose cube is 343 then
10. (7a + b) has the value (A*) 98 (B) – 49 (C) – 98 (D) 49 [Sol. the cube root of 343 are the roots of x3 – 343 = 0 (x – 7)(x2 + 7x + 49) = 0 where a = 7 and b = 49  7a + b = 98 Ans.] Direction for Q.36 to Q.40. Let k̂ 3 j ˆ 2 î A     and k̂ 5 j ˆ 4 î 3 B     Q.363(i)/vec The value of the scalar 2 2 ) B · A ( | B A |       is equal to (A) 8 (B) 10 7 (C*) 7 10 (D) 64 [Sol. 2 2 | b | | a |   = 50 · 14 = 700 = 7 10 Ans] Q.373(ii)/vec Equationofalinepassingthroughthepointwithpositionvector j ˆ 3 î 2  andorthogonal totheplane containingthevectors A  and B  ,is (A*) k̂ j ˆ ) 3 2 ( î ) 2 ( r          (B) k̂ j ˆ ) 3 2 ( î ) 2 ( r          (C) k̂ j ˆ ) 3 2 ( î r         (D) none [Sol. A  × B  = 5 4 3 3 2 1 k̂ ĵ î = k̂ ) 6 4 ( j ˆ ) 9 5 ( î ) 12 10 (      = k̂ 2 j ˆ 4 î 2    = ) k̂ j ˆ 2 î ( 2    Here ) k̂ j ˆ 2 î ( j ˆ 3 î 2 r        = k̂ j ˆ ) 2 3 ( î ) 2 ( r          Ans.] Q.383(iii)/vec Equation of aplane containingthe point with position vector ) k̂ j ˆ î (   and parallel tothe vectors A  and B  , is (A) x + 2y + z = 0 (B) x – 2y – z – 2 = 0 (C*) x – 2y + z – 4 = 0 (D) 2x + y + z – 1 = 0 [Sol. k̂ j ˆ 2 î n     k̂ j ˆ î a     0 n · ) a r (      ) k̂ ĵ 2 î ( · r    = n · a   = 4 x – 2y + z = 4 ] Q.393(iv)/vec Volume ofthe tetrahedron whose 3coterminous edges are thevectors A  , B  and k̂ 4 j ˆ î 2 C     is (A) 1 (B*) 4/3 (C) 8/3 (D) 8 [Sol. ] c b a [ 6 1    = 4 1 2 5 4 3 3 2 1 6 1  = 6 1 [1(–16 – 5) – 2(–12 – 10) + 3(3 – 8)] = 6 1 [–21 + 44 – 15] = 6 8 = 3 4 ]
11. Q.403(v)/vec Vector component of A  perpendicular to the vector B  is given by (A*) 2 B ) B A ( B       (B) 2 B ) B A ( A       (C) 2 A ) B A ( B       (D) 2 A ) B A ( A       [Sol. B B B · A A x 2                  (A) ] Select the correct alternatives : (More than one are correct) Q.41501/vec If a, b, c are different real numbers anda i b j ck      ; bi c j a k      & ci a j bk      are position vectors of threenon-collinear pointsA, B &C then: (A*) centroid oftriangleABC is   a b c i j k     3    (B*)    i j k   is equallyinclinedto the three vectors (C*) perpendicular from the origin totheplane of triangleABC meet at centroid (D*)triangleABCisanequilateral triangle. Q.42504/vec The vectors    a b c , , are of the same length & pairwise form equal angles. If  a i j     &  b j k    , the pv's of  c can be : (A*) (1, 0, 1) (B)         4 3 1 3 4 3 , , (C) 1 3 4 3 1 3 , ,        (D*)         1 3 4 3 1 3 , , [Hint: Let  c x i y j zk       x2 + y2 + z2 = 2  (1) now       a b b c c a . . .    I = y + z = x + y  (2)  z = x y = 1  x put z and y in terms of x in (1) to get x and then get yand z ] Q.43512/complex Which of the following locii of z on the complex plane represents a pair of straight lines? (A*) Re z2 = 0 (B*) Im z2 = 0 (C) z + z = 0 (D) z1= zi [Hint: C  negative real axis ; D  perpendicular bisector of the line joining (0, 1) & (1, 0) ] Q.44506/vec If    a b c , , &  d are linearlyindependent set of vectors & K a K b K c K d 1 2 3 4        = 0 then : (A*) K1 + K2 + K3 + K4 = 0 (B*) K1 + K3 = K2 + K4 = 0 (C*) K1 + K4 = K2 + K3 = 0 (D) none of these [Hint: k a k b k c k d 1 2 3 4        = 0     a b c d , , , arelinearlyindependent  k1 = k2 = k3 = k4 = 0 ] Q.45507/vec If    a b c , , &  d are the pv's of the pointsA, B, C & D respectively in three dimensional space & satisfythe relation3 2 2     a b c d    =0, then : (A*)A, B, C & D are coplanar (B) the line joiningthe points B& D divides the line joining the pointA& C in the ratio 2 : 1. (C*) theline joiningthe pointsA& C divides the line joiningthepoints B &Din the ratio 1 : 1 (D*) the four vectors    a b c , , &  d arelinearlydependent.
12. [Hint: 3 4   a c  = 2 2 4   b d  =   b d  2 HencelinejoiningA& C intersect linejoiningB& C ] Q.46519/complex If z3  i z2  2 i z  2 = 0 then z can be equal to : (A) 1  i (B*) i (C*) 1 + i (D*)  1  i [Hint: (z  i) (z2  2 i) = 0  z = i or z2 = 2 i = 2 ei /2  z = 1 + i or  1  i ] Q.47509/vec If  a &  b are two non collinear unit vectors &  a ,  b , x  a  y  b form a triangle, then : (A*) x =  1 ; y = 1 &   a +  b  = 2 cos   a b            2 (B*) x =  1 ; y = 1 & cos   a b        +   a +  b  cos      a a b ,          =  1 (C)   a +  b  =  2 cot   a b            2 cos   a b            2 & x =  1 , y = 1 (D) none [Hint: ,  a b & xa yb    form a triangle hence,     a b xa yb    = 0  (x + 1)  a + (1  y)  b = 0 Since &  a b are collinear  x =  1 & y = 1 Also   a b  2 = 2 + 2 .  a b = 2 (1 + cos ).   a b  = 2 cos  2 = 2 cos   a b            2  A A Also cos  =  b̂ â b̂ â . â ) (   =  1   cos    a b (where is the angle between   a &   a b  )    a b  cos  =  (1 + cos )    a b  cos  + cos =  1 ] Q.48510/vec The lines with vector equations are ;  r1 =         3 6 4 3 2    ~  i j i j k  and  r2 =         2 7 4    ~  i j i j k  are such that : (A) theyare coplanar (B*) theydo not intersect (C*) theyare skew (D*) the angle between them is tan1 (3/7) Q.49523/complex Given a,b, x, y R then which of the following statement(s) hold good? (A*) (a + ib) (x + iy)–1 = a – ib  x2 + y2 = 1 (B*) (1–ix) (1 + ix)–1 = a – ib  a2 + b2 = 1 (C*) (a + ib) (a – ib)–1 = x – iy  |x + iy| = 1 (D*) (y – ix) (a + ib)–1 = y + ix |a – ib| = 1 [Hint: Modulus ofacomplex number, whichis theratiooftwoconjugates is unity. e.g. inA, ib a ib a   = x + iy  a ib a ib   = |x + iy|  x2 + y2 = 1 ]
13. Q.50514/vec The acute angle that the vector 2 2    i j k   makes with the plane contained by the two vectors 2 3    i j k   and    i j k   2 is given by: (A) cos–1 1 3       (B*) sin–1 1 3       (C) tan–1   2 (D*) cot–1   2 [Hint:  n1 =   a b  =    2 3 2       i j k i j k      =   5    i j k    n1 =    i j k   3  v i j k    2 2      v i j k    2 2 3    cos   2        = sin  = .  v n  1 3 ] Q.51518/vec The volume of a right triangular prism ABCA1B1C1 is equal to 3. If the position vectors of the vertices of the baseABC areA(1, 0, 1) ; B(2,0, 0) and C(0, 1, 0) the position vectors of the vertexA1 canbe: (A*) (2, 2, 2) (B) (0, 2, 0) (C) (0,  2, 2) (D*) (0,  2, 0) [Hint: knowing the volume of the prism we find its altitude H = (AA1) = 6 and designating the vertex A1 (x1, y1, z1) relate the co-ordinates of the vector AA1  = (x  1, y, z  1) and its length.We get the other equation from the contition AA1  perpendicular toAC  ] OR compute ± AB AC AB AC       =  n  6  n = AA A1 = ±      i j k   2 = (x1  1)  i + (y1  1)  j + (z1  1)  k Compare to get at the possible coordinates ofA ] Q.52528/complex If xr = CiS  2r       for 1  r  n r, n  N then : (A*) Limit n   Re xr r n         1 =  1 (B) Limit n   Re xr r n         1 = 0 (C) Limit n   Im xr r n         1 = 1 (D*) Limit n   Im xr r n         1 = 0 Q.53524/vec If a line has a vector equation ,    r i j i j     2 6 3      then which of the following statements holds good ? (A)the lineis parallel to 2 6   i j  (B*) theline passes through thepoint 3 3   i j  (C*) the line passes through thepoint   i j  9 (D*) theline is parallel toxyplane
14. [Hint: Line is parallel to   i j  3  D Alsoput  r i j 1 3 3     for which  = 1 and  r i j 1 9     for which  =  1  B & C ] Q.54525/vec If    a b c , , are non-zero, non-collinear vectors such that a vector  p = a b cos     2      a b c and a vector  q = a c cos           a c b then  p +  q is (A) parallel to  a (B*) perpendicular to  a (C*) coplanar with   b c & (D) coplanar with a  and c  [Sol. c ) 2 ( cos b a p       where is the angle betweena  andb  and b ) cos( c a q       where  is the angle between a  and c  now q p    = b cos c a c ) cos b a (      = b ) c . a ( c ) b . a (        = ) b c ( a       B and C ] Q.55539/complexThegreatestvalueofthemodulusofofthecomplexnumber 'z' satisfyingtheequality z z  1 = 1 is (A)   1 5 2 (B*) 3 5 2  (C) 3 5 2  (D*) 5 1 2  SUBJECTIVE: Q.190/5 Let j ˆ î 3 a    and j ˆ 2 3 î 2 1 b    and b ) 3 q ( a x 2       , b q a p y       . If y x    , then express p as a function of q, say p = f (q), (p  0 & q  0) and find the intervals of monotonicity of f (q). [Sol.                ĵ 2 3 î 2 1 ) 3 q ( ĵ î 3 x 2  = ĵ ) 3 q ( 2 3 1 î 2 3 q 3 2 2                                     ĵ 2 3 î 2 1 q ĵ î 3 p y  y · x   = 0 gives p = 4 ) 3 q ( q 3  Ans. dq dp = 4 1 [3q2 – 3] > 0 q2 – 1 > 0 q > 1 or q < – 1 and decreasing in q  (–1, 1), q  0 Ans. ]
15. Q.225/3 Using onlythe limit theorems 1 x x n Lim 1 x   l = 1 and x 1 e Lim x 0 x   = 1. Evaluate 1 x x n x x Lim x 1 x     l . [Ans. – 2] [Sol. 1 x x n x x Lim x 1 x     l l = 1 x x n e e Lim x n x n x 1 x     l l l = 1 x x n x n x n x · ) x n x n (x ] 1 e [ · e Lim x n x n x x n 1 x        l l l l l l l l = (1) (1) · ) 1 x x n ( ) 1 x ( ) 1 x )( 1 x ( x n Lim 1 x       l l = (1) (1) (1) · 1 x x n ) 1 x ( Lim 2 1 x     l put x = 1 + h, as x  1, h  0 = h ) h 1 ( n h Lim 2 0 h    l put ln (1 + h) = y  1 + h = ey = ) 1 e ( y ) 1 e ( Lim y 2 y 0 y     = 1 e y y Lim · y 1 e Lim y 2 0 y 2 y 0 y              = – (1) 1 e y y Lim y 2 0 y    l = – 2 and 2 y 0 y y 1 y e Lim    = 2 1 Ans. ] Q.392/5 The three vectors  a i j k    4 2    ;  b i j k    2   and  c i k   2  are all drawn from the point with p.v. j ˆ î . Findthe equationof theplanecontainingtheir endpoint in scalar dot product form. [ Ans.   2 2    . i j k r    = 3 ] Q.4222/3 (cos cos ) 2 1 2 1 2 2 n n x x dx     z   where n N  [Sol. I = 2 1 2 1 2 0 2 (cos ) ( cos ) x x dx n  z  as f is even = 2 2 1 2 0 2 (cos ) . sin x x dx n z  = 2 2 1 2 0 1 t dt n z . when cosx = t = 2 2 2 1 4 2 1 2 1 2 0 1 . n t n n  L N M M O Q P P    ]
16. Q.597/5 Let points P, Q & R have position vectors, 1 r  = 3i  2j  k; 2 r  = i + 3j + 4k & 3 r  = 2i+j2k respectively, relative toan origin O. Find the distance of Pfrom the plane OQR. [REE ’90, 6] [Ans:3 units] [Sol. 2 1 2 4 3 1 k̂ ĵ î r r n 3 2        ) 6 1 ( k̂ ) 8 2 ( j ˆ ) 4 6 ( î        = k̂ 5 j ˆ 10 î 10    3 k̂ ĵ 2 î 2 n̂     d = n on OP of ojection Pr  = 3 ) k̂ ĵ 2 î 2 ( )· k̂ ĵ 2 î 3 (     = 3 1 4 6   = 3 units] Q.6228/3 Evaluate:     3 1 dx ) 3 x )( 2 x )( 1 x ( [Ans. 1/2] [Sol. I =     3 1 dx ) 2 x )( x 3 )( 1 x ( let x = cos2 + 3 sin2 dx = 2 sin 2 d x – 1 = 2 sin2; 3 – x = 2 cos2 and x – 2 = cos2 + 3 sin2 – 2 = 2 sin2 – 1 = – cos 2 I =        2 0 2 d 2 sin 2 2 cos · cos 2 · sin 2 =        2 0 2 2 d 2 cos 2 sin 2 · cos · sin 4 =      2 0 3 d 2 cos 2 sin 2 put 2 = t I =     0 3 2 dt t cos t sin 2 =   2 0 3 dt ) t cos · t (sin 2 put sin t = y I =  1 0 3 dy y 2 = 1 0 4 4 y · 2 = 2 1 Ans. ]
17. Q.798/5 Given that vectors    A B C , , form a triangle such that    A B C   , find a,b,c,d such that the area of the triangle is5 6 where  A = ai + bj + ck;  B= di + 3j + 4k &  C =3i + j  2k. [Ans: (8, 4, 2, 11) or (8, 4, 2, 5)] [ REE ’90, 6 ] [Sol.    A B C   ai bj ck d i j k    ( ) ( ) ( )         3 3 1 4 2 = ( )   d i j k    3 4 2 Hence d + 3 = a; b = 4 and c = 2 again| |   B x C = 5 6 | | | |   B C 2 2 – ( )   B C  2 = 150 (25 + d2)14 – (3d + 3 – 8)2 = 150 14(25 + d2) – (3d – 5)2 = 150 now proceed to get two values of d ] Q.8246/3                       1 n 0 k n 1 k n k n dx x n 1 k n k x n Lim [Ans. 8  ] [Sol. Let        dx ) x )( x ( where  = n k ;  = n 1 k  x =  cos2 +  sin2 dx = ( – )2 sin  cos  x –  = ( – )sin2 I = 2( – )2      2 0 2 2 d ) cos (sin =        2 0 2 2 d 2 sin 2 ) ( put 2 = t I =      0 2 2 dt t sin 4 ) ( =      2 0 2 2 dt t sin · 2 · 4 ) ( =     8 ) ( 2 = 2 ) ( 8     = 2 n 1 · 8  which is independent of k.  l =       1 n 0 k 2 n n 1 · 8 · n Lim =       1 n 0 k n ) 1 ( 8 n 1 Lim = n · n 8 Lim n    = 8  Ans. ]
18. Q.9114/5 Find the distance of the point P(i + j + k) from the plane L which passes through the three points A(2i + j + k), B(i +2j + k), C(i+ j + 2k).Also find thepvofthefoot of the perpendicularfrom Pon the plane L. [ REE ’92 , 6 + 6 ] [Ans :       3 4 , 3 4 , 3 4 , 3 1 ] [Sol. k̂ j ˆ î 0 a     k̂ 0 j ˆ î b     0 1 1 1 1 0 k̂ ĵ î b a       = ) 1 ( k̂ ) 1 ( j ˆ ) 1 ( î    n k̂ j ˆ î b a         (say) c k̂ 0 ĵ î 0 BP         PN Projection of c  on n  = n n . c    = 1 1 1 1    = 3 1 Now equation of a line through P and | | n  is ) k̂ j ˆ î ( k̂ j ˆ î r         = ] k̂ j ˆ î [ 1     Let the position vector of N = (1+), (1+) , (1+) k̂ ĵ î ) 1 ( AN         Now b , a   and  AN must be copalnar       1 0 1 1 1 1 0 = 0 1[] + 1[ +  – 1] = 0 3l = 1   = 1/3  Position vector of N       3 4 , 3 4 , 3 4 ] Q.10 Evaluate: (a)   dx x cos x sin x cos x sin 3 4 4 , x         2 , 0 ; (b)   dx x cos x sin x cos x sin 3 4 4 [Sol.(a)I =   dx x cos x sin x cos x sin 3 4 4 , x         4 , 0 =   dx x cos x sin x tan 1 x cos 3 4 2 =   dx x sin x tan 1 x cos 3 4 =   dx x ec cos · x cot · x cot x cot 1 2 2 4
19. put cot2x = t  2 cot x · cosec2x dx = – dt I = –   dt t t 1 2 1 2 put 1 + t2 = y2  t dt = y dy I = –  dy t y · y 2 1 2 = –     dy 1 y 1 1 y 2 1 2 2 = –             1 y dy dy 2 1 2 = C – 2 y – 1 y 1 y n 4 1   l = C – 1 1 t 1 1 t n 4 1 2 t 1 2 2 2       l where t = cot2x Ans(a). (b) I =   dx x cos x sin x cos x sin 3 4 4 =   dx x cos x sin x cot 1 x sin 3 4 2 =   dx x sec · x tan · x tan x tan 1 2 2 4 put tan2x = t =   dt t t 1 2 1 2  1 1 t 1 1 t n 4 1 2 t 1 2 2 2       l + C, where t = tan2x Ans(b). ] Q.11115/5 Findtheequationofthestraight linewhich passes through thepoint withposition vector a  ,meets the line c t b r      andis parallel to theplane 1 n . r    . [Sol. Suppose the required line intersects the given line at P with p.v.( c t b    ). As the line l is | | to the plane 1 n . r    . Hence 0 n . AP    n · c t ) a b ( ] [       = 0 n . c n . ) b a ( t         Henceequationofthelineis ) c t a b ( a r                       c n . c n . ) b a ( a b a r                         c n . c n . ) b a ( ) b a ( a r           Ans ] Q.12 Integrate:   x sin x cos dx 3 3 . [Ans. 2 [tan–1(sin x + cos x) + 2 2 1 ln x cos x sin 2 x cos x sin 2     + C] [Sol. I =   x sin x cos dx 3 3 =    ) x cos x sin 1 )( x sin x (cos dx =     ) x 2 sin 2 ( ) x sin x (cos dx ) x sin x (cos 2 2 = 2     ) x 2 sin 2 )( x 2 sin 1 ( dx ) x sin x (cos
20. I =          2 2 ) x cos x (sin 1 ) x cos x (sin 2 dx ) x sin x (cos put sin x + cos x = t  (cos x – sin x) dx = dt hence I =    ) t 1 )( t 2 ( dt 2 2 =       dt ) t 1 )( t 2 ( ) t 1 ( ) t 2 ( 2 2 2 2 =      2 2 t 2 dt t 1 dt = tan–1(t) + 2 2 1 ln t 2 t 2   + C = 2 [tan–1(sin x + cos x) + 2 2 1 ln x cos x sin 2 x cos x sin 2     + C Ans. ] Q.13147/5 Find the equation of the line passing through the point (1, 4, 3) which is perpendicular to both of the lines 2 1 x  = 1 3 y  = 4 2 z  and 3 2 x  = 2 4 y  = 2 1 z   Also find all points on this linethe square of whose distance from (1, 4,3) is 357. [Ans. 1 3 z 16 4 y 10 1 x       , ; (–9, 20, 4) ; (11, –12, 2) ] [Sol. Equationoftheline passingthrough(1, 4, 3) c 3 z b 4 y a 1 x      ....(1) since (1) is perpendicular to 2 1 x  = 1 3 y  = 4 2 z  and 3 2 x  = 2 4 y  = 2 1 z   hence 2a + b + 4c = 0 and 3a + 2b – 2c = 0  3 4 c 4 12 b 8 2 a        1 c 16 b 10 a    hence theequation of thelines is 1 3 z 16 4 y 10 1 x       ....(2) Ans. now any point P on (2) can be taken as 1 – 10 ; 16 + 4 ;  + 3 distance of P from Q (1, 4, 3) (10)2 + (16)2 + 2 = 357 (100 + 256 + 1)2 = 357  = 1 or – 1 Hence Q is (–9, 20, 4) or (11, – 12, 2) Ans.]
21. Q.14 1 n n 2 2 n 2 n 1 n n Lim               [Ans. e–1] [Sol. L = ) 1 n 1 n n ( 1 n n 2 Lim 2 2 n e        = el, where l =                      n ) n 1 ( n n 1 n n 2 Lim 2 2 n =                      ) 1 n ( n n Lim · n n 1 n 1 1 2 n Lim 2 n n = 2 ·                 ) 1 n ( n n ) 1 n ( ) n n ( Lim 2 2 2 n (rationalisation) = 2 · 1 n n n 1 n 2 n n n Lim 2 2 2 n          = 2 ·                          n 1 1 n 1 1 n ) 1 n ( Lim n = 2 ·                          n 1 1 n 1 1 n n 1 1 n Lim n = – 2       2 1 = – 1  L = e–1 ans. ] Q.15151/5 If z-axis be vertical, find the equation of the line of greatest slope through the point (2, –1, 0) on the plane 2x + 3y – 4z = 1. [Sol. Equationof thelineofgreatest slope c z b 1 y a 2 x     where 2a + 3b – 4c = 0 ....(1) nowequation ofthehorizontal plane is z = 0 i.e. 0 · x + 0 · y + 1 · z = 0 nowavectoralongthelineofintersection ofgiven planeandhorizontalplaneis 4 3 2 1 0 0 k̂ ĵ î v    = – ( j ˆ 2 î 3  ) = k̂ 0 j ˆ 2 î 3   since the lineof greatest slope is also perpendicular to the vector v  hence – 3a + 2b + 0 · c = 0 ....(2) from (1) and (2) 2a + 3b – 4c = 0 – 3a + 2b + 0 · c = 0 9 4 c 12 b 8 0 a      13 c 12 b 8 a    equation ofthe line of greatest slope = 13 z 12 1 y 8 2 x     ]
22. Q.16 Let I =    2 0 dx x sin b x cos a x cos and J =    2 0 dx x sin b x cos a x sin , where a > 0 and b > 0. Compute the values of Iand J. [Sol. aI + bJ = 2  ....(1) and bI – aJ =     2 0 dx x sin b x cos a x sin a x cos b  bI – aJ = ln   2 0 x sin b x cos a    bI – aJ = ln       a b .....(2) from (1) and (2) a2I + abJ = 2 a b2I – abJ = b ln  a b ———————— I =                  a b n b 2 a b a 1 2 2 l Ans. again abI + b2I = 2 b and abI – a2J = a ln   a b subtract ————————— J =                  a b n a 2 b b a 1 2 2 l Ans. Alternatively: convert a cos x + b sin x into a single cosine say cos(x + f) and put x – f = t ]