MATHEMATICS

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 ]

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 :

(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

(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

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

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

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

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

(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
]

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.

[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 ]

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

[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. ]

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



]

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. ]

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. ]

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

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

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.]

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




]

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 ]