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[233] Mathematics for I I T-J EE (Hints/Solutions)
1. , and b c b c

are three non-coplanar vectors , hence
any vector ( ) r

in space can be represented by

( ) ,
| | | | | |
r b c b c
b c b c

= + +

where , ,
are the projections of on , r b c

and
b c

respectively.
projection of
.
on
| |
b
r b
b

= =

Similarly ,
. ( ).
and
| | | |
r c b c r
c b c


= =

Now ,
2
.( )
( . ) ( . ) ( )
| |
r b c
r r b b r c c b c
b c

= + +

( | | | | 1) b c = =

2
.( )
( . ) ( . ) ( )
| |
a b c
a b b a c c b c a
b c

+ + =

required vector a =

2. Let point P represents the origin
2
( )
2
1 1
a b
b a
SX
PX
XM

| |
+
+
|
\ .
= =
+

2
2
a b
PM

+
=
`

)

( )
1 1
PX u u
PX a b
XR u
= = +
+

...(ii)
PX

from (i) and (ii) is equal

( 1) 1
2
( )
1 1
a b
u
a b
u

| |
+ +
|
\ .
+ =
+ +


( 1) 1
2

| |
= +
|
\ .
4 =
1 4
4 5
XM SM
SX SX
= =
4
5
k =
3. | | | | | | 1 a b c = = =

. 0, . . cos a b a c b c = = =

( ) ( ) c p a b q a b = + +

. . cos . a c b c p p = = =

Now ,
2
2
| | ( ) ( ) c p a b q a b = + +

2 2
1 ( | | 1) p p q a b = + + =

2 2
1 2cos cos2 q = =
(cos2 ) 0 s
3
4 4

s s
4. Applying vector triple product

( )

( )

( ) ( )

( )
3 . . . 0 a ak k a j j ai i i j k + + + + =



( )

( )

( )

( )
2 . . . a i j k a ai i a j j ak k = + + = + +

3 3
or
2 2 2 3
i j k
a a
+ +
= =

5. x

lies in plane of and a b

and it is normal to
( )
. . 0 b xb=

Let
( ) ( )
x b a b =

( ) ( ) ( )
. . x b b a b a b =


( )
3 5 6 x i j k = + +

Now ,
1
. 7
2
a x = =


( )
1
3 5 6
2
x i j k = + +

Chapter No -29 ( Vectors )

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[234] Mathematics for I I T-J EE (Hints/Solutions)
6. ( ) ( ) ( ) ( ) ( ) ( ) { }
a b r c b c r a c a r b + + =

AddingthethreeV.T.P.
a b c r a b r c
b c a r b c r a
c a b r c a r b
( (

( (

`


( (
)



{ }
3 a b c r a b r c b c r a c a r b ( ( ( ( + +


{ }
3 a b c r a b c r
( (
=


2 a b c r ( =


7. Volume a b b c c a
(
= + + +



2 a b c
(
=


2| | 2 = =
8.
2 2 2
a b b c c a + +

{ }
2 2 2
2 2 . . . a b c ab bc ca
| |
= + + + +
|
\ .

{ }
2 2 2 2
2(3) a b c a b c = + + + + +

2
9 a b c = + +

maximumvalue =9
( )
0 a b c + + >

9.
( ) ( )
. 0 a d b c AD BC =

( ) ( )
. 0 b d c a BD AC =

D
is intersection point of altitudes
D
is orthocentre of
ABC A
10. ( )
a b c

represents the vector in plane of and b c

,
and the vector is normal to a

( )
( )
a b c
n
a b c
| |

|
=
|
+ |
\ .


Required unit vector

3
10
j k
| |

= |
|
\ .
11.
( ) ( ) ( ) ( )
. a b c b c b c +

( ) { } ( )
. a b c b c b c b c a b c
( (
= + +


( ) { } ( )
0 . a b c b c b c
(
= +


{ }
2 2
. a b c b c
(
=


( )
0 1 b c = = =

12. and r a b r c d = =

( ) ( )
r a c b d =

( ) ( )
. 0 a c b d =

. . . . 0 ab ad cb cd + =

( )
. . 0 . . 0 ad cb ab cd + = = =

13. Let d xa yb = +

( )
.
1
3
xa yb c
c
+
=

( ) ( )
. .
1
3
x a c y b c
c
+
=

2 1 or (2 1) x y y x = =
(2 1) d xa x b = +


(3 1) (3 2 ) (3 1) d x i x j x k = + +

for

1, 2 2 x d i j k = = + +

14.
1
2
. ba
b b a
a
=

( ) 1
2
.
. . .
ba
b a ba aa
a
=

2
1. 0 . b a aa a
| |
= =
|
\ .

and a b

are orthogonal vectors.
Now ,
( )
1
1 1 1 1 1
2
.
. . 0 .
b c
c b cb b b
b
=

1 1 1 1 . 0 or . 0 c b b c = =

15.
( )
. a b c

represents the volume of a parallelepiped
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[235] Mathematics for I I T-J EE (Hints/Solutions)
and if it equals to , a b c

then it represents the
volume of cuboid.
. . . 0 ab bc ca = = =

16. Let normal vector of plane determined by vectors

1 and be i i j n +


and for the another plane let the
normal vector be 2 n

As shown in figure , a

is normal to both the vectors

( ) 1 2 . . and i e n n

( ) 1 2 Let a n n =

( ) ( )

( )

( ) ( ) { }
a i i j i j i k = + +

( )
a i j =

Let angle between a

and

2 2 i j k +

be
( )
( )
2
(1 2) 1
cos
2
2 9

+
= =
4

=
17.

( )

( )

( )

( )
. . . 1 b a b a a a b a b a a = =

( )

( )
. b a b a a b a =

( )

( )
3 and a b a b a

are orthogonal vectors,
hence the triangle is right-angled
Let the internal angles be 90 , , 90

( )

( )
3
tan
a b
a b a

( ) ( )
3
tan
sin90
a b
a b a
a b a

tan 3 60 = =
internal angles are 30 , 60 , 90 .
18.
( )
. U V W U V W
(
=


cos U V W U V W
(
=


( )
min
1 U V W V W U
(
= =

59 =
19. Let

w xa yv = +


( )
. . wa x y v a = +

cos cos x y = ...(i)
Similarly ,

( )
. . wv x a v y = +

cos2 ( cos )x y = + ...(ii)
from (i) and (ii) ,

( )

2 . w v a v a =

20. In , ABC BC AC AB A =

BA CA BC = +

2e e f
BA
e e f
= +

e f
BA
e f
| |
|
= +
|
\ .

Now ,
2 2
. .
.
e e f f
BC BA
e f
| |
|
=
|
|
\ .



. 0 BC BA =

90 ABC Z =
cos2 cos2 2cos( ).cos( ) A C A C A C + = +
0 ( 90 ) A C = + =
cos2 cos2 cos2 (cos2 0) A B C B + + = +
1 ( 2 180 ) B = =
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[236] Mathematics for I I T-J EE (Hints/Solutions)
21.
( ) ( )
. 1 a b c d =

( ) ( )
1 and || a b c d a b c d = =

, , and arecoplanar a b c d

1
.
2
a c=

angle between and a c

is
60
1 a b =

angle between and a b

is 90 , and
similarly angle between and c d

is 90.
As shown in figure , and b d

are orthogonal to
and a c

respectively
, b d

are non-parallel.
22. Let ; 0 a b c
(
= =


1 2 6
, ,..... can be or P P P
i k
j l
P P
P P


+
`

)
can not attain the value of 1
23. For cyclic quadrilateral ABCD , A C Z + Z = and
. B D Z + Z =
tan( ) 0 tan tan 0 A C A C + = + =
sin sin
0
cos cos
A C
A C
+ =
( ) ( )
0
. .
AB AD CB CD
ABAD CBCD

+ =


( ) ( ) ( ) ( )
0
. .
a b b d d a b c c a d b
b a d a b c d c

+ + + +

+ =
`


)


24.
( ) ( )
. . r a b c a c b =

( )
r b c a =

( )
. 0 and . 0 r a r b c = =

25. Diagonal vectors are and a b a b +

( ) ( )
.
cos
a b a b
a b a b

+
=
+ +


1
cos
2
=

=
4
Now , angle between diagonals can be
3
4
26. . 0 a b x R > e

( 1) ( 1) 0 x x x a x R + + + > e
2
2 1 0 x x a x R + + > e
0 2 D a < >
27.
( ) ( ) { }
. . a b a c d

. a b c a d
(
=


( )
. a d a b c
(
=


28. 1 a b = =

( ) ( ) ( ) 1 . . . u a a b b b b a a b b = =

( ) 1 u b a b =

1. 0 u b =

2 u a b =

1 sin90 u b a b =

1 2 u u =

( ) 2 1 2 1 or . . 0 u u u b u b = + =

29. Given conditions imply that r

is orthogonal to
, , a b c

.
, , a b c

are coplanar
0 a b c
(
=


30. Let the vertex points be ( ) ( ) ( )
, , A a B b C c

and
( )
D d

Now ,
1 1 AA AB AC + =

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[237] Mathematics for I I T-J EE (Hints/Solutions)
( )
2 2
b c c d
a a c a
| | | |
+ +
+ =
| |
| |
\ . \ .


( )
2
2 2
b d
c a c a + + =


( )
2
2 2 2
a c b d c a
c a c a
| |
+ + +
+ = =
|
|
\ .

( )
3 3
2 2
c a c a =

3
2
=
31.
from figure , cos CD a =

( )
. a b
b
b

=
`

)

( )
2
. ab
b
b

=
`

)

Now , CD DB CB + =

CB CD =


( )
2
2
. b a a b b
a
b

= =

32. 0 a b =

and a b

are collinear vectors
| |
2
2
1 cos
sin
sin
|sin | |cos |
cos
1
x
x
x
x x
e x
xe
e

+
= =
| | ( )
sin
cos
1 1 |sin | |cos | 1
x
x
x x x R
xe
= = + = e
sin
cos 0
x
x xe =
Let
sin
( ) cos
x
f x xe x =
( )
sin sin
'( ) . cos sin
x x
f x e xe x x = + +
'( ) 0 0,
2
f x x
| |
> e
|
\ .
Now , (0). (1)
2 2
e
f f
| | | |
=
| |
\ . \ .
(0). 0 and ( )
2
f f f x
| |
<
|
\ .
is strictly increasing
in
0,
2
| |
|
\ .
Unique value of ' ' x exists in 0,
2
| |
|
\ .
Similarly ,
3
'( ) 0 ,
2
f x x


| |
> e
|
\ .
( )
3 3
( ) 1
2 2
f f
e


| | | |
=
| |
\ . \ .
( )
3
0
2
f f

| |
>
|
\ .
and ( ) f x is increasing in
3
,
2

| |

|
\ .
no value of
' ' x
exists in
3
,
2

| |

|
\ .
.
33.

( )
2 2
cos sin (2cos sin ) . OP t t t t a b = + +

( )
1 (sin2) . OP t a b = +

maximum value of

( )
1 . OP a b = +

( )
. 0 a b >

( )
1 . L a b = +

for 2 90 t =
Now , if

45 ,
2
a b
t OP
| |
+
= = |
|
\ .

2
2
a b a b
n
a b a b
+ +
= =
+ +

a b
n
a b
+
=
+

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[238] Mathematics for I I T-J EE (Hints/Solutions)
34.
( )
2
2
2
1 1
. 1 1 0
1 1
a b c

= =

1 1 2 3
R R R R + +
2 2 2
2
2
2 2 2
1 1 0
1 1

( )
2 2
2
1 1 1
2 1 1 0
1 1

Apply
1 1 2 2 2 3
and C C C C C C
( )
2 2 2
2 2
0 0 1
2 1 1 1 0
0 1


+ =
+
( )( )
2
2 2
2 1 0 + =
2 =
35. . 0 >

( ) ( )
2
3 3 ( ) 0 a b c ab bc ac + + + >
( )
2 2 2
(2 3 ) 0 a b c ab bc ac + + + + + >
( )
2 2 2
3 2
a b c
ab bc ac

| |
+ +
s
|
|
+ +
\ .
for scalene triangle ,
2
2
a b c < ...(i)
2
2
b c a < ...(ii)
2
2
c a b < ...(iii)
Adding (i) , (ii) and (iii)
( )
( )
2 2 2
2
a b c
ab bc ac
+ +
<
+ +
( ) 3 2 2 <
4
3
<
Now , exhaustive set of values of
4
is ,
3

| |

|
\ .
36. If
( ) ( )
and a b c a b c

are perpendicular to each
other , then :
( ) ( ) { } ( ) ( ) { }
. . . . . 0 a c b a b c a c b b c a =

( ) ( )( )
{ }
2
. . . 0 a c b a c b c =

( )
. a c

is not necessarily zero
Statement (1) is false and statement (2) is true.
37.
( ) ( )
r a r b a b b a + = +

( )
0 r a b + =

( )
and r a b +

are collinear
( )
r a b = +


( )
1
2 2
9
r i j k = +

Statements (1) and (2) are true and the explanation is

appropriate.
38. Given equation can be written as :
2 2
3 2 0 p u v w pq u v w q u v w ( ( ( + =


( )
2 2
3 2 0 0 p pq q u v w
(
+ = =

Disc =124 <0

( ) 0 , p q p q R = = e
Statements (1) and (2) are true and the reasoning is
appropriate.
39.
( )
r b r a = +

2 2
r r a b =

( ) ( )
2 2 2
2 . r r a r r a b + =

( )
2 2
1 . 0 r r r a + = =

1 2
of
2 2
r =

Statement (1) is true.

As per statement (2) , 2r b a b = +

2r b a b =

( )
r r b a b + =

( ) ( )
r r a a b r b r a + = = +

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[239] Mathematics for I I T-J EE (Hints/Solutions)
( )
r a b r = +

( )
r a b r = +

is not true because r

and b r +

are not
perpendicular to each other
Statement (2) is false.
40.
( ) ( )
r a b a c =

r a a c b b a c a
( (
=


r a b c a
(
=


and r a

are collinear
( )
0 a b c
(
=

and r a

are linearly dependent
Statement (1) is true.
Now . . r b a b c a b
(
=


. r b

is not essentially zero.
Similarly . r c

is also not essentially zero.
Statement (2) is false.
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[240] Mathematics for I I T-J EE (Hints/Solutions)
1.
Let
1
CD
DA

=
position vector of

4 1 2 2 1
1 1 1
D i j k


+ | | | | | |
= + +
| | |
+ + +
\ . \ . \ .

Now ,

5 6 3 (2 1)
1 1 ( 1)
DB i j k


+ + | | | |
= +
| |
+ + + \ . \ .

. 0 . 0 ADBD DBDC = =

25 18 9 2 1 0 =
15 20 =
3
4
=
p . r . of

13 2 10
7 7 7
D i j k = +

169 4 100
49
OD
+ +
=

39
7

2. Area of
1
2
CDB DB DC A =


1
600 420 60
98
i j k =

150 6
Area Squareunits.
49
=
3.
.
cos( )
BC BD
DBC
BC BD
Z =


2 10
7
=
1
2 10
cos
7
DBC

| |
Z =
|
|
\ .
4.
as shown in figure , PQ RQ =

p r p p r r + = +

p r =

OP OR PQ RQ = = =

OPQR is rhombus
3 2
5 5
PQR OPQ

Z = Z =
SR r p p = =

r

2 2
2
2 . p r p p + =

( )
r p =

2
2
2
3
cos
5
r p
p

3 2
2cos or 2cos
5 5

=
Now ,

2 5 1 5 1
1 2cos 1 2
5 4 2
PS
QR

| |
+
= + = + =
|
|
\ .

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[241] Mathematics for I I T-J EE (Hints/Solutions)
5. Area of polygon PQRST =OPQR +OTS +2(ORS)
( ) ( )
2
2
r p r p p r

= + +

2
1
2
r p




= +
`

)

2
2 2
2
r p


+
=
`

)

2 2
( 1) 1 ( 2cos72 1) 1
2 2 2 2

= + = +
2
1 5 1 1 5 5
1
2 2 2 4
| | | |
+
= + + =
| |
| |
\ . \ .
5 5
4

| |
+
=
|
|
\ .
6.
( ) 0
1
5
OC OP OQ OR OS OT = + + + +

( )
1
5
p p r r p r = + + + + +

( ) ( )
1
2
5
p r = + +

( )( )
1
2 2cos72
5
p r = +

2 5 1
1
5 4
p r
| |

= +
|
|
\ .

( )
5 5
10
p r

= +

7. I f 1 2 3 , , e e e

is given set of non-coplanar
vectors and
1 2 3
1 ;
. , then , ,
0;
m n
m n
e f f f f
m n
=
=

=

represent the reciprocal system of vectors.
1 2 3
1 2 3
1 e e e f f f
( (
=


Now ,
1 2 3
1 2 3
16 9 e e e f f f
( (
+


1 2 3
1 2 3
9
16 e e e
e e e
(
= +

(



By AM GM > property
minimumvalue 2 16 9 =

24 =
8. 1 2 3 2 e e e
(
=


1 2 3
2 f f f
(
=


4 =
given quadratic equation can be written as :
2
(2)(4)(3) ( ) (2)(3) 0
2 2
x x

| | | |
+ + =
| |
\ . \ .
2
(12 ) ( ) 3 0 x x + + =
Now ,
2
( ) 4(12)(3) D =

2
( ) 144( ) =

2
(4) 144(4) =
0 D <
9.
2
1 2 3 e e e
(
=


2
1 2 3
f f f
(
=


1 =
2
1 2 3
2
1 2 3
1
e e e
e e e

(
+ = +

(



By AM GM > property
2 + >
Now , sin cos x x + = +
sin cos 2 x x + >
Above inequality is not valid.
( )
1 sin cos 2 s + s
10. and a b

are non-collinear unit vectors
. 1 and 0 ab a b = =

Now
( )
2 2 a b a b + + =

Squaring both sides
( )
2
1 1 4 2 . 4 a b a b + + + =

( ) ( )
2
2 4 1 . 2 . 4 a b a b
| |
+ + =
|
\ .

( )
{ }
2 2 2 2
. a b a b a b =

( ) ( ) ( )
. 1 2 . 1 0 a b a b + =

1
. ; . 1
2
a b a b = =

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[242] Mathematics for I I T-J EE (Hints/Solutions)
( )
2 2
2 . a b a b a b = +

3 a b =

( )
2 2 2 2
. a b a b a b = +


1 3
1
4 4
= =
3
2
a b =

Now, 2
a b
a b

11. ( ) ( ) ( )
1 1 1 2 2 2 3 3 3
6 a b c a b c a b c + + + + + + + + =
for maximumvolume , V a b c =

( )
3
2 2 2 2
1
r r r
r
V a b c
=
= + +
[
( ) ( ) ( )
1 1 1 2 2 2 3 3 3
3
a b c a b c a b c + + + + + + + +
>

( ) ( )
1/3
3
1
r r r
r
a b c AM GM
=


+ + >
`

)
[

( )( )( ) { }
3
1 1 1 2 2 2 3 3 3
6
3
a b c a b c a b c
| |
+ + + + + + s
|
\ .
( )( )( )
1 1 1 2 2 2 3 3 3
8 a b c a b c a b c + + + + + + s ...(i)
( ) ( )
2 2 2 2
1 1 1 1 1 1 1 1 1 1 1 1
2 a b c a b c ab bc c a + + = + + + + +
( )
2 2 2 2
1 1 1 1 1 1
a b c a b c + + > + + ...(ii)
( ) , , 0
r r r
a b c >
Similarly , ( )
2 2 2 2
2 2 2 2 2 2
a b c a b c + + > + + ...(iii)
( )
2
2 2 2
3 3 3 3 3 3
a b c a b c + + > + + ...(iv)
from (ii) , (iii) and (iv)
( ) ( ) ( )
2 2 2
1 1 1 2 2 2 3 3 3
. a b c a b c a b c + + + + + + >

( )
3
2 2 2
1
r r r
r
a b c
=
+ +
[
( )( )( ) { }
2
2
1 1 1 2 2 2 3 3 3
V a b c a b c a b c s + + + + + +
Using Inequality (i)
2
64 V s
8 V s
maximum volume=8
{
1 1 1 1 1 1
8exists, for 0, V ab bc ca = + + =
}
2 2 2 2 2 2 2 2 2 2 2 2
0 and 0 a b b c c a a b b c c a + + = + + =
12.
( ) ( ) ( )
. . a i a a a i a i a =



( ) ( )

( )

( ) { }
2
. . . b a i j k a i a a j a a k a = + + + +



( )

( )
2
. . . b a i j k a ai a j ak = + + + +



( )

( ) ( )
2
. 0 b a i j k a i j k = + + + + =

{ }
2 4
3 b a =

2
4
3
b
a


=
`

)

13. 1and 0, 0 a a b a c = = =

( ) ( ) { } ( )
. 2 P a b a b c a b c = +

{ } ( )
. 2 P b c a c a b c = +

( ) ( ) { }
. .2 P b c a a c b =

( )
3 . P a b c =

3 P a b c s

3 P b c s

( )
9 3 P b c s =

14. , ,
u v v w w u
x y z
u v v w w u
+ + +
= = =
+ + +



1 1 12cos 2cos
2
u v

+ = + + =

2cos and cos
2 2
r
v w w u

+ = + =

, and
2cos 2cos 2cos
2 2 2
u v v w w u
x y z

+ + +
= = =


Now ,
x y y z z x
(



( ) ( ) { }
. x y y z z x =

{ }
0 . x y z y z x
(
=


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[243] Mathematics for I I T-J EE (Hints/Solutions)
2
x y z
(
=


2
2 2 2
64cos .cos .cos
2 2 2
u v v w w u
x y y z z x

(
+ + +

(
=



{ }
2
2 2 2
2
64cos cos .cos
2 2 2
u v w
x y y z z x

(

(
=



1
2
2
2 2 2
sec .sec .sec
2 2 2
4
u v w
x y y z z x

(

=
`
(



)


15. (a) Area of
1 1 1
2 2 2
ABC a b b c c a A = = =
a b b c c a = =
(b) In regular tetrahedron all the edges are of equal
length and the angle between two adjacent edges is
60
2
1
. . . (edgelength)
2
a b b c c a = = =

Area of all the faces is same
a b b c c a = =

(c) , a b c b c a = =

c a b =

, , a b c

forms an orthogonal system of
vectors having equal magnitude
. . . 0 a b b c c a = = =

2
a b b c c a a = = =

(d) a b c + =

1
.
2
a b c ab + = =

similarly ,
1
. .
2
b c c a = =

1
. . .
2
a b b c c a = = =

3
. . .
2
a b b c c a + + =

3
2
a b b c c a = = =