Vector Algebra Questions and Answers

VECTOR ALGEBRA JEE ADVANCED


VECTOR ALGEBRA
WORKEDOUT EXAMPLES 2 1 1
  1,   ;    ,  
3 3 3
W.E-1:Let PN be the Perpendicular from the point
W.E-3:If a and b are unit vectors then greatest and
P 1, 2,3 to xy-plane. If OP makes an angle 
with positive direction of the Z-axis and ON least values of a  b  a  b are
makes an angle  with the positive direction of A. 2,-2 B. 4, 4 2
x-axis, where O is the Origin (  and  are
C. 2 2, 2 D. 2 2, 2 2
acute angles) then
Sol: (C)
2 1
A. sin  sin   B. Cos cos   Given a  b  1
14 14

C. tan  
5
D. tan   2
 
Let a, b   then
3 2 2 2
Sol: (A, C, D) a  b  a  b  2a.b
Let OP  r then = 2  2 cos 
r sin  cos   1......(1) 2
= 4 cos
r sin  sin   2......(2) 2
r cos   3......(3) 
 a  b  2 cos
On squaring and adding, we get 2
2
r  14  r   14 
Using (1),(2),(3) and (4) we get required result. Similarly, a  b  2sin
2
W.E-2:In a four-Dimensional space where unit vector
  
along axes are i, j , k and l and a1 , a 2 , a 3 , a 4  a  b  a  b  2  cos  sin 
 2 2
are four non-zero vectors such that no vector
can be expressed as linear combination of others 0 
Clearly   0,180     0,900 
and 2
  1  a1 a2   a2 a3  r a3 a4 2a2  a3 a4  o  greatest value= 2 2 , least value= 2(1)  2
Then W.E-4:Let ABCD be a tetrahedron with AB=41,
BC=36, CA=7, DA=18, DB=27, DC=13. If
A.   1 B.    ‘d’ is the distance between the midpoints of edges
C. r  D.   1/ 3 AB and CD, then
Sol: (A, B) A. The last digit of d 2 is 7
 1  a1 a2  a2 a3 r a3 a4 2a2 a3 a4 o B. d 2  137
C. The last digit of d 2 is 6
 1 a1 12 a2   1 a3   a4 0
D. d 2  126
a1 , a 2 , a 3 , a 4 are linearly independent Sol: (A, B)
   1  0 .............(1) Take D as the origin DA  a , DB  b , DC  c
1      2  0.......(2)
a 18, b 27, c 13, a b 41, b c 36, c a 7
    1  0.....(3)
2
    0....(4) 2 a b c 
By solving (1), (2), (3) and (4), we get d   
 2 2

MATHS BY Er.ROHIT SIR 119

, JEE ADVANCED VECTOR ALGEBRA
2 2 2 2
 4d 2  a  b  c  2a .b  2a .c  2b .c Now, b  c  144
2 2 2 2 2
a  b  a  b  2a .b  b  c  2b.c  144
2 2 2
b  c  b  c  2b .c  b.c  72
2 2 2 1 b
c  a  c  a  2c .a Since AE : EB  1: 3  AE  AB 
4 4
using the above results we get d 2  137
b
W.E-5:Let aˆ , bˆ, cˆ be unit vectors such that CE  AE  AC   c
4
aˆ  bˆ  cˆ  0 and x, y, z be distinct integers then
CE.CA
x aˆ  y bˆ  zcˆ cannot be equal to cos  
CE C A
A)1 B) 2 C.2 D.3
b 
sol: (A,B)   c  c
4 
 
a, b, c form an equilateral triangle and angle cos   
b 
2 c c
between each pair is 4
3
2
 xa  yb  zc  x 2  y 2  z 2  xy  yz  zx b.c
2
c 
 4 3 7
b 8
1 2 2 2
c c

  x  y    y  z    z  x   4
2
1
1  1  4

W.E-7:If the distance of the point B i  2 j  3k from 
Minimum value  the line which is passing through
2
=3 
A 4i  2 j  2k  and Parallel to the vector
 xa  yb  zc  3
c  2i  3 j  6k is  then  6   4   2  1 is
A.1001 B.1101 C.1111 D.1011
W.E-6:In an Isosceless triangle ABC, AB  BC  8 .
Sol: (D)
A point E divides AB internally in the ratio 1 :3 Given B  1, 2,3
then the angle between CE and CA where
A   4, 2, 2  and
CA  12 is
c  2i  3 j  6 k
 7 
1 3 7

cos 1
  cos  
A. B.  8 
 8   
1
 7 
1 3 7

C. cos  4  D. cos  4  A
    c C
Sol: (B)
Let ‘A’ be the fixed point and AB  b, AC  c AB  c
Then Then    10
c
b  8 , c  8 , b  c  12
  6   4   2  1  1000  100  10  1  1111

120 MATHS BY Er.ROHIT SIR

,VECTOR ALGEBRA JEE ADVANCED

W.E-8: a  i  k , b  i  j and c  i  2 j  3k      
a b  b c  c a
 
be three given vectors. if r is a vector such p
  
that r  b  c  b and r.a  0 then the value
3 p / 2   p / 2  p
 
p
of r.b is 
Sol: (9) 3p
  3
Given r  b  c  b p

Taking cross product on both sides with a W.E-10:If the planes x  cy  bz  0 , cx  y  az  0
and bx  ay  z  0 Pass through a line then the
   
 r b a  cb a value of a 2  b 2  c 2  2abc is
Sol: (1)
  r.a  b   a.b  r   c.a  b   a.b  c Given plane are
x  cy  bz  0.......(1)
 0  r  4b  c
cx  y  az  0.......(2)
 r  4b  c bx  ay  z  0.......(3)
 r  3i  6 j  3k Equation of plane passing through the line of
intersection of planes (1) and (2) may be taken
 r.b  3  6  9 as
W.E-9:Let O be an interior point of ABC such that  x  cy  bz     cx  y  az   0.........(4)
   
OA  2OB  3OC  0, Then the ratio of the Now, Planes (3) and (4) are same
area of ABC to the area of AOC is 1  c   c    b  a
  
Sol: (3) b a 1
      By Eliminating , we get
Area of ABC 1/ 2  a  b  b  c  c  a
   a 2  b 2  c 2  2abc  1
Area of AOC 1/ 2  a  b
 W.E-11:ABCD is a square of unit side. It is folded along
Now a  2 b  3 c  0 the diagonal AC, So that the plane ABC is
     Perpendicular to the plane ACD. The shortest
cross with b , a  b  3c  b  0 distance between AB and CD is
     1 3 2
 a , 2a  b  3a  c  0
A) 3 B) C. D.
    3 2 3
 a b  3 b c   Sol: (D)

  3     C
Hence, a  b   c  a   3 b  c
2
  D
  
Let  c  a   p. Them
N
 
  3p   p
a b  b c 
2 2 A B
Hence, the ratio is Let Square ABCD intially lies in xy-plane with A
lying at Origin , AB along x-axis and AD along
y-axis then AB  BC  CD  AD  1

MATHS BY Er.ROHIT SIR 121

, JEE ADVANCED VECTOR ALGEBRA
Let ' N ' be the foot of peependicular from ‘B’ to Now shortest dist an ce  d
1
AC then BN   m l n  m  l m n
2   d
  d
Let B be new position of B after folding along
' l nm ab sin
diagonal AC then Co-ordinates of various points
in 3D are  l m n   abd sin 
A =(0,0,0), B =(1,0,0) ,C=(1,1,0),D =(0,1,0)
1 1
1 1 1  1 1   volume = l m n  abd sin
B'   , ,  , N   , ,0 6 6
2 2 2 2 2  W.E-13:Let a  i  2 j  2k , b  2i  3 j  6k and

Now AB line equation is
c  4i  4 j  7 k . A vector which is equally
1 1 1  inclined to these three vectors is perpendicular
r t i j k
2 2 2  to the plane passing through the points pa, qb

CD line equation is r   i  j   s  i  and rc . If p,q,r are the least possible positive
 shortest distance integers, then the value of p  q  r is
1 1 0 A)19 B)37 C)53 D)111
Sol: (B)
1 1 1 Let r be required vector then
2 2 2
= 
a  c b d 
 1 0 0 2  r, a    r, b    r, c 
 
bd 1
 k j
1 3 cos  r , a   cos  r , b   cos  r , c 
2 2
x2y 2z 2x3y 6z 4x4y 7z
W.E-12:The length of two opposite edges of a     k say
tetrahedron are a,b and their shortest distance 3 7 9
is d and angle between them is  then volume of  x  2 y  2 z  3k ..........(1)
tetrahedron is 2 x  3 y  6 z  7 k ...........(2)
1 2 4 x  4 y  7 z  9k ...........(3)
A. abd sin  B. abd sin 
6 6 by solving above equations we get
1 x : y : z  3:5:7
C. d sin  D. abd cos 
6  r  3i  5 j  7 k
Sol: (A)
Since r is perpendicular to the plane passing
C
through the points pa, pb, rc

O A
  
 r. pa  pb  0, r. qb  rc  0 
 p  r.a   q  r.b  and q  r.b   r  r.c 
B  p  27   q  63  r  81
Let OA  l , OB  m , OC  n then
 p q r
The Equation of line OA is r  tl .............(1)   
 21 9 7
and Equation of line BC is  least possible value of p  q  r  37
W.E-14:Let the position vector of the orthocentre of
 
r  m  s n  m ................(2)
ABC be r , then which of the following
Since OA, BC are opposite edges statement(s) is/are correct ( Given position


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