Exam (elaborations)
Vector Algebra Questions and Answers
- Module
- MATHEMATICS VECTORS
- Institution
- The University Of Edinburgh (ED)
Vector Algebra Questions and Answers pt1
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VECTOR ALGEBRA JEE ADVANCEDVECTOR 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 12 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 cb 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 nm 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
bd 1
k j
1 3 cos r , a cos r , b cos r , c
2 2
x2y 2z 2x3y 6z 4x4y 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
122 MATHS BY Er.ROHIT SIR
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