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Examen

Vector Algebra Questions and Answers pt3

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Vector Algebra Questions and Answers pt3

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Subido en
6 de abril de 2024
Número de páginas
26
Escrito en
2023/2024
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JEE ADVANCED - VOL - I VECTOR ALGEBRA
LEVEL-VI  11
(A) n  ,nZ (B) 2n  ,nZ
6 6
SINGLE ANSWER QUESTIONS
 11
(C) 2n  , n  Z (D) 2n  ,nZ
 ^ ^ ^ 6 6
1. The unit vector c if  i  j  k bisects the 
5. If b is vector whose initial point divides the
 ^ ^
^
angle between vectors c and 3 i  4 j is ^
join of 5 i and 5 j in the ratio k:1 and whose
1 ^ ^ ^
 
(A)  11 i  10 j  2 k  terminal point is the origin and b  37 ,
15  
then k lies in the interval
1 ^ ^ ^

(B)  11i  10 j  2 k   1   1 
15   (A)  6,  (B)  , 6   ,  
 6 6 
1 ^ ^ ^

(C) 11 i  10 j  2 k   1 
15   (C)  0, 6 (D)  6, 
 5
1 ^ ^ ^

(D) 11 i  10 j  2 k  6. Let x 2  3 y 2  3 be the equation of an ellipse
15  
2. A man travelling east at 8km/h finds that the in the xy-plane. A and B are two points whose
wind seems to blow directly from the north. ^ ^

On doubling the speed, he finds that it position vectors are  3 i and  3 i  2 k
Narayana Junior Colleges




appears to come from the north-east. The then the position vector of a point P on the
velocity of the wind is ellipse such that APB   / 4 is




(A)   450 , v  8 2 (B)   450 , v  8 2
(C)   900 , v  4 2 (D)   600 , v  8 2
3. ABCD is a parallelogram . If L and M be
the middle points of BC and CD,
 
respectively express AL and AM in terms ^ ^ ^ 
     (A)  j (B)  i  j 

of AB and AD . If AL  AM  K . AC then  
the value of K is ^ 
(A) 1/2 (B) 3/2 (C) 2 (D) 1 (C)  i (D)  k
 
4. If the vectors a and b are linearly  ^ ^ ^  ^ ^ ^
7. If vectors a  3 i  j  2 k , b   i  3 j  4 k
independent satisfying
   ^ ^ ^
   
3 tan   1 a  3 sec   2 b  0 , then and c  4 i  2 j  6 k constitute the sides of
the most general values of  are ABC , then the length of the median

bisecting the vector c

160 Narayana Junior Colleges

,VECTOR ALGEBRA JEE ADVANCED - VOL - I
A) Equilateral triangle B) Isosceles triangle
14
(A) 2 (B) (C) 74 (D) 6 C) Scalene triangle D) Straight line
2 13. Any plane cuts the sides AB, BC, CD, DA
8. Let AC be an arc of a circle, subtending a of a quadrilateral in P, Q, R, S respectively,
right angle at the centre O. The point B
divides the arc AC in the ratio 1:2 . If AP BQ CR
and if  1 ,  2 ,  3 ,
     PB QC RD
OA  a and OB  b , then calculate OC in
  DS
terms of a and b  4 . Then 1234 
SA
A) 1 B) 2 C) 3 D) 4
14. Points X and Y are taken on the sides QR
and RS respectively of a parallelogram
PQRS. So that QX  4 XR and RY  4YS the
PZ
    line XY cuts the line PR at Z then 
(A) 2b  3 a (B) 2b  3 a ZR
    1 21 21 16
(C) 2b  3 a (D) 3b  2 a
A) B) C) D)
9. ABC is a triangle and O any point in the same 4 5 4 3
plane. AO, BO and CO meet the sides BC, 15. Let OABCD is a pentagon in which the side
CA and AB respectively at points D, E and OA and CB are parallel and the sides OD
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OD OE OF and AB are parallel and OA : CB  2 :1 and
F then  
AD BE CF OD : AB  1: 3 . If the diagonal OC and AD
(A) 1 (B) 1/2 (C) 1/3 (D) 2 OX
10. In triangle ABC, A  300 , H is the meet at X then 
XC
orthocenter and D is the mid-point of BC.
Segment HD is produced to T such that 1 1 2 2
HD=DT. The length AT is equal to A) B) C) D)
3 5 3 5
4 1 16. Let a  i  j , b  j  k . If ‘m’ be slope of
A) 2BC B) 3BC C) BC D) BC
3 3
11. The value of  so that the points P, Q, R tangent to the curve y  f  x  at x  1
and S on the sides OA, OB, OC and AB of a
regular tetrahedron OABC are coplanar  3x  4   4
where f   x  2  x  R   
OP 1 OQ 1 OR 1  3x  4   3
when  ,  ,  and
OA 3 OB 2 OC 3 then the vector of magnitude “m” along the

OS
  then bisector of angle between a and b is
AB
1 3 3 2
A)  
2
B)   1 A)
2

i  2 j  k B) 3
i  j  2k 
C)   0 D) for no value of 
12. In triangle ABC, AD and AD' are the 2 i jk
bisectors of the angle A meeting BC in D
C)
3 3

i  2 j  k D) 2
and D ' respectively. A ' is the mid point of 17. If the resultant of two forces is equal in
DD ' ; B ' and C ' are the points on CA and magnitude to one of the components and
perpendicular to it in direction , find the other
AB similarly obtained then A ', B ', C ' forms
component using the vector method
Narayana Junior Colleges 161

, JEE ADVANCED - VOL - I VECTOR ALGEBRA
  
  
a  b .c  
A) 2b  a  2 2 c
c
  

  a  b .c  
B) b  a  2 c
c
  
(A) 1350 (B) 1200 (C) 900 (D) 450 

   a  b .c  


C) b  a   c 2  a D) None
    
18. If the vector   tan  , 1, 2 sin 2  and
b
22. The lar distance from A 1, 4, 2  from the
 
  segment BC where B   2,1, 2  ,
  3 
c   tan  , tan  ,   C  0, 5,1
   are orthogonal
 sin 
 2  3 26 26 3
A) B) C) D) 26
 7 7 7
and a vector a  1,3,sin 2  makes an obtuse
angle with the z-axis then the value of  is 23. If a, b, c are unit vectors such that
1
A)    4n  1   tan 2
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a  2b  3c  3  2 2 , angle between a
B)    4n  2    tan 1 2
and b is  , angle between a and c is 
C)    4n  1   tan 1 2
and angle between b and c varies in
D)    4n  2    tan 1 2
19. Image of the point P with position vector   2 
 2 , 3  then the greatest value of
7iˆ  ˆj  2kˆ in the line whose vector equation
 4 cos   6 cos 
 
is r  9iˆ  5 ˆj  5kˆ   iˆ  3 ˆj  5kˆ has the
position vector A) 2 2  5 B) 2 2  5
A) 9iˆ  5 ˆj  2kˆ B) 9iˆ  5 ˆj  2kˆ C) 2 2  5 D) 42
C) 9i  5 ˆj  2kˆ D) 9iˆ  5 ˆj  2kˆ 24. Let P, Q, R and S be the points on the plane
20. Foot of the perpendicular from the point with position vectors –2 ˆi  ˆj, 4i,3i
ˆ ˆ  3jˆ and
 
P  a  to the line r  b  tc 3iˆ  2ˆj respectively. The quadrilateral
   PQRS must be a [IIT JEE 2010]
 
 a  b .c    a.c   (A) parallelogram, which is neither a rhombus
A) b   2 c B) b    2  c nor a rectangle
c c 
(B) square
   (C) rectangle, but not a square
 
  b  c .c   (D) rhombus, but not a square
C) a   c 2  c D) None 25. Two adjacent sides of a parallelogram ABCD
  
 are given by AB  2iˆ  10ˆj  11kˆ and
21. The reflection of the point P  a  with 
 AD   ˆi  2ˆj  2kˆ . The side AD is rotated by
respect to the line r  b  tc
an acute angle a in the plane of the
162 Narayana Junior Colleges
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