SET WISE SOLUTION /TC–2102/12/06/2017
Target IIT-JEE 2019
+1 TEST (VECTORS) SET WISE Solution
Date: 12.06.2017 (Test Code :2102)
Q.No.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
Set A
Set B
Set C
Set D
Q.29/B
Q.29/D
Q.20/B
Q.16/D
Q.5/B
Q.6/D
Q.21/D
Q.3/D
Q.12/B
Q.17/C
Q.10/A
Q.9/C
Q.7/C
Q.2/B
Q.12/A
Q.30/B
Q.6/B
Q.12/D
Q.7/C
Q.24/D
Q.28/B
Q.23/A
Q.27/A
Q.11/D
Q.27/D
Q.25/D
Q.4/B
Q.1/C
Q.20/B
Q.21/C
Q.18/D
Q.25/D
Q.13/B
Q.10/A
Q.5/C
Q.5/D
Q.21/A
Q.1/A
Q.16/C
Q.19/C
Q.2/D
Q.5/A
Q.6/D
Q.21/B
Q.26/A
Q.7/C
Q.15/A
Q.20/A
Q.10/D
Q.27/C
Q.19/C
Q.28/A
Q.19/D
Q.19/D
Q.13/C
Q.17/B
Q.18/B
Q.11/A
Q.9/B
Q.23/A
Q.1/C
Q.13/C
Q.23/B
Q.2/B
Q.16/D
Q.30/D
Q.2/D
Q.8/B
Q.17/C
Q.4/A
Q.25/C
Q.13/B
Q.3/A
Q.16/B
Q.1/A
Q.7/A
Q.15/C
Q.24/B
Q.22/B
Q.4/C
Q.11/A
Q.3/C
Q.11/D
Q.27/C
Q.30/A
Q.15/C
Q.17/B
Q.14/B
Q.14/D
Q.26/A
Q.28/B
Q.12/D
Q.8/C
Q.14/B
Q.8/A
Q.26/A
Q.22/B
Q.28/A
Q.26/D
Q.10/B
Q.24/D
Q.8/D
Q.14/C
Q.29/C
Q.25/C
Q.22/C
Q.3/C
Q.22/A
Q.9/C
Q.18/B
Q.30/C
Q.18/D
Q.23/B
Q.9/A
Q.29/A
Q.15/D
Q.4/A
Q.20/B
Q.24/C
Q.6/A
Page #1
SET WISE SOLUTION /TC–2102/12/06/2017
  
  
a  b  c  0a  b  c
1.
3.
4.
5.
R = 10 N

b 120o

c

a
 


We can find that A B  0 . So A  r B

Horizontal component = 10 sin 30 = 5N

A–BRA+B
27.
6.

c
acute angle 
d
28.


 
A  B is negative because angle b/w A and B is
7.
11.
2
R = (3P) + (2P) + 2  3P  2 P cos 
….(1)
On doubling the first force
(2R)2 = (6P)2 + (2P)2 + 2  6P  2Pcos
2
2
2
4R = 40P + 24P cos …(2)

From (i) & (2)
cos  = –1/2
 = 120

22. Resultant of given vectors is
= (4  8)iˆ  [3  8]jˆ  12iˆ  5jˆ
Dot product of this vector is Zero with the vectors
given is each option. Hence all vectors are
perpendicular to this vector
23. Sum of three angles should be 3600.



B = 20 N
A + B = 16
B2 = R2 + A2
….(1)
and
….(2)
Given R  8 3 N
Solve to get A = 2N, B = 14 N
Coordinates of G : (O, a, a)
a
Coordinates of center of line DC:  a, , 0 
2


a
a

 (a  0)iˆ    a  ˆj  (0  a)kˆ  aiˆ  ˆj  akˆ
2
2


2
 R2 = 13P2 + 12P2cos 
R
 cos 60 o
B
10 1

B 2
Displacement
obtuse (120).
2
Target IIT-JEE 2019

Magnitude =
29.
30.
2
9a 2 3a
a
a     a2 

4
2
2
2
 
Area of parallelogram = | A  B |
 1
ˆ  1 ˆi  1 ˆj  1 kˆ
a  [2iˆ  2ˆj  k]
4
2 2
4
2
2
2

1 1 1
| a |          1
2 2 4

So, a is not a unit vector

 
1
Let b  ˆi  ˆj  kˆ we can see that a . b  0
2
 r 
So a  b
 7 7 7
7 1 1 1  7 
Let c  ˆi  ˆj  kˆ   ˆi  ˆj  kˆ   a
4 4 8
2 2


So c is parallel to a
2
4 
2
24. OC  OA  AB BC
= 30ˆj  20iˆ  20(ˆi)  20(ˆj)  10ˆj
25. Unit vector in the direction of velocity = unit vector
along the given vector v̂ 
26.
 4î  3 ĵ
(4) 2  (3) 2
 
4 3 
î  ĵ  = 12iˆ  9ˆj m/s
v | v | v̂ = 15 
5 
 5
B
60o
R
90o
A
Page #2