23

ELEMENTS OF VECTORS

PREVIOUS EAMCET BITS

1. Two persons A and B are located in X-Y plane at the points (0, 0) and (0, 10) respectively. (The distances are measured in M.K.S units). At a time t = 0, they start moving simultaneously with velocities 1

Aˆv 2jms−= and 1

Bˆv 2i ms−= respectively. The time after which A and B are at their

closest distance is [EAMCET 2009 E]

1) 2.5 s 2) 4s 3) 1 s 4) 10 s2

Ans: 1 Sol: Resultant velocity = 2 2 1

A BV V 2 2ms−+ =

Resultant displacement = 10 2m The time after which both A and B are at their closest distance is

10 2displacement 2time 2.5s

velocity 2 2= = =

2. The component of vector x y zˆ ˆ ˆA a i a j a k= + + along the direction of ˆ ˆi j− is [EAMCET 2008 E]

1) x y za a a− + 2) x ya a− 3) x ya a2

− 4) ( )x y za a a+ +

Ans: 3 Sol: Component of x y z

ˆ ˆ ˆa i a j a k+ + along the direction of ˆ ˆi j− is

= ( ) ( )

( ) ( )x y z

2 2

ˆ ˆ ˆ ˆ ˆa i a j a k . i j

1 1

+ + −

+ −= x ya a

2−

Formula : If A and B are two vectors then the (i) component of A a lo ng B is

A.BA cosB

θ =

(ii) component of B along A is A BBcosA−

θ =

3. Velocity and acceleration vectors of charged particle moving perpendicular to the direction of a magnetic field at a given time are ˆ ˆv 2i cj= + and ˆ ˆa 3i 4 j= + respectively, then the value of ‘c’ is [EAMCET 2007 E]

1) 3 2) 1.5 3) – 1.5 4) –3 Ans: 3 Sol: In the problem it is given that acceleration and velocity are perpendicular. ∴ We know that dot product between two perpendicular vectors is zero. ∴ a.v 0= ( ) ( )ˆ ˆ ˆ ˆ3i 4 j . 2i cj 0∴ + + =

Elements of Vectors

24

66 4c 0 c 1.54

−∴ + = ⇒ = = −

4. When a man is standing, rain drops appear to him falling at 60° from the horizontal from his front side. When he is travelling at 5 km per hour on a horizontal road they appear to him falling at 30° , from the horizontal from his front side. The actual speed of the rain in (in km per hour) [EAMCET 2006E]

1) 3 2) 4 3) 5 4) 6 Ans: 3 Sol: From parallelogram law of vectors

R

m R

Qsin 1 V sin 60tanP Qcos V V cos603

θ °α = ⇒ =

+ θ + °

Where Vr is the velocity of rain and Vm is the velocity of man

∴ R

R

3V1 213 5 V2

×⇒ =

+ ×

On solving VR = 4 kmph 5. At a given instant of time the position vector of a particle moving in a circle with a velocity

ˆ ˆ ˆ ˆ ˆ ˆ3i 4 j 5k is i 9 j 8k− + + − . Its angular velocity at that time is [EAMCET 2005 E]

1) ( )ˆ ˆ ˆ13i 29j 31k

146

− − 2)

( )ˆ ˆ ˆ13i 29 j 31k

146

− − 3)

( )ˆ ˆ ˆ13i 29 j 31k

146

+ − 4)

( )ˆ ˆ ˆ13i 29 j 31k

146

+ +

Ans: 2 Sol: From the relation v r= ω× v should be perpendicular to ω and r . So check the condition v. 0 , .r 0ω = ω = [since the dot product of any two perpendicular vectors is zero] On verifying the above conditions the answer is 2 option 6. At a given instant of time two particles are having the position vectors 4 i 4 j 7k− + metres and

2 i 2 j 5k+ + meters respectively. If the velocity of the first particle be 10.4 i ms− , the velocity of second particle in metre per second if they collide after 10 sec is [EAMCET 2004E ]

1) 16 i j k3

⎛ ⎞− +⎜ ⎟⎝ ⎠

2) 10.6 i j k3

⎛ ⎞− +⎜ ⎟⎝ ⎠

3) 16 i j k3

⎛ ⎞+ +⎜ ⎟⎝ ⎠

4) 10.6 i j k3

⎛ ⎞+ −⎜ ⎟⎝ ⎠

Ans : 2 Sol: As both the particles are colliding the distances travelled are same 1 1 2 2s u t s u t+ = +

( ) ( ) ( ) ( )2ˆ ˆ ˆ ˆ ˆ ˆ ˆ4i 4 j 7k 0.4i 2i 2 j 5k u 10⇒ − + + = + + +

2ˆ ˆ ˆ ˆ ˆ ˆ ˆ4i 4 j 7k 4i 2i 2 j 5k 10u⇒ − + + = + + +

2ˆ ˆ ˆ6i 6 j 2k 10u⇒ − + =

2kˆ ˆu 0.6 i j3

⎛ ⎞⇒ = − +⎜ ⎟⎜ ⎟

⎝ ⎠

Elements of Vectors

25

7. Two particles having position vectors ( )1r 3i 5 j= + metre and ( )2r 5i 3 j= − − metres are moving

with velocities ( )1r 4i 3 j= + m/s and ( )2r ai 7 j= + m/s. If they collide after 2 seconds. The value

of ‘a’ is [EAMCET 2003] 1) 2 2) 4 3) 6 4) 8 Ans: 4 Sol: If two particles collide after a time of ‘t’ then they travel equal distances ∴ 1 1 2 2r v t r v t+ = +

( ) ( ) ( ) ( )ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ3i 5j 4i 3j 2 5i 3j ai 4 j 2+ + + = − + + − +

On solving a = 8 8. A proton of velocity ( )3i 2 j+ ms–1 enters a field of magnetic induction ( )2 j 3k+ Tesla. The

acceleration produced in the proton in ms–2 is (Specific charge of proton = 8 10.96 10 Ckg− −× ) [EAMCET 2002]

1) ( )8 ˆ ˆ ˆ0.96 10 6i 4k 9j× + + 2) ( )8 ˆ ˆ ˆ0.96 10 6i 9j 4k× − −

3) ( )8 ˆ ˆ ˆ0.96 10 i j k× − − 4) ( )8 ˆ ˆ ˆ0.96 10 5i 9 j 4k× − −

Ans: 2 Sol: From the relation F q V B⎡ ⎤= ×⎣ ⎦

ma q V B⎡ ⎤= ×⎣ ⎦

qa V Bm

⎡ ⎤∴ = ×⎣ ⎦

∴ ( )ˆ ˆ ˆi j k

ˆ ˆ ˆV B 3 2 0 6i 9 j 4k2 0 3

× = = − −

∴ 8 ˆ ˆ ˆa 0.96 10 6i 9j 4k⎡ ⎤= × − −⎣ ⎦

9. An electron moves with speed 2 x 105 m/s along the positive x-direction in the presence of a magnetic induction B i 4 j 3k= + − (in tesla). The magnitude of the force experienced by the electron in Newtons is (Charge on the electron = 191.6 10 C−× ) [EAMCET 2001 E]

1) 131.18 10−× 2) 131.28 10−× 3) 131.6 10−× 4) 131.72 10−× Ans: 3 Sol: F q V B⎡ ⎤= ×⎣ ⎦

( ) ( )19 5 ˆ ˆ ˆ ˆF 1.6 10 2 10 i i 4 j 3k− ⎡ ⎤= × × × + −⎣ ⎦

( )19 5 5ˆ ˆ1.6 10 8 10 k 6 10 j− ⎡ ⎤= × × + ×⎣ ⎦

19 6 13F 1.6 10 10 1.6 10 N− −∴ = × × = ×

10. The displacement r of a charge Q in an electric field 1 2 3E e i e j e k= + + is r a i b j= + . The work done is [EAMCET 2000 E]

1) ( )1 2Q ae be+ 2) ( ) ( )2 21 2Q ae be+ 3) ( ) 2 2

1 2Q e e a b+ + 4) ( ) 2 21 2Q e e a b+ +

Elements of Vectors

26

Ans: 1 Sol: Work done = F.S Eq.S q E.S⎡ ⎤= = ⎣ ⎦

( )1 2 3ˆ ˆ ˆ ˆ ˆQ e i e j e k ai bj⎡ ⎤= + + +⎣ ⎦

[ ] [ ]1 2 1 2Q e a e b Q ae be= + = +

MEDICAL 11. Two persons A and B are located in X-Y plane at the points (0, 0) and (0, 10) respectively. (The

distances are measured in M.K.S units). At a time t = 0, they start moving simultaneously with velocities 1

Aˆv 2jms−= and 1

Bˆv 2i ms−= respectively. The time after which A and B are at their

closest distance is [EAMCET 2009 M]

1) 2.5 s 2) 4s 3) 1 s 4) 10 s2

Ans: 1 Sol: Resultant velocity = 2 2 1

A BV V 2 2ms−+ =

Resultant displacement = 10 2m The time after which both A and B are at their closest distance is

10 2displacement 2time 2.5s

velocity 2 2= = =

12. Three forces ( ) ( )ˆ ˆ ˆ ˆ ˆ ˆA i j k , B 2i j 3k= + + = + + and C acting on a body to keep it in equilibrium.

Then C is [EAMCET 2008 M] 1) ( )ˆ ˆ3i 4k− + 2) ( )ˆ ˆ4i 3k− + 3) ˆ ˆ3i 4 j+ 4) ˆ ˆ2i 3k+

Ans: 1 Sol: If three forces are acting on a body and keep it in equilibrium then 1 2 3F F F 0+ + =

∴ A B C 0+ + = On solving ( )ˆ ˆC 3i 4k= − +

13. Given two vectors ˆ ˆ ˆA i 2j 3k= + − and ˆ ˆ ˆB 4i 2j 6k= − + . The angel made by ( )A B+ with X-axis is [EAMCET 2007 M]

1) 30° 2) 45° 3) 60° 4) 90° Ans: 2 Sol: ˆ ˆA B 3i 3k+ = + and unit vector along x-axis is i

∴ ( ) ( )

2 2 2

ˆ ˆ ˆ3i 3k . i 3 1cos 453 2 23 3 1

+θ = = = = °

+

14. Of the vectors given below, the parallel vectors are , ˆ ˆA 6i 8j= + , ˆ ˆB 210i 280j= + , ˆ ˆC 5.1i 6.8j= + , ˆ ˆ ˆD 3.6 i 8j 48k= + + [EAMCET 2006 M] 1) A and B 2) A and C 3) A and D 4) C and D Ans: 2

Elements of Vectors

27

Sol: Two vectors A and B are said to be parallel if AB

= constant

∴ ( )17 17ˆ ˆ ˆ ˆC 5.1i 6.8j 6i 8j A20 20

⎡ ⎤= + = + =⎣ ⎦

A and C∴ are parallel 15. Angle (in rad) made by the vector ˆ ˆ3 i j+ with the x-axis [EAMCET 2005 M]

1) 6π 2)

4π 3)

3π 4)

2π

Ans: 1 Sol: Angle made by the vector x y

ˆ ˆA A i A j= + with x- axis is

y

x

A 1tanA 63

πα = = ⇒ α =

16. Two vector Q which has a magnitude of 8 is added to the vector P which lies along the x-axis. The resultant of these two vectors is a third vector R which lies along the y-axis and has a magnitude twice that of P . The magnitude of P is [EAMCET 2004 M]

1) 65

2) 85

3) 125

4) 165

Ans: 2 Sol: In the problem Q = 8 units ………(1) [ ]R 2P given= …………...(2)

Since P is along x-axis and R is along y-axis ∴ P and R are perpendicular vectors ∴ 2 2 2Q P R= + ……………(3)

Sub (1) and (2) in (3) 8P5

=

17. A stationary body of mass 3 kg explodes into three equal pieces. Two of the pieces fly off at right angles to each other, one with a velocity 2 im / s and the other with a velocity 3 jm / s . If the explosion takes place in 10–5 sec. The average force acting on the third piece in Newton is [EAMCET 2003 M]

1) ( ) 52i 3 j 10−+ × 2) ( ) 52i 3 j 10− + × 3) ( ) 53 j 2 i 10+ × 4) ( ) 52i 3 j 10−+ ×

Ans: 2 Sol: According to the law of conservation of momentum ( )1 2 3 3 1 2P P P 0 P P P+ + = ⇒ = − +

( )ˆ ˆ2i 3j= − +

nddpF Newtons II lawdt

⎡ ⎤= ⎣ ⎦

( ) ( ) 5

5

ˆ ˆ2i 3jˆ ˆ2i 3j 10

10−

− += = − + ×

Elements of Vectors

28

18. A boat which has a speed of 13 kmph in still water crosses a river of width 1 km along the shortest possible path in 12 minutes. The velocity of the river water in kmph is [EAMCET 2002M]

1) 12 2) 10 3) 8 4) 6 Ans: 1

Sol: Time taken to cross the river in shortest possible path is = 2 2b r

width of riverv v−

r2r

12 1 v 12kmph60 169 v

= ⇒ =−

19. The unit vector parallel to the resultant of the vectors A 4i 3 j 6k and B i 3 j 8k= + + = − + − is [EAMCET 2000 M]

1) 1 3i 6 j 2k7

⎡ ⎤+ −⎣ ⎦ 2) 1 3i 6 j 2k7

⎡ ⎤+ +⎣ ⎦ 3) 1 3i 6 j 2k49

⎡ ⎤+ +⎣ ⎦ 4) 1 3i 6 j 2k49

⎡ ⎤+ −⎣ ⎦

ans: 1

Sol: The unit vectors parallel to the resultant of A and B is ( )A B

A B

+

+

Resultant of A and B A B= + ˆ ˆ ˆA B 3i 6j 2k+ = + −

∴ Unit vector = ˆ ˆ ˆA B 3i 6j 2k

7A B+ + −

=+

= 1 ˆ ˆ ˆ3i 6j 2k7

⎡ ⎤+ −⎣ ⎦