JEE Previous Year Questions With Solutions Of Vector Analysis Part - 2 | IIT JEE | AIEEE | NEET | JEE Main & Advance | AIIMS | WBJEE | KCET | PMT | OTHER JE EXAMS

It is very important to have the idea of any examination to get success in that. The best way to have that idea is to have a look at previous year Questions of that examination. That's why we have collected some previous year Questions with Solutions of JEE Main, JEE Advance, NEET, IIT JEE, AIEEE, AIIMS, WBJEE, EAMCET, Karnataka CET, CPMT, Kerala CET, MP PMT and Other State JE Exams so that you can get clear pattern and level of question that will be asked in forthcoming JEE Main, JEE Advance, NEET, IIT JEE, AIEEE, AIIMS, WBJEE, EAMCET, Karnataka CET, CPMT, Kerala CET, MP PMT and Other State JE Exams.

JEE Previous Year's Questions With Solutions

1. If A x B = B x A, then the angle between A and B is: (AIEEE 2004)

(A) π/2

(B) π/3

(C) π

(D) π/4. 

Sol: (C)

Given: A x B = B x A

Or, (A x B) – (B x A) = 0

Or, (A x B) + (A x B) = 0 [same magnitude & opposite direction]

Or, 2 (A x B) = 0

Or, 2AB Sin θ n = 0

| A| and |B| are non-zero value.

Therefore it is possible if Sin θ = 0 → θ = 0 or π.

2. A force F = 5i + 3j + 2k N is applied over a particle which displaces it from its origin to the point r = 2i – j m. The work done on the particle in joules is: (AIEEE 2004)

(A) + 7

(B) – 7

(C) + 10

(D) + 13. 

Sol: (A)

Displacement, s = (2i – j) – (0i + 0j) = 2i – j m.

Force, F = 5i + 3j + 2k N.

So, work done, W = F.s

Or, W = (5i + 3j + 2k). (2i – j)

Or, W = + 7 J.

3. A particle has an initial velocity of 3i + 4j and acceleration of 0.4i + 0.3j. Its speed after 10s is: (AIEEE 2009)

(A) 7√2 units

(B) 7 units

(C) 8.5 units

(D) 10 units.   

Sol: (D)

Initial velocity, u = 3i + 4j

Magnitude of initial velocity, u = 5 units.

Acceleration, a = 0.4i + 0.3j.

Magnitude of acceleration, a = 0.5 units.

Time, t = 10 s.

We have, v = u + at

Or, v = 5 + 0.5 * 10 = 10 units.

4. A sphere is rolling without slipping on a fixed horizontal plane surface. In the figure, A is the point of contact, B is the centre of the sphere and C is its topmost point. Then (IIT JEE 2009)

(A) V_C – V_A = 2 (V_A – V_C)

(B) V_C – V_B = V_B – V_A

(C) |V_C – V_A| = 2 |V_B – V_C|

(D) |V_C – V_A| = 4 |V_B|. 

Sol: (C)

Let ω be the angular velocity of rotation of sphere V_A, V_B, V_C be the velocities at A, B and C respectively.

Then V_A = 0; V_B = ωr; V_C = 2ωr.

|V_C – V_A| = 2ωr and |V_B – V_C| = ωr.

Therefore, |V_C – V_A| = 2 |V_B – V_C|.

5. Which of the following is a vector quantity? (WBJEE 2007)

(A) Temperature

(B) Flux density

(C) Magnetic field intensity

(D) Time.  

Sol: (C)

Magnetic field intensity is a vector quantity.

6. The angle subtended by the vector A = 4i + 3j + 12k with the x-axis is: (WBJEE 2008)

(A) Sin^-1 (3/13)

(B) Sin^-1 (4/13)

(C) Cos^-1 (3/13)

(D) Cos^-1 (4/13).  

Sol: (D)

A = 4i + 3j + 12k and B = i (x- axis)

→ A.B = (4i + 3j + 12k). (i)

Or, A.B = 4

Magnitude of A = 13

Magnitude of B = 1

We have, Cos θ = (A.B)/AB

Or, Cos θ = 4/13

Or, θ = Cos^-1 (4/13).   

7. A and B are two vectors given by A = 2i + 3j and B = i + j. The magnitude of the components of A along B is: (WBJEE 2009, 2011)

(A) 5/√2

(B) 3/√2

(C) 7/√2

(D) 1/√2.  

Sol: (D)

The magnitude of the components of A along B is

= (A.B)/|B|

= 5/√2.

8. Given, C = A x B and D = B x A. what is the angle between C and D? (WBJEE 2009)

(A) 30⁰

(B) 60⁰

(C) 90⁰

(D) 180⁰.  

Sol: (D)

Since A x B = – B x A [same magnitude & opposite direction]

Or, C = – D 

Therefore, the angle between C and D is 180⁰.

9. The value of ‘λ’ for which the two vectors a = 5i + λj + k and b = i − 2j + k are perpendicular to each other is: (WBJEE 2010)

(A) 2

(B) – 2

(C) 3

(D) – 3

Sol: (C)

a.b = 0 (since two vectors perpendicular to each other)

→ (5i + λj + k).(i − 2j + k) = 0

Or, 5 − 2λ + 1 = 0

Or, λ = 3. 

10. If a + b = c and a + b = c, then the angle included between a and b is: (WBJEE 2010)

(A) 90⁰

(B) 180⁰

(C) 120⁰

(D) Zero

Sol: (D)

Given: a + b = c and a + b = c

Now, c = √ (a² + b² + 2ab cos θ)

Or, (a + b) = √ (a² + b² + 2ab cos θ)

Or, a² + b² + 2ab = a² + b² + 2ab cos θ

Or, cos θ = 1

Or, θ = 0⁰.

11. Three vectors satisfy the relation A.B and A.C = 0, then A is parallel to: (Karnataka CET 2003)

(A) B x C

(B) B

(C) C

(D) B.C

Sol: (A)

Given: A.B = 0, A.C = 0.

→ A is perpendicular to B and C.

And, B x C is perpendicular to the plane containing B and C.

Hence, A is parallel to B x C.

12. Two equal forces (F each) act a point inclined to each other at an angle of 120⁰. The magnitude of their resultant is: (Karnataka CET 2004)

(A) F

(B) F/2

(C) F/4

(D) 2F

Sol: (A)

Resultant = √ (F² + F² + 2.F.F Cos 120)

Resultant = √ (F²) = F.

13. If a and b are two vectors then the value of (a + b) x (a – b) is: (BHU 2002)

(A) (a x b)

(B) – 2 (b x a)

(C) 2 (b x a)

(D) (b x a)

Sol: (C)

Given: (a + b) x (a – b)

= a x a – a x b + b x a – b x b 

= 2 (b x a) [since, a x a = b x b = 0; – a x b = b x a].

14. If |A x B| = √3 (A.B), then the value of |A + B| is: (CBSE 2004)

(A) (A² + B² + AB/√3)^1/2

(B) (A + B)

(C) (A² + B² + √3AB)^1/2

(D) (A² + B² + AB)^1/2

Sol: (C)

Given: |A x B| = √3 (A.B)

Or, AB Sin θ = √3AB Cos θ

Or, Tan θ = √3 → θ = 60⁰.

Now, |A + B| = (A² + B² + 2AB Cos θ)^1/2

Or, |A + B| = (A² + B² + 2AB Cos 60)^1/2

Or, |A + B| = (A² + B² + AB/√3)^1/2.

15. If the scalar and vector products of two vectors A and B are equal in magnitude, then the angle between the two vectors is: (J & K CET 2009)

(A) 180⁰

(B) 90⁰

(C) 360⁰

(D) 45⁰.

Sol: (D)

Given: |A x B| = (A.B)

Or, AB Sin θ = AB Cos θ

Or, Tan θ = 1 → θ = 45⁰.

16. The component of vector A = a_xi + a_yj + a_zk along the direction of (i – j) is: (EAMCET 2008)

(A) (a_x + a_y)

(B) (a_x – a_y)/√2

(C) (a_x – a_y + a_z)
(D) (a_x + a_y + a_z).

Sol: (B)

Unit vector of (i – j) = (i – j)/|(i – j)| = 1/√2 (i – j).

Component of A along (i – j) = A. ((i – j)/|(i – j)|)

Component of A along (i – j) = (a_x – a_y)/√2.

17. Given A = 2i + 3j and B = i + j. The component of vector A along vector B is: (WBJEE 2011)

(A) 1/√2

(B) 3/√2

(C) 5/√2

(D) 7/√2

Sol: (C)

Component of A along B = A. (B/|B|)

Component of A along B = 5/√2.

18. If A = B + C and A, B, C have scalar magnitudes of 5, 4, 3 units respectively then the angle between A and C is: (WBJEE 2012)

(A) cos^-1 (3/5)

(B) cos^-1 (4/5)

(C) π/2

(D) sin^-1 (3/5)

Sol: (A)

Given: A = B + C

Or, 5² = 4² + 3² + 2 * 4 * 3 cos θ

Or, angle between B and C is = 90⁰.

And A is the resultant of B + C.

So, the angle between A and C is = cos^-1 (3/5).

19. Two vectors are given by A = i + 2 j + 2k and B = 3i + 6j + 2k. Another vector C has the same magnitude as B but has the same direction as A. Then which of the following vectors represents C? (WBJEE 2013)

(A) (7/3) (i + 2j + 2k)

(B) (7/3) (i – 2j + 2k)

(C) (7/9) (i – 2j + 2k)

(D) (9/7) (i + 2j + 2k)

Sol: (A)

Magnitude of B = √ (3² + 6² + 2²) = 7.

Unit vectors of A = A/| A| = (i + 2 j + 2k)/3.

So, vector C = (7/3) (i + 2 j + 2k).

Discussion - If you have any Query or Feedback comment below.

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