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.

Tagged With: Previous year Chapter-wise JEE Questions Solutions, Chapter-wise previous year solutions of IIT JEE, AIEEE, Other JEE Exams, HC Verma Solutions, Concepts ofPhysics Volume 1, Concepts of Physics Volume 2, HCVERMA SOLUTION (CHAPTERWISE), JEE Main and advance questions and solutions.

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JEE Previous Year Questions With Solutions Of Vector Analysis Part - 1 | 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. The magnitudes of vectors A, B and C are 3, 4 and 5 units respectively. If A + B = C, the angle between A and B is: (CBSE 1990, CPMT 1997, UPSEAT 1999)

(A) π/2

(B) π/4

(C) Cos^{– 1} 0.6

(D) Tan^{– 1} (7/5) 

Sol: (A)

Magnitude and direction of A + B same as the magnitude and direction of C.

Therefore, √ (A² + B² + 2AB Cos θ) = C

Or, 3² + 4² + 24 Cos θ = 5²

Or, Cos θ = 0/24 = 0

Or, θ = π/2.

2. The angle between A and B is θ. The value of triple product A. (B x A) is: (CBSE 1991)

(A) 0

(B) A²B cos θ

(C) A²B

(D) A²B sin θ

Sol: (A)

Since (B x A) is perpendicular to vector A.

Therefore, A. (B x A) = 0.

3. What is the torque due to a force F = 2i – 3j + 4k N acting at a point r = 3i + 2j + 3k about the origin? (CBSE 1995, 1997, Karnataka CET 1998)

(A) 6i – 6j + 12k

(B) – 6i + 6j + 12k

(C) – 17i – 6j + 13k

(D) 17i – 6j – 13k

Sol: (D)        

We have, torque = r x F

Or, torque = (3i + 2j + 3k) x (2i – 3j + 4k)

Or, torque = 17i – 6j – 13k. 

4. A force vector applied on a mass is represented as F = 6i – 8j + 10k N and accelerates with 1 m/s². What will be the mass of the body in kg? (CBSE 1996)

(A) 20

(B) 2√10

(C) 10

(D) 10√2

Sol: (D)

Magnitude of force, F = √ (6² + (– 8)² + 10²) = 10√2 N.

We have, F = ma

Or, m = F/a = 10√2/1 = 10√2 kg.

5. What is the value of linear velocity, if ω = 3i – 4j + k and r = 5i – 6j + 6k? (CBSE 1999)

(A) 4i – 13j + 6k

(B) 6i + 2j – 3k

(C) – 18i – 13j + 2k

(D) 6i – 2j + 8k

Sol: (C)

We have, v = r x ω

Or, v = (5i – 6j + 6k) x (3i – 4j + k)

Or, v = – 18i – 13j + 2k. 

6. The angle between two vectors – 2i + 3j + k and i + 2j – 4k is: (EAMCET 1990, EAMCET 1995, CPMT 1997)

(A) 90⁰

(B) 180⁰

(C) 0⁰

(D) None of the above.

Sol: (A)

Let, A = – 2i + 3j + k and B = i + 2j – 4k.

→ A.B = (– 2i + 3j + k).( i + 2j – 4k)

Or, A.B = (– 2 + 6 – 4) = 0

Magnitude of A = √14

Magnitude of B = √21

We have, Cos θ = (A.B)/AB

Or, Cos θ = 0

Or, θ = 90⁰.

7. If A and B are perpendicular vectors and vector A = 5i + 7j – 3k and B = 2i + 2j – ak. The value of ‘a’ is: (EAMCET 1991)

(A) 2

(B) – 7

(C) – 8

(D) 8.

Sol: (C)

A.B = 0 (since two vectors perpendicular to each other)

→ (5i + 7j – 3k).( 2i + 2j – ak) = 0

Or, 10 + 14 + 3a = 0

Or, a = – 8.

8. If 0.5i + 0.8j + ck is a unit vector, then c is: (EAMCET 1994)

(A) √0.89

(B) 0.2

(C) 0.3

(D) √0.11.

Sol: (D)

Magnitude of a unit vector is one.

Therefore, √ (0.5² + 0.8² + c²) = 1

Or, 0.5² + 0.8² + c² = 1

Or, c = √0.11. 

9. A vector is represented by 3i + j + 2k. Its length in X-Y plane is: (EAMCET 1994)

(A) 2

(B) √14

(C) √10

(D) √5.

Sol: (C)

The projection of vector in X-Y plane is 3i + j.

So the length of vector in X-Y is √ (3² + 1²) = √10.

10. Two forces F₁ = 5i + 10j – 20k and F₂ = 10i – 5j – 15k act on a single point. The angle between F₁ and F₂ is nearly: (AMU 1995)

(A) 90⁰

(B) 60⁰

(C) 45⁰

(D) 30⁰.

Sol: (C)

→ F₁. F₂ = (5i + 10j – 20k).(10i – 5j – 15k)

Or, F₁. F₂ = (50 – 50 + 300) = 300

Magnitude of F₁ = √525

Magnitude of F₂ = √350

We have, F₁. F₂ = F₁ F₂ Cos θ

Or, 300 = 428 Cos θ

Or, Cos θ = 0.7 → θ = 45⁰.

11. Two adjacent sides of a parallelogram are represented by the two vectors i + 2j + 3k and 3i – 2j + k. What is the area of the parallelogram? (AMU 1997)

(A) 8√3

(B) 8

(C) 3√8

(D) 192.

Sol: (A)

We have,

Area of parallelogram = |a x b|

So, A = |( i + 2j + 3k) x (3i – 2j + k)|

Or, A = |8 i + 8j – 8k |

Or, A = 8√3.

12. Magnitude of vector which comes on addition of two vectors, 6 i + 7j and 3 i + 4j is: (BHU 2000)

(A) √13.2

(B) √136

(C) √160

(D) √202.

Sol: (D)

Vector, V = (6 i + 7j) + (3 i + 4j)

Or, V = 9 i + 11j.

Magnitude of V = √ (9² + 11²) = √202.

13. The position vectors of radius are 2 i + j + k and 2 i – 3j + k while those of linear momentum are 2i + 3j – k. Then the angular momentum is: (BHU 1997)

(A) 4 i – 8j

(B) – 4 i – 8k

(C) 2 i – 4k

(D) 2 i – 4j + 2k.

Sol: (B)

Radius vector, r = (2 i + j + k) – (2 i – 3j + k) = 4j.

Linear momentum, p = 2 i + 3j – k.

We have, angular momentum, L = r x p

Or, L = (4j) x (2 i + 3j – k) = – 4 i – 8k.

14. A particle moves from position r₁ = 3 i + 2j – 6k to position r₂ = 14 i + 13j + 9k under the action of force 4i + j + 3k N. The work done will be: (Punjab PMT 2002, 2003)

(A) 50 J

(B) 75 J

(C) 100 J

(D) 200 J.

Sol: (C)

Displacement, s = (14 i + 13j + 9k) – (3 i + 2j – 6k) = 11 i + 11j + 15k.

Force, F = 4i + j + 3k N.

So, work done, W = F.s 

Or, W = (4i + j + 3k). (11 i + 11j + 15k) 

Or, W = 100 J.

15. If the vectors P = ai + aj + 3k and Q = ai – 2j – k are perpendicular to each other, then the positive value of a is: (AIIMS 2002)

(A) 0

(B) 3

(C) 2

(D) 1.

Sol: (B)

Since P and Q are perpendicular to each other.

→ P.Q = 0

Or, (ai + aj + 3k).( ai – 2j – k) = 0

Or, a² – 2a – 3 = 0

Or, a = – 1, 3.

So, the positive value of a is 3.

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

(A) A + B

(B) (A² + B² + AB/√3)^1/2

(C) (A² + B² + AB)^1/2  

(D) (A² + B² + √3 AB)^1/2.

Sol: (C)

Given: |A x B| = √3 A.B

Or, AB sin θ = √3 AB cos θ

Or, tan θ = √3

Or, θ = 60⁰.

Therefore, |A + B| = (A² + B² + 2AB cos 60⁰)^1/2  

Or, |A + B| = (A² + B² + AB)^1/2.

17. When A.B = – |A|.|B|, then: (Odisha JEE 2003)

(A) A and B act in the same direction

(B) A and B can act in any direction

(C) A and B act in the opposite direction

(D) A and B are perpendicular to each other.

Sol: (C)

Given: A.B = – |A|.|B|

Or, |A|.|B| cos θ = – |A|.|B|

Or, cos θ = – 1

Or, θ = 180⁰.

Therefore, A and B act in the opposite direction.

18. With respect to a rectangular Cartesian coordinate system, three vectors are expressed as a = 4i – j; b = – 3i + 2j; c = – k where i, j, k are unit vectors, along the x, y, z-axis respectively. The unit vector r along the direction of sum of these vector is: (Kerala CET 2003)

(A) r = 1/√3 (i + j – k)

(B) r = 1/3 (i – j + k)

(C) r = 1/√2 (i + j – k)

(D) r = 1/2 (i – j + k)

(E) r = 1/3 (i – j + k).

Sol: (A)

Sum of three vectors = a + b + c = i + j – k.

Magnitude of sum of three vectors = √ [1² + 1² + (– 1)²] = √3.

So, the unit vector r = 1/√3 (i + j – k).

19. If |A + B| = |A| + |B|, then the angle between A and B will be: (CBSE 2001)

(A) 60⁰

(B) 90⁰

(C) 120⁰

(D) 0⁰.

Sol: (D)

Given: |A + B| = |A| + |B|

Or, √ (|A|² + |B|² + 2|A||B| cos θ) = |A| + |B|

Or, |A|² + |B|² + 2|A||B| cos θ = |A|² + |B|² + 2|A||B|

Or, cos θ = 1 → θ = 0⁰.

20. The unit vector parallel to the resultant of the vectors A = 4i + 3j + 6k and B = – i + 3j – 8k is: (EAMCET 2000)

(A) 1/49 (3i + 6j – 2k)

(B) 1/49 (3i – 6j + 2k)

(C) 1/7 (3i + 6j + 2k)

(D) 1/7 (3i + 6j – 2k).

Sol: (D)

Resultant of A and B = A + B

Or, A + B = 3i + 6j – 2k

And, | A + B | = √ [3² + 6² + (– 2)²] = √49 = 7.

The unit vector parallel to the resultant = (A + B)/ (| A + B |).

Therefore, unit vector = 1/7 (3i + 6j – 2k).

21. A body of mass 5 kg starts from the origin with an initial velocity u = 30i + 40j m/s. If a constant force – 6i – 5j N acts on the body, the time in which the y-component of the velocity becomes zero is: (EAMCET 2000)

(A) 20 s

(B) 40 s

(C) 80 s

(D) 5 s.

Sol: (B)

Given: mass, m = 5 kg, y-component of initial velocity, u_y = 40 m/s, y-component of force = – 5 N, y-component of acceleration, a_y = – 5/5 = – 1 m/s², v_y = 0.

We have, v_y – u_y = a_yt

Or, 0 – 40 = – 1 * t

Or, t = 40 s.

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

Tagged With: Previous year Chapter-wise JEE Questions Solutions, Chapter-wise previous year solutions of IIT JEE, AIEEE, Other JEE Exams, HC Verma Solutions, Concepts ofPhysics Volume 1, Concepts of Physics Volume 2, HCVERMA SOLUTION (CHAPTERWISE), JEE Main and advance questions and solutions.

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