mcqs.eprepare.org

Blog

Vector Practice Test 5 ( Vector Product )

0%

1

Report a question

What’s wrong with this question?
You cannot submit an empty report. Please add some details.

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
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50

Vector Practice Test 5 ( Vector Product )

Type Your Name

1 / 50

If A = 2i and B = 3j, then A × B is

A) 6k

B) −6k

C) 6i

D) 6j

Standard unit vectors follow i × j = k.
Here A × B = (2i) × (3j)
= 6 (i × j).
So A × B = 6k.

2 / 50

If we have two vectors A and B and |A| ≠ |B| and |A × B| = |A · B|, then:

A) A & B are perpendicular

B) A & B are parallel

C) A and B are anti-parallel

D) A is inclined to B at an angle of 45°

Explanation:

3 / 50

What is the torque of a force F = (2î – 3ĵ + 4k̂) newton acting at a point r = (3î + 2ĵ + 3k̂) metre about the origin?
(Given: τ = r × F)

A) 6î – 6ĵ + 12k̂

B) 17î – 6ĵ – 13k̂

C) –6î + 6ĵ – 12k̂

D) –17î + 6ĵ – 13k̂

Explanation:

4 / 50

The vector product of two parallel non-zero vectors is

A) Maximum

B) Equal to |A||B|

C) Zero vector

D) A unit vector

For parallel vectors, the angle between them is 0°.
Sine of 0° is zero.
So |A × B| = |A||B| sin0° = 0, meaning the cross product is the zero vector.

5 / 50

If A = 1î + 2ĵ + 3k̂ and B = 2î + 4ĵ + 6k̂, then A × B = ?

A) 0

B) (–3î + 6ĵ – 3k̂)

C) (3î – 6ĵ + 3k̂)

D) (1î + 2ĵ + 3k̂)

Explanation:

B is 2·A (parallel to A), so A × B = 0 for any parallel vectors.

6 / 50

Which one of the following statements about vector product is correct?

A) Result of A × B is always a scalar

B) Result of A × B is always a vector

C) Result of A × B is zero only when A and B are perpendicular

D) Result of A × B is independent of angle between A and B

Cross product is defined to give a new vector quantity.
This new vector is perpendicular to the plane containing the original two vectors.
So the result of A × B is always a vector, not a scalar.

7 / 50

The vector product (cross product) of two vectors A and B is defined as a vector whose magnitude is

A) |A||B| cosθ

B) |A||B| sinθ

C) |A| / |B|

D) |A| + |B|

For vectors A and B with angle θ between them, the cross product magnitude is given by |A × B|.
By definition, |A × B| = |A||B| sinθ.
So the magnitude depends on both magnitudes and sine of the included angle.

8 / 50

The scalar and vector product of two vectors, a = (3î – 4ĵ + 5k̂) and b = (–2î + ĵ – 3k̂) is equal to:

A) –25 & (7î – ĵ – 5k̂)

B) 25 & (–7î + ĵ – 5k̂)

C) 0 & (–7î + ĵ + 3k̂)

D) –25 & (–7î + ĵ + 5k̂)

Explanation:

9 / 50

If A × B = 0, which of the following must be true?

A) A and B are perpendicular

B) A and B are parallel or one vector is zero

C) Angle between A and B is 90°

D) Vectors have equal magnitudes

Explanation:

Cross product is zero if vectors are parallel (θ = 0° or 180°) or if either vector is a zero vector.
Because sin0° = sin180° = 0.

10 / 50

If A is a non-zero vector and λ is a scalar, then A × (λA) equals

A) λA²

B) Zero vector

C) A

D) λA

Vector A and λA have the same or opposite direction depending on sign of λ.
So they are parallel or antiparallel vectors.
Cross product of parallel vectors is zero, hence A × (λA) is zero vector.

11 / 50

If A × B = 0 for non-zero vectors A and B, then

A) A and B must be perpendicular

B) A and B must be parallel or antiparallel

C) A and B must have equal magnitudes

D) A and B must be unit vectors

Cross product magnitude is |A × B| = |A||B| sinθ.
If A and B are non-zero and A × B = 0, then sinθ must be zero.
Sine becomes zero at θ = 0° or 180°, meaning the vectors are parallel or antiparallel.

12 / 50

The direction of A × B is

A) Parallel to A

B) Parallel to B

C) Along A + B

D) Perpendicular to the plane containing A and B

Cross product produces a vector that is normal to the plane of the two vectors.
This new vector is at right angle to both A and B.
So the direction of A × B is perpendicular to the plane containing A and B.

13 / 50

The vector product of two vectors is maximum when the angle between them is

A) 0°

B) 45°

C) 90°

D) 180°

Magnitude of cross product is |A × B| = |A||B| sinθ.
This is maximum when sinθ is maximum.
Sine has maximum value 1 at θ = 90°, so cross product is maximum at 90°.

14 / 50

The magnitude of torque τ = r × F is equal to

A) rF cosθ

B) rF sinθ

C) r + F

D) r − F

Torque magnitude is |τ| = |r × F|.
By cross product definition, |r × F| = r F sinθ.
Here θ is the angle between r and F.

15 / 50

A and B are two vectors and θ is the angle between them. If
|A × B| = √3 (A · B),
then the value of θ will be:

A) 60°

B) 45°

C) 30°

D) 90°

Explanation:

|A × B| = |A| |B| sin θ
A · B = |A| |B| cos θ

Given:
|A| |B| sin θ = √3 (|A| |B| cos θ)
⇒ sin θ = √3 cos θ
⇒ tan θ = √3
⇒ θ = 60°

16 / 50

If the angle between vectors A and B is θ, the value of the product
(B × A) · A is equal to:

A) zero

B) BA²sin θcos θ

C) BA² cos θ

D) BA² sin θ

Explanation:

The vector (B × A) is perpendicular to both B and A.
So when we take the **dot product** of (B × A) with A, the result is:

(B × A) · A = 0

Reason: dot product of two perpendicular vectors is zero.

17 / 50

If i, j and k are unit vectors along x, y and z axes respectively, then j × i is

A) k

B) −k

C) i

D) −j

We know that i × j = k in right-handed system.
Reversing the order gives j × i = −(i × j).
So j × i = −k.

18 / 50

If A and B are two non-zero vectors such that |A × B| = |A||B|, then the angle between them is

A) 0°

B) 45°

C) 90°

D) 180°

From cross product magnitude, |A × B| = |A||B| sinθ.
Given |A × B| equals |A||B|, so sinθ must be 1.
Sine becomes 1 at θ = 90°, so the vectors are perpendicular.

19 / 50

The SI unit of the vector product of force and position vector (r × F) is the unit of

A) Linear momentum

B) Torque

C) Energy per unit time

D) Pressure

The vector r × F represents torque or moment of force.
Force has unit newton and position vector has unit metre.
So unit of torque is newton metre (N m).

20 / 50

Which of the following correctly expresses that vector product is not associative?

A) (A × B) × C ≠ A × (B × C) in general

B) (A × B) × C = A × (B × C) always

C) A × (B × C) is always zero

D) Cross product cannot be taken for three vectors

Associative law would require (A × B) × C to equal A × (B × C).
For cross product these two expressions usually give different vectors.
Therefore vector product does not satisfy associative law.

21 / 50

Which of the following combinations of two vectors will always give the largest possible magnitude of cross product |A × B| for fixed |A| and |B|?

A) Vectors parallel

B) Vectors perpendicular

C) Vectors forming 45° between them

D) Vectors forming 135° between them

Magnitude |A × B| = |A||B| sinθ depends on sine of angle between them.
It is maximum when sinθ is maximum.
Sine function reaches its maximum value 1 at 90°, so perpendicular vectors give largest |A × B|.

22 / 50

If A and B are two position vectors of adjacent vertices of a triangle, then ½|A × B| represents

A) Perimeter of the triangle

B) Area of the triangle

C) Circumference of a circle

D) Volume of a prism

Magnitude |A × B| gives area of parallelogram formed by A and B.
A triangle formed by the same two vectors has half of this area.
So ½|A × B| represents area of the triangle formed by A and B.

23 / 50

If |A| = 5, |B| = 7 and the angle between A and B is 90°, then |A × B| = ?

A) 35

B) 12

C) 0

D) √74

Explanation:

|A × B| = |A| |B| sin θ
= 5 × 7 × sin 90°
= 35.

24 / 50

The angle between vectors ( A x B ) & (B X A )

A) Zero

B) π

C) π/4

D) π/2

Explanation:

A × B = –(B × A)
So, the two vectors are equal in magnitude but opposite in direction,
hence the angle between them is π radians (180°).

25 / 50

For two vectors A and B, which of the following is correct?

A) |A × B| = |A||B| cosθ

B) |A × B| = |A||B| sinθ

C) |A × B| = |A||B| tanθ

D) |A × B| = |A||B| / sinθ

Magnitude of vector product depends on sine of angle between vectors.
By definition, |A × B| = |A||B| sinθ.
So the cross product magnitude is proportional to sine of the included angle.

26 / 50

Given are two vectors, A = (2î – 5ĵ + 2k̂) and B = (4î – 10ĵ + ck̂).
What should be the value of c so that vector A and B become parallel to each other?

A) 1

B) 2

C) 3

D) 4

Explanation:

27 / 50

Given vectors A = 2i + 3j + k and B = i – j + 4k, what is the vector A × B?

A) 14i + 7j – 7k

B) 13i – 7j – 5k

C) 11i + 10j + 1k

D) 0

Explanation:

Calculate cross product using determinant:

28 / 50

Cross product of two vectors A and B represents which geometrical quantity in magnitude?

A) Area of a square built on A

B) Area of rectangle formed by A and B

C) Area of parallelogram formed by A and B

D) Volume of a cube built on A

Magnitude |A × B| equals |A||B| sinθ.
This is equal to the area of a parallelogram having sides A and B.
So cross product magnitude represents area of the parallelogram formed by two vectors.

29 / 50

Which of the following option is not true, if vector A = 3î + 4ĵ and B = 6î + 8ĵ, where A and B are magnitudes of A and B?

A) A × B = 0

B) A/B=1/2

C) A . B = 48

D) A=5

Explanation:

A · B = 3×6 + 4×8
= 18 + 32 = 50 ❌ (not 48)

30 / 50

In three dimensions, the cross product of A = (Ax, Ay, Az) and B = (Bx, By, Bz) is given symbolically by

A) A × B = (Ax Bx, Ay By, Az Bz)

B) A × B = determinant with i, j, k in first row and components in next rows

C) A × B = Ax Bx + Ay By + Az Bz

D) A × B = (Ax + Bx, Ay + By, Az + Bz)

Vector product in components is written using a 3 × 3 determinant.
First row contains unit vectors i, j, k.
Second and third rows contain components of A and B respectively.
Expanding this determinant gives the components of A × B.

31 / 50

If A, B and C are coplanar vectors, then the vector A × B is

A) Parallel to C

B) Perpendicular to C

C) Perpendicular to the plane containing A and B (and hence C)

D) Zero only

If A, B and C are coplanar, they all lie in the same plane.
Vector A × B is perpendicular to the plane containing A and B.
Since C is also in that plane, A × B is perpendicular to C as well.

32 / 50

The direction of the vector A × B is given by:

A) Left-hand rule

B) Right-hand rule

C) Vector addition rule

D) None of the above

Explanation:

The direction of the cross product vector is given by the right-hand rule: if fingers point along A and curl towards B, then the thumb points in the direction of A × B.

33 / 50

The right-hand rule for A × B states that

A) Fingers point along A, curl towards B, thumb gives direction of A × B

B) Fingers point along B, curl towards A, thumb gives direction of A × B

C) Thumb points along A, fingers along B, palm gives A × B

D) Thumb points opposite to A × B always

In the right-hand rule, you first point the fingers of your right hand along A.
Then curl the fingers towards B through the smaller angle between them.
Your extended thumb then points in the direction of A × B.

34 / 50

The value of the unit vector which is perpendicular to both
A = î + 2ĵ + 3k̂ and
B = î – 2ĵ – 3k̂
is equal to:

A) (î + 2ĵ + 3k̂) / 6

B) (6ĵ – 4k̂) / √52

C) (6ĵ + 4k̂) / √52

D) (2î – ĵ) / √5

Explanation:

Its magnitude = √(6² + 4²)
= √(36 + 16) = √52

35 / 50

If A, B and C are mutually perpendicular unit vectors, then (A × B) · C equals

A) 1

B) 2

C) −1

D) zero

Take A, B and C as i, j and k in a right-handed coordinate system.
We know that i × j = k and k is a unit vector.
Dot product of k with itself is 1.
So (A × B) · C equals 1.

36 / 50

The cross product of two vectors A and B is:

A) A scalar quantity

B) A vector perpendicular to both A and B

C) A vector parallel to A

D) Zero vector always

Explanation:

The cross product A × B produces a vector that is perpendicular to both A and B following the right-hand rule. Its magnitude equals |A||B|sinθ, where θ is the angle between A and B.

37 / 50

The vector product î × k̂ equals:

A) î

B) –ĵ

C) ĵ

D) k̂

Explanation:

Using right-hand rule / cyclic order î → ĵ → k̂, we have î × k̂ = –(k̂ × î) = –ĵ.

38 / 50

A student calculates the vector product of two vectors and gets a scalar answer. This indicates that

A) The vectors were parallel

B) The vectors were perpendicular

C) He mistakenly used the formula of dot product instead of cross product

D) His answer is correct for all cases

Cross product of two vectors should give a vector quantity.
Dot product gives a scalar result.
If the answer is scalar while calculating cross product, it means dot product formula was used by mistake.

39 / 50

Which algebraic property is NOT satisfied by vector product?

A) Distributive law over addition

B) Associative law for scalar multiplication

C) Commutative law A × B = B × A

D) Existence of zero result for parallel vectors

Cross product satisfies A × (B + C) = A × B + A × C, so distributive law holds.
It also obeys (kA) × B = k(A × B), so scalar multiplication behaves well.
But it is not commutative, because B × A = −(A × B).
So the commutative law does not hold for vector product.

40 / 50

Which of the following is distributive over vector addition?

A) A × (B + C) = A × B + A × C

B) A × (B + C) = A × B − A × C

C) A × (B + C) = B × C

D) A × (B + C) = 0

The cross product obeys distributive law over addition.
So A × (B + C) expands as A × B plus A × C.
Therefore A × (B + C) = A × B + A × C.

41 / 50

Which of the following is true about the vector product of two vectors?

A) It is commutative, A × B = B × A

B) It is anti-commutative, A × B = −B × A

C) It is associative with respect to scalar multiplication only

D) It always gives a scalar result

Swapping the order in a cross product changes the sign of the result.
So B × A equals −(A × B).
This property is called anti-commutativity of the vector product.

42 / 50

Torque τ acting on a particle is defined as

A) τ = F × r

B) τ = r × F

C) τ = r · F

D) τ = F · r

Torque measures turning effect of a force about a point.
By definition, torque is given by cross product of position vector r and force F.
So τ = r × F.

43 / 50

If A = 3i + 4j and B = i, then A × B is

A) 4k

B) −4k

C) 3k

D) −3k

Compute A × B = (3i + 4j) × i.
This equals 3(i × i) + 4(j × i).
i × i is zero and j × i = −k.
So A × B = 0 + 4(−k) = −4k.

44 / 50

If A × B is a zero vector, then which statement is correct?

A) Either A or B is zero or they are parallel

B) A and B must be perpendicular non-zero vectors

C) A and B must have equal magnitudes only

D) A and B must form 60° angle

Cross product magnitude is |A × B| = |A||B| sinθ.
If |A × B| is zero, then either |A| or |B| is zero or sinθ is zero.
Sinθ is zero when vectors are parallel or antiparallel.
So cross product is zero if one vector is zero or the vectors are parallel.

45 / 50

The cross product of two vectors is closely related to which physical quantity in magnetic field problems?

A) Electric potential

B) Magnetic flux

C) Magnetic force on a moving charge

D) Electrostatic force

Magnetic force on a moving charge is given by F = q v × B.
Here v is velocity vector and B is magnetic field vector.
This force is a vector product of charge velocity and magnetic field.

46 / 50

If two vectors a = 2î + ĵ and b = 3î + 2ĵ, then |a × b| = ?

A) 1

B) √65

C) 8

D) 4

Explanation:

47 / 50

The linear velocity of a rotating body is given by
v = ω × r, where ω is the angular velocity and r is the radius vector.
If ω = î − 2ĵ + 2k̂ and r = 4ĵ − 3k̂ then the value of |v| will be:

A) √29 units

B) 31 units

C) √37 units

D) √41 units

Explanation:

= î·[(-2)(-3) – (2·4)]
– ĵ·[1·(-3) – (2·0)]
+ k̂·[1·4 – (-2·0)]

= î·(6 – 8)
– ĵ·(-3 – 0)
+ k̂·(4 – 0)

= (–2)î + 3ĵ + 4k̂

|v| = √[(-2)² + 3² + 4²]
= √[4 + 9 + 16]
= √29

48 / 50

If two vectors A and B are perpendicular and have magnitudes 4 and 6 respectively, what is the magnitude of A × B?

A) 0

B) 10

C) 24

D) 12

Explanation:

Magnitude of cross product = |A||B|sinθ
Since vectors are perpendicular, θ = 90°, sin90° = 1.
So magnitude
= 4 × 6 × 1 = 24.

49 / 50

If for two vectors A and B, A × B = 0, then the vectors:

A) are perpendicular to each other.

B) are parallel to each other.

C) act at an angle of 60°.

D) act at an angle of 30°.

Explanation:

Two vectors will be parallel to each other.

50 / 50

What is the magnitude of the cross product A × B when A = 5 units, B = 3 units, and the angle between them is 90°?

A) 15

B) 8

C) 0

D) 2

Explanation:

Magnitude of cross product = |A||B|sinθ
= 5 × 3 × sin90°
= 15 × 1 = 15.

Your score is
The average score is 0%
0%

December 5, 2025
Vectors Practice Test 3 ( Resultant of Vector)

0%

1

Report a question

What’s wrong with this question?
You cannot submit an empty report. Please add some details.

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
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57

Vectors Practice Test 3 ( Resultant of Vector)

Type Your Name

1 / 57

Given A = 3i + 4j – 5k and B = 1i – 2j + 2k, the magnitude of R = A – 2B is:

A) √82

B) √100

C) √146

D) √164

Explanation:

Vector 2B
= 2 × (1 i – 2 j + 2 k)
= (2 i – 4 j + 4 k)

Resultant R
= A – 2B
= (3 i + 4 j – 5 k) – (2 i – 4 j + 4 k)
= (3 – 2) i + (4 – (-4)) j + (-5 – 4) k
= 1 i + 8 j – 9 k

|R| = √(1² + 8² + (-9)²)
= √(1 + 64 + 81)
= √146

2 / 57

Vectors A = 4i − 2j and B = −i + 5j act at a point. The direction of the resultant R = A + B, measured from the positive x-axis, is:

A) 30°

B) 45°

C) 60°

D) 90°

Explanation:

Add components:
Rₓ = 4 + (−1) = 3;
Rᵧ = −2 + 5 = 3.
Thus
tanθ = Rᵧ/Rₓ
= 3/3 = 1,
giving θ = 45°.

3 / 57

Three mutually perpendicular vectors of magnitudes 2, 3 and 6 units act simultaneously at a point. The magnitude of their resultant is:

A) √11

B) 7

C) √7

D) 49

Explanation:

Since the vectors are at right angles, square the magnitudes and sum:
R² = 2² + 3² + 6²
= 4 + 9 + 36
= 49, hence
R = √49 = 7.

4 / 57

Two vectors of magnitudes 5 units and 7 units act with an angle of 60° between them. The magnitude of their resultant is:

A) √45

B) √71

C) √109

D) √114

Explanation:

By the parallelogram law,
R² = 5² + 7² + 2·5·7·cos 60°.
Since cos 60° = 0.5,
R² = 25 + 49 + 35
= 109,
so R = √109.

5 / 57

Consider vectors A = 3i + 4j and B = 2i − j. The magnitude of the resultant R = A + B is:

A) √19

B) √34

C) √29

D) √14

Explanation:

First compute the components:
Rₓ = 3 + 2 = 5, Rᵧ
= 4 + (−1) = 3.
The magnitude is
√(Rₓ² + Rᵧ²)
= √(5² + 3²)
= √(25 + 9)
= √34.

6 / 57

For the figure –

A) A+B=C

B) B+C=A

C) C+A=B

D) A+B+C=0

Explanation:

Use head to tail rule:

7 / 57

If the angle between the unit vectors â and b̂ is 60°, then |â − b̂| is:

A) 0

B) 1

C) 2

D) 4

Explanation:

8 / 57

What displacement must be added to the displacement
25 î − 6 ĵ m to give a displacement of 7.0 m pointing in the x-direction?

A) 18 î − 6 ĵ

B) 32 î − 13 ĵ

C) −18 î + 6 ĵ

D) −25 î + 13 ĵ

Explanation:

What displacement must be added to the displacement
25 î − 6 ĵ m to give a displacement of 7.0 m pointing in the x-direction?

Solution:

Let the required displacement vector be:
B = x î + y ĵ

We are given:
A = 25 î − 6 ĵ
Resultant R = 7 î + 0 ĵ

We know:
A + B = R

So:
(25 + x) î + (−6 + y) ĵ = 7 î + 0 ĵ

Now equate components:

i-component:
25 + x = 7 → x = 7 − 25 = −18

j-component:
−6 + y = 0 → y = 6

So the required displacement is:
B = −18 î + 6 ĵ

9 / 57

Two forces, F1 and F2 are acting on a body. One force is doubled of the other force and the resultant is equal to the greater force. Then the angle between the two forces is –

A) cos⁻¹(1/2)

B) cos⁻¹(−1/2)

C) cos⁻¹(−1/4)

D) cos⁻¹(1/4)

Explanation:

10 / 57

A truck travelling due north with 20m/s turns towards west and travels at the same speed. Then the change in velocity is –

A) 40 m/s north-west

B) 20√2 m/s north-west

C) 40 m/s south-west

D) 20√2 m/s south-west

Explanation:

11 / 57

The three vectors OA, OB and OC have the same magnitude R. Then the sum of these vectors have magnitude –

A) R

B) √2 R

C) 3R

D) ( 1 + √ 2 )R

Explanation:

12 / 57

A force of 6 N and another of 8 N can be applied together to produce the effect of a single force of –

A) 1 N

B) 11 N

C) 15 N

D) 20 N

Explanation:

To find the resultant of two forces, we use vector addition. The resultant depends on the angle between them.

Maximum resultant (when both forces act in the same direction, angle = 0°):
R = 6 + 8 = 14 N

Minimum resultant (when both forces act in opposite directions, angle = 180°):
R = |8 – 6| = 2 N

Therefore, the resultant of 6 N and 8 N can be any value between 2 N and 14 N depending on the angle.

13 / 57

Which of the sets given below may represent the magnitude of resultant of three vectors adding to zero?

A) 2, 4, 8

B) 4, 8, 16

C) 1, 2, 1

D) 0.5, 1, 2

Explanation:

Largest ≤ sum of other two

If three vectors add to zero, they must form a triangle. According to the triangle inequality, the sum of the magnitudes of any two vectors must be greater than or equal to the third.

Let’s check each option:

Option (1): 2, 4, 8
→ 2 + 4 = 6 < 8 → ❌ Cannot form triangle
Invalid

Option (2): 4, 8, 16
→ 4 + 8 = 12 < 16 → ❌ Cannot form triangle
Invalid

Option (3): 1, 2, 1
→ 1 + 2 = 3 > 1 ✅
→ 2 + 1 = 3 > 1 ✅
→ 1 + 1 = 2 = 2 ✅
→ This is a degenerate triangle (straight line) → Resultant = 0

14 / 57

In the given figure

A) Angle between A and B is 110∘

B) Angle between C and D is 60∘

C) Angle between B and C is 110∘

D) Angle between B and C is 70∘

Explanation:

15 / 57

Two time-dependent vectors are defined as

A = cos t i + sin t j
B = cos (t⁄2) i + sin (t⁄2) j

At what smallest positive value of t (in radians) are A and B perpendicular?

A) π⁄2

B) π

C) 3π⁄2

D) 2π

Explanation:

Orthogonality ⇒ A·B = 0

A·B = cos t · cos (t⁄2) + sin t · sin (t⁄2)
= cos [t − t⁄2]
= cos (t⁄2).

Set cos (t⁄2) = 0
⇒ t⁄2 = π⁄2 + nπ
⇒ t = π + 2nπ.
Smallest positive solution: t = π.

16 / 57

If the magnitude of the sum of two vectors is equal to the magnitude of the difference of the two vectors, the angle between these vectors is:

A) 0∘

B) 90∘

C) 45∘

D) 180∘

Explanation:

17 / 57

Two forces are such that the sum of their magnitudes is 18N and their resultant is perpendicular to the smaller force and the magnitude of the resultant is 12N. Then the magnitudes of the forces will be:

A) 4 N and 14 N

B) 5 N and 13 N

C) 6 N and 12 N

D) 7 N and 11 N

Explanation:

Video Explanation:

<span data-mce-type="bookmark" style="display: inline-block; width: 0px; overflow: hidden; line-height: 0;" class="mce_SELRES_start"></span><br />

18 / 57

Two forces of magnitude F have a resultant of the same magnitude F. The angle between the two forces is

A) 45°

B) 120°

C) 150°

D) 60°

Explanation:

19 / 57

If the angle between the two forces increases, the magnitude of their resultant:

A) Decreases

B) Increases

C) Remains unchanged

D) First decreases, then increases

Explanation:

The angle between two vectors can never be greater than 180^o.

So, on increasing the θ, the magnitude of resultant vectors decreases.

20 / 57

if we have two vectors P and Q and P=Q then which of the following is not correct

A) P̂ = Q̂

B) |P| = |Q|

C) P Q̂ = Q P̂

D) P + Q = P̂ + Q̂

Explanation:

The sum of the vectors cannot be equal to the sum of their unit vectors.

Detailed Explanation:

P = Q implies both vectors have the same magnitude and direction.

Therefore P̂ = Q̂ (unit vectors match) and |P| = |Q|, so options A and B are true.

In option C, since Q̂ = P̂, both sides reduce to P P̂, so it is also true.

Option 4 equates 2P (because P + Q = 2P) with 2 P̂ (sum of two unit vectors). Unless |P| = 1, those vectors are not equal, so statement D is generally false.

21 / 57

If A = B + C and the magnitudes of A, B, C are 5, 4, 3 units respectively, the angle between A and C is:

A) cos⁻¹ (3⁄5)

B) cos⁻¹ (4⁄5)

C) sin⁻¹ (3⁄4)

D) π⁄2

Explanation:

22 / 57

Two forces of the same magnitude are acting on a body in the East and North directions, respectively. If the body remains in equilibrium, then the third force should be applied in the direction of:

A) North-East

B) North-West

C) South-West

D) South-East

Explanation:

The resultant of these two forces would lie in the north-east direction. To balance this resultant, the third force should be in a south-west direction.

23 / 57

For the given figure, which of the following is true?

A) B=A+C

B) A=B+C

C) C=A+B

D) All of these

Explanation:

24 / 57

If the magnitude of the sum of two vectors equals the magnitude of their difference, the angle between the vectors is:

A) 90°

B) 45°

C) 180°

D) 0°

Explanation:

Let A and B be the vectors.
Given |A + B| = |A − B|.

Square both sides:

|A + B|² = |A − B|²
⇒ (A + B)·(A + B) = (A − B)·(A − B)
⇒ |A|² + |B|² + 2 A·B = |A|² + |B|² − 2 A·B

Cancel the common terms |A|² + |B|²:

2 A·B = −2 A·B
⇒ 4 A·B = 0
⇒ A·B = 0

A zero dot product means the vectors are perpendicular, hence the angle between them is 90°.

25 / 57

There are two force vectors, one of 5 N and the other of 12 N. At what angle should the two vectors be added so that the resultant has a magnitude of 17 N, 7 N, and 13 N respectively?

A) 0°, 180° and 90°

B) 0°, 90° and 180°

C) 0°, 90° and 90°

D) 180°, 0° and 90°

Explanation:

For 17 N both the vector should be parallel i.e., the angle between them should be zero.

For 7 N both the vectors should be antiparallel i.e., the angle between them should be
180°.

For 13 N both the vectors should be perpendicular to each other i.e., the angle between them should be
90°.

26 / 57

If | v₁ + v₂ | = | v₁ − v₂ | and v₁, v₂ are non-zero vectors, then:

A) v₁ is parallel to v₂

B) v₁ = v₂

C) v₁ and v₂ are mutually perpendicular

D) | v₁ | = | v₂ |

Explanation:

Hint: Use Resultant of vectors.
Step: Find the relation between v1 and v2

27 / 57

Two forces A and B have a resultant R₁. If B is doubled, the new resultant R₂ is perpendicular to A. Then

A) R₁ = A

B) R₁ = B

C) R₂ = A

D) R₂ = B

Explanation:

Original resultant: R1 = A + B
New resultant after doubling B: R2 = A + 2B

Given that R2 is perpendicular to A:
A · R2 = 0
⟹ A · (A + 2B) = 0

Expand the dot product:
A·A + 2 A·B = 0
⟹ A² + 2ABcosθ = 0
⟹ ABcosθ = −A² / 2 … (✱)

Find |R1|²:
|R1|² = (A + B) · (A + B)
= A² + B² + 2ABcosθ

Substitute (✱):
|R1|² = A² + B² − A² = B²

Therefore |R1| = |B|, so R1 equals B in magnitude and direction.

28 / 57

Three equal forces of magnitude F act at a point and are directed at 120° to each other in the same plane. The magnitude of their resultant is

A) F

B) 3F

C) F√3

D) zero

Explanation:
Draw the three equal forces at 120° to each other head to tail.
They form a closed equilateral triangle or hexagon-like figure when arranged properly.
A closed polygon of vectors indicates that their vector sum is zero.
So the resultant of the three equal forces is zero.

29 / 57

Two equal forces act on a body and their resultant is at 30° to one of them. The angle between the two forces is

A) 30°

B) 60°

C) 90°

D) 120°

Explanation:
For two equal vectors, the resultant lies along the bisector of the angle between them.
If resultant is at 30° to one force, it is also at 30° to the other force.
So the total angle between the two forces is 30° + 30° = 60°.

30 / 57

A boat moves with velocity 15 m/s east and a river current of 8 m/s north acts on it. The magnitude of the resultant velocity is

A) 7 m/s

B) 17 m/s

C) 23 m/s

D) √289 m/s

Explanation:
Boat velocity and current velocity are at right angles to each other.
Take east as x-direction and north as y-direction.
Resultant speed R = √(15² + 8²).
R = √(225 + 64).
R = √289 = 17 m/s.

31 / 57

If two vectors of magnitudes 4 units and 9 units act in the same direction, their resultant has magnitude

A) 5 units

B) 13 units

C) 36 units

D) √97 units

Explanation:
When vectors act in the same direction, their magnitudes add directly.
So resultant magnitude is 4 units + 9 units.
That gives a resultant of 13 units.

32 / 57

Which statement about the resultant of two non-zero vectors is correct?

A) The resultant can be zero if vectors are equal and opposite

B) The resultant is always greater than each vector

C) The resultant is always equal to the difference of magnitudes

D) The resultant is always perpendicular to both vectors

Explanation:
Two non-zero vectors can add to give zero only when they cancel each other.
Cancellation requires equal magnitudes and opposite directions.
In that special case the resultant is zero even though each vector is non-zero.

33 / 57

A cyclist travels 5 km east and then 12 km north. The magnitude of his resultant displacement is

A) 7 km

B) 10 km

C) 13 km

D) 17 km

Explanation:
Take east as x-direction and north as y-direction.
The two displacements are at right angles.
Resultant displacement R = √(5² + 12²).
R = √(25 + 144).
R = √169 = 13 km.

34 / 57

If the magnitude of the sum of two vectors equals the magnitude of their difference, the angle between the vectors is

A) 0°

B) 45°

C) 90°

D) 180°

Explanation:
Let the vectors be A and B with angle θ between them.
Given |A + B| = |A − B| so their squares are also equal.
Using |A ± B|² = A² + B² ± 2AB cosθ
gives A² + B² + 2AB cosθ
= A² + B² − 2AB cosθ.
This simplifies to 4AB cosθ = 0 so cosθ = 0.
Therefore the angle between the vectors is 90°.

35 / 57

Two forces each of 20 N act at 90° to each other. The magnitude of their resultant is

A) 20 N

B) 20√2 N

C) 40 N

D) 10√2 N

Explanation:
The forces are equal and perpendicular.
Using Pythagoras, R = √(20² + 20²).
R = √(400 + 400).
R = √800 = 20√2 N.

36 / 57

For two forces P and Q, the magnitude of the resultant lies between

A) P − Q and P + Q

B) 0 and P + Q

C) |P − Q| and P + Q

D) 0 and P − Q

Explanation:
Maximum resultant occurs for forces in the same direction and equals P + Q.
Minimum resultant occurs for forces in opposite directions and equals |P − Q|.
So the resultant must always lie between |P − Q| and P + Q.

37 / 57

If two forces 7 N and 24 N act at right angles, the magnitude of their resultant is

A) 17 N

B) 25 N

C) 31 N

D) √50 N

Explanation:
Use R = √(A² + B²) for perpendicular forces.
R = √(7² + 24²).
R = √(49 + 576).
R = √625 = 25 N.

38 / 57

Two forces P and Q act at an angle θ. The magnitude of their resultant is equal to zero. This condition is satisfied when

A) P = Q and θ = 0°

B) P = Q and θ = 180°

C) P > Q and θ = 90°

D) P < Q and θ = 0°

Explanation:
Zero resultant means the forces cancel each other exactly.
For that, they must have equal magnitudes and opposite directions.
Equal magnitudes give P = Q and opposite directions correspond to θ = 180°.

39 / 57

A force of 10 N is resolved into two perpendicular components of 6 N and 8 N. When these components are added vectorially, the resultant is

A) 2 N

B) 6 N

C) 8 N

D) 10 N

Explanation:
The components are at right angles to each other.
Resultant of perpendicular components is found using R = √(6² + 8²).
R = √(36 + 64).
R = √100 = 10 N.
So the resultant equals the original 10 N force.

40 / 57

Three forces acting on a body are represented in magnitude and direction by three sides of a triangle taken in order. The resultant of these forces is

A) Maximum

B) Minimum but not zero

C) Equal to one side

D) Zero

Explanation:
When forces are represented by sides of a triangle taken in order, the head of the last meets the tail of the first.
This means the path closes back on itself.
A closed polygon for force vectors indicates that their vector sum is zero.
So the resultant of the three forces is zero.

41 / 57

If two equal forces act at an angle of 60° and their resultant is 10 N, the magnitude of each force is

A) 5 N

B) 10 N

C) 10√3 N

D) 10/√3 N

Explanation:
Let each force have magnitude F and angle between them be 60°.
Using R² = F² + F² + 2F² cos60°
gives R² = 2F² + 2F²(1/2).
This simplifies to R² = 3F².
Given R = 10 N,
so 10² = 3F².
Thus 100 = 3F²
and F² = 100 / 3,
so F = 10 / √3

42 / 57

The direction of the resultant of two non-parallel forces is along

A) The direction of the larger force only

B) The direction of the smaller force only

C) The diagonal of the parallelogram constructed on the two forces

D) Always exactly midway between the two forces

Explanation:
For two forces represented as adjacent sides of a parallelogram, the resultant is shown by the diagonal.
This diagonal is drawn from the common tail of the two forces.
Therefore the direction of the resultant is along that diagonal.

43 / 57

If the resultant of two forces is equal to the larger force, the angle between them is

A) 0°

B) 60°

C) 90°

D) 180°

Explanation:
Let the forces be P and Q with P > Q and angle between them be θ.
Given that resultant equals the larger force P.
Condition R = P means the smaller force Q must cancel part of P.
This is possible only when they act in opposite directions, so θ = 180°.

44 / 57

Two forces of 5 N and 12 N act at 90° to each other. The magnitude of their resultant is

A) 7 N

B) 10 N

C) 13 N

D) 17 N

Explanation:
The forces are perpendicular, so use R = √(A² + B²).
R = √(5² + 12²).
R = √(25 + 144).
R = √169 = 13 N.

45 / 57

If the angle between two equal forces is 120°, then the magnitude of their resultant is equal to

A) Zero

B) Half of either force

C) Either force

D) Twice either force

Explanation:
Let each force have magnitude F and angle between them be 120°.
Using R² = F² + F² + 2F² cos120°
gives R² = 2F² + 2F²(−1/2).
That simplifies to
R² = 2F² − F² = F².
So R = F, which equals the magnitude of either force.

46 / 57

The magnitude R of the resultant of two forces P and Q acting at angle θ is given by

A) R = P + Q

B) R = √(P² + Q² + 2PQ cosθ)

C) R = √(P² + Q² − 2PQ cosθ)

D) R = PQ sinθ

Explanation:
When two vectors act at any angle θ, we use the law of cosines.
This gives R² = P² + Q² + 2PQ cosθ.
Taking square root on both sides gives
R = √(P² + Q² + 2PQ cosθ).

47 / 57

Two forces P and Q have a resultant of magnitude R. The resultant will be minimum when the angle between P and Q is

A) 0°

B) 60°

C) 90°

D) 180°

Explanation:
Minimum resultant occurs when one force tends to cancel the other.
This happens when they act in exactly opposite directions.
Opposite directions correspond to an angle of 180° between P and Q.

48 / 57

Two forces P and Q have a resultant of magnitude R. The resultant will be maximum when the angle between P and Q is

A) 0°

B) 60°

C) 90°

D) 180°

Explanation:
Resultant depends on both magnitudes and angle between the forces.
The cosine formula shows that R is largest when cosθ is maximum.
cosθ is maximum at θ = 0°, meaning both forces act in the same direction.

49 / 57

The minimum possible value of the resultant of two forces of magnitudes 6 N and 8 N is

A) 0 N

B) 2 N

C) 4 N

D) 14 N

Explanation:
Minimum resultant occurs when the forces act in opposite directions.
In that case the effective magnitude is the difference of their magnitudes.
So minimum resultant is |8 N − 6 N| = 2 N.

50 / 57

The maximum possible value of the resultant of two forces of magnitudes 6 N and 8 N is

A) 2 N

B) 10 N

C) 14 N

D) 48 N

Explanation:
Maximum resultant occurs when forces act in the same direction.
In that situation their magnitudes simply add.
So maximum resultant is 6 N + 8 N = 14 N.

51 / 57

Two forces P and Q acting at a right angle have a resultant R. Which relation is correct?

A) R² = P² + Q²

B) R = P + Q

C) R = P − Q

D) R² = P² − Q²

Explanation:
For two vectors at 90° the law of cosines reduces to Pythagoras.
That gives R² = P² + Q².
So the correct relation for the resultant is R² = P² + Q².

52 / 57

Two forces P and P act at right angles. The magnitude of their resultant is

A) P

B) P√2

C) 2P

D) P / √2

Explanation:
The forces are equal and perpendicular to each other.
For perpendicular vectors, resultant is R = √(P² + P²).
R = √(2P²).
R = P√2.

53 / 57

Two forces of 10 N and 10 N act at 60° to each other. The magnitude of their resultant is

A) 10 N

B) 10√3 N

C) 20 N

D) 5√3 N

Explanation:
Both forces have magnitude 10 N and the angle between them is 60°.
Use R = √(A² + B² + 2AB cosθ).
R² = 10² + 10² + 2(10)(10) cos60°.
cos60° = 1/2, so R² = 100 + 100 + 200(1/2).
R² = 200 + 100 = 300, so R = √300 = 10√3 N.

54 / 57

Two forces of equal magnitude F act at an angle of 180° to each other. The magnitude of the resultant is

A) F

B) 2F

C) F / 2

D) zero

Explanation:
When two equal vectors are in opposite directions, their effects cancel.
Mathematically they have the same magnitude but opposite directions.
Adding them gives a net vector of zero magnitude.
So the resultant is zero.

55 / 57

Two equal forces of 5 N act at an angle of 120° between them. The magnitude of the resultant is

A) 0 N

B) 5 N

C) 5√3 N

D) 10 N

Explanation:
Use the formula R = √(A² + B² + 2AB cosθ) for any two vectors.
Here A = B = 5 N and θ = 120°.
cos120° is equal to −1/2.
So R² = 25 + 25 + 2(5)(5)(−1/2).
R² = 50 − 25
= 25, so R = 5 N.

56 / 57

Two forces 6 N and 8 N act at right angles. The magnitude of their resultant is

A) 2 N

B) 10 N

C) 12 N

D) 14 N

Explanation:
The forces are mutually perpendicular.
Resultant of perpendicular vectors is given by R = √(A² + B²).
Here R = √(6² + 8²).
R = √(36 + 64).
R = √100 = 10 N.

57 / 57

Two forces of 3 N east and 4 N north act simultaneously on a body. The magnitude of the resultant force is

A) 1 N

B) 5 N

C) 7 N

D) 12 N

Explanation:
The two forces are at right angles along east and north.
For perpendicular vectors, the resultant magnitude is found by Pythagoras.
R = √(3² + 4²).
R = √(9 + 16).
R = √25 = 5 N.

Your score is
The average score is 2%
0%

December 5, 2025
Vector Practice Test 4 ( Scalar Product )
0%

1

Report a question

What’s wrong with this question?
You cannot submit an empty report. Please add some details.

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
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65

Vector Practice Test 4 ( Scalar Product )

Type Your Name

1 / 65

The angle between the two vectors (–2i + 3j + k) and (i + 2j – 4k) is:

A) 0°

B) 90°

C) 180°

D) 45°

Explanation:

If the dot product is zero the angle will be 90° and if the dot product is 1 the angle will be 0°

2 / 65

If A · B = |A||B|, then the angle between A and B is

A) 0°

B) 45°

C) 90°

D) 180°

Explanation:
From dot product formula, A · B = |A||B| cosθ.
Given A · B equals |A||B|, so cosθ must be 1.
Cosθ is 1 when θ = 0°, so the vectors are in the same direction.

3 / 65

If the dot product of two vectors is negative, the angle between them lies between:

A) 0° – 90°

B) 90° – 180°

C) 0° – 45°

D) 45° – 90°

Explanation:

cosθ < 0 in this range ⇒ A·B < 0

4 / 65

If A · B > 0, then the angle between A and B is

A) Acute

B) Right angle

C) Obtuse

D) Exactly 180°

Explanation:
Sign of the dot product depends on cosθ.
A · B > 0 means cosθ is positive.
Cosθ is positive for angles less than 90°, so the angle is acute.

5 / 65

If A · B = A · C and A ≠ 0, then

A) B = C

B) B – C is perpendicular to A

C) B = –C

D) A is perpendicular to B + C

Explanation:

Start from the given equality:
A · B = A · C

Move all terms to one side:
A · B – A · C = 0

Factor out A using the distributive property of the dot product:
A · (B – C) = 0

A dot product is zero precisely when the two vectors are perpendicular.
Therefore, A is perpendicular to the vector B – C (equivalently, B – C ⟂ A).

6 / 65

The scalar product of two non-zero vectors is zero. Which of the following must be true?

A) Vectors are collinear

B) Vectors are equal in magnitude

C) Vectors are perpendicular

D) Vectors lie in different planes

Explanation:

A·B = |A||B|cosθ.
If A·B = 0 and both A, B ≠ 0,
then
cosθ = 0
⇒ θ = 90°

7 / 65

The scalar product of two unit vectors is 1. The angle between them is:

A) 0°

B) 45°

C) 90°

D) 180°

Explanation:

A·B = |A||B|cosθ
= 1×1×cosθ = 1
⇒ cosθ = 1
⇒ θ = 0°

8 / 65

The scalar product (dot product) of two vectors A and B is defined as

A) A × B

B) |A||B| sinθ

C) |A||B| cosθ

D) |A| / |B|

Explanation:
The dot product is defined using magnitudes of vectors and the angle between them.
If θ is the angle between A and B, then A · B = |A||B| cosθ.
It is called a scalar product because its result is a single number, not a vector.

9 / 65

What is the scalar product of vector A = 2i + 3j and B = -i + 4j?

A) 10

B) 2

C) 6

D) 4

Explanation:

A·B = (2)(-1) + (3)(4)
= -2 + 12 = 10

10 / 65

If two vectors are perpendicular, then their scalar product is:

A) Maximum

B) Minimum

C) Zero

D) One

Explanation:

Scalar product is given by A·B = |A||B|cosθ.
If vectors are perpendicular, θ = 90°, and cos(90°) = 0,
⇒ A·B = 0

11 / 65

Let a = 2î + ĵ – k̂, b = î – 3ĵ + k̂, and c = î + ĵ + k̂. If a · (b + λc) = 0, then λ = ?

A) 0

B) 1

C) -1

D) 2

Explanation:

a · (b + λc)
= a · b + λ(a · c)
= (–2) + λ(2)= 0
⇒ λ = 1

12 / 65

Which of the following statements is incorrect about scalar product?

A) It depends on the angle between vectors

B) It gives a scalar result

C) It is non-commutative

D) It is distributive

Explanation:

This is false. Scalar product is
commutative
⇒ A·B = B·A.

13 / 65

Which of the following statements about scalar product is correct?

A) It is commutative but not distributive over addition

B) It is distributive over addition but not commutative

C) It is both commutative and distributive over vector addition

D) It is neither commutative nor distributive

Explanation:
Commutative property means A · B = B · A, which is true for dot product.
Distributive property means
A · (B + C) = A · B + A · C,
which also holds.
So scalar product is both commutative and distributive over vector addition.

14 / 65

. If |A| = 5, |B| = 12, and A · B = 30, then the angle between A and B is:

A) 0°

B) 60°

C) 90°

D) 120°

Explanation:

A · B = |A||B|cosθ
⇒ 30 = 5×12×cosθ
⇒ cosθ = 0.5
⇒ θ = 60°

15 / 65

Two vectors of equal magnitude P have a scalar product P². The angle between them is

A) 0°

B) 60°

C) 90°

D) 120°

Explanation:
For equal vectors A and B with magnitude P, dot product is A · B = P² cosθ.
Given A · B = P², so P² cosθ = P².
Dividing by P² (non-zero) gives cosθ = 1 and θ = 0°.

16 / 65

The scalar product of vector A with itself is:

A) A

B) 1

C) |A|

D) |A|²

Explanation:

A·A = |A||A|cos0°
= |A|²

17 / 65

If A·B = 0 and neither A nor B is zero, then:

A) A and B are parallel

B) A and B are anti-parallel

C) A and B are perpendicular

D) Angle between them is 0°

Explanation:

A·B = |A||B|cosθ = 0
implies
cosθ = 0,
so θ = 90°

18 / 65

If the scalar projection of vector A on vector B is maximum, then angle between them is:

A) 0°

B) 45°

C) 90°

D) 180°

Explanation:

Scalar projection = |A||cosθ|.
Max when cosθ = 1
⇒ θ = 0°

19 / 65

If the angle between two vectors is obtuse (>90°), then their dot product is:

A) Positive

B) Zero

C) Negative

D) Undefined

Explanation:

cosθ < 0 for obtuse angles ⇒ A·B = |A||B|cosθ becomes negative.

20 / 65

Two vectors A and B are such that |A| = 4, |B| = 4 and A · B = 8. The angle between them is

A) 0°

B) 30°

C) 60°

D) 90°

Explanation:
Apply dot product formula A · B = |A||B| cosθ.
Here 8 = 4 × 4 × cosθ.
So 8 = 16 cosθ and
cosθ = 8 / 16
= 1 / 2.
Angle whose cosine is 1/2 is 60°,
so θ = 60°.

21 / 65

The scalar product of a vector A with itself, A · A, is equal to

A) Zero

B) |A|

C) |A|²

D) 2|A|

Explanation:
In A · A, the angle between the two A vectors is zero.
So A · A = |A||A| cos0°.
Cos0° is 1,
so A · A = |A|².

22 / 65

If vector A = xî + 3ĵ and vector B = 2î + yĵ, and A · B = 0, then the relation between x and y is:

A) xy = –6

B) 2x + 3y = 0

C) x = y

D) x + y = 0

Explanation:

A · B = (x)(2) + (3)(y)
= 0 ⇒ 2x + 3y = 0

23 / 65

A particle moves from position null to (11i + 11j + 15k) due to a uniform force of (4i + j + 3k) N. If the displacement is in m, then the work done will be:

A) 100 J

B) 200 J

C) 300 J

D) 250 J

Explanation:

24 / 65

If two equal vectors of magnitude P have scalar product zero, their angle of separation is

A) 0°

B) 60°

C) 90°

D) 180°

Explanation:
Dot product for equal vectors is P² cosθ.
Given P² cosθ = 0 and P ≠ 0, so cosθ must be zero.
Cosθ is zero at θ = 90°, hence they are perpendicular.

25 / 65

If A · B < 0, then the angle between A and B is

A) Acute

B) Right angle

C) Obtuse (between 90° and 180°)

D) Zero

Explanation:
Dot product A · B = |A||B| cosθ.
Negative dot product means cosθ is negative.
Cosθ is negative when θ lies between 90° and 180°, so the angle is obtuse.

26 / 65

If angle between two vectors is acute, then their dot product is:

A) Zero

B) Negative

C) Positive

D) Depends on magnitude only

Explanation:

• Acute angle (0° – 90°)
→dot product > 0 (positive)
____________
• Right angle (90°)
→dot product = 0
_____________
• Obtuse angle (90° – 180°)
→dot product < 0 (negative)

27 / 65

A and B are two vectors given by A = 2i + 3j and B = i + j. The component of A parallel to B is:

A) (2i − j)

B) (5/2)(i − j)

C) (5/2)(i + j)

D) (3i − 2j)/2

Explanation:

A = 2i + 3j
B = i + j

We need to find the component of A parallel to B (i.e., projection of A on B).

Formula:
Projection of A on B = [(A · B) / |B|²] × B

Step 1: Find dot product A · B
A · B = (2)(1) + (3)(1) = 5

Step 2: Find |B|²
|B|² = 1² + 1² = 2

Step 3: Apply formula
Projection = (5 / 2)(i + j)

28 / 65

The scalar (dot) product of two vectors is a:

A) Vector

B) Scalar

C) Tensor

D) Complex number

Explanation:

Dot product results in a scalar quantity, not a vector.
A·B = |A||B|cosθ is a number, not a direction-based quantity.

29 / 65

For any vectors A and B, (A + B) · (A + B) equals:

A) |A|² + |B|²

B) (A · B)²

C) |A|² + |B|² + 2(A · B)

D) |A – B|²

Explanation:

(A + B)² = A · A + B · B + 2A · B

30 / 65

If the magnitude of a vector A is 10 units, then A · A is

A) 10

B) 20

C) 100

D) zero

Explanation:
Dot product of a vector with itself is equal to square of its magnitude.
So A · A = |A|².
Here |A| = 10,
so A · A = 10²
= 100.

31 / 65

Which of the following is true about scalar product?

A) Commutative: A·B = B·A

B) Distributive: A·(B + C) = A·B + A·C

C) A·A = |A|²

D) All of the above

Explanation:

All listed properties are standard for scalar products.

32 / 65

What is the scalar product of two vectors inclined at 120°, both having magnitude 10?

A) 100

B) -50

C) 50

D) -100

Explanation:

A·B = |A||B|cos(120°)
= 10×10×cos(120°)
= 100×(-0.5)
= -50

33 / 65

The magnitude of the resultant of two vectors of magnitude 3 units and 4 units is 1 unit. What is the value of their dot product?

A) -12 units

B) -7 units

C) -1 units

D) 0

Explanation:

34 / 65

If two non-zero vectors are perpendicular, then their scalar product is

A) Positive

B) Negative

C) Zero

D) Equal to |A||B|

Explanation:
For perpendicular vectors, the angle between them is 90°.
Cos90° is equal to zero.
So A · B = |A||B| cos90° = 0 for non-zero perpendicular vectors.

35 / 65

For vectors A = Ai + Bj and B = Ci + Dj in a plane, the scalar product A · B equals

A) AC + BD

B) AD + BC

C) A C − B D

D) A / C + B / D

Explanation:
Dot product in two dimensions is found by multiplying corresponding components and adding them.
So A · B = A × C for i-components plus B × D for j-components.
Thus A · B
= AC + BD.

36 / 65

The dot product of two vectors depends on

A) Only the magnitudes of the vectors

B) Only the angle between them

C) Magnitudes and the angle between them

D) Only the direction of one vector

Explanation:
Dot product formula is A · B = |A||B| cosθ.
This involves magnitude of A, magnitude of B and cos of the angle between them.
So the dot product depends on both magnitudes and the angle between the vectors.

37 / 65

If A = 3i + 4j and B = 4i – 3j, find the angle between them using scalar product.

A) 90°

B) 180°

C) 45°

D) 0°

Explanation:

A·B = (3)(4) + (4)(–3)
= 12 – 12 = 0
⇒ angle= 90°

38 / 65

Which of the following is a correct expression for work done by a constant force F when it causes displacement s making angle θ with the force?

A) W = F s

B) W = F s cosθ

C) W = F s sinθ

D) W = F / s

Explanation:
Work is defined as the dot product of force and displacement.
So W = F · s.
Using the scalar product definition, W = |F||s| cosθ
= F s cosθ.

39 / 65

Which of the following vector is perpendicular to the vector A = 2i + 3j + 4k

A) i + j + k

B) 4i + 3j – 2k

C) i – 3j + k

D) i + 2j – 2k

Explanation:

40 / 65

If A · B = 0 and |A| = 3, |B| = 4, the angle between A and B is:

A) 0°

B) 45°

C) 90°

D) 180°

Explanation:

A · B = |A||B|cosθ
= 0 ⇒ cosθ
= 0
⇒ θ = 90°

41 / 65

The value of (A + B) · (A – B) is:

A) |A|² + |B|²

B) |A|² – |B|²

C) 2(A · B)

D) |A + B| · |A – B|

Explanation:

(A + B) · (A – B)
= A · A – B · B
= |A|² – |B|²

42 / 65

If A · B = −|A||B|, then A and B are

A) Parallel in same direction

B) Perpendicular

C) Antiparallel (opposite directions)

D) Equal and perpendicular

Explanation:
Dot product A · B = |A||B| cosθ.
Given A · B = −|A||B|, so cosθ = −1.
Cosθ is −1 at θ = 180°, so the vectors are opposite in direction.

43 / 65

If A = aî + bĵ + ck̂ and A · A = 49, then which of the following is true?

A) a + b + c = 7

B) a² + b² + c² = 49

C) a – b + c = 0

D) 3

Explanation:

For any vector A = a î + b ĵ + c k̂, the dot product with itself is
A · A = a² + b² + c².
The MCQs states A · A = 49,
so a² + b² + c² = 49.

44 / 65

If |A| = 5, |B| = 10, and angle between them is 60°, then A·B = ?

A) 50

B) 25

C) 100

D) 75

Explanation:

A·B = |A||B|cosθ
= 5×10×cos(60°)
= 50×0.5 = 25

45 / 65

If A = aî + bĵ and B = bî – aĵ, then A · B is:

A) 0

B) a² + b²

C) a² – b²

D) 2ab

Explanation:

A · B = a·b + b·(–a)
= ab – ab
= 0

46 / 65

Which one of the following is a scalar quantity that can be written as a dot product of two vectors?

A) Torque

B) Momentum

C) Work

D) Angular momentum

Explanation:
Work is equal to the dot product of force and displacement.
Torque and angular momentum are vector quantities written using cross products.
Momentum is also a vector, not a scalar.

47 / 65

The component of a vector A along the direction of a unit vector n̂ is given by

A) A · n̂

B) A × n̂

C) |A||n̂| sinθ

D) |A||n̂|

Explanation:
Component of A along a direction is the projection of A on that direction.
When n̂ is a unit vector along that direction, projection is A · n̂.
This value is a scalar giving how much of A lies along n̂.

48 / 65

When a vector A is resolved into components parallel and perpendicular to a unit vector n̂, the component parallel to n̂ is

A) A − (A · n̂) n̂

B) (A · n̂) n̂

C) A × n̂

D) n̂ / (A · n̂)

Explanation:
Component parallel to n̂ has the same direction as n̂.
Its magnitude is given by the projection A · n̂.
So the parallel component vector is (A · n̂) n̂.

49 / 65

The scalar product of two non-zero vectors is zero. Which statement is true?

A) The vectors must be parallel

B) The vectors must be perpendicular

C) The vectors must have equal magnitudes

D) The vectors must be zero vectors

Explanation:
For non-zero vectors A and B, A · B = |A||B| cosθ.
Given A · B = 0, so cosθ must be zero.
Cosθ is zero when θ = 90°, meaning the vectors are perpendicular.

50 / 65

In terms of components, if A = Ax i + Ay j + Az k and B = Bx i + By j + Bz k, then A · B equals

A) Ax Bx + Ay By + Az Bz

B) Ax By + Ay Bx + Az Bz

C) Ax Bx − Ay By − Az Bz

D) Ax + Ay + Az + Bx + By + Bz

Explanation:
Dot product in three dimensions is obtained by multiplying corresponding components and adding them.
So A · B = Ax Bx for x-components plus Ay By for y-components plus Az Bz for z-components.
Thus A · B = Ax Bx + Ay By + Az Bz.

51 / 65

The SI unit of the scalar product of force and displacement is

A) Newton

B) Joule

C) Pascal

D) Watt

Explanation:
Dot product of force and displacement gives work done.
Unit of force is newton and unit of displacement is metre.
So unit of work is newton metre, which is called joule.

52 / 65

The angle which the vector A = 2i + 3j makes with the y-axis, where i and j are unit vectors along x- and y-axis, respectively, is:

A) cos⁻¹(3/5)

B) cos⁻¹(2/3)

C) tan⁻¹(2/3)

D) sin⁻¹(2/3)

Explanation:

Magnitude of A
|A| = √(2² + 3²) = √13.

Angle θ with the y-axis:

Adjacent side = y-component = 3

Opposite side = x-component = 2

Therefore tan θ = (opposite)/(adjacent) = 2 ⁄ 3.

Hence θ = tan⁻¹ ( 2 ⁄ 3 ).

53 / 65

Let A = 2î – 3ĵ + k̂, B = –î + 4ĵ + 2k̂. Find A · B.

A) 1

B) –4

C) –3

D) -12

Explanation:

A · B = (2)(–1) + (–3)(4) + (1)(2)
= –2 – 12 + 2
= –12

54 / 65

If A = 5i and B = 5j, then A · B is

A) 5

B) 10

C) 25

D) zero

Explanation:
Vector A lies along the x-axis and vector B lies along the y-axis.
These directions are perpendicular to each other.
So A · B = |A||B| cos90° = 0.

55 / 65

If A = 2i − 3j and B = 4i + j, then A · B is

A) 5

B) 7

C) 11

D) 14

Explanation:
Dot product is A · B = Ax Bx + Ay By.
Here Ax = 2, Ay = −3,
Bx = 4 and By = 1.
So A · B = 2 × 4 + (−3) × 1
= 8 − 3
= 5.

56 / 65

If A = 2i + j − k, B = i + 2j + 3k, and C = 6i − 2j − 6k, then the angle between (A + B) and C will be:

A) 30°

B) 45°

C) 60°

D) 90°

Explanation:

Step 1: Find A + B

Add components of A and B:

i-component: 2 + 1 = 3

j-component: 1 + 2 = 3

k-component: -1 + 3 = 2

So,
A + B = 3i + 3j + 2k

First, calculate the dot product:

(A + B) · C = (3)(6) + (3)(-2) + (2)(-6)
= 18 − 6 − 12 = 0

as the dot product is zero angle will be 90°

57 / 65

Two vectors A and B satisfy A·B = |A||B|. The angle between them is:

A) 0°

B) 45°

C) 90°

D) 180°

Explanation:

A·B = |A||B|cosθ.
If A·B = |A||B|
⇒ cosθ = 1
⇒ θ = 0°

58 / 65

The vector sum of two forces is perpendicular to their vector difference. In that case, the forces:

A) are not equal to each other in magnitude.

B) cannot be predicted.

C) are equal to each other.

D) are equal to each other in magnitude.

Explanation:

Let the two force vectors be F₁ and F₂.
Given: (F₁ + F₂) is perpendicular to (F₁ − F₂)
⇒ (F₁ + F₂) · (F₁ − F₂) = 0

Now expand the dot product:
F₁ · F₁ − F₁ · F₂ + F₂ · F₁ − F₂ · F₂ = 0
⇒ |F₁|² − |F₂|² = 0
⇒ |F₁| = |F₂|

59 / 65

Let vector A = aî + bĵ. The angle between A and the x-axis is 60°. Then the value of b/a is:

A) √3

B) 1/√3

C) 1

D) 0

Explanation:

Vector components
A = a î + b ĵ
• x-component = a
• y-component = b

60 / 65

If A · B = 15, |A| = 5 and |B| = 3, then the angle between A and B is

A) 0°

B) 30°

C) 60°

D) 90°

Explanation:
Use A · B = |A||B| cosθ.
Here 15 = 5 × 3 × cosθ.
So 15 = 15 cosθ and cosθ = 1.
Angle whose cosine is 1 is 0°, so vectors are in same direction.

61 / 65

Which of the following is always true for the scalar product of two vectors A and B?

A) A · B is always positive

B) A · B is always negative

C) A · B can be positive, negative or zero

D) A · B is always zero

Explanation:
Dot product A · B = |A||B| cosθ.
Cosθ can be positive, negative or zero depending on θ.
Therefore the scalar product can be positive, negative or zero.

62 / 65

If the angle between two vectors A and B is 60° and |A| = 4, |B| = 5, then A · B is

A) 10

B) 5

C) 8

D) 20

Explanation:
Use dot product formula
A · B = |A||B| cosθ.
Here A · B = 4 × 5 × cos60°.
Cos60° is 1/2, so A · B
= 20 × 1/2
= 10.

63 / 65

If A · B = 0 and A × B ≠ 0, then A and B are

A) Parallel non-zero vectors

B) Perpendicular non-zero vectors

C) Zero vectors

D) Same vectors

Explanation:
A · B = 0 indicates that vectors are perpendicular or one is zero.
A × B ≠ 0 means their cross product has non-zero magnitude, so neither is zero and they are not parallel.
Together these conditions show that A and B are perpendicular non-zero vectors.

64 / 65

If A = 3i + 4j and B = i + 2j, then A · B is

A) 5

B) 11

C) 14

D) 18

Explanation:
Dot product in component form is A · B = Ax Bx + Ay By.
Here Ax = 3, Ay = 4,
Bx = 1 and By = 2.
So A · B = 3 × 1 + 4 × 2
= 3 + 8
= 11.

65 / 65

If A · B = 0 for non-zero vectors A and B, then the angle between them is

A) 0°

B) 45°

C) 90°

D) 180°

Explanation:
Dot product formula is
A · B = |A||B| cosθ.
Given A · B = 0 and both vectors are non-zero, so cosθ must be zero.
Cosθ is zero at θ = 90°, hence the vectors are perpendicular.

Your score is
The average score is 0%
0%
December 5, 2025
Vector Practice Test 2 ( Resolution of Vector )

0%

1

Report a question

What’s wrong with this question?
You cannot submit an empty report. Please add some details.

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
31
32
33
34
35
36
37
38

Vector Practice Test 2 ( Resolution of Vector )

Type Your Name

1 / 38

Which of the following is a correct statement about unit vector along a given direction?

A) It has magnitude zero

B) It has magnitude one and shows direction of a vector

C) It can have any magnitude greater than one

D) It is always along positive x-axis

Explanation:
A unit vector is defined as a vector whose magnitude is exactly one.
It is used only to indicate direction and not strength.
Therefore it has magnitude one and points along a specific direction.

2 / 38

If a vector lies along the x-axis, its y-component is

A) Equal to its x-component

B) Greater than its x-component

C) Zero

D) Negative of its x-component

Explanation:
A vector along the x-axis makes angle 0° with that axis.
Its entire magnitude lies along x, so projection on y-axis is zero.
Hence for a vector along the x-axis the y-component vanishes.

3 / 38

A 10 N force acts at 0° with the x-axis. Its components are

A) 10 N along x and 0 N along y

B) 0 N along x and 10 N along y

C) 5 N along x and 5 N along y

D) 10 N along both x and y

Explanation:
Angle 0° means the force lies exactly along the positive x-axis.
So the whole magnitude is in the x-direction and none in the y-direction.
Therefore components are 10 N along x and 0 N along y.

4 / 38

A vector of magnitude F making angle θ with the x-axis has an x-component equal to

A) F sinθ

B) F cosθ

C) F tanθ

D) F / cosθ

Explanation:
The x-component is the projection of the vector on the x-axis.
In a right triangle, adjacent side to angle θ is given by magnitude times cosθ.
So the x-component of the vector is F cosθ.

5 / 38

If a vector lies along the y-axis, its x-component is

A) Equal to its y-component

B) Zero

C) Negative of its y-component

D) Greater than its y-component

Explanation:
A vector along the y-axis makes angle 90° with the x-axis.
The projection of such a vector on the x-axis is zero because cos90° = 0.
So the x-component of a purely vertical vector is zero.

6 / 38

A force of 25 N acts at 180° to the x-axis. What is the magnitude of its x-component?

A) 0 N

B) 12.5 N

C) 25 N

D) –25 N

Explanation:

The x-component is
Fₓ = F·cos θ
= 25·cos 180°
= 25·(–1) = –25 N,
whose magnitude is 25 N.

7 / 38

A force vector F makes angle 30° with the positive x-axis. Its y-component is

A) F / 2

B) F √3 / 2

C) F

D) Zero

Explanation:
The y-component of a vector at angle 30° is F sin30°.
Sin30° has value 1 / 2.
Therefore the y-component of the force is F / 2.

8 / 38

If a vector has magnitude 13 units and components 5 units along x and 12 units along y, then

A) The components are incorrect

B) The components are correct

C) The components are too small to form this vector

D) The vector must point exactly along x-axis

Explanation:
For perpendicular components, magnitude of the vector is √(Ax² + Ay²).
Using given components,
√(5² + 12²)
= √(25 + 144)
= √169
= 13.
So the components are consistent with the given vector magnitude.

9 / 38

A force of 20 N acts at 60° to the x-axis. What is the magnitude of its x-component?

A) 5 N

B) 10 N

C) 15 N

D) 20 N

Explanation:

The x-component is F_x = F·cos θ
= 20·cos 60°
= 20·0.5
= 10 N.

10 / 38

Resolving a vector into components is especially useful because

A) Scalars are always easier to draw than vectors

B) Components along chosen axes can be added and subtracted separately

C) It changes the magnitude of the original vector

D) It removes the direction information entirely

Explanation:
When vectors are written as components along axes, each component behaves like a simple scalar quantity.
Components along the same axis can be added or subtracted directly.
This simplifies calculation of resultant vectors while preserving full vector information.

11 / 38

Which statement is true when a vector is resolved into components along coordinate axes?

A) The sum of magnitudes of components always equals the magnitude of the vector

B) Each component is always greater than the original vector

C) The components may be smaller but together reproduce the vector exactly

D) Resolution destroys the original direction information

Explanation:
Components along axes are usually smaller than the original vector in magnitude.
However, when combined vectorially, they give back the same original vector.
So resolution does not change the vector, it just expresses it in another convenient form.

12 / 38

Resolution of a vector means

A) Adding two or more vectors

B) Breaking a vector into two or more components

C) Subtracting a scalar from a vector

D) Converting a vector into a scalar

Explanation:
Many problems become simpler when a vector is expressed in terms of simpler parts.
This process of splitting a single vector into parts is called resolution of a vector.
Usually these parts are chosen along convenient directions such as coordinate axes.
So resolution means breaking a vector into components, not adding vectors.

13 / 38

A force of 10 N is acting along the x-axis. Which of the following is the magnitude of its y-component?

A) 1 N

B) 10 N

C) 5 N

D) 0 N

Explanation:

A force entirely along x has no y-component ⇒ 0 N.

14 / 38

A vector of magnitude 20 units makes an angle 60° with the positive x-axis. Its x-component is

A) 10 units

B) 20 units

C) 20 sin60°

D) 5 units

Explanation:
The x-component is given by magnitude multiplied by cos of the angle with x-axis.
Here it is 20 cos60°.

=20 (1/2)
=10 units

15 / 38

In two-dimensional motion, how many independent rectangular components does a vector have?

A) One

B) Two

C) Three

D) Four

Explanation:
Two-dimensional motion takes place in a plane with two perpendicular axes.
A vector in this plane can be completely described by its x- and y-components.
So it has two independent rectangular components.

16 / 38

A child pulls a box with a force of 200N at an angle of 60∘ above the horizontal. Then the horizontal and vertical components of the force will be:

A) 100N, 175 N

B) 86.6 N, 100N

C) 100 N, 86.6 N

D) 100 N, 0 N

Explanation:

17 / 38

Which of the following statements about components is correct?

A) The number of components of a vector depends on its magnitude

B) The original vector is the vector sum of its components

C) The components of a vector are always larger than the vector itself

D) Components cannot be negative

Explanation:
When a vector is resolved, it is expressed as the sum of simpler vectors along chosen directions.
These simpler vectors are called components of the original vector.
If the resolution is correct, their vector sum reproduces exactly the original vector in both magnitude and direction.

18 / 38

A vector A can be written in terms of unit vectors i and j as A = Ax i + Ay j. Here Ax and Ay represent

A) Magnitudes of two separate vectors with no relation to A

B) Scalar components of A along x- and y-axes

C) Angles between A and coordinate axes

D) Unit vectors along some arbitrary directions

Explanation:
Unit vectors i and j point along x- and y-axes respectively.
Ax and Ay are scalar multipliers that give the length of A along each axis.
Hence Ax and Ay are the scalar components of A along x and y directions.

19 / 38

For a vector of magnitude F at angle 45° with x-axis, its x- and y-components are

A) F and F

B) F / √2 and F / √2

C) F √2 and 0

D) 0 and F √2

Explanation:
At 45°, both cosine and sine have the same value 1 / √2.
So x-component is
F cos45°
= F / √2.
Similarly y-component is
F sin45°
= F / √2.
Thus both components have equal magnitude F / √2.

20 / 38

If the components of a vector are Ax = –4 units and Ay = –3 units, the magnitude of the vector is

A) 1 unit

B) 5 units

C) 7 units

D) 12 units

Explanation:
Magnitude of a vector with components Ax and Ay is √(Ax² + Ay²).
Here magnitude is
√((-4)² + (-3)²).
That equals √(16 + 9)
= √25
= 5 units.

21 / 38

A force of 10 N is acting along the x-axis. Which of the following is the magnitude of its y-component?

A) 1 N

B) 5 N

C) 15 N

D) 0 N

Explanation:

A force along the x-axis makes
θ=0° with the x-axis, so
F_y = F·sin θ
= 10·sin 0°
= 10·0 = 0 N.

22 / 38

A force of 8 N acts at 45° to the x-axis. What is the magnitude of its x-component?

A) 4 N

B) 4√2 N

C) 8 N

D) 0 N

Explanation:

Fₓ = F·cos θ
= 8·cos 45°
= 8·(√2/2)
= 4√2 N.

23 / 38

A 10 N force acts at 90° with the x-axis. Its components are

A) 10 N along x and 0 N along y

B) 0 N along x and 10 N along y

C) 5 N along x and 5 N along y

D) 10 N along both x and y

Explanation:
Angle 90° means the force lies along the positive y-axis.
So there is no horizontal component and full magnitude in vertical direction.
Thus components are 0 N along x and 10 N along y.

24 / 38

Which one of the following is a necessary condition for resolving a vector into two components along given directions?

A) The two directions must be parallel

B) The two directions must be perpendicular

C) The two directions must not be collinear

D) The angle between them must be 180°

Explanation:
A vector can be resolved along any two non-collinear directions.
Non-collinear means the two directions are not along the same straight line.
Perpendicular directions are a convenient special case, but not strictly necessary.

25 / 38

If a vector has magnitude F and its y-component is zero, then the angle it makes with the positive x-axis can be

A) 0° or 180°

B) 90° or 270°

C) 45° only

D) 135° only

Explanation:
Zero y-component means the vector lies along the x-axis.
Along positive x-axis the angle is 0°, and along negative x-axis it is 180°.
So for zero y-component, the angle with positive x-axis is either 0° or 180°.

26 / 38

When resolving a weight vector of magnitude W acting vertically downward into components along horizontal and vertical directions, the horizontal component is

A) W

B) 2W

C) 4W

D) Zero

Explanation:
Weight acts along the vertical direction.
Horizontal direction is perpendicular to the direction of weight.
Projection of a vector on a perpendicular direction is zero.
So horizontal component of weight is zero.

27 / 38

A force vector F makes angle 30° with the positive x-axis. Its x-component is

A) F / 2

B) F √3 / 2

C) F

D) Zero

Explanation:
For a vector at angle 30°, x-component is F cos30°.
cos30° is √3 / 2.
So the x-component of the force is F √3 / 2.

28 / 38

If a vector makes an angle θ with the positive x-axis and its x-component is negative while y-component is positive, the vector lies in

A) First quadrant

B) Second quadrant

C) Third quadrant

D) Fourth quadrant

Explanation:
A negative x-component points towards the left.
A positive y-component points upwards.
Together they place the vector in the second quadrant of the x–y plane.

29 / 38

A vector in the xy-plane has components Ax = 8 units and Ay = 0 units. This vector

A) Is purely horizontal

B) Is purely vertical

C) Has equal horizontal and vertical parts

D) Has no direction

Explanation:
A non-zero Ax with Ay equal to zero means there is only horizontal part.
The vector lies completely along the x-axis.
So it is a purely horizontal vector with no vertical component.

30 / 38

A vector of magnitude F making angle θ with the x-axis has a y-component equal to

A) F cosθ

B) F sinθ

C) F / sinθ

D) F tanθ

Explanation:
The y-component is obtained by projecting the vector on the y-axis.
In the right triangle formed, the opposite side to angle θ is F sinθ.
Therefore the y-component of the vector is F sinθ.

31 / 38

The most useful components of a vector in plane problems are usually taken along

A) Any two random directions

B) Mutually perpendicular axes

C) Parallel lines only

D) Circular paths

Explanation:
In two-dimensional motion we often use x- and y-axes at right angles to each other.
Expressing a vector along these perpendicular axes gives two independent components.
These perpendicular components can be handled with simple algebra and Pythagoras.

32 / 38

A force of 15 N acts at 90° to the x-axis. What is the magnitude of its x-component?

A) 0 N

B) 7.5 N

C) 15 N

D) 30 N

Explanation:

Fₓ = F·cos θ
= 15·cos 90°
= 15·0 = 0 N.

33 / 38

The magnitude of a vector whose components are Ax and Ay along two perpendicular axes is

A) Ax + Ay

B) Ax – Ay

C) √(Ax² + Ay²)

D) Ax Ay

Explanation:
Components Ax and Ay act like the legs of a right-angled triangle.
The original vector is represented by the hypotenuse of this triangle.
By Pythagoras, magnitude of the vector is
√(Ax² + Ay²).

34 / 38

If the components of a vector along x- and y-axes are equal and positive, the vector makes an angle

A) 0° with x-axis

B) 30° with x-axis

C) 45° with x-axis

D) 90° with x-axis

Explanation:
Equal positive components mean Ax = Ay and both are greater than zero.
Then
tanθ = Ay / Ax = 1.
So θ = tan⁻¹(1) = 45°,
measured from the x-axis towards the y-axis.

35 / 38

The process of resolving a force at an oblique angle into horizontal and vertical components uses

A) Only sine function for both components

B) Only cosine function for both components

C) Both sine and cosine functions

D) No trigonometric functions

Explanation:
Horizontal component is given by F cosθ when θ is measured from the horizontal.
Vertical component is F sinθ with the same angle definition.
So resolution into horizontal and vertical components uses both sine and cosine.

36 / 38

The direction θ of a vector with components Ax and Ay (Ax > 0) relative to x-axis is given by

A) θ = tan(Ax / Ay)

B) θ = tan(Ay / Ax)

C) θ = tan⁻¹(Ay / Ax)

D) θ = tan⁻¹(Ax Ay)

Explanation:
In the right triangle for components, tanθ equals opposite over adjacent.
Here opposite side is Ay and adjacent side is Ax.
So
tanθ = Ay / Ax
and θ = tan⁻¹(Ay / Ax)
when Ax is positive.

37 / 38

If a vector has components Ax = 0 and Ay < 0, its direction is

A) Along positive x-axis

B) Along negative x-axis

C) Along positive y-axis

D) Along negative y-axis

Explanation:
Component Ax = 0 means there is no horizontal part of the vector.
Ay < 0 means the vertical component points downward. So the vector lies along the negative y-axis.

38 / 38

A force of 12 N acts at 30° to the x-axis. What is the magnitude of its y-component?

A) 3 N

B) 6 N

C) 10 N

D) 12 N

Explanation:

Fᵧ = F·sin θ
= 12·sin 30°
= 12·0.5
= 6 N.

Your score is
The average score is 0%
0%

December 5, 2025
Vector Practice Test 1 ( Addition of Vector )
0%

1

Report a question

What’s wrong with this question?
You cannot submit an empty report. Please add some details.

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
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74

Vector Practice Test 1 ( Addition of Vector )

Type Your Name

1 / 74

Which statement about vector addition is correct?

A) Vector addition depends on the choice of origin.

B) Vector addition ignores direction and adds only magnitudes.

C) Vector addition takes both magnitude and direction into account.

D) Vector addition is valid only in one dimension.

Explanation:
A vector is defined by its magnitude and direction together.
When vectors are added, both their sizes and directions are combined.
So vector addition always takes magnitude and direction into account.

2 / 74

A vector of magnitude 13 units has components 5 units and 12 units along two perpendicular axes. These two components are

A) Added scalarly to get 17

B) Added by Pythagoras to give the original vector

C) Multiplied to get resultant 60

D) Subtracted to get resultant 7

Explanation:
The components are mutually perpendicular.
Magnitude of the original vector is found by
R = √(5² + 12²).
R = √(25 + 144)
= √169
= 13 units.
So the components combine by the Pythagorean relation to give the original vector.

3 / 74

A force of 10 N is resolved into two perpendicular components 6 N and 8 N. When these two components are added back vectorially, the resultant is

A) 2 N

B) 10 N

C) 14 N

D) 48 N

Explanation:
The components are at right angles, so
R = √(6² + 8²).
R = √(36 + 64).
R = √100 = 10 N.
So the components recombine to give the original 10 N.

4 / 74

If three vectors acting on a particle add to give zero resultant, the vectors can be represented graphically as

A) Three arrows starting from a common point and diverging

B) Three arrows placed head-to-tail forming a closed triangle

C) Three arrows parallel to each other

D) Three arrows forming an open polygon

Explanation:
Zero resultant requires that the final head comes back to the initial tail.
When three vectors are arranged head-to-tail and close to form a triangle, this condition is met.
Thus three vectors giving zero sum are represented as a closed triangle.

5 / 74

When resolving a vector into components and adding vectors by components, the resultant vector’s components along x- and y-axes are obtained by

A) Subtracting corresponding components of all vectors

B) Adding corresponding components of all vectors

C) Multiplying all components together

D) Dividing each component by its magnitude

Explanation:
Each vector is first split into its x- and y-components.
All x-components are combined algebraically to get the resultant x-component.
Similarly all y-components are combined to get the resultant y-component.
These two components together define the resultant vector.

6 / 74

Two vectors A and B are represented by two adjacent sides of a parallelogram. The diagonal passing through their common point represents

A) A – B

B) B – A

C) A + B

D) 2(A + B)

Explanation:
In the parallelogram law, A and B are drawn from the same point as adjacent sides.
A parallelogram is completed on these two vectors.
The diagonal from the common tail gives the vector sum A + B.

7 / 74

A particle is acted upon by three displacement vectors that form three sides of a triangle taken in order. The net displacement of the particle is

A) Equal to the longest side

B) Equal to the shortest side

C) Zero

D) Equal to the perimeter of the triangle

Explanation:
When vectors are arranged head-to-tail and form a closed triangle, start and end points coincide.
This means there is no overall change in position.
So the resultant or net displacement is zero.

8 / 74

The polygon law of vector addition is useful for

A) Adding any number of vectors one by one

B) Adding only two vectors at a time

C) Subtracting vectors only

D) Multiplying a vector by a scalar

Explanation:
Polygon law is an extension of the triangle law.
Vectors are placed head-to-tail in sequence to form a polygon.
The side joining the first tail and the last head gives the resultant.
This law allows addition of any number of vectors together.

9 / 74

Which of the following cannot be a possible resultant when two vectors of magnitudes 6 units and 10 units are added?

A) 2 units

B) 4 units

C) 10 units

D) 16 units

Explanation:
Resultant magnitude must lie between |10 – 6| and 10 + 6.
So it must be between 4 and 16 inclusive.
Values 4, 10 and 16 fall in this range and are possible.
A magnitude of 2 units is less than 4, so it is not possible.

10 / 74

Vector addition is

A) Commutative and associative

B) Commutative but not associative

C) Associative but not commutative

D) Neither commutative nor associative

Explanation:
Commutative property means A + B = B + A, which holds for vectors.
Associative property means
(A + B) + C = A + (B + C),
which also holds.
Therefore vector addition is both commutative and associative.

11 / 74

A displacement of 3 km north and 4 km east gives a resultant direction

A) Exactly towards north

B) Exactly towards east

C) North of east

D) East of north

Explanation:
The components are 4 km east and 3 km north.
Resultant lies between north and east.
Larger component is along east, so direction is closer to east.
This is described as a direction east of north.

12 / 74

For two vectors A and B, if |A + B| = |A – B|, then the angle between A and B is

A) 0°

B) 45°

C) 90°

D) 180°

Explanation:
Use the identities
|A ± B|² = A² + B² ± 2AB cosθ.
Given |A + B|² = |A – B|²,
the 2AB cosθ terms must cancel.
That is possible only if cosθ = 0.
So the angle between A and B is 90°.

13 / 74

Two vectors of magnitudes 5 and 5 act at a right angle. The magnitude of their resultant is

A) 5

B) 5√2

C) 10

Explanation:
The two vectors are equal and perpendicular.
For perpendicular vectors,
R = √(A² + B²).
Here R = √(5² + 5²).
R = √(25 + 25)
= √50.
So the magnitude of the resultant is 5√2.

14 / 74

A person walks 6 km east and then 8 km west. The magnitude of his resultant displacement is

A) 0 km

B) 2 km

C) 14 km

D) 10 km

Explanation:
Take east as positive direction and west as negative.
Net displacement = 6 – 8 = –2 km.
Negative sign shows direction is towards west.
Magnitude of displacement is 2 km.

15 / 74

When two vectors are added and the resultant has magnitude zero, the polygon representing these two vectors must be

A) A right-angled triangle

B) A straight line segment with overlapping arrows

C) A closed triangle with zero area

D) A square

Explanation:
Zero resultant means net effect of the two vectors is nothing.
Graphically, one vector must exactly cancel the other.
This appears as two equal arrows on the same line pointing in opposite directions.
So they overlap on a straight line segment with opposite arrows.

16 / 74

For two vectors of magnitudes 7 and 24, the resultant has minimum possible magnitude when the angle between them is

A) 0°

B) 45°

C) 90°

D) 180°

Explanation:
Minimum resultant occurs when vectors act in opposite directions.
Opposite directions correspond to an angle of 180° between them.
So for magnitudes 7 and 24, the least resultant is obtained at 180°.

17 / 74

The minimum possible magnitude of the resultant of two vectors A and B is

A) |A – B|

B) |A + B|

C) AB

D) A² + B²

Explanation:
Resultant is smallest when the vectors point in opposite directions.
Then the effective magnitude is the difference of their magnitudes.
Thus the minimum possible resultant is ||A| – |B||, written as |A – B|.

18 / 74

If R is the resultant of two vectors A and B, then the maximum possible value of R is

A) |A – B|

B) |A + B|

C) A² + B²

D) AB

Explanation:
Resultant is largest when the vectors point in the same direction.
In that case their magnitudes simply add.
So maximum R is |A| + |B|,
which we write as |A + B|.

19 / 74

In the triangle law of vector addition, the two vectors are represented by two sides of a triangle taken in order. The third side taken in the opposite order represents

A) Their difference

B) Their scalar sum

C) Their resultant

D) A zero vector

Explanation:
First vector is drawn from a point, and the second is drawn from the head of the first.
These two sides in order represent the two vectors being added.
The side joining the tail of the first to the head of the second gives the sum.
Thus the third side of the triangle taken in opposite order represents the resultant.

20 / 74

The graphical method that uses a parallelogram to find the sum of two vectors is called

A) Component method

B) Triangle law of vector addition

C) Parallelogram law of vector addition

D) Polygon law

Explanation:
Two vectors are drawn from a common point as adjacent sides of a parallelogram.
A parallelogram is then completed on these two sides.
The diagonal passing through the common tail gives the sum of the vectors.
This construction is known as the parallelogram law of vector addition.

21 / 74

A body is acted upon by two forces 5 N north and 12 N east. The magnitude of the resultant force is

A) 7 N

B) 13 N

C) 17 N

D) 25 N

Explanation:
Forces towards north and east are at right angles.
So we use the Pythagorean relation for perpendicular vectors.
R = √(5² + 12²).
R = √(25 + 144) = √169.
Thus the resultant force has magnitude 13 N.

22 / 74

The condition for two non-zero vectors A and B to give a zero resultant (A + B = 0) is

A) They must be perpendicular and equal in magnitude

B) They must be parallel and have different magnitudes

C) They must be antiparallel and equal in magnitude

D) They must be parallel and perpendicular at the same time

Explanation:
A + B = 0 means that the two vectors cancel each other exactly.
For this to happen, one must be equal in length but opposite in direction to the other.
Geometrically they are collinear but point in opposite directions.
So they are antiparallel with equal magnitudes.

23 / 74

Two vectors 10 N and 6 N act in opposite directions along the same line. The magnitude of their resultant is

A) 4 N

B) 6 N

C) 10 N

D) 16 N

Explanation:
Opposite collinear vectors subtract in magnitude.
Take the larger force direction as positive.
Net magnitude is |10 – 6|.
So the resultant force has magnitude 4 N.

24 / 74

Two vectors A and B are collinear and in the same direction. Their resultant has magnitude

A) |A – B|

B) A + B

C) A – B only

D) √(A² + B²)

Explanation:
Collinear vectors in the same direction simply add in magnitude.
The total effect is obtained by adding their lengths.
So the magnitude of the resultant is |A| + |B|.
This is written as A + B when we talk about magnitudes only.

25 / 74

When two vectors A and B are added head-to-tail, the resultant vector

A) Starts from head of A and ends at head of B

B) Starts from tail of A and ends at head of B

C) Starts from head of B and ends at head of A

D) Starts from tail of B and ends at tail of A

Explanation:
In the head-to-tail method, the tail of B is placed at the head of A.
The free tail is at the start of A and the free head is at the end of B.
The resultant vector is drawn from the free tail to the free head.
So it starts from tail of A and ends at head of B.

26 / 74

The angle between two equal vectors whose resultant has the same magnitude as any one of them is

A) 0°

B) 60°

C) 90°

D) 120°

Explanation:
Let each vector have magnitude A and angle between them be θ.
Resultant magnitude is R = √(A² + A² + 2A² cosθ).
Condition is R = A,
so A² = 2A²(1 + cosθ).
Dividing by A² gives
1 = 2(1 + cosθ),
so 1/2 = 1 + cosθ.
Thus cosθ = –1/2
and θ = 120°.

27 / 74

Two equal vectors of magnitude 5 units act at 60° to each other. The magnitude of their resultant is

A) 2.5 units

B) 5 units

C) 5√3 units

D) 10 units

Explanation:
For any two vectors A and B with angle θ, resultant is R = √(A² + B² + 2AB cosθ).
Here A = B = 5 units and θ = 60°.
So R² = 25 + 25 + 2(5)(5) cos60°.
cos60° = 1/2,
so R² = 50 + 50/2
= 50 + 25 = 75.
Therefore R = √75
= 5√3 units.

28 / 74

Two vectors 6 N and 8 N act at right angles. The magnitude of their resultant is

A) 2 N

B) 10 N

C) 14 N

D) 48 N

Explanation:
The two forces are mutually perpendicular.
For such vectors, resultant R = √(A² + B²).
Here R = √(6² + 8²).
R = √(36 + 64) = √100.
So the magnitude of the resultant is 10 N.

29 / 74

Two displacement vectors 4 m east and 3 m north act on a body. The magnitude of the resultant displacement is

A) 1 m

B) 5 m

C) 7 m

D) 12 m

Explanation:
The two displacements are at right angles to each other.
For perpendicular vectors we use Pythagoras to find the resultant.
R = √(4² + 3²).
R = √(16 + 9) = √25.
So the magnitude of the resultant displacement is 5 m.

30 / 74

For any nonzero scalar k, the vector kA is ______ to A.

A) parallel

B) perpendicular

C) identical

D) zero

Explanation:

Scalar multiples of A always lie along the same line as A, so kA is parallel to A.

31 / 74

The magnitude of the vector 3A is

A) 3|A|

B) |A|

C) |A|/3

D) 0

Explanation:

|3A| = |3|·|A| = 3·|A|

32 / 74

If a vector is multiplied by a scalar k then the resultant may be

A) 0

B) kA

C) –kA

D) all of these

Explanation:

• If k=0, kA = 0 (the zero vector).
• If k>0, kA points in the same direction as A.
• If k<0, kA points opposite to A (–kA).

33 / 74

Two vectors are said to be collinear if the angle between them is

A) 0° or 180°

B) 90°

C) 45°

D) undefined

Explanation:

Collinear vectors lie along the same line, so they point in exactly the same (0°) or exactly opposite (180°) direction

34 / 74

The magnitude of the cross product A × B is given by

A) A + B

B) A B cos θ

C) A B sin θ

D) |A – B|

Explanation:

|A × B| = A B sin θ, with θ the angle between A and B.

35 / 74

The vector having zero magnitude and arbitrary direction is called

A) zero vector

B) null vector

C) unit vector

D) both a & b

Explanation:

A “zero vector” has zero length and its direction is undefined; it is also called the “null vector.”

36 / 74

The cross product A × B of two parallel vectors is:

A) AB

B) 0

C) A – B

D) a vector perpendicular to both

Explanation:

|A × B| = AB sin θ, and sin 0° = 0, so the result is the zero vector.

37 / 74

The dot product A · B of two perpendicular vectors is:

A) AB

B) 0

C) A + B

D) √(A² + B²)

Explanation:

A · B = AB cos θ, and cos 90° = 0, so the result is 0.

38 / 74

A unit vector is defined as a vector whose magnitude is:

A) 0

B) 1

C) equal to the magnitude of the original vector

D) greater than 1

Explanation:

A unit vector has been scaled so its length is exactly 1.

39 / 74

Adding any vector A to the zero vector 0 gives:

A) A

B) 0

C) 2A

D) –A

Explanation:

A + 0 = A by definition of the zero vector.

40 / 74

If two perpendicular vectors of magnitudes A and B are added, the magnitude of the resultant R is:

A) A + B

B) A² + B²

C) √(A² + B²)

D) |A – B|

Explanation:

By Pythagoras’ theorem (since the angle is 90°), |R| = √(A² + B²).

41 / 74

Vector addition is commutative. That means:

A) A + B ≠ B + A

B) (A + B) + C = A + (B + C)

C) A + B = B + A

D) A × B = B × A

Explanation:

“Commutative” means you can swap the order of the vectors without changing the resultant: A + B = B + A.

42 / 74

Addition of vectors obey the law of _____

A) head-to-tail rule

B) simple addition

C) cross product

D) dot product

Explanation:

Graphical vector addition is done by placing the tail of one vector at the head of the other (head-to-tail). This is neither a scalar “simple addition” nor a product operation.

43 / 74

Two forces each of 10 N act at angles of 0° and 90° to the x-axis. What is the magnitude of their resultant?

A) 10 N

B) 14.14 N

C) 20 N

D) 0 N

Explanation:

F₁ at 0° has components (10, 0),
F₂ at 90° has (0, 10).
Resultant R = (10 + 0, 0 + 10)
= (10, 10),
|R| = √(10² + 10²)
= √200
≈ 14.14 N

44 / 74

Unit vector along the vector 4î + 3ĵ is _____

A) (4î + 3ĵ)/5

B) 4î + 3ĵ

C) (4î + 3ĵ)/6

D) (4î + 3ĵ)/10

Explanation:

Unit vector along 4î + 3ĵ , is obtained by dividing the present vector by its magnitude.

The magnitude of the given vector is 5. Hence, the required unit vector is (4î + 3ĵ)/5.

45 / 74

When two vectors in the same direction are added, the magnitude of resulting vector is equal to _______

A) Sum of magnitudes of the vectors

B) Difference of magnitudes of the vectors

C) Product of magnitudes of the vectors

D) Sum of the roots of magnitudes of the vectors

Explanation:

Consider the graphical representation of these two vectors. When one vector is added to the other in the same direction, the lengths will be added.

The resultant vector will bear the resultant length. Length is the magnitude of the vector. Hence the magnitudes add to give the magnitude of the resultant vector.

46 / 74

Following sets of three forces act on a body. Whose resultant cannot be zero

A) 10, 10, 20

B) 10, 20, 23

C) 10, 20, 40

D) 10, 10, 10

Explanation:

For three forces to have zero resultant, their magnitudes must satisfy the triangle inequality.

(each pair’s sum ≥ the third).

Here 10 + 20 = 30 < 40, so they can’t form a closed triangle and cannot balance to zero.

47 / 74

When subtracting B = 4î + 2ĵ from A = î – 6ĵ, the resultant A – B is:

A) –3î – 8ĵ

B) 5î + 4ĵ

C) –3î + 8ĵ

D) 3î – 8ĵ

Explanation:

A – B = (î – 6ĵ) – (4î + 2ĵ)
= î – 6ĵ – 4î – 2ĵ
= (1 – 4)î + (–6 – 2)ĵ
= –3î – 8ĵ

48 / 74

If u and v are perpendicular unit vectors, then |u – v| equals:

A) 0

B) 1

C) √2

D) 2

|u| = |v| = 1 (unit vectors)
Angle between u and v = 90° (perpendicular)

Use the formula:
|u − v|² = |u|² + |v|² − 2|u||v|cosθ

= 1² + 1² − 2(1)(1)cos(90°)
= 1 + 1 − 0
= 2

Therefore,
|u − v| = √2

49 / 74

A plane flies 300 km due north and then 400 km due west. Its resultant displacement is:

A) 500 km northwest

B) 700 km

C) 100 km

D) 250 km

R = √(300² + 400²)
= 500 km.

50 / 74

Two vectors of equal magnitude V with angle 60° to each other. The magnitude of their resultant is:

A) V

B) V√2

C) V√3

D) 2V sin 30°

R = √(V² + V² + 2·V·V·cos 60°)
= √(2V² + 2V²·½)
= √(2V² + V²)
= √(3V²)
= V√3

51 / 74

If A = 7î + 24ĵ, the magnitude of A is:

A) 17

B) 25

C) 31

D) 49

|A| = √(7² + 24²)
= √(49 + 576)
= √625 = 25.

52 / 74

Two equal forces $P$ each act at a point with an angle of 120° between them. The magnitude of their resultant is:

A) P

B) 0

C) P/√2

D) P√2

We know:
cos(120°) = -1/2

So,
R = √[2P² + 2P²·(-1/2)]
R = √[2P² – P²]
R = √[P²]
R = P

53 / 74

If the sum of two unit vectors is a unit vector, then the magnitude of their difference and the angle between them are, respectively:

A) √2, 60°

B) √3, 60°

C) √2, 120°

D) √3, 120°

Let the two unit vectors be A and B.
So,
|A| = |B| = 1
|A + B| = 1

Use the formula:
|A + B|² = |A|² + |B|² + 2|A||B|cosθ

⇒ 1² = 1² + 1² + 2(1)(1)cosθ
⇒ 1 = 1 + 1 + 2cosθ
⇒ 1 = 2 + 2cosθ
⇒ 2cosθ = -1
⇒ cosθ = -1/2
⇒ θ = 120°

Now, find |A − B|:
|A − B|² = |A|² + |B|² − 2|A||B|cosθ
= 1 + 1 − 2(1)(1)(cos120°)
= 2 − 2(−1/2)
= 2 + 1 = 3
⇒ |A − B| = √3

54 / 74

At what angle should the two force vectors 5 N and 12 N be added to get a resultant of 13 N?

A) 0°

B) 30°

C) 90°

D) 180°

For resultant we use below formula:

55 / 74

Two equal forces act at right angles on a point. The magnitude of their resultant is:

A) F

B) 2F

C) F/√2

D) F√2

56 / 74

A particle moves 5 m east then 12 m north. Its resultant displacement from the start is:

A) 13 m

B) 7 m

C) 17 m

D) 6 m

57 / 74

The angle between vector Q and the resultant of (2Q + 2P) and (2Q – 2P)

A) tan⁻¹(P/Q)

B) 0°

C) tan⁻¹((2Q–2P)/(2Q+2P))

D) tan⁻¹(2Q/P)

(2Q + 2P) + (2Q – 2P) = 4Q,
which is parallel to Q, so the angle between them is 0°.

58 / 74

Two perpendicular forces have magnitudes A and A/2. The magnitude of their resultant is:

A) 5A/2

B) (√5 A)/4

C) (A√5 )/2

D) (A²√5)/2

59 / 74

When A = 2î + 3ĵ + 2k̂ is subtracted from B, the result is 2ĵ. The magnitude of B is:

A) 3

B) √33

C) √6

D) √5

B – A = 2ĵ
⇒ B = A + 2ĵ
= 2î + (3+2)ĵ + 2k̂
= 2î + 5ĵ + 2k̂
⇒ |B| = √(4+25+4)
= √33.

60 / 74

A vector in the xy-plane makes an angle of 30° with the y-axis. Its y-component is 2√3. The magnitude of its x-component is:

A) √3

B) 2

C) 6

D) 1/√3

61 / 74

Given A = (2, 3) and B = (–2, –3), the graphical sum A + B forms:

A) A triangle

B) A straight line

C) A closed loop

D) A parallelogram

Plot A, then attach B head-to-tail; you return to the origin.

A and B are equal in magnitude but opposite in direction.

Their head-to-tail connection completes the path back to the start, you will get a straight line.

62 / 74

Which property of vector addition states A + B = B + A?

A) Associative

B) Commutative

C) Distributive

D) Identity

“Commutative” means order doesn’t matter.

Graphically, placing A then B, or B then A, yields the same head-to-tail resultant.

Magnitude and direction are identical either way.

This holds for all vectors in a vector space.

63 / 74

Which is not part of adding vectors by rectangular components?

A) Resolve each vector into î and ĵ

B) Sum corresponding components

C) Draw the resultant by the parallelogram law

D) Multiply x- and y-components together

Steps:
(1) resolve into components,
(2) add all x’s and all y’s,
(3) recombine into R.

Parallelogram law is an equivalent graphical check.

At no point do you multiply components.

Option (d) is irrelevant to vector addition.

64 / 74

Two vectors A and B add to a resultant purely along the x-axis if:

A) Aᵧ + Bᵧ = 0

B) Aₓ + Bₓ = 0

C) Aᵧ = Bᵧ = 0

D) Aₓ and Bₓ are both positive

Pure x-axis resultant means no net vertical component.

Component addition: Rᵧ = Aᵧ + Bᵧ must be zero.

Horizontal sum Aₓ + Bₓ can be nonzero.

Hence only condition (a) guarantees R is along x.

65 / 74

If A = 8î + 6ĵ and B = –8î + 6ĵ, the resultant A + B lies:

A) along +y-axis

B) along +x-axis

C) in the 1st quadrant

D) in the 2nd quadrant

Sum x-components: 8 + (–8) = 0
→ no horizontal part.

Sum y-components: 6 + 6 = 12
→ positive vertical part.

Zero x and positive y
⇒ a vector straight up along +y.

66 / 74

Which method adds two vectors by drawing them from the same origin to form a parallelogram?

A) Triangle law

B) Parallelogram law

C) Pythagorean law

D) Commutative law

Parallelogram law uses the two vectors as adjacent sides.

The diagonal from the common origin is the resultant.

It decomposes into perpendicular components on x and y.

This visual proof also shows commutativity of addition.

67 / 74

A = 4î + 3ĵ and B = –2î + 6ĵ. What is the angle θ of R from the +x-axis?

A) tan⁻¹(3/4)

B) tan⁻¹(9/2)

C) tan⁻¹(2/9)

D) tan⁻¹(3/6)

Rₓ = 4 + (–2)
Rₓ = 2,
R_y = 3 + 6
R_y = 9.

θ = tan⁻¹(R_y/R_x)
= tan⁻¹(9/2).

Compute components first, then use tan⁻¹.

This yields the correct direction of R.

68 / 74

A = 3î – 4ĵ, B = –5î + 2ĵ. The resultant A + B is:

A) –2î – 2ĵ

B) –2î + 6ĵ

C) 8î – 2ĵ

D) 2î – 2ĵ

Sum x-components:
3 + (–5)
= –2
→ –2î.

Sum y-components:
–4 + 2
= –2
→ –2ĵ.

R = –2î + (–2ĵ) by component-wise addition.

Vector closure ensures the result is itself a vector.

69 / 74

If A_x+ B_x= 0 but A_y + B_y ≠ 0, then A + B

A) lies along the x-axis

B) lies along the y-axis

C) is zero

D) is undefined

Zero x-sum means no horizontal part; nonzero y-sum gives a pure vertical resultant.

70 / 74

Two equal-magnitude vectors at right angles are added. The resultant’s magnitude is V√2 because

A) you multiply one component by √2

B) the dot product equals zero

C) Pythagoras’ theorem on the two equal legs

D) vectors become scalar when perpendicular

If each component= V,
then R = Vî + Vĵ
use Pythagoras’ theorem
⇒ |R|
= √(V² + V²)
= V√2.

71 / 74

A = 5î + 0ĵ and B = 0î + 12ĵ. The magnitude of A + B is

A) 7

B) 13

C) 17

D) 60

R = (5+0)î + (0+12)ĵ
= 5î + 12ĵ.
Magnitude = √(5² + 12²) = 13.

72 / 74

Subtracting 2î + 7ĵ from î + ĵ gives ______

A) -î – 6ĵ

B) 3î + 8ĵ

C) î + 6

D) 7ĵ

Subtract component-wise:
(î+ĵ) – (2î+7ĵ)
= (1–2)î + (1–7)ĵ
X-component:
1 – 2
= –1
→ –î
Y-component:
1 – 7 = –6 → –6ĵ
Resultant: –î –6ĵ

73 / 74

ding 2î + 7ĵ and î + ĵ gives ______

A) 3î + 8ĵ

B) î + 35ĵ

C) î + 8ĵ

D) 2î + 7ĵ

Add x-components:
2 + 1 = 3
Add y-components:
7 + 1 = 8
Result: 3î + 8ĵ

74 / 74

On adding two vectors we get _____

A) A vector

B) A scalar

C) A number

D) An operation

On adding two vectors we get a vector.
Vector addition preserves both magnitude and direction.
Use the head-to-tail or parallelogram law to find the resultant.
Vectors are closed under addition, so the result is always another vector.

Your score is
The average score is 0%
0%
December 5, 2025
Buffer Solution

0%

1

Report a question

What’s wrong with this question?
You cannot submit an empty report. Please add some details.

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
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92

Thank you for your interest in Eprepare Academy Tests! These quizzes are only available to registered students of our Academy. If you are a registered student, please log in with your account information to access the quizzes.

If you are not a registered student, unfortunately you will not be able to take the quizzes at this time. We apologize for any inconvenience this may cause. If you have any questions or need further assistance, please don’t hesitate to contact us.

Username or Email Address
Password
Remember Me

December 2, 2025
Common Ion Effect

0%

1

Report a question

What’s wrong with this question?
You cannot submit an empty report. Please add some details.

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
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86

Thank you for your interest in Eprepare Academy Tests! These quizzes are only available to registered students of our Academy. If you are a registered student, please log in with your account information to access the quizzes.

If you are not a registered student, unfortunately you will not be able to take the quizzes at this time. We apologize for any inconvenience this may cause. If you have any questions or need further assistance, please don’t hesitate to contact us.

Username or Email Address
Password
Remember Me

December 2, 2025
Solubility Product Ksp Part-2

0%

1

Report a question

What’s wrong with this question?
You cannot submit an empty report. Please add some details.

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
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50

Thank you for your interest in Eprepare Academy Tests! These quizzes are only available to registered students of our Academy. If you are a registered student, please log in with your account information to access the quizzes.

If you are not a registered student, unfortunately you will not be able to take the quizzes at this time. We apologize for any inconvenience this may cause. If you have any questions or need further assistance, please don’t hesitate to contact us.

Username or Email Address
Password
Remember Me

December 2, 2025
Solubility Product Ksp Part-1

0%

1

Report a question

What’s wrong with this question?
You cannot submit an empty report. Please add some details.

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
31
32
33
34
35
36
37
38
39
40

Thank you for your interest in Eprepare Academy Tests! These quizzes are only available to registered students of our Academy. If you are a registered student, please log in with your account information to access the quizzes.

If you are not a registered student, unfortunately you will not be able to take the quizzes at this time. We apologize for any inconvenience this may cause. If you have any questions or need further assistance, please don’t hesitate to contact us.

Username or Email Address
Password
Remember Me

December 2, 2025
Haber’s Process

0%

1

Report a question

What’s wrong with this question?
You cannot submit an empty report. Please add some details.

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
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56

Thank you for your interest in Eprepare Academy Tests! These quizzes are only available to registered students of our Academy. If you are a registered student, please log in with your account information to access the quizzes.

If you are not a registered student, unfortunately you will not be able to take the quizzes at this time. We apologize for any inconvenience this may cause. If you have any questions or need further assistance, please don’t hesitate to contact us.

Username or Email Address
Password
Remember Me

December 2, 2025