Vectors Practice Test 3 ( Resultant of Vector)
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1 / 27
Given A = 3i + 4j – 5k and B = 1i – 2j + 2k, the magnitude of R = A – 2B is:
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 / 27
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:
Add components: Rₓ = 4 + (−1) = 3; Rᵧ = −2 + 5 = 3. Thus tanθ = Rᵧ/Rₓ = 3/3 = 1, giving θ = 45°.
3 / 27
Three mutually perpendicular vectors of magnitudes 2, 3 and 6 units act simultaneously at a point. The magnitude of their resultant is:
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 / 27
Two vectors of magnitudes 5 units and 7 units act with an angle of 60° between them. The magnitude of their resultant is:
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 / 27
Consider vectors A = 3i + 4j and B = 2i − j. The magnitude of the resultant R = A + B is:
First compute the components: Rₓ = 3 + 2 = 5, Rᵧ = 4 + (−1) = 3. The magnitude is√(Rₓ² + Rᵧ²) = √(5² + 3²) = √(25 + 9) = √34.
6 / 27
For the figure -
Use head to tail rule:
7 / 27
If the angle between the unit vectors â and b̂ is 60°, then |â − b̂| is:
8 / 27
What displacement must be added to the displacement25 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 ĵResultant 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 / 27
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 -
10 / 27
A truck travelling due north with 20m/s turns towards west and travels at the same speed. Then the change in velocity is -
11 / 27
The three vectors OA, OB and OC have the same magnitude R. Then the sum of these vectors have magnitude -
12 / 27
A force of 6 N and another of 8 N can be applied together to produce the effect of a single force of -
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 / 27
Which of the sets given below may represent the magnitude of resultant of three vectors adding to zero?
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 triangleInvalid
Option (2): 4, 8, 16→ 4 + 8 = 12 < 16 → ❌ Cannot form triangleInvalid
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 / 27
In the given figure
15 / 27
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?
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 / 27
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:
17 / 27
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:
Video Explanation:
18 / 27
Two forces of magnitude F have a resultant of the same magnitude F. The angle between the two forces is
19 / 27
If the angle between the two forces increases, the magnitude of their resultant:
The angle between two vectors can never be greater than 180o.
So, on increasing the θ, the magnitude of resultant vectors decreases.
20 / 27
if we have two vectors P and Q and P=Q then which of the following is not correct
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 / 27
If A = B + C and the magnitudes of A, B, C are 5, 4, 3 units respectively, the angle between A and C is:
22 / 27
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:
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 / 27
For the given figure, which of the following is true?
24 / 27
If the magnitude of the sum of two vectors equals the magnitude of their difference, the angle between the vectors is:
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 / 27
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?
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 / 27
If | v₁ + v₂ | = | v₁ − v₂ | and v₁, v₂ are non-zero vectors, then:
Hint: Use Resultant of vectors. Step: Find the relation between v1 and v2
27 / 27
Two forces A and B have a resultant R₁. If B is doubled, the new resultant R₂ is perpendicular to A. Then
Original resultant: R1 = A + BNew 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.
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