Vector Practice Test 5 ( Vector Product )
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1 / 21
If two vectors A and B are perpendicular and have magnitudes 4 and 6 respectively, what is the magnitude of A × B?
Explanation:
Magnitude of cross product = |A||B|sinθSince vectors are perpendicular, θ = 90°, sin90° = 1.So magnitude = 4 × 6 × 1 = 24.
2 / 21
Given vectors A = 2i + 3j + k and B = i - j + 4k, what is the vector A × B?
Calculate cross product using determinant:
3 / 21
The direction of the vector A × B is given by:
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.
4 / 21
If A × B = 0, which of the following must be true?
Cross product is zero if vectors are parallel (θ = 0° or 180°) or if either vector is a zero vector. Because sin0° = sin180° = 0.
5 / 21
What is the magnitude of the cross product A × B when A = 5 units, B = 3 units, and the angle between them is 90°?
Magnitude of cross product = |A||B|sinθ = 5 × 3 × sin90° = 15 × 1 = 15.
6 / 21
The cross product of two vectors A and B is:
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.
7 / 21
If A = 1î + 2ĵ + 3k̂ and B = 2î + 4ĵ + 6k̂, then A × B = ?
B is 2·A (parallel to A), so A × B = 0 for any parallel vectors.
8 / 21
The vector product î × k̂ equals:
Using right-hand rule / cyclic order î → ĵ → k̂, we have î × k̂ = –(k̂ × î) = –ĵ.
9 / 21
If |A| = 5, |B| = 7 and the angle between A and B is 90°, then |A × B| = ?
|A × B| = |A| |B| sin θ = 5 × 7 × sin 90° = 35.
10 / 21
If the angle between vectors A and B is θ, the value of the product (B × A) · A is equal to:
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.
11 / 21
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)
12 / 21
The value of the unit vector which is perpendicular to both A = î + 2ĵ + 3k̂ and B = î – 2ĵ – 3k̂ is equal to:
Its magnitude = √(6² + 4²) = √(36 + 16) = √52
13 / 21
The angle between vectors ( A x B ) & (B X A )
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°).
14 / 21
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?
15 / 21
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 · B = 3×6 + 4×8 = 18 + 32 = 50 ❌ (not 48)
16 / 21
The scalar and vector product of two vectors, a = (3î – 4ĵ + 5k̂) and b = (–2î + ĵ – 3k̂) is equal to:
17 / 21
If we have two vectors A and B and |A| ≠ |B| and |A × B| = |A · B|, then:
18 / 21
If two vectors a = 2î + ĵ and b = 3î + 2ĵ, then |a × b| = ?
19 / 21
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:
= 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
20 / 21
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 × 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°
21 / 21
If for two vectors A and B, A × B = 0, then the vectors:
Two vectors will be parallel to each other.
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