QCC Notes
CLASS 11 · PHYSICSJEE MAIN × NEETहिंदी
§ 6.2NCERT Class 11 · Physics · Chapter 6

Vector Product & Angular Velocity

QCC Notes · Class 11 Physics NCERT-aligned Updated 26 Sep 2026 9 min read

Translational kinematics relies predominantly on scalar (dot) products for work and power. In contrast, rotational mechanics is fundamentally governed by vector (cross) products, which describe directional twist, perpendicular relationships, and axial vectors like angular velocity, torque, and angular momentum.

1. Definition and Properties of the Vector (Cross) Product

The vector product of two vectors A and B, denoted as A × B, is a vector C whose magnitude is proportional to the area spanned by the vectors and whose direction is perpendicular to both:

A × B = (A B sin θ) n̂   (0 ≤ θ ≤ π)

where A and B are vector magnitudes, θ is the angle between them, and n̂ is a unit vector perpendicular to the plane containing A and B, oriented according to the Right-Hand Thumb Rule.

Key Mathematical Laws of Cross Products

  • Anti-Commutative Property: A × B = −(B × A). Order of multiplication cannot be inverted without introducing a negative sign.
  • Distributive over Addition: A × (B + C) = (A × B) + (A × C).
  • Self Cross Product: The cross product of any vector with itself (or with any collinear/parallel vector, θ = 0 or π) is the zero vector: A × A = 0.
  • Orthogonal Unit Vectors:
    • î × î = ĵ × ĵ = k̂ × k̂ = 0
    • Cyclic clockwise: î × ĵ = k̂,   ĵ × k̂ = î,   k̂ × î = ĵ
    • Anticyclic counter-clockwise: ĵ × î = −k̂,   k̂ × ĵ = −î,   î × k̂ = −ĵ

Cartesian Determinant Formula

For vectors given in rectangular components A = Axî + Ayĵ + Azk̂ and B = Bxî + Byĵ + Bzk̂:

A × B = | î     ĵ     k̂ |
| Ax   Ay   Az | = î(AyBz − AzBy) − ĵ(AxBz − AzBx) + k̂(AxBy − AyBx)
| Bx   By   Bz |

2. Geometric Applications: Areas of Parallelograms & Triangles

Geometric Figure Vector Inputs Area Formula
Parallelogram (Adjacent Sides) Adjacent sides A and B Area = |A × B| = A B sin θ
Parallelogram (Diagonals) Diagonals d1 and d2 Area = ½ |d1 × d2|
Triangle Two adjacent sides A and B Area = ½ |A × B|

3. Angular Velocity and Its Relation with Linear Velocity

When a rigid body rotates about a fixed axis, every constituent particle executes circular motion in a plane perpendicular to the axis, with its center on the axis.

  • Angular Velocity (ω): The rate of angular displacement, ω = dθ/dt (measured in rad/s).
  • Axial Vector Character: Angular velocity ω is an axial vector directed along the axis of rotation following the right-hand grip rule (fingers curl along rotation, thumb points along ω).
  • Relation to Linear Velocity: For a particle with position vector r measured from any point on the rotation axis:
v = ω × r   (Linear speed v = ω r sin φ = ω r⊥)
NTA TRAP: Never write v = r × ω! Because the vector product is anti-commutative, r × ω = −v, which reverses the physical direction of linear velocity. The strict canonical order is v = ω × r.

General Acceleration in Circular & Curved Motion

Differentiating v = ω × r with respect to time:

a = dv/dt = (dω/dt × r) + (ω × dr/dt) = (α × r) + (ω × v)

This resolves acceleration into two mutually perpendicular physical components:

  1. Tangential Acceleration: at = α × r (magnitude at = α r, changes speed).
  2. Centripetal (Radial) Acceleration: ac = ω × v (magnitude ac = ω² r = v²/r, directed radially inward toward the rotation center, changes direction).

Total linear acceleration magnitude: a = √(at² + ac²) = √((α r)² + (ω² r)²).

4. Visualising the Right-Hand Rule and Vector Geometry

Figure 6.2 — Right-Hand Rule for Vector Product & Angular Velocity

^ C = A x B (Along axis of rotation: ω) | | Thumb points along normal unit vector n̂ | +---+---+ / | / / | / / | / B (Fingers curl from A toward B) / | / / / |/ / +---------O-----+---------> A \ / \ Plane of / \ A and B/ +-------+
Exam Read-out: Notice that C is perpendicular to BOTH A and B simultaneously: C · A = 0 and C · B = 0.

Step-by-Step Solved Numerical Examples

Solved Example 1: Evaluating Cross Product & Area of Triangle

Problem: Given two vectors A = 2î + 3ĵ − k̂ and B = î − 2ĵ + 2k̂. Find (1) the cross product A × B, (2) the unit vector perpendicular to both, and (3) the area of the triangle formed by A and B.

A × B = 4î − 5ĵ − 7k̂ · Area = √90 / 2 ≈ 4.74 sq. units
Step 1: Compute Cross Product via Determinant:
A × B = î[(3)(2) − (−1)(−2)] − ĵ[(2)(2) − (−1)(1)] + k̂[(2)(−2) − (3)(1)]
= î[6 − 2] − ĵ[4 + 1] + k̂[−4 − 3] = 4î − 5ĵ − 7k̂.

Step 2: Magnitude:
|A × B| = √(4² + (−5)² + (−7)²) = √(16 + 25 + 49) = √90 ≈ 9.487.

Step 3: Unit Perpendicular Vector:
n̂ = (A × B) / |A × B| = (4î − 5ĵ − 7k̂) / √90.

Step 4: Area of Triangle:
Area = ½ |A × B| = ½ √90 = 3√10 / 2 ≈ 4.74 sq. units.
Solved Example 2: Linear Velocity from Angular Velocity and Position Vector

Problem: A particle rotates with an angular velocity ω = 3î − 4ĵ + k̂ rad/s. At a certain instant, its position vector with respect to the center of rotation is r = 5î − 6ĵ + 6k̂ m. Calculate the linear velocity vector of the particle.

v = −18î − 13ĵ + 2k̂ m/s
Step 1: Apply Vector Relation v = ω × r:
| î     ĵ     k̂ |
| 3   −4    1 |
| 5   −6    6 |

Step 2: Expand Determinant:
vx = (−4)(6) − (1)(−6) = −24 + 6 = −18
vy = −[(3)(6) − (1)(5)] = −[18 − 5] = −13
vz = (3)(−6) − (−4)(5) = −18 + 20 = +2

Therefore, v = −18î − 13ĵ + 2k̂ m/s.
Verification: v · ω = (−18)(3) + (−13)(−4) + (2)(1) = −54 + 52 + 2 = 0 (velocity is orthogonal to angular velocity!).

5. Solved NEET & JEE Practice Questions

Q1. If the magnitude of the vector product of two vectors is equal to √3 times their scalar product, the angle between the vectors is:

A. 30°
B. 45°
C. 60°
D. 90°

Answer: C
Explanation: Given |A × B| = √3 (A · B).
AB sin θ = √3 AB cos θ ⇒ tan θ = √3 ⇒ θ = 60° (π/3 rad).
Q2. Which of the following statements is INCORRECT regarding vector cross product?

A. A × B = −(B × A)
B. A × A = 0
C. (A × B) · A = 0
D. A × (B × C) = (A × B) × C

Answer: D
Explanation: The vector triple product is NOT associative: A × (B × C) ≠ (A × B) × C. Statements A, B, and C are true properties of the cross product.
Q3. A body is rotating with uniform angular velocity ω about a fixed axis. The linear velocity of a particle at distance r from the axis is v. If the distance from the axis is doubled while angular velocity is halved, the linear velocity becomes:

A. v
B. 2v
C. v / 2
D. 4v

Answer: A
Explanation: Linear speed v = ω r. New speed v' = (ω/2)(2r) = v. The linear speed remains unchanged.
Q4. A particle moves in a circle of radius 0.5 m with an angular speed increasing at a constant rate of 2 rad/s². If at t = 0 it starts from rest, its total linear acceleration at t = 2 s is:

A. 1 m/s²
B. 8 m/s²
C. √65 m/s²
D. 9 m/s²

Answer: C
Explanation: Given r = 0.5 m, α = 2 rad/s², ω0 = 0.
At t = 2 s: ω = α t = (2)(2) = 4 rad/s.
Tangential acceleration at = α r = (2)(0.5) = 1 m/s².
Centripetal acceleration ac = ω² r = (4²)(0.5) = (16)(0.5) = 8 m/s².
Total acceleration a = √(at² + ac²) = √(1² + 8²) = √65 m/s² ≈ 8.06 m/s².

6. Frequently Asked Questions (FAQs)

Why is angular velocity defined as an axial vector rather than a polar vector?
A polar vector (like displacement, velocity, force) has a clear point of application and changes sign under spatial inversion (parity transformation: r → −r implies v → −v). In contrast, angular velocity describes rotation about a line (axis); its direction is defined by a mathematical convention (right-hand screw rule). Under spatial inversion, rotation sense appears inverted, so the cross product ω = (r × v)/r² remains unchanged in direction. Hence ω is a pseudovector (axial vector).
How can we test if three vectors A, B, and C are coplanar?
Three vectors are coplanar if and only if their scalar triple product is zero: A · (B × C) = 0. Geometrically, the scalar triple product represents the volume of the parallelepiped spanned by the three vectors; if they lie in the same plane, the volume is zero.
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