Home › Physics › Mechanics › Vector Product of Vectors

Mechanics

Vector Product of Vectors

The vector product (or cross product) of two vectors yields a third vector whose magnitude is proportional to the product of their magnitudes and the sine of the angle between them. Its direction is perpendicular to the plane containing the constituent vectors, following a right-handed screw convention.

UP TGT ScienceUP PGT PhysicsStudy Notes + Practice

Scope and prerequisites

TGT core

Definition of cross product, geometric interpretation as area of parallelogram and triangle, right-hand rule, scalar components in Cartesian coordinates, unit vectors cross products (i x j = k), condition for collinearity and parallelism.

PGT extension

Lagrange's identity, vector triple product, applications to rotational mechanics including torque, angular momentum, central force motion, and Coriolis force.

Prerequisites

Scalar and vector quantities
Resolution of vectors into Cartesian components
Right-hand coordinate systems

Definition and Direction of Vector Product

The vector product of two vectors A and B is defined as a vector $C = A x B = AB \sin(\theta)\hat{n}$, where θ is the smaller angle between A and B (0 ≤ θ ≤ π) and n^ is a unit vector perpendicular to the plane containing A and B.
The direction of n^ is given by the right-hand thumb rule: if fingers curl from vector A to vector B, the outstretched thumb points in the direction of A x B.
The cross product is anticommutative: A x $B = -(B x A)$.

Cartesian Component Form

In terms of orthogonal unit vectors i^, j^, k^, if $A = A_{x}\hat{i} + A_{y}\hat{j} + A_{z}\hat{k}$ and $B = B_{x}\hat{i} + B_{y}\hat{j} + B_{z}\hat{k}$, the cross product is evaluated using a determinant.
The expanded expression is A x $B = (A_{y}B_{z} - A_{z}B_{y})\hat{i} - (A_{x}B_{z} - A_{z}B_{x})\hat{j} + (A_{x}B_{y} - A_{y}B_{x})\hat{k}$.
Cyclic relations for unit vectors: i^ x j^ = k^, j^ x k^ = i^, k^ x i^ = j^; and self cross products vanish: i^ x i^ = 0.

Geometric Interpretation

The magnitude of the vector product |A x B| represents the area of the parallelogram formed with vectors A and B as adjacent sides.
The area of a triangle formed by vectors A and B as two adjacent sides is given by (1/2)|A x B|.
If A x B = 0 for non-zero vectors A and B, the vectors are collinear or parallel, indicating $\theta = 0 or \pi$.

Applications in Mechanics

Torque (τ) is defined as the cross product of position vector r and force F: τ = r x F.
Angular momentum (L) of a particle is given by L = r x p = r x (mv).
Velocity in a rigid body rotating with angular velocity ω about a fixed axis is v = ω x r.

Derivations

Expansion of Cross Product in Cartesian Coordinates

Conditions: Vectors defined in a right-handed orthogonal Cartesian coordinate system

Write vectors A and B in component form: $A = A_{x}\hat{i} + A_{y}\hat{j} + A_{z}\hat{k}$ and $B = B_{x}\hat{i} + B_{y}\hat{j} + B_{z}\hat{k}$.
Apply distributive property of cross product over addition: A x $B = (A_{x}\hat{i} + A_{y}\hat{j} + A_{z}\hat{k}) x (B_{x}\hat{i} + B_{y}\hat{j} + B_{z}\hat{k})$.
Expand terms using cross products of unit vectors (e.g., i^ x i^ = 0, i^ x j^ = k^, j^ x i^ = -k^).
Group coefficients of i^, j^, and k^ to form the determinant structure.

Result: A x $B = \hat{i}(A_{y}B_{z} - A_{z}B_{y}) + \hat{j}(A_{z}B_{x} - A_{x}B_{z}) + \hat{k}(A_{x}B_{y} - A_{y}B_{x})$

Lagrange's Identity for Vector Products

Conditions: Any two vectors A and B in 3D Euclidean space

Consider the vector identity involving dot and cross products: |A x B|² + (A · B)².
Substitute magnitudes: (AB sin(θ))² + (AB cos(θ))².
Factor out A²B² to obtain A²B²(sin²(θ) + cos²(θ)).
Simplify using trigonometric identity sin²(θ) + cos²(θ) = 1.

Result: |A x B|² = A²B² - (A · B)²

Formula sheet

Formula / law	Equation	Variables and units	Conditions
Definition of Vector Product	$$A \times B = AB \sin(\theta) \hat{n}$$	A, B = magnitudes; \theta = angle between vectors; \hat{n} = unit normal vector Product of individual units	0 \le \theta \le \pi
Anticommutative Property	$$A \times B = - (B \times A)$$	A, B = vectors None	Always valid
Determinant Expansion	$$A \times B = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{vmatrix}$$	A_x, A_y, A_z and B_x, B_y, $B_z = Cartesian components$ Product of individual units	Cartesian coordinate system
Magnitude of Vector Product	$$\|A \times B\| = \sqrt{(A_yB_z - A_zB_y)^2 + (A_zB_x - A_xB_z)^2 + (A_xB_y - A_yB_x)^2}$$	Cartesian components of A and B Product of individual units	Cartesian coordinate system
Area of Parallelogram	$$Area = \|A \times B\|$$	A, B = adjacent side vectors square units	Non-collinear vectors
Area of Triangle	$$Area = \frac{1}{2} \|A \times B\|$$	A, B = adjacent side vectors forming the triangle square units	Non-collinear vectors
Torque Definition	$$\tau = r \times F$$	r = position vector, F = force vector N m	Point of application relative to reference axis
Angular Momentum Definition	$$L = r \times p = r \times (m v)$$	r = position vector, p = linear momentum, m = mass, v = velocity kg m^2 s^-1	Point particle about origin

Graphs and experiments

Right-Hand Rule Orientation Diagram

Draw: Two vectors A and B originating from a common point, with angle θ between them, and a perpendicular vector A x B pointing upwards according to the right-hand screw rule.

Interpret: Visualizes the spatial orientation of the cross product vector relative to the plane of the operands.

Exam use: Tested in conceptual MCQs regarding the direction of torque or magnetic force.

Geometric Area of Parallelogram

Draw: A parallelogram with sides A and B, height $h = B \sin(\theta)$, illustrating that the magnitude of the cross product equals base times height.

Interpret: Establishes the physical equivalence between the cross product magnitude and planar area.

Exam use: Direct application questions for calculating triangle or parallelogram areas using coordinate vectors.

Torque Vector Representation Experiment/Diagram

Draw: A wrench of length r acted upon by force F at an angle, showing the resulting torque vector τ pointing perpendicularly out of the plane of rotation.

Interpret: Demonstrates how rotational effect is maximized when force is perpendicular to the position vector.

Exam use: Core mechanics problems involving rotational equilibrium and axis of rotation.

Solved numericals

TGT core

Find the cross product of two vectors $A = 2\hat{i} - 3\hat{j} + \hat{k}$ and $B = \hat{i} + 2\hat{j} - 4\hat{k}$.

Set up the determinant: A x $B = \hat{i}( (-3)(-4) - (1)(2) ) - \hat{j}( (2)(-4) - (1)(1) ) + \hat{k}( (2)(2) - (-3)(1) )$.
Calculate i^ component: 12 - 2 = 10.
Calculate j^ component: -8 - $1 = -9$, so -(-9)j^ = 9j^.
Calculate k^ component: 4 - (-3) = 7.
Combine terms: 10i^ + 9j^ + 7k^.

Answer: 10i^ + 9j^ + 7k^

TGT core

Calculate the area of a triangle whose adjacent sides are represented by vectors $A = 3\hat{i} + 4\hat{j}$ and $B = -\hat{i} + 2\hat{j}$.

Compute the cross product A x $B = (3\hat{i} + 4\hat{j}) x (-\hat{i} + 2\hat{j})$.
Using unit vector cross products: 3(2)(i^ x j^) + 4(-1)(j^ x $\hat{i}) = 6\hat{k} - 4(-\hat{k}) = 6\hat{k} + 4\hat{k} = 10\hat{k}$.
Find the magnitude: |A x B| = |10$k^| = 10$.
Calculate area of triangl$e = (1/2) |A x B| = (1/2)(10) = 5 square units$.

Answer: 5 sq. units

PGT extension

A force $F = (2\hat{i} + 3\hat{j} - \hat{k}) N acts at a point whose position vector$ is $r = (\hat{i} - \hat{j} + 2\hat{k}) m$. Determine the torque acting about the origin.

Use the torque formula: $\tau = r x F$.
Set up the determinant with r = [1, -1, 2] and F = [2, 3, -1].
Calculate i^ component: (-1)(-1) - (2)(3) = 1 - $6 = -5$.
Calculate j^ component: (1)(-1) - (2)(2) = -1 - $4 = -5$, so -(-5)j^ = 5j^.
Calculate k^ component: (1)(3) - (-1)(2) = 3 - (-2) = 5.
Combine components to get torque vector.

Answer: (-5i^ + 5j^ + 5k^) N m

Assumptions and limitations

The cross product is exclusively defined in 3-dimensional (and 7-dimensional) Euclidean space; it does not have a standard direct definition for 2D planes except by treating 2D vectors as 3D vectors with zero z-components.
The right-hand convention must be strictly maintained; switching to a left-hand coordinate system inverts the sign of the vector product.

Common misconceptions

Common error: Assuming the vector product is commutative, so A x B = B x A.

Correct: The vector product is anticommutative: A x $B = -(B x A)$.

Reversing the order of operands reverses the direction of the right-hand thumb rule.

Common error: Believing that the cross product of two non-zero vectors can yield a scalar.

Correct: The vector product always results in a vector quantity.

By definition, the cross product includes a perpendicular unit vector n^.

Common error: Thinking that |A x B| is simply equal to AB without regard to the angle.

Correct: |A x B| = AB sin(θ), which equals AB only when the vectors are perpendicular ($\theta = 90°)$.

The sine factor accounts for the projection of one vector perpendicular to the other.

Quick revision

A x $B = AB \sin(\theta)\hat{n}$ and is perpendicular to both A and B.
Definition of Vector Product: A \times B = AB \sin(\theta) \hat{n}
Anticommutative Property: A \times $B = - (B \times A)$

Must remember

The vector product of two vectors $A$ and $B$ is given by $A \times B = AB \sin(\theta)\hat{n}$, which results in a vector quantity perpendicular to both $A$ and $B$.
The cross product is anticommutative, meaning $A \times B = -(B \times A)$.
The self cross product of any vector is zero, i.e., $A \times A = 0$, and for orthogonal unit vectors, $\hat{i} \times \hat{j} = \hat{k}$, $\hat{j} \times \hat{k} = \hat{i}$, and $\hat{k} \times \hat{i} = \hat{j}$.
The magnitude $|A \times B|$ represents the area of the parallelogram formed by vectors $A$ and $B$, while $\frac{1}{2}|A \times B|$ gives the area of the triangle.
If $A \times B = 0$ for non-zero vectors, the vectors are parallel or collinear, implying $\theta = 0$ or $\theta = \pi$.
In mechanics, torque is defined as $\tau = r \times F$ and angular momentum is defined as $L = r \times p = r \times (mv)$.

Glossary

Vector ProductA multiplicative operation between two vectors producing a third vector perpendicular to the plane of the two original vectors.

Anticommutative PropertyThe property where reversing the order of operands negates the result: $A \times B = -(B \times A)$.

Unit Normal VectorA vector of magnitude 1 pointing perpendicularly away from the plane containing the multiplying vectors, denoted by $\hat{n}$.

TorqueThe rotational equivalent of linear force, defined as the cross product of position vector and force vector: $\tau = r \times F$.

Angular MomentumThe rotational momentum of a particle, defined as the cross product of position vector and linear momentum: $L = r \times p$.

Scalar Product vs. Vector Product

Feature	Scalar (Dot) Product	Vector (Cross) Product
Notation	$A \cdot B$	$A \times B$
Result Type	Scalar	Vector
Formula	$AB \cos(\theta)$	$AB \sin(\theta)\hat{n}$
Commutativity	Commutative ($A \cdot B = B \cdot A$)	Anticommutative ($A \times B = -(B \times A)$)
Parallel Vectors ($0^\circ$)	Maximum ($AB$)	Zero ($0$)

Practice Quiz

Choose an option to reveal the explanation.

1. What is the result of the cross product of two parallel vectors $A$ and $B$ where $\theta = 0^\circ$?

Explanation: Since $|A \times B| = AB \sin(0^\circ) = 0$, the cross product of parallel vectors vanishes, yielding a zero vector.

2. Which of the following relations correctly represents the anticommutative property of the vector product?

Explanation: The cross product is anticommutative, meaning that interchanging the vectors reverses the sign: $A \times B = -(B \times A)$.

3. What is the cross product of unit vector $\hat{i}$ with itself, i.e., $\hat{i} \times \hat{i}$?

Explanation: The angle between a vector and itself is $0^\circ$. Therefore, $\hat{i} \times \hat{i} = (1)(1)\sin(0^\circ)\hat{n} = 0$.

4. Which cyclic relation among orthogonal unit vectors is correct?

Explanation: Following the right-hand cyclic order, $\hat{i} \times \hat{j} = \hat{k}$, $\hat{j} \times \hat{k} = \hat{i}$, and $\hat{k} \times \hat{i} = \hat{j}$.

5. What does the magnitude $|A \times B|$ geometrically represent when $A$ and $B$ are adjacent sides of a geometric figure?

Explanation: The magnitude of the cross product of two vectors represents the area of the parallelogram formed by taking those vectors as adjacent sides.

6. If vector $A = 2\hat{i} + 3\hat{j}$ and vector $B = \hat{i} - \hat{j}$, what is $A \times B$?

Explanation: Using determinant expansion: $(2(-1) - (3)(1))\hat{k} = (-2 - 3)\hat{k} = -5\hat{k}$.

7. What is the area of a triangle formed by vectors $A$ and $B$ as two adjacent sides?

Explanation: The area of a triangle formed by two adjacent vectors is half the area of the corresponding parallelogram, given by $\frac{1}{2}|A \times B|$.

8. Torque ($\tau$) about a reference point is defined as the cross product of which two vectors?

Explanation: Torque is defined as $\tau = r \times F$, where $r$ is the position vector and $F$ is the applied force vector.

9. What is the SI unit of angular momentum ($L = r \times p$)?

Explanation: Angular momentum has dimensions of position ($m$) times linear momentum ($kg \cdot m \cdot s^{-1}$), yielding $kg \cdot m^2 \cdot s^{-1}$.

10. Which rule is used to determine the direction of the unit vector $\hat{n}$ in the vector product?

Explanation: The right-hand thumb rule determines the direction of the cross product vector: curling fingers from $A$ to $B$ points the thumb in the direction of $A \times B$.

11. If the angle between two non-zero vectors $A$ and $B$ is $90^\circ$, what is the magnitude of their vector product?

Explanation: When $\theta = 90^\circ$, $\sin(90^\circ) = 1$, so $|A \times B| = AB \sin(90^\circ) = AB$.

12. What is the velocity $v$ of a particle in a rigid body rotating with angular velocity $\omega$ about a fixed axis at position $r$?

Explanation: Linear velocity in rotational motion is related to angular velocity via the cross product: $v = \omega \times r$.

13. Which of the following statements is INCORRECT regarding the vector product?

Explanation: The vector product is anticommutative ($A \times B = -(B \times A)$), not commutative. Thus, the statement claiming it is commutative is incorrect.

14. If $A = 3\hat{i} - 4\hat{j}$, what is the magnitude of the vector product of $A$ with unit vector $\hat{k}$?

Explanation: $A \times \hat{k} = (3\hat{i} - 4\hat{j}) \times \hat{k} = 3(\hat{i} \times \hat{k}) - 4(\hat{j} \times \hat{k}) = 3(-\hat{j}) - 4(-\hat{i}) = 4\hat{i} - 3\hat{j}$. The magnitude is $\sqrt{4^2 + (-3)^2} = 5$.

15. What condition must be satisfied by two non-zero vectors $A$ and $B$ for their vector product to be zero?

Explanation: Since $|A \times B| = AB \sin(\theta)$, the cross product is zero when $\sin(\theta) = 0$, which occurs when $\theta = 0^\circ$ or $180^\circ$ (parallel or collinear).

PYQ Patterns

These are pattern-based practice questions. Authentic year labels are only used when verified from a source.

1. If $A = 2\hat{i} + 2\hat{j} - \hat{k}$ and $B = 6\hat{i} - 3\hat{j} + 2\hat{k}$, what is the angle between vectors $A$ and $B$ derived via vector product and dot product consistency?

Explanation: Evaluating components or cross product yields a non-zero cross product and non-zero dot product, but precise angle values require evaluating inverse trigonometric ratios matching vector magnitudes.

2. The position vector of a particle is given by $r = (3\hat{i} + 2\hat{j} + \hat{k})$ m and linear momentum is $p = (3\hat{i} - 2\hat{j} + 2\hat{k})$ kg m s$^{-1}$. What is the angular momentum $L$ about the origin?

Explanation: Using $L = r \times p$: determinant gives $L_x = (2)(2) - (1)(-2) = 6$, $L_y = -[(3)(2) - (1)(3)] = -3$, $L_z = (3)(-2) - (2)(3) = -12$. Thus $L = 6\hat{i} - 3\hat{j} - 12\hat{k}$.

3. What is the unit vector perpendicular to both vectors $A = \hat{i} + \hat{j} + \hat{k}$ and $B = \hat{i} + \hat{j}$?

Explanation: $A \times B = (1+1)$ components calculated via determinant yields $-\hat{i} + \hat{j}$. Dividing by its magnitude $\sqrt{2}$ gives $\frac{-\hat{i} + \hat{j}}{\sqrt{2}}$ or $\frac{\hat{i} - \hat{j}}{\sqrt{2}}$ depending on orientation.

4. A force $F = (2\hat{i} - 3\hat{j} + \hat{k})$ N is applied at a point with position vector $r = (3\hat{i} + 2\hat{j} - 2\hat{k})$ m. What is the magnitude of the torque about the origin?

Explanation: $\tau = r \times F = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & -2 \\ 2 & -3 & 1 \end{vmatrix} = (2(-1)-(-6))\hat{i} - (3(-1)-(-4))\hat{j} + (-9-4)\hat{k} = 4\hat{i} - 1\hat{j} - 13\hat{k}$. Magnitude = $\sqrt{4^2 + (-1)^2 + (-13)^2} = \sqrt{16 + 1 + 169} = \sqrt{186}$. Let's recompute: $2(1) - (-2)(-3) = 2 - 6 = -4$; $-[3(1) - (-2)(2)] = -(3+4) = -7$; $3(-3) - 2(2) = -9 - 4 = -13$. Thus $-4\hat{i} - 7\hat{j} - 13\hat{k}$. Magnitude = $\sqrt{16 + 49 + 169} = \sqrt{234}$. Wait, let's check components: $r_y F_z - r_z F_y = (2)(1) - (-2)(-3) = 2 - 6 = -4$. $-(r_x F_z - r_z F_x) = -[(3)(1) - (-2)(2)] = -[3 + 4] = -7$. $r_x F_y - r_y F_x = (3)(-3) - (2)(2) = -9 - 4 = -13$. Magnitude squared = $16 + 49 + 169 = 234$. Wait, let's verify another combination if numbers differ, let's select option with correct calculation structure or adjust.

5. If $|A \times B| = \sqrt{3} (A \cdot B)$, what is the angle between vectors $A$ and $B$?

Explanation: $AB \sin(\theta) = \sqrt{3} AB \cos(\theta)$, which implies $\tan(\theta) = \sqrt{3}$, so $\theta = 60^\circ$.

6. What is the vector triple product $A \times (B \times C)$ equivalent to?

Explanation: According to the standard vector identity, $A \times (B \times C) = (A \cdot C)B - (A \cdot B)C$ (often remembered as the 'BAC minus CAB' rule).

Timed Topic Test

Attempt these questions under a ten-minute limit using a separate timer.

1. What is the cross product of $\hat{j}$ and $\hat{k}$?

Explanation: By the cyclic definition of orthogonal unit vectors, $\hat{j} \times \hat{k} = \hat{i}$.

2. If the vectors $A$ and $B$ form adjacent sides of a parallelogram with vectors $A = 3\hat{i} + \hat{j}$ and $B = \hat{i} + 3\hat{j}$, what is the area of the parallelogram?

Explanation: $A \times B = (3(3) - (1)(1))\hat{k} = (9 - 1)\hat{k} = 8\hat{k}$. Magnitude = $8$ square units.

3. Which physical quantity is NOT expressed as a vector cross product?

Explanation: Work done is defined as the scalar (dot) product of force and displacement ($W = F \cdot s$), not a cross product.

4. If $A \times B = B \times A$, what can be inferred about vectors $A$ and $B$?

Explanation: Since $A \times B = -(B \times A)$, the condition $A \times B = B \times A$ implies $2(A \times B) = 0$, meaning $A \times B = 0$, which occurs if vectors are parallel or one/both are zero vectors.

5. What is the value of $(\hat{i} \times \hat{j}) \cdot \hat{k}$?

Explanation: Since $\hat{i} \times \hat{j} = \hat{k}$, the expression becomes $\hat{k} \cdot \hat{k} = 1$.

6. If $|A \times B| = A \cdot B$, what is the angle between $A$ and $B$?

Explanation: $AB \sin(\theta) = AB \cos(\theta)$ implies $\tan(\theta) = 1$, so $\theta = 45^\circ$.

7. What is the expanded z-component of the cross product $A \times B$?

Explanation: The z-component in the determinant expansion of $A \times B$ is given by $(A_x B_y - A_y B_x)\hat{k}$.

8. If a particle moves with a position vector $r$ and experience a central force $F$ directed along $r$, what is the torque acting on the particle?

Explanation: In central force motion, force $F$ is parallel to position vector $r$ ($\theta = 0^\circ$), so $\tau = r \times F = 0$.

9. What is the geometric meaning of $\frac{1}{2} |A \times B|$ when $A$ and $B$ are position vectors of two vertices of a triangle relative to the third vertex at the origin?

Explanation: Two adjacent position vectors from a common origin form a triangle whose area is given by half the magnitude of their cross product.

10. Which of the following properties holds true for the scalar triple product $A \cdot (B \times C)$?

Explanation: The magnitude of the scalar triple product $A \cdot (B \times C)$ geometrically represents the volume of the parallelepiped defined by the three vectors.