Physics Lab
Class XI/Chapter 4: Motion in a Plane/Cross Product (Vector Product)

Cross Product (Vector Product)

The cross product turns two vectors into a third vector perpendicular to both. It encodes rotational quantities — torque, angular momentum, magnetic force.

Concept

Definition

For vectors A⃗\vec{A} and B⃗\vec{B} with angle θ\theta between them: A⃗×B⃗=ABsin⁡θ n^,\vec{A}\times\vec{B} = AB\sin\theta\,\hat{n}, where n^\hat{n} is a unit vector perpendicular to the plane of A⃗\vec{A} and B⃗\vec{B}, given by the right-hand rule.

Right-Hand Rule

Point your fingers along A⃗\vec{A}, curl toward B⃗\vec{B}; the thumb points along A⃗×B⃗\vec{A}\times\vec{B}.

Component Form (Determinant)

A⃗×B⃗=∣i^j^k^AxAyAzBxByBz∣\vec{A}\times\vec{B} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\A_x&A_y&A_z\\B_x&B_y&B_z\end{vmatrix}

Expanding: A⃗×B⃗=(AyBz−AzBy)i^−(AxBz−AzBx)j^+(AxBy−AyBx)k^.\vec{A}\times\vec{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}.

Identities of Unit Vectors

CrossResult
i^×j^\hat{i}\times\hat{j}k^\hat{k}
j^×k^\hat{j}\times\hat{k}i^\hat{i}
k^×i^\hat{k}\times\hat{i}j^\hat{j}
j^×i^\hat{j}\times\hat{i}−k^-\hat{k}
i^×i^\hat{i}\times\hat{i}0⃗\vec{0}

The cyclic pattern: i→j→k→ii \to j \to k \to i gives positive results.

Properties

PropertyStatement
Anti-commutativeA⃗×B⃗=−B⃗×A⃗\vec{A}\times\vec{B} = -\vec{B}\times\vec{A}
DistributiveA⃗×(B⃗+C⃗)=A⃗×B⃗+A⃗×C⃗\vec{A}\times(\vec{B}+\vec{C}) = \vec{A}\times\vec{B} + \vec{A}\times\vec{C}
Self-cross zeroA⃗×A⃗=0⃗\vec{A}\times\vec{A} = \vec{0}
Parallel testA⃗×B⃗=0⃗\vec{A}\times\vec{B} = \vec{0} iff parallel/antiparallel

Geometric Meaning

∣A⃗×B⃗∣|\vec{A}\times\vec{B}| = area of the parallelogram formed by A⃗\vec{A} and B⃗\vec{B}.

Applications

  • Torque: τ⃗=r⃗×F⃗\vec{\tau} = \vec{r}\times\vec{F}.
  • Angular momentum: L⃗=r⃗×p⃗\vec{L} = \vec{r}\times\vec{p}.
  • Magnetic force on a charge: F⃗=qv⃗×B⃗\vec{F} = q\vec{v}\times\vec{B}.
  • Area vector for a planar loop.

Worked Example

Q1: A⃗=2i^+3j^\vec{A} = 2\hat{i} + 3\hat{j}, B⃗=i^−j^+2k^\vec{B} = \hat{i} - \hat{j} + 2\hat{k}. Find A⃗×B⃗\vec{A}\times\vec{B}.

Solution: A⃗×B⃗=∣i^j^k^2301−12∣\vec{A}\times\vec{B} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&3&0\\1&-1&2\end{vmatrix}

  • i^\hat{i}-component: 3(2)−0(−1)=63(2) - 0(-1) = 6.
  • j^\hat{j}-component: −(2(2)−0(1))=−4-(2(2) - 0(1)) = -4.
  • k^\hat{k}-component: 2(−1)−3(1)=−52(-1) - 3(1) = -5.

So A⃗×B⃗=6i^−4j^−5k^\vec{A}\times\vec{B} = 6\hat{i} - 4\hat{j} - 5\hat{k}.

Q2: Torque from F⃗=3i^\vec{F} = 3\hat{i} N applied at r⃗=2j^\vec{r} = 2\hat{j} m from origin: τ⃗=r⃗×F⃗=2j^×3i^=6(j^×i^)=−6k^ N⋅m.\vec{\tau} = \vec{r}\times\vec{F} = 2\hat{j}\times 3\hat{i} = 6(\hat{j}\times\hat{i}) = -6\hat{k}\,\text{N·m}.

Common Confusions

  • "A⃗×B⃗=B⃗×A⃗\vec{A}\times\vec{B} = \vec{B}\times\vec{A}." — Wrong, it changes sign: anti-commutative.
  • "A⃗×B⃗=0\vec{A}\times\vec{B} = 0 implies A⃗=0⃗\vec{A} = \vec{0}." — Could also mean A⃗∥B⃗\vec{A}\parallel\vec{B}.
  • "Cross product gives a scalar." — No, it gives a vector.
  • Wrong sign on j^\hat{j} in the determinant expansion — always remember the alternating signs +,−,++, -, +.

Key Takeaways

  • A⃗×B⃗=ABsin⁡θ n^\vec{A}\times\vec{B} = AB\sin\theta\,\hat{n}, direction by right-hand rule.
  • Magnitude equals the area of the parallelogram spanned by the two vectors.
  • Anti-commutative; distributive; self-cross is zero.
  • Computed via the 3×33\times 3 determinant in components.
  • Cornerstone of rotational and magnetic physics.

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