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Cross Product

Formulas & Definitions
Core Definition

The cross product (also known as the vector product) is an operation defined exclusively for vectors in three-dimensional space (R3\mathbb{R}^3). Unlike the dot product, the cross product produces a new vector that is strictly orthogonal (perpendicular) to both original vectors:

  1. Determinant Form (3×33 \times 3 Matrix):
    u⃗×v⃗=∣i^j^k^u1u2u3v1v2v3∣=(u2v3−u3v2)i^−(u1v3−u3v1)j^+(u1v2−u2v1)k^\vec{u} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix} = (u_2 v_3 - u_3 v_2)\hat{i} - (u_1 v_3 - u_3 v_1)\hat{j} + (u_1 v_2 - u_2 v_1)\hat{k}
  2. Geometric Magnitude & Sine Form:
    ∣u⃗×v⃗∣=∣u⃗∣∣v⃗∣sin⁡θ|\vec{u} \times \vec{v}| = |\vec{u}| |\vec{v}| \sin\theta
    where θ\theta is the angle between u⃗\vec{u} and v⃗\vec{v} (0≤θ≤π0 \le \theta \le \pi).

Concept Explanation

Direction and the Right-Hand Rule

The direction of u⃗×v⃗\vec{u} \times \vec{v} is determined by the Right-Hand Rule:

  1. Point the index finger of your right hand along vector u⃗\vec{u}.
  2. Curl your middle fingers toward vector v⃗\vec{v}.
  3. Your thumb points in the exact direction of u⃗×v⃗\vec{u} \times \vec{v} (orthogonal to the plane containing u⃗\vec{u} and v⃗\vec{v}).
Common Student Trap: Dot vs. Cross Product Output
  • Dot Product (u⃗⋅v⃗\vec{u} \cdot \vec{v}): Produces a SCALAR (a single real number).
  • Cross Product (u⃗×v⃗\vec{u} \times \vec{v}): Produces a VECTOR (has components ⟨cx,cy,cz⟩\langle c_x, c_y, c_z \rangle or cxi^+cyj^+czk^c_x\hat{i} + c_y\hat{j} + c_z\hat{k}).

Never return a scalar for a cross product (unless asked specifically for its magnitude ∣u⃗×v⃗∣|\vec{u} \times \vec{v}|) and never return a vector for a dot product!

Key Algebraic Properties

  1. Anti-Commutativity: u⃗×v⃗=−(v⃗×u⃗)\vec{u} \times \vec{v} = -(\vec{v} \times \vec{u}) (reversing order reverses the spatial vector direction).
  2. Parallel Vectors Test: If u⃗\vec{u} and v⃗\vec{v} are parallel (θ=0\theta = 0 or π\pi), sin⁡θ=0  ⟹  u⃗×v⃗=0⃗\sin\theta = 0 \implies \vec{u} \times \vec{v} = \vec{0} (the zero vector).
  3. Distributive Law: u⃗×(v⃗+w⃗)=u⃗×v⃗+u⃗×w⃗\vec{u} \times (\vec{v} + \vec{w}) = \vec{u} \times \vec{v} + \vec{u} \times \vec{w}.
  4. Scalar Multiplication: (cu⃗)×v⃗=c(u⃗×v⃗)=u⃗×(cv⃗)(c\vec{u}) \times \vec{v} = c(\vec{u} \times \vec{v}) = \vec{u} \times (c\vec{v}).

Geometric Interpretation: Parallelogram Area

The magnitude ∣u⃗×v⃗∣|\vec{u} \times \vec{v}| represents the geometric area of the parallelogram formed by u⃗\vec{u} and v⃗\vec{v}. Consequently, the area of a triangle with vertices defined by these vectors is:

Areatriangle=12∣u⃗×v⃗∣\text{Area}_{\text{triangle}} = \frac{1}{2} |\vec{u} \times \vec{v}|

Engineering Applications

Example Problems

Example 1: 3D Cross Product & Orthogonal Unit Vector

Compute a⃗×b⃗\vec{a} \times \vec{b} for a⃗=⟨1,2,3⟩\vec{a} = \langle 1, 2, 3 \rangle and b⃗=⟨4,5,6⟩\vec{b} = \langle 4, 5, 6 \rangle, then find a unit vector orthogonal to both.

Solution
  1. Set Up Determinant:
    a⃗×b⃗=∣i^j^k^123456∣\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 4 & 5 & 6 \end{vmatrix}
  2. Expand Along Top Row:
    a⃗×b⃗=i^(2⋅6−3⋅5)−j^(1⋅6−3⋅4)+k^(1⋅5−2⋅4)\vec{a} \times \vec{b} = \hat{i}(2\cdot 6 - 3\cdot 5) - \hat{j}(1\cdot 6 - 3\cdot 4) + \hat{k}(1\cdot 5 - 2\cdot 4)
    =i^(12−15)−j^(6−12)+k^(5−8)=−3i^+6j^−3k^=⟨−3,6,−3⟩= \hat{i}(12 - 15) - \hat{j}(6 - 12) + \hat{k}(5 - 8) = -3\hat{i} + 6\hat{j} - 3\hat{k} = \langle -3, 6, -3 \rangle
  3. Find Orthogonal Unit Vector:
    ∣a⃗×b⃗∣=(−3)2+62+(−3)2=9+36+9=54=36|\vec{a} \times \vec{b}| = \sqrt{(-3)^2 + 6^2 + (-3)^2} = \sqrt{9 + 36 + 9} = \sqrt{54} = 3\sqrt{6}
    u^⊥=⟨−3,6,−3⟩36=⟨−16,26,−16⟩\hat{u}_{\perp} = \frac{\langle -3, 6, -3 \rangle}{3\sqrt{6}} = \left\langle -\frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}}, -\frac{1}{\sqrt{6}} \right\rangle

Connections & References