Cross product — formula, geometric meaning (area), right-hand rule

WHAT is the cross product?

This lives only in $\mathbb{R}^3$ (the cross product as a vector is special to 3 dimensions).

WHY perpendicular and WHY area? (derive from scratch)

We want a recipe that turns two vectors into a third with the three properties above. Let's derive the coordinate formula instead of memorizing it.

Step 1 — Demand perpendicularity. Write $\vec c = \vec a \times \vec b = (c_1,c_2,c_3)$. We require $\vec c\cdot\vec a = 0$ and $\vec c\cdot\vec b=0$: $a_1c_1 + a_2c_2 + a_3c_3 = 0, \qquad b_1c_1 + b_2c_2 + b_3c_3 = 0.$ Why this step? Property (1) says $\vec c$ is orthogonal to both inputs — that's two linear equations in the three unknowns $c_1,c_2,c_3$. Two equations, three unknowns → a whole line of solutions, all pointing the same direction.

Step 2 — Solve the system. Treat it as eliminating variables. From the two equations the solution (up to a scalar) is $\vec c = \lambda\,(a_2b_3 - a_3b_2,\ \ a_3b_1 - a_1b_3,\ \ a_1b_2 - a_2b_1).$ Why this step? Plug this back in: $\vec a\cdot\vec c = a_1(a_2b_3-a_3b_2)+a_2(a_3b_1-a_1b_3)+a_3(a_1b_2-a_2b_1)=0$ — every term cancels. So any such $\vec c$ is automatically perpendicular. We just need to fix the scale $\lambda$.

Step 3 — Fix the scale using the area condition. Choose $\lambda = 1$. We now prove this choice gives magnitude $|\vec a||\vec b|\sin\theta$. Use the algebraic identity (Lagrange's identity): $|\vec a\times\vec b|^2 = |\vec a|^2|\vec b|^2 - (\vec a\cdot\vec b)^2.$ Since $\vec a\cdot\vec b = |\vec a||\vec b|\cos\theta$, $|\vec a\times\vec b|^2 = |\vec a|^2|\vec b|^2(1-\cos^2\theta) = |\vec a|^2|\vec b|^2\sin^2\theta.$ $\boxed{\ |\vec a\times\vec b| = |\vec a||\vec b|\sin\theta\ }$ Why this step? The base of the parallelogram is $|\vec a|$ and its height is $|\vec b|\sin\theta$ (drop a perpendicular from the tip of $\vec b$). Base × height = area = $|\vec a||\vec b|\sin\theta$. So $\lambda=1$ is exactly the right scale.

The right-hand rule (direction)

Point the fingers of your right hand along $\vec a$, curl them toward $\vec b$ (through angle $\theta$); your thumb points along $\vec a\times\vec b$.

Consequences:

• $\hat i\times\hat j=\hat k,\quad \hat j\times\hat k=\hat i,\quad \hat k\times\hat i=\hat j$ (cyclic order $i\to j\to k\to i$).
• Anticommutative: $\vec b\times\vec a = -(\vec a\times\vec b)$ — swap hand direction, flip thumb.
• $\vec a\times\vec a = \vec 0$ (parallel vectors → $\sin 0 =0$ → zero area).

Worked examples

Forecast-then-Verify

Common mistakes (steel-manned)

80/20 — the 20% that gives 80%

1. Output is a vector ⊥ to both inputs.
2. Length = area $=|\vec a||\vec b|\sin\theta$; triangle area is half.
3. Right-hand rule gives direction; swapping order flips the sign.
4. Compute via the $3\times3$ determinant (watch the middle sign).

Connections

• Dot Product — partner operation; $|\vec a\times\vec b|^2 + (\vec a\cdot\vec b)^2 = |\vec a|^2|\vec b|^2$
• Determinants — the formula is a $3\times3$ determinant; signs come from cofactors
• Scalar Triple Product — $\vec a\cdot(\vec b\times\vec c)$ = signed volume of a parallelepiped
• Area and Volume — parallelogram/triangle area, oriented area
• Orthogonality — why perpendicularity gives two linear equations
• Torque and Angular Momentum — physics applications of $\vec r\times\vec F$

Flashcards

Concept Map

Hinglish (regional understanding)