Q.If $\vec{a}, \vec{b}, \vec{c}$ are three Vectors and $\vec{a} + \vec{b} + \vec{c} = 0$ prove that $\vec{a}\times\vec{b} = \vec{b}\times\vec{c} = \vec{c}\times\vec{a}$

The Distributive Property for the Cross Product

You already know distribution from arithmetic: $a(b+c) = ab + ac$ — multiplication "spreads" over addition. The cross product obeys the same kind of rule, with one important twist: it is not commutative, so the order of the vectors must be respected.

Intuition

Put two vectors $\vec{u}$ and $\vec{v}$ at a common point; their sum $\vec{u}+\vec{v}$ is the diagonal of the parallelogram they form. Crossing a third vector $\vec{w}$ with this sum, $\vec{w}\times(\vec{u}+\vec{v})$, gives the same result as crossing $\vec{w}$ with each piece separately and adding. This works because the cross product is bilinear — its geometry (parallelogram area, right-hand rule) is linear in each argument.

The Precise Statement

For any vectors $\vec{a},\vec{b},\vec{c}$ in $\mathbb{R}^3$ and any scalar $k$:

1. Left distributivity: $\vec{a}\times(\vec{b}+\vec{c}) = \vec{a}\times\vec{b} + \vec{a}\times\vec{c}$
2. Right distributivity: $(\vec{b}+\vec{c})\times\vec{a} = \vec{b}\times\vec{a} + \vec{c}\times\vec{a}$
3. Scalar multiplication: $(k\vec{a})\times\vec{b} = k(\vec{a}\times\vec{b}) = \vec{a}\times(k\vec{b})$

Why It Matters

Distributivity lets you expand a product of sums term by term, exactly like FOIL in algebra:

Every term keeps its left–right order intact. This is the routine behind expanding cross products in proofs (areas of triangles, testing collinearity, deriving vector identities). …