Proof of (A + B) × (A − B) = −2(A × B)

For vectors $A$ and $B$ in three-dimensional space, the identity follows from two basic properties of the cross product: it is distributive over addition, and reversing the order of its inputs reverses its sign. Expanding the left-hand side leaves two equal terms, each equal to $-(A \times B)$.

Here, the notation $A X B$ is interpreted as $A \times B$, the vector cross product.

Proof using cross-product rules

The cross product is distributive in each input, so expand the expression one term at a time:

$$\begin{aligned} (A+B) \times (A-B) &= A \times A - A \times B + B \times A - B \times B. \end{aligned}$$

A vector crossed with itself is the zero vector: $A \times A=\mathbf{0}$ and $B \times B=\mathbf{0}$. Also, the cross product is antisymmetric:

$$B \times A = -(A \times B).$$

Substitute these facts into the expansion:

$$\begin{aligned} (A+B) \times (A-B) &= \mathbf{0} - A \times B - A \times B - \mathbf{0} \\ &= -2(A \times B). \end{aligned}$$

That proves the identity. The negative sign is essential: it comes from reversing the order in $B \times A$.

Why the cross-product rules apply

For vectors in $\mathbb{R}^3$, the cross product is bilinear, meaning it distributes over vector addition and scalar multiplication. It is also antisymmetric, so swapping its inputs changes the sign. In particular, swapping a vector with itself gives

$$A \times A = -(A \times A),$$

which implies $A \times A=\mathbf{0}$. These properties make the expansion proof shorter than computing the cross product from coordinates.

Worked numerical example

Take

$$A=(1,2,0), \qquad B=(0,1,3).$$

First calculate the cross product on the right:

$$A \times B = (6,-3,1),$$

so

$$-2(A \times B)=(-12,6,-2).$$

Now compute the left-hand side directly. The two vectors in the cross product are

$$A+B=(1,3,3), \qquad A-B=(1,1,-3).$$

Their cross product is

$$\begin{aligned} (A+B) \times (A-B) &= (3(-3)-3(1),\;3(1)-1(-3),\;1(1)-3(1)) \\ &= (-12,6,-2). \end{aligned}$$

Both sides agree, as the identity predicts.

Common mistakes

• Dropping the minus sign: The term $B \times A$ is $-(A \times B)$, not $A \times B$.
• Treating the cross product like ordinary multiplication: It is not commutative. In general, $A \times B \ne B \times A$.
• Forgetting the self-products: Both $A \times A$ and $B \times B$ are zero vectors, so they disappear from the expansion.
• Confusing cross and dot products: This identity concerns the vector cross product in three dimensions. A dot product has different properties and produces a scalar.

The same proof works if $A$ and $B$ are vector-valued expressions, provided the cross product is defined for them and the usual distributive and antisymmetric rules apply.