Cross Product of Vector with Itself is Zero/Proof 2

Theorem

Let $\mathbf x$ be a vector in a vector space of $3$ dimensions:

$\mathbf x = x_i \mathbf i + x_j \mathbf j + x_k \mathbf k$

Then:

$\mathbf x \times \mathbf x = \mathbf 0$

where $\times$ denotes vector cross product.

Proof

By definition, a vector is parallel to itself.

The result follows from Cross Product of Parallel Vectors.

$\blacksquare$

Sources

• 1957: D.E. Rutherford: Vector Methods (9th ed.) ... (previous) ... (next): Chapter $\text I$: Vector Algebra: $\S 3$