Cross Product of Vector with Itself is Zero/Proof 2
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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: $\S 3$