Cross Product of Elements of Standard Ordered Basis

Theorem
Let $\tuple {\mathbf i, \mathbf j, \mathbf k}$ be the standard ordered basis of Cartesian $3$-space $S$.

Then:
 * $\mathbf i \times \mathbf i = \mathbf j \times \mathbf j = \mathbf k \times \mathbf k = 0$

and:

where $\times$ denotes the dot product.

Proof
From Cross Product of Vector with Itself is Zero:
 * $\mathbf i \times \mathbf i = \mathbf j \times \mathbf j = \mathbf k \times \mathbf k = 0$

Then we can take the definition of cross product:


 * $\mathbf a \times \mathbf b = \begin {vmatrix}

\mathbf i & \mathbf j & \mathbf k \\ a_i & a_j & a_k \\ b_i & b_j & b_k \\ \end {vmatrix} = \mathbf a \times \mathbf b = \paren {a_j b_k - a_k b_j} \mathbf i - \paren {a_i b_k - a_k b_i} \mathbf j + \paren {a_i b_j - a_j b_i} \mathbf k$

and note that:

Hence:

The remaining identities follow from Vector Cross Product is Anticommutative: