Cross Product of Elements of Standard Ordered Basis

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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:

\(\ds \mathbf i \times \mathbf j\) \(=\) \(\, \ds \mathbf k \, \) \(\, \ds = \, \) \(\ds -\mathbf j \times \mathbf i\)
\(\ds \mathbf j \times \mathbf k\) \(=\) \(\, \ds \mathbf i \, \) \(\, \ds = \, \) \(\ds -\mathbf k \times \mathbf j\)
\(\ds \mathbf k \times \mathbf i\) \(=\) \(\, \ds \mathbf j \, \) \(\, \ds = \, \) \(\ds -\mathbf i \times \mathbf k\)

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:

\(\ds \mathbf i\) \(=\) \(\ds 1 \mathbf i + 0 \mathbf j + 0 \mathbf k\)
\(\ds \mathbf j\) \(=\) \(\ds 0 \mathbf i + 1 \mathbf j + 0 \mathbf k\)
\(\ds \mathbf k\) \(=\) \(\ds 0 \mathbf i + 0 \mathbf j + 1 \mathbf k\)


Hence:

\(\ds \mathbf i \times \mathbf j\) \(=\) \(\ds \paren {0 \cdot 0 - 0 \cdot 1} \mathbf i - \paren {1 \cdot 0 - 0 \cdot 0} \mathbf j + \paren {1 \cdot 1 - 0 \cdot 0} \mathbf k\) \(\ds = \mathbf k\)
\(\ds \mathbf j \times \mathbf k\) \(=\) \(\ds \paren {1 \cdot 1 - 0 \cdot 0} \mathbf i - \paren {0 \cdot 1 - 0 \cdot 0} \mathbf j + \paren {0 \cdot 0 - 1 \cdot 0} \mathbf k\) \(\ds = \mathbf i\)
\(\ds \mathbf k \times \mathbf i\) \(=\) \(\ds \paren {0 \cdot 0 - 1 \cdot 0} \mathbf i - \paren {0 \cdot 0 - 1 \cdot 1} \mathbf j + \paren {0 \cdot 0 - 0 \cdot 1} \mathbf k\) \(\ds = \mathbf j\)


The remaining identities follow from Vector Cross Product is Anticommutative:

\(\ds \mathbf i \times \mathbf j\) \(=\) \(\ds -\mathbf j \times \mathbf i\)
\(\ds \mathbf j \times \mathbf k\) \(=\) \(\ds -\mathbf k \times \mathbf j\)
\(\ds \mathbf k \times \mathbf i\) \(=\) \(\ds -\mathbf i \times \mathbf k\)

$\blacksquare$


Sources

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