Editing Cross product (section)

=== Coordinate notation ===
[[File:3D Vector.svg|thumb|right|[[Standard basis]] vectors '''i''', '''j''', '''k''' and [[vector component]]s of '''a''', denoted here '''a'''<sub>x</sub>, '''a'''<sub>y</sub>, '''a'''<sub>z</sub>]]
If <math>(\mathbf{\color{blue}{i}}, \mathbf{\color{red}{j}}, \mathbf{\color{green}{k}})</math> is a positively oriented orthonormal basis, the basis vectors satisfy the following equalities<ref name=":1" />
:<math>\begin{alignat}{2}
  \mathbf{\color{blue}{i}}&\times\mathbf{\color{red}{j}} &&= \mathbf{\color{green}{k}}\\
  \mathbf{\color{red}{j}}&\times\mathbf{\color{green}{k}} &&= \mathbf{\color{blue}{i}}\\
  \mathbf{\color{green}{k}}&\times\mathbf{\color{blue}{i}} &&= \mathbf{\color{red}{j}}
\end{alignat}</math>
A [[mnemonic]] for these formulas is that they can be deduced from any other of them by a [[cyclic permutation]] of the basis vectors. This mnemonic applies also to many formulas given in this article.

The [[anticommutativity]] of the cross product, implies that 
:<math>\begin{alignat}{2}
  \mathbf{\color{red}{j}}&\times\mathbf{\color{blue}{i}} &&= -\mathbf{\color{green}{k}}\\
  \mathbf{\color{green}{k}}&\times\mathbf{\color{red}{j}} &&= -\mathbf{\color{blue}{i}}\\
  \mathbf{\color{blue}{i}}&\times\mathbf{\color{green}{k}} &&= -\mathbf{\color{red}{j}}
\end{alignat}</math>

The anticommutativity of the cross product (and the obvious lack of linear independence) also implies that
:<math>\mathbf{\color{blue}{i}}\times\mathbf{\color{blue}{i}} = \mathbf{\color{red}{j}}\times\mathbf{\color{red}{j}} = \mathbf{\color{green}{k}}\times\mathbf{\color{green}{k}} = \mathbf{0}</math> (the [[zero vector]]).

These equalities, together with the [[distributivity]] and [[linearity]] of the cross product (though neither follows easily from the definition given above), are sufficient to determine the cross product of any two vectors '''a''' and '''b'''. Each vector can be defined as the sum of three orthogonal components parallel to the standard basis vectors:
:<math>\begin{alignat}{3}
  \mathbf{a} &= a_1\mathbf{\color{blue}{i}} &&+ a_2\mathbf{\color{red}{j}} &&+ a_3\mathbf{\color{green}{k}} \\
  \mathbf{b} &= b_1\mathbf{\color{blue}{i}} &&+ b_2\mathbf{\color{red}{j}} &&+ b_3\mathbf{\color{green}{k}}
\end{alignat}</math>

Their cross product {{nowrap|1='''a''' × '''b'''}} can be expanded using distributivity:
:<math> \begin{align}
 \mathbf{a}\times\mathbf{b} = {} &(a_1\mathbf{\color{blue}{i}} + a_2\mathbf{\color{red}{j}} + a_3\mathbf{\color{green}{k}}) \times (b_1\mathbf{\color{blue}{i}} + b_2\mathbf{\color{red}{j}} + b_3\mathbf{\color{green}{k}})\\
                            = {} &a_1b_1(\mathbf{\color{blue}{i}} \times \mathbf{\color{blue}{i}}) + a_1b_2(\mathbf{\color{blue}{i}} \times \mathbf{\color{red}{j}}) + a_1b_3(\mathbf{\color{blue}{i}} \times \mathbf{\color{green}{k}}) + {}\\
                                 &a_2b_1(\mathbf{\color{red}{j}} \times \mathbf{\color{blue}{i}}) + a_2b_2(\mathbf{\color{red}{j}} \times \mathbf{\color{red}{j}}) + a_2b_3(\mathbf{\color{red}{j}} \times \mathbf{\color{green}{k}}) + {}\\
                                 &a_3b_1(\mathbf{\color{green}{k}} \times \mathbf{\color{blue}{i}}) + a_3b_2(\mathbf{\color{green}{k}} \times \mathbf{\color{red}{j}}) + a_3b_3(\mathbf{\color{green}{k}} \times \mathbf{\color{green}{k}})\\
\end{align}</math>

This can be interpreted as the decomposition of {{nowrap|1='''a''' × '''b'''}} into the sum of nine simpler cross products involving vectors aligned with '''i''', '''j''', or '''k'''. Each one of these nine cross products operates on two vectors that are easy to handle as they are either parallel or orthogonal to each other. From this decomposition, by using the above-mentioned [[#Coordinate notation|equalities]] and collecting similar terms, we obtain:
:<math>\begin{align}
 \mathbf{a}\times\mathbf{b} = {} &\quad\ a_1b_1\mathbf{0} + a_1b_2\mathbf{\color{green}{k}} - a_1b_3\mathbf{\color{red}{j}} \\
                                 &- a_2b_1\mathbf{\color{green}{k}} + a_2b_2\mathbf{0} + a_2b_3\mathbf{\color{blue}{i}} \\
                                 &+ a_3b_1\mathbf{\color{red}{j}}\ - a_3b_2\mathbf{\color{blue}{i}}\ + a_3b_3\mathbf{0} \\
                            = {} &(a_2b_3 - a_3b_2)\mathbf{\color{blue}{i}} + (a_3b_1 - a_1b_3)\mathbf{\color{red}{j}} + (a_1b_2 - a_2b_1)\mathbf{\color{green}{k}}\\
\end{align}</math>

meaning that the three [[scalar component]]s of the resulting vector '''s''' = ''s''<sub>1</sub>'''i''' + ''s''<sub>2</sub>'''j''' + ''s''<sub>3</sub>'''k''' = {{nowrap|1='''a''' × '''b'''}} are
:<math>\begin{align}
  s_1 &= a_2b_3-a_3b_2\\
  s_2 &= a_3b_1-a_1b_3\\
  s_3 &= a_1b_2-a_2b_1
\end{align}</math>

Using [[column vector]]s, we can represent the same result as follows:
:<math>\begin{bmatrix}s_1\\s_2\\s_3\end{bmatrix}=\begin{bmatrix}a_2b_3-a_3b_2\\a_3b_1-a_1b_3\\a_1b_2-a_2b_1\end{bmatrix}</math>