K^3/Cross product/Properties/Fact/Proof

Proof

(1) is clear from the definition.

(2). We have

${}{\begin{aligned}{\left(a{\begin{pmatrix}x_{1}\\x_{2}\\x_{3}\end{pmatrix}}+b{\begin{pmatrix}y_{1}\\y_{2}\\y_{3}\end{pmatrix}}\right)}\times {\begin{pmatrix}z_{1}\\z_{2}\\z_{3}\end{pmatrix}}&={\begin{pmatrix}ax_{1}+by_{1}\\ax_{2}+by_{2}\\ax_{3}+by_{3}\end{pmatrix}}\times {\begin{pmatrix}z_{1}\\z_{2}\\z_{3}\end{pmatrix}}\\&={\begin{pmatrix}(ax_{2}+by_{2})z_{3}-(ax_{3}+by_{3})z_{2}\\-(ax_{1}+by_{1})z_{3}+(ax_{3}+by_{3})z_{1}\\(ax_{1}+by_{1})z_{2}-(ax_{2}+by_{2})z_{1}\end{pmatrix}}\\&={\begin{pmatrix}ax_{2}z_{3}-ax_{3}z_{2}\\-ax_{1}z_{3}+ax_{3}z_{1}\\ax_{1}z_{2}-ax_{2}z_{1}\end{pmatrix}}+{\begin{pmatrix}by_{2}z_{3}-by_{3}z_{2}\\-by_{1}z_{3}+by_{1}z_{3}\\by_{1}z_{2}-by_{2}z_{1}\end{pmatrix}}\\&=a{\begin{pmatrix}x_{2}z_{3}-x_{3}z_{2}\\-x_{1}z_{3}+x_{3}z_{1}\\x_{1}z_{2}-x_{2}z_{1}\end{pmatrix}}+b{\begin{pmatrix}y_{2}z_{3}-y_{3}z_{2}\\-y_{1}z_{3}+y_{1}z_{3}\\y_{1}z_{2}-y_{2}z_{1}\end{pmatrix}}\\&=a(x\times z)+b(y\times z).\end{aligned}}$

The second equation follows from this and from (1).

(3). If ${}x$ and ${}y$ are linearly dependent, then we can write ${}x=cy$ (or the other way round). In this case,

${}{\begin{pmatrix}cy_{1}\\cy_{2}\\cy_{3}\end{pmatrix}}\times {\begin{pmatrix}y_{1}\\y_{2}\\y_{3}\end{pmatrix}}={\begin{pmatrix}cy_{2}y_{3}-cy_{2}y_{3}\\-cy_{1}y_{3}+cy_{3}y_{1}\\cy_{1}y_{2}-cy_{2}y_{1}\end{pmatrix}}=0\,.$

If the cross product is ${}0$, then all entries of the vectors ${}{\begin{pmatrix}x_{2}y_{3}-x_{3}y_{2}\\-x_{1}y_{3}+x_{3}y_{1}\\x_{1}y_{2}-x_{2}y_{1}\end{pmatrix}}$ equal ${}0$. For example, let ${}y_{1}\neq 0$. From ${}x_{1}=0$, we can deduce directly

${}x_{2}=x_{3}=0\,,$

and ${}x$ is the zero vector. So suppose that ${}x_{1}\neq 0$. Then ${}y_{2}={\frac {y_{1}}{x_{1}}}x_{2}$ and ${}y_{3}={\frac {y_{1}}{x_{1}}}x_{3}$; therefore, we get

${}y={\frac {y_{1}}{x_{1}}}x\,.$

(4). See exercise.

(5). We have

${}{\begin{aligned}\left\langle x\times y,z\right\rangle &=\left\langle {\begin{pmatrix}x_{2}y_{3}-x_{3}y_{2}\\-x_{1}y_{3}+x_{3}y_{1}\\x_{1}y_{2}-x_{2}y_{1}\end{pmatrix}},{\begin{pmatrix}z_{1}\\z_{2}\\z_{3}\end{pmatrix}}\right\rangle \\&=z_{1}x_{2}y_{3}-z_{1}x_{3}y_{2}-z_{2}x_{1}y_{3}+z_{2}x_{3}y_{1}+z_{3}x_{1}y_{2}-z_{3}x_{2}y_{1}.\end{aligned}}$

This coincides with the determinant, due to Sarrus.

(6) follows from (5).

To fact