Cross product/K^3/Introduction/Section

A special feature of ${}\mathbb {R} ^{3}$ is the so-called cross product. This assigns, to two given vectors, a vector that is orthogonal to them.

Definition

Let ${}K$ be a field. The operation on ${}K^{3}$, defined by

${}x\times y={\begin{pmatrix}x_{1}\\x_{2}\\x_{3}\end{pmatrix}}\times {\begin{pmatrix}y_{1}\\y_{2}\\y_{3}\end{pmatrix}}:={\begin{pmatrix}x_{2}y_{3}-x_{3}y_{2}\\-x_{1}y_{3}+x_{3}y_{1}\\x_{1}y_{2}-x_{2}y_{1}\end{pmatrix}}\,,$

is called the

cross product.

The cross product is also called the vector product. To remember this formula, one might think

${}x\times y=\det {\begin{pmatrix}e_{1}&x_{1}&y_{1}\\e_{2}&x_{2}&y_{2}\\e_{3}&x_{3}&y_{3}\end{pmatrix}}\,,$

where ${}e_{1},e_{2},e_{3}$ are the standard vectors, and where we expand formally with respect to the first column. In this way, the cross product is defined with respect to the standard basis.

Example

The cross product of the vectors ${}{\begin{pmatrix}5\\8\\-2\end{pmatrix}},{\begin{pmatrix}3\\4\\1\end{pmatrix}}\in \mathbb {R} ^{3}$ is

${}{\begin{pmatrix}5\\8\\-2\end{pmatrix}}\times {\begin{pmatrix}3\\4\\1\end{pmatrix}}={\begin{pmatrix}8\cdot 1-4\cdot (-2)\\-5\cdot 1-2\cdot 3\\5\cdot 4-8\cdot 3\end{pmatrix}}={\begin{pmatrix}16\\-11\\-4\end{pmatrix}}\,.$

Lemma

The cross product on ${}K^{3}$ fulfills the following properties (where ${}x,y,z\in K^{3}$ and

${}a,b\in K$).

1. We have

   ${}x\times y=-(y\times x)\,.$

2. We have

   ${}(ax+by)\times z=a(x\times z)+b(y\times z)\,$

   and

   ${}z\times (ax+by)=a(z\times x)+b(z\times y)\,.$

3. We have

   ${}x\times y=0\,$

   if and only if ${}x$ and ${}y$ are linearly dependent.

4. We have

   ${}x\times (y\times z)+y\times (z\times x)+z\times (x\times y)=0\,.$

5. We have

   ${}\left\langle x\times y,z\right\rangle =\det(x\,,y\,,z)\,,$

   where ${}\left\langle -,-\right\rangle$ denotes the formal evaluation in the sense of the standard inner product.

6. We have

   ${}\left\langle x,x\times y\right\rangle =0=\left\langle y,x\times y\right\rangle \,,$

   where ${}\left\langle -,-\right\rangle$ denotes the formal evaluation in the sense of the standard inner product.

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).

$\Box$

The expression ${}\left\langle x\times y,z\right\rangle$ from (5), that is, the determinant of the three vectors, considered as a column vector, is also called triple product.

Lemma

Let ${}u_{1},u_{2},u_{3}$ be an orthonormal basis of ${}\mathbb {R} ^{3}$ with

${}\det {\left(u_{1},\,u_{2},\,u_{3}\right)}=1\,.$

Then the

cross product ${}x\times y$ can be computed with the coordinates of ${}x$ and ${}y$ with respect to this basis (and the formula from

Definition).

Proof

Let

${}x=c_{1}u_{1}+c_{2}u_{2}+c_{3}u_{3}\,$

and

${}y=d_{1}u_{1}+d_{2}u_{2}+d_{3}u_{3}\,.$

Due to fact  (2), we have

${}x\times y={\left(c_{1}u_{1}+c_{2}u_{2}+c_{3}u_{3}\right)}\times {\left(d_{1}u_{1}+d_{2}u_{2}+d_{3}u_{3}\right)}=\sum _{1\leq i,j\leq 3}c_{i}d_{j}{\left(u_{i}\times u_{j}\right)}\,.$

Due to fact  (3), we have

${}u_{i}\times u_{i}=0\,,$

and, because of fact  (1), we have

${}u_{i}\times u_{j}=-u_{j}\times u_{i}\,.$

According to fact  (6), ${}u_{1}\times u_{2}$ is perpendicular to ${}u_{1}$ and to ${}u_{2}$; therefore,

${}u_{1}\times u_{2}=\lambda u_{3}\,$

with some ${}\lambda \in \mathbb {R}$, as this orthogonality condition defines a line. Because of fact  (5) and the condition, we get

${}\lambda =\left\langle \lambda u_{3},u_{3}\right\rangle =\left\langle u_{1}\times u_{2},u_{3}\right\rangle =\det {\left(u_{1},\,u_{2},\,u_{3}\right)}=1\,;$

hence,

${}u_{1}\times u_{2}=u_{3}\,.$

Using fact  (3), we obtain ${}u_{1}\times u_{3}=-u_{2}$ and ${}u_{2}\times u_{3}=u_{1}$. Altogether we get

${}{\begin{aligned}x\times y&=\sum _{1\leq i,j\leq 3}c_{i}d_{j}{\left(u_{i}\times u_{j}\right)}\\&=\sum _{i<j}(c_{i}d_{j}-c_{j}d_{i}){\left(u_{i}\times u_{j}\right)}\\&=(c_{1}d_{2}-c_{2}d_{1})u_{3}-(c_{1}d_{3}-c_{3}d_{1})u_{2}+(c_{2}d_{3}-c_{3}d_{2})u_{1},\end{aligned}}$

and this is the claim.

$\Box$