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\begin{document}
\centerline{Math 201 lecture for Wednesday, Week 13}
\bigskip

\centerline{\textbf{Cross product}}  
\bigskip

Let~$v_1,\dots,v_{n-1}$ be a set of~$n-1$ vectors in~$\R^n$.  Define the function

\begin{align*}
  \chi\colon\R^n&\to\R\\
  x&\mapsto\det(x,v_1,\dots,v_{n-1}).
\end{align*}
where we think of the determinant as a function of the
rows~$x,v_1,\dots,v_{n-1}$ of a matrix, as usual.  The~$1\times n$ matrix
representing~$\chi$ has the form~$(a_1\cdots a_n)$. We define the {\em cross
product} to be the row vector
\[
  v_1\times\cdots\times v_{n-1}:=(a_1,\dots.,a_n).
\]
The mapping~$\chi$ is just dot product with the cross product:
\begin{center}
  \begin{tikzpicture}
    \node at (0,0) {$\displaystyle
  \chi(x) =(a_1\cdots a_n)\left(\begin{array}{c}
      x_1\\\vdots\\x_n
  \end{array} \right)
  =(a_1,\dots,a_n)\cdot x=(v_1\times\cdots\times v_{n-1})\cdot x.
    $};
    \node[brown] at (-2.6,-1.3) {matrix multiplication};
    \node[brown] at (0.55,-1.3) {dot product};
  \end{tikzpicture}
\end{center}
\medskip

{\bf Theorem.} (Properties of the cross product.)
\begin{enumerate}[itemsep=0pt]
  \item\label{13wedit1} The cross product is a multilinear alternating function of~$v_1,\dots,v_{n-1}$.
  \item\label{13wedit2} Swapping~$v_i$ with~$v_j$ for~$i\neq j$ changes the sign of the cross
    product.
  \item\label{13wedit3} Adding a scalar multiple of~$v_i$ to~$v_j$ for some~$i\neq j$ does not
    change the cross product.
  \item\label{13wedit4} The cross product is orthogonal to the subspace spanned
    by~$v_1,\dots,v_{n-1}$.
  \item\label{13wedit5} The length of the cross product is the volume of the parallelepiped
    spanned by~$v_1,\dots,v_{n-1}$.
  \item\label{13wedit6} Given~$w\in\R^n$, the volume of the parallelepiped spanned by~$w$
    and~$v_1,\dots,v_{n-1}$ is~$|w\cdot(v_1\times\cdots\times v_{n-1})|$.
  \item\label{13wedit7} Let~$A$ be the~$(n-1)\times n$ matrix with
    rows~$v_1,\dots,v_{n-1}$, and let~$A^{(j)}$ be the~$(n-1)\times(n-1)$ matrix
    formed by removing the~$j$-th column of~$A$.  Then
    \[
      v_1\times\cdots\times
      v_{n-1}
      =
      \left(\det(A^{(1)}),-\det(A^{(2)}),\det(A^{(3)}),\dots,(-1)^{n-1}\det(A^{(n)})\right).
    \]
\end{enumerate}
\smallskip

{\bf Proof.} Properties~\ref{13wedit1}--\ref{13wedit3} follow immediately from
the properties of $\det(x,v_1,\dots,v_{n-1})$.  For property~\ref{13wedit4}, note that
\[
  (v_1\times\cdots\times v_{n-1})\cdot v_i=\det(v_i,v_1,\dots,v_{n-1})=0
\]
since~$v_i$ is a repeated row.  

For property~\ref{13wedit5}, let~$P$ be the parallelepiped
spanned by~$v_1,\dots,v_{n-1}$, and let~$Q$ be the parallelepiped spanned
by~$v_1\times\cdots\times v_{n-1}$ and $v_1,\dots,v_{n-1}$.  Since~$v_1\times\cdots\times
v_{n-1}$ is perpendicular to~$P$, the volume of~$Q$ is given by the volume of
the base,~$P$, times the height~$\lVert v_1\times\cdots\times v_{n-1}\rVert$:
\begin{equation}\label{12wed1}
  \vol(Q)=\lVert v_1\times\cdots\times v_{n-1}\rVert\vol(P).
\end{equation}
The volume of~$Q$ is the absolute value of the determinant of its spanning
vectors.  Therefore,
\begin{align*}
  \vol(Q)&= |\det(v_1\times\cdots\times v_{n-1},v_1,\dots,v_{n-1})|\\
  &=|\chi(v_1\times\cdots\times v_{n-1},v_1,\dots,v_{n-1})|\\
  &= (v_1\times\cdots\times v_{n-1})\cdot(v_1\times\cdots\times v_{n-1})\\
  &= \lVert v_1\times\cdots\times v_{n-1}\rVert^2.
\end{align*}
Combining this with equation~\eqref{12wed1} yields the result:
\[
  \lVert v_1\times\cdots\times v_{n-1}\rVert=\vol(P).
\]

For property~\ref{13wedit6}, note that
\[
  |w\cdot(v_1\times\cdots\times v_{n-1})|=|\det(w,v_1,\dots,v_{n-1})|,
\]
which gives the volume of the parallelepiped in question.

Property~\ref{13wedit7} follows by expanding the determinant defining~$\chi$ along its first
row:
\begin{align*}
  \chi(x)&= \det(x,v_1,\dots,v_{n-1})\\
  &= \det(A^{(1)}x_1-\det(A^{(2)})x_2+\dots+(-1)^{n-1}\det(A^{(n)})x_n\\
    &=
  (\det(A^{(1)},-\det(A^{(2)}),\dots,(-1)^{n-1}\det(A^{(n)}))\cdot
    (x_1,\dots,x_n).
\end{align*}
\hfill$\Box$
\medskip

{\bf The cross product in~$\R^3$.} The cross product is most well-known in the
case~$n=3$.  Here, we have vectors~$x=(x_1,x_2,x_3)$ and~$y=(y_1,y_2,y_3)$.  The cross product is
\[
  x\times y=(x_2y_3-x_3y_2,x_3y_1-x_1y_3,x_1y_2-x_2y_1)\in\R^3.
\]
The usual mnemonic is
\[
  x\times y = 
  \det
    \left(\begin{array}{ccc}
	\mathbf{i}&\mathbf{j}&\mathbf{k}\\
	x_1&x_2&x_3\\
	y_1&y_2&y_3
\end{array} \right)
=(x_2y_3-x_3y_2)\mathbf{i}-(x_1y_3-x_3y_1)\mathbf{j}+(x_1y_2-x_2y_1)\mathbf{k},
\]
where~$\mathbf{i}=e_1=(1,0,0)$, $\mathbf{j}=e_2=(0,1,0)$,
and~$\mathbf{k}=e_3=(0,0,1)$. We get exactly the formula given by
part~\ref{13wedit7} of the
Theorem.  The above is only a mnemonic since we have not defined a determinant in the
case where the entries are vectors of various dimensions.  

The cross product here is perpendicular to the parallelogram spanned
by~$x$ and~$y$, and its length is
\[
  \lVert x\times y\rVert=\lVert x\rVert\lVert y\rVert\sin(\theta)
\]
where~$\theta$ is the angle between~$x$ and~$y$.  This last formula gives the
area of the parallelogram spanned by~$x$ and~$y$:
\begin{center}
  \begin{tikzpicture}[scale=1.4]
    \draw[fill=gray!10] (0,0)--(3,1)--(4,3)--(1,2)--(0,0);
    \draw[->,thick] (0,0)--(3,1);
    \draw[->,thick] (0,0)--(1,2);
    \node at (0.5,0.45) {$\theta$};
    \node at (1.5,0.0) {$\lVert x\rVert$};
    \node at (3.1,0.6) {$x$};
    \node at (0.6,2) {$y$};
    \draw[dashed] (1,2)--(1.45,0.41);
    \node at (2.0,1.4) {$\lVert y\rVert\sin(\theta)$};
  \end{tikzpicture}
\end{center}
\medskip

{\bf Example.} Find an equation for the plane through the
points~$p=(1,2,3)$,~$q=(1,0,-2)$, and~$r=(0,7,2)$.  
\smallskip

{\sc solution:} To find a vector perpendicular to the plane, we take the cross
product of~$q-p$ and~$r-p$.  Below is a picture that illustrates the geometry
(with no attempt to get the actual coordinates correct!).
The sides of the base parallelogram are spanned by the vectors~$q-p$ and~$r-p$.
\begin{center}
  \begin{tikzpicture}[scale=0.6]
    \draw[->] (0,0)--(4,1);
    \draw[->] (0,0)--(1.2,3);
    \draw[->] (0,0)--(-3,4);
    \node at (0,-0.5) {$p$};
    \node at (4.3,0.7) {$q$};
    \node at (0.9,3.2) {$r$};
    \node at (-6,3.4) {$(q-p)\times(r-p)$};
  \end{tikzpicture}
\end{center}
\medskip

Compute:
\begin{align*}
  (q-p)\times(r-p)
  &= (0,-2,-5)\times(-1,5,-1)\\[10pt]
  &= \det
  \left(\begin{array}{rrr}
	\mathbf{i}&\mathbf{j}&\mathbf{k}\\
	0&-2&-5\\
	-1&5&-1
    \end{array} \right)\\[10pt]
    &= 27\,\mathbf{i}+5\,\mathbf{j}-2\,\mathbf{k}\\[8pt]
    &= (27,5,-2).
\end{align*}
To double-check, note that the cross product is perpendicular to~$q-p$
and~$r-p$:
\[
  (0,-2,-5)\cdot(27,5,-2)=0\quad\text{and}\quad
  (-1,5,-1)\cdot(27,5,-2)=0.
\]
The set of all points~$(x,y,z)$ perpendicular to the cross product is the plane
defined by
\[
  (27,5,-2)\cdot(x,y,z)=0,
\]
i.e., the plane with equation
\[
27x+5y-2z=0.
\]
This plane passes through the origin,~$(0,0,0)$.
We want the translation of this plane that passes through~$p$.
(It will automatically then pass through~$q$ and~$r$. So we could choose
  either~$q$ or~$r$ for this requirement, instead.)  The equation of this
  translated plane will have the form
\[
27x+5y-2z=c.
\]
for some constant~$c$.  Plug in~$p$ (or~$q$ or~$r$) to solve for~$c$:
\[
  c= 27(1)+5(2)-2(3)=31.
\]
So the equation of the plane is
\[
  27x+5y-2z=31.
\]
(Check that the equation is satisfied by~$p$,~$q$, and~$r$!)
\medskip

{\em Parametric equation of the plane.}  As we saw earlier in the semester,
we can parametrize this plane by
\begin{align*}
  f(s,t) &=  p + s(q-p)+t(r-p)\\
  &= (1,2,3)+s(0,-2,-5)+t(-1,5,-1)\\
  &= (1-t,2-2s+5t,3-5s-t).
\end{align*}
Thus, we get the function:
\begin{align*}
  f\colon\R^2&\to\R^3\\
  (s,t)&\mapsto(1-t,2-2s+5t,3-5s-t).
\end{align*}
The image of~$f$ is the plane passing through~$p$, $q$, and~$r$.  One may check
that if we let
\[
  x=1-t,\quad y=2-2s+5t,\quad z=3-5s-t,
\]
then $27x+5y-2z=31$, i.e., the point satisfies the equation for the plane.
 
\end{document}  
