\section{The Cross Product}
\subsection{Definitions and Basic Properties} We introduce a foundational definition for the study of vector geometry in $\mathbb{R}^3$.


\begin{definition}\label{13defn1}
Given two vectors $\mathbf{u}$ and $\mathbf{v}$ in $\mathbb{R}^3$, the \textbf{cross product} $\mathbf{u} \times \mathbf{v}$ is the unique vector in $\mathbb{R}^3$ such that:
\begin{enumerate}[(1)]
\item{$\mathbf{u} \times \mathbf{v}$ is perpendicular to both ${\bf u}$ and ${\bf v}$;}

\smallskip

\item{$|\mathbf{u} \times \mathbf{v}| = |\mathbf{u}||\mathbf{v}| \sin \theta$, where $\theta$ is the angle between $\mathbf{u}$ and $\mathbf{v}$.}

\medskip

\item{The triple $(\mathbf{u},\mathbf{v},\mathbf{u} \times \mathbf{v})$ forms a \textbf{right-handed triad} -- essentially, it's oriented the same way as the $(\ii,\jj,\kk)$ triple, going counterclockwise from $\mathbf{u}$ to $\mathbf{v}$ to $\mathbf{u} \times \mathbf{v}$. Your textbook may have other (different but equivalent) descriptions.}
\end{enumerate}
\end{definition}

\medskip

We first check that this determines $\mathbf{u} \times \mathbf{v}$ uniquely. Assume first that $\mathbf{u}$ and $\mathbf{v}$ are not parallel. Then (1) says that $\mathbf{u} \times \mathbf{v}$ is perpendicular to the plane spanned by $\mathbf{u}$ and $\mathbf{v}$, so this determines the direction of $\mathbf{u} \times \mathbf{v}$ up to the choice of sign. That choice is then determined by (3). Then (2) gives the magnitude of $\mathbf{u} \times \mathbf{v}$, and we are done.

Now, if $\mathbf{u}$ and $\mathbf{v}$ are parallel, there are many vectors that are perpendicular to both of them. However, in that case the angle between  $\mathbf{u}$ and $\mathbf{v}$ is either $0$ or $\pi$, so that $\sin\theta=0$ and $|\mathbf{u} \times \mathbf{v}|=0$. Therefore $\mathbf{u} \times \mathbf{v}$ is the zero vector in this case.



The following theorem tells us how to calculate the cross product.

\begin{theorem}\label{13thm1}
If $\mathbf{u} := u_1 \mathbf{i} + u_2 \mathbf{j} + u_3 \mathbf{k}$ and $\mathbf{v} := v_1 \mathbf{i} + v_2 \mathbf{j} + v_3 \mathbf{k}$, the cross product is given by
\begin{align*}
\mathbf{u} \times \mathbf{v} &=
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
u_1 & u_2 & u_3 \\
v_1 & v_2 & v_3
\end{vmatrix}
=\begin{vmatrix}
 u_2 & u_3 \\
 v_2 & v_3
\end{vmatrix}\mathbf{i}
- \begin{vmatrix}
 u_1 & u_3 \\
 v_1 & v_3
\end{vmatrix}\mathbf{j}
+ \begin{vmatrix}
 u_1 & u_2 \\
 v_1 & v_2
\end{vmatrix}\mathbf{k}
\\
%\end{equation*}
%
%\begin{equation*}
&=
(u_2v_3 - u_3v_2) \mathbf{i} - (u_1v_3 - u_3v_1) \mathbf{j} + (u_1v_2 - u_2v_1) \mathbf{k}
\end{align*}

\end{theorem}

The formal $3 \times 3$ determinant (with vectors in the first row!) is evaluated as indicated above. We first expand the $3\times 3$ determinant in the first row (as if the vectors ${\bf i}$, ${\bf j}$, ${\bf k}$ were just numbers) -- you will see this in your linear algebra class. Then, for each of the remaining $2\times 2$ determinants we 
use the fact that $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc$ for any $a,b,c,d \in \mathbb{R}$. The easiest way to remember the formula for the cross product is to memorize the $3 \times 3$ determinant formula.

\begin{proof}[Sketch of Proof of Theorem \ref{13thm1}]
Let $\mathbf{w}$ be the vector on the left-hand side of the equality in Theorem \ref{13thm1}. We need to verify the conditions $(1)$, $(2)$ and $(3)$ from Definition \ref{13defn1}. We work in order.

\begin{itemize}
\item{To check $(1)$, we need to check that the dot products $\mathbf{u} \cdot \mathbf{w}$ and $\mathbf{v} \cdot \mathbf{w}$ are zero. Write $\mathbf{u} \cdot \mathbf{w}=u_1w_1+u_2w_2+u_3w_3$, and then substitute the components of ${\bf w}$.
If you like using linear algebra, you may notice that
$$
\mathbf{u} \times \mathbf{w} =
\begin{vmatrix}
u_1& u_2& u_3 \\
u_1 & u_2 & u_3 \\
v_1 & v_2 & v_3
\end{vmatrix}
$$
and the last determinant is zero because two of the rows are identical. Or you could actually calculate the dot product in terms of the components of ${\bf u}$ and ${\bf v}$ and simplify to get 0.
}

\medskip

\item{To check $(2)$,  we first rewrite the scalar $|\mathbf{u} \times \mathbf{v}|$ in a form that will be easier to use.  Write,
\begin{align*}
|\mathbf{u} \times \mathbf{v}|^2 & = |\mathbf{u}|^2 |\mathbf{v}|^2 \sin^2 \theta \\[2ex]
\quad & = |\mathbf{u}|^2 |\mathbf{v}|^2 (1 - \cos^2 \theta) \\[2ex]
\quad & = |\mathbf{u}|^2 |\mathbf{v}|^2 \bigg( 1 - \big(\frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}||\mathbf{v}|} \big)^2 \bigg) \\[2ex]
\quad & = |\mathbf{u}|^2 |\mathbf{v}|^2 - (\mathbf{u} \cdot \mathbf{v})^2
\end{align*}
Now, you need a calculation to evaluate both this quantity and the quantity
$$
|\mathbf{w}|^2 =
\begin{vmatrix}
u_2 & u_3 \\
v_2 & v_3
\end{vmatrix}^2
+
\begin{vmatrix}
u_1 & u_3 \\
v_1 & v_3
\end{vmatrix}^2
+
\begin{vmatrix}
u_1 & u_2 \\
v_1 & v_2
\end{vmatrix}^2 
$$
and check that they are equal. We omit the calculation because it is a bit long, but it should be completely straightforward from here.
}
\item{To prove $(3)$, we have to verify that the triad $\mathbf{u}, \mathbf{v},\mathbf{w}$ is right-handed. To do this, we will set up our coordinate system so that the calculations are easy. It turns out that what works best is to make the $x$-axis parallel to $\uu$, then rotate the coordinate system so that $\vv$ lies in the $xy$-plane and has a positive $y$-component. (If the $y$-component of $v$ is zero for all such rotations, then $\uu$ and $\vv$ are parallel and the cross product is zero.) This means that $\uu=u_1\ii$ and $\vv=v_1\ii+v_2\jj$ with $u_1>0$ and $v_2>0$. But then $w_3=u_1v_2$ is positive, so the right-handed property holds.}

\end{itemize}
\end{proof}

\begin{remark}
The previous is \textbf{not a complete proof}! However, it should give you a good start. What is missing is, first, you need to complete the missing calculations. Second, and this is a little bit more delicate, you need to verify that (3) does not depend on the placement of $\uu$ and $\vv$ in the coordinate system. 
An important idea here is that the cross product is invariant under rigid motions: if $R$ is a rotation or a translation or a composition of both, then $(R\uu)\times (R\vv)=R(\uu\times\vv)$. This follows from the geometric definition of $\uu\times\vv$. To complete the proof, one should check that the coordinate formula in 
Theorem \ref{13thm1} is also invariant under rigid motions. This can be done using the invariance of determinants under orthogonal transformations in linear algebra.
\end{remark}


We list a few basic properties of the cross product.

\begin{proposition}
For vectors $\mathbf{u}$, $\mathbf{v}$ and $\mathbf{w}$ in $\mathbb{R}^3$, the cross product satisfies the following,
\begin{enumerate}[(1)]
\item{$\mathbf{u} \times \mathbf{u} = \mathbf{0}$}

\medskip

\item{$\mathbf{u} \times \mathbf{v} = - (\mathbf{v} \times \mathbf{u})$}

\medskip

\item{$(c \mathbf{u}) \times \mathbf{v} = c (\mathbf{u} \times \mathbf{v})$}

\medskip

\item{$(\mathbf{u} + \mathbf{v}) \times \mathbf{w} = \mathbf{u} \times \mathbf{w} + \mathbf{v} \times \mathbf{w}$}



\end{enumerate}
\end{proposition}


\medskip

\subsection{Examples} We calculate a few cross products.

\begin{example}
Recall that $\mathbf{i} = \langle 1, 0, 0 \rangle$, $\mathbf{j} = \langle 0,1,0 \rangle$ and $\mathbf{k} = \langle 0,0,1\rangle$.  Let us practice with the following calculations.
\begin{enumerate}[(1)]
\item{
$$
\mathbf{i} \times \mathbf{j} =
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
1 & 0 & 0 \\
0 & 1 & 0
\end{vmatrix}
=
0 \mathbf{i} + 0 \mathbf{j} + 1 \mathbf{k} = \mathbf{k}
$$
}

\medskip

\item{Now perform the necessary calculations to show that $\mathbf{j} \times \mathbf{i} = - \mathbf{k}$.  In general, one always has $\mathbf{u} \times \mathbf{v} = - \mathbf{v} \times \mathbf{u}$.
}

\medskip

\item{Let $\mathbf{u} = 3 \mathbf{i} + \mathbf{k}$ and $\mathbf{v} = \mathbf{i} - \mathbf{j} + 2 \mathbf{k}$. Then,
\begin{align*}
\mathbf{u} \times \mathbf{v} & =
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
3 & 0 & 1 \\
1 & - 1 & 2
\end{vmatrix}
=
(0+1) \mathbf{i} - (6 - 1) \mathbf{j} + (-3 -0) \mathbf{k} \\[1ex]
\quad & = \mathbf{i} - 5 \mathbf{j} - 3 \mathbf{k}
\end{align*}}
\end{enumerate}
\end{example}




\textbf{Warning}! It is entirely possible that the cross product $(\mathbf{u} \times \mathbf{v}) \times \mathbf{w}$ might \textbf{not be equal} to the cross product $\mathbf{u} \times (\mathbf{v} \times \mathbf{w})$. In other words, the cross product operation $\times$ is \textbf{not} associative! For example,
$$
(\mathbf{i} \times \mathbf{j}) \times \mathbf{j} = \mathbf{k} \times \mathbf{j} = - \mathbf{i},
$$
and yet
$$
\mathbf{i} \times (\mathbf{j} \times \mathbf{j}) = \mathbf{i} \times \mathbf{0} = \mathbf{0}.
$$



\subsection{Applications}
The cross product is used to calculate areas and volumes of certain shapes in $\mathbb{R}^3$.  We provide some applications below.

\begin{application}[Areas of a Triangle and Parallelogram]
Suppose that $\mathbf{u}$ and $\mathbf{v}$ in $\mathbb{R}^3$ make angle $\theta$ with one another; see the figure below.

\begin{minipage}{.8\textwidth}
\begin{figure}[H]\label{13fig1}
\centering
\includegraphics[width= .5\textwidth]{13fig1.pdf}
\end{figure}
\end{minipage}


The vectors $\mathbf{u}$ and $\mathbf{v}$ form two sides of a triangle.  Calculating this area using trigonometry gives,
$$
AREA_{TRI} = \frac{1}{2} |\mathbf{u}| |\mathbf{v}| \sin \theta = \frac{1}{2} |\mathbf{u} \times \mathbf{v}|.
$$
Similarly,  we can calculate the area of the parallelogram; see the figure below.
\begin{center}
\begin{minipage}{.8\textwidth}
\begin{figure}[H]\label{13fig2}
\centering
\includegraphics[width= .65\textwidth]{13fig2.pdf}
\end{figure}
\end{minipage}
\end{center}
This area is 
$$
AREA_{PAR} = 2 (AREA_{TRI}) = | \mathbf{u} \times \mathbf{v}|
$$
\end{application}

The cross product also gives a vector perpendicular to a plane. This will be used extensively in what follows.

\begin{application}[Vector perpendicular to a plane]
Let $\mathbf{u}$ and $\mathbf{v}$ be (non-parallel) vectors in $\mathbb{R}^3$.  Consider any $2$-dimensional plane parallel to both $\mathbf{u}$ and $\mathbf{v}$ in $\mathbb{R}^3$.  Then $\mathbf{w} = \mathbf{u} \times \mathbf{v}$ is perpendicular to the plane.
\end{application}

\begin{application}[Volume of a Parallelepiped]
Suppose $\mathbf{u}$, $\mathbf{v}$ and $\mathbf{w}$ are three non-parallel vectors in $\mathbb{R}^3$. We may view these vectors as one corner of a box with parallelogram faces.  This shape is called a \textbf{parallelepiped}. See the figure below.


\begin{minipage}{.85\textwidth}
\begin{figure}[H]
\centering
\includegraphics[width= .85\textwidth]{13fig3.png}
\end{figure}
\end{minipage}


In order to calculate the area of this shape (the parallelepiped), we need to calculate the \textbf{area of the base} and the \textbf{height}.  We know that,
$$
AREA_{BASE} = |\mathbf{u} \times \mathbf{v}|
$$
and we can calculate the height as follows,
\begin{align*}
HEIGHT & = |(\textrm{scalar projection of } \mathbf{w} \textrm{ on } \mathbf{u} \times \mathbf{v})| = \bigg|\frac{\mathbf{w} \cdot (\mathbf{u} \times \mathbf{v})}{|\mathbf{u} \times \mathbf{v}|} \bigg|,
\end{align*}
where the outer $|\cdot |$ denote the absolute value.  Note that, by \textbf{height}, we mean the height of $\mathbf{w}$ above the plane spanned by $\mathbf{u}$ and $ \mathbf{v}$.  See the figure below.


\begin{minipage}{.8\textwidth}
\begin{figure}[H]\label{13fig4}
\centering
\includegraphics[width= .85\textwidth]{13fig4.pdf}
\end{figure}
\end{minipage}


Now putting together everything so far, we see that the volume of the parallelepiped is given by
$$
VOL_{PAR} = (AREA_{BASE})(HEIGHT) = |\mathbf{u} \times \mathbf{v}| \cdot \bigg|\frac{\mathbf{w} \cdot (\mathbf{u} \times \mathbf{v})}{|\mathbf{u} \times \mathbf{v}|} \bigg| = |\mathbf{w} \cdot (\mathbf{u} \times \mathbf{v})|
$$
You can check yourself that the order of $\mathbf{u},\mathbf{v}$ and $\mathbf{w}$ does not matter in calculating the area of the parallelepiped.
\end{application}

The expression $\mathbf{w} \cdot (\mathbf{u} \times \mathbf{v})$ is sometimes called the \textbf{scalar triple product} of $\mathbf{w}, \mathbf{u}$ and $\mathbf{v}$. To evaluate it, we should use the component representations of $\mathbf{u} \times \mathbf{v}$. That is,
\begin{align*}
\mathbf{w} \cdot (\mathbf{u} \times \mathbf{v}) & =
w_1
\begin{vmatrix}
u_2 & u_3 \\
v_2 & v_3
\end{vmatrix}
-
w_2
\begin{vmatrix}
u_1 & u_3 \\
v_1 & v_3
\end{vmatrix}
+
w_3
\begin{vmatrix}
u_1 & u_2 \\
v_1 & v_2
\end{vmatrix}
\\[2ex]
\quad &=
\begin{vmatrix}
w_1 & w_2 & w_3 \\
u_1 & u_2 & u_3 \\
v_1 & v_2 & v_3
\end{vmatrix}
\end{align*}
This is not the last time you will see this $3 \times 3$ determinant appear in this course. More to come.


\section*{Worked Problems}
We have some worked problems related to the cross product and its relation to areas and volumes.

\begin{enumerate}[1.]
\item{{\it Find the area of the triangle in} $\mathbb{R}^3$ {\it with vertices } $(0,0,0), (3,0,1), (1,-1,2)$.

\bigskip

Let $P = (3,0,1)$ and $Q = (1,-1,2)$.  Then $\vec{OP} = 3 \mathbf{i} + \mathbf{k}$ and $\vec{OQ} = \mathbf{i} - \mathbf{j} + 2 \mathbf{k}$. These are the same vectors that appeared in Example 1.3.4(3). So, the area of the triangle $OPQ$ is given by
$$
\frac{1}{2}| \vec{OP} \times \vec{OQ}| = \frac{1}{2} |\mathbf{i} - 5 \mathbf{j} - 3 \mathbf{k}| = \frac{1}{2} \sqrt{1 + 25 + 9} = \frac{\sqrt{35}}{2}.
$$
}

\bigskip

\item{{\it Find the volume of the parallelepiped spanned by the vectors} $\mathbf{u} = \langle 1, 0, -1 \rangle$, $\mathbf{v} = \langle 1,3,3 \rangle$ and $\mathbf{w} = \langle 0,1,-2 \rangle$.

\bigskip

The volume of the parallelepiped is given by $|\mathbf{w} \cdot (\mathbf{u} \times \mathbf{v})|$. The scalar product is given by
\begin{align*}
\mathbf{w} \cdot (\mathbf{u} \times \mathbf{v}) 
& =
\begin{vmatrix}
0 & 1 & -2 \\
1 & 0 & -1 \\
1 & 3 & 3
\end{vmatrix}
=
-1
\begin{vmatrix}
1 & -1 \\
1 & 3
\end{vmatrix}
-2
\begin{vmatrix}
1 & 0 \\
1 & 3
\end{vmatrix}
\\[2ex]
\quad & = -1(3+1) -2(3-0) = -10.
\end{align*}

\noindent

Recall: volumes cannot be negative, so the volume is $|-10| = 10$.
}


\bigskip


\end{enumerate}


\section*{Practice Problems}
\begin{enumerate}[1.]
\item
{ Give an example of three non-zero vectors ${\bf u}$, ${\bf v}$, ${\bf w}$ such that ${\bf u}\cdot( {\bf v}\times {\bf w})=4$. }

\bigskip

\item Give an example of three non-zero vectors ${\bf u}$, ${\bf v}$, ${\bf w}$, such that ${\bf v}\times {\bf w}\neq 0$ but ${\bf u}\cdot( {\bf v}\times {\bf w})=0$.

\bigskip

\item {Find the area of the triangle in $\RR^3$ with vertices $(0,1,1)$, $(2,2,1)$, $(1,0,3)$.}

\bigskip



\item { 
Find the volume of the parallelepiped in ${\mathbb R}^3$ spanned by the vectors 
$\ii-\jj+4\kk$, $3\jj-\kk$, $2\ii+\jj$.}
\end{enumerate}





