# 2.4 The cross product  (Page 5/16)

 Page 5 / 16

## The triple scalar product

Because the cross product of two vectors is a vector, it is possible to combine the dot product and the cross product. The dot product of a vector with the cross product of two other vectors is called the triple scalar product because the result is a scalar.

## Definition

The triple scalar product    of vectors $\text{u},$ $\text{v},$ and $\text{w}$ is $\text{u}·\left(\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{w}\right).$

## Calculating a triple scalar product

The triple scalar product of vectors $\text{u}={u}_{1}\text{i}+{u}_{2}\text{j}+{u}_{3}\text{k},$ $\text{v}={v}_{1}\text{i}+{v}_{2}\text{j}+{v}_{3}\text{k},$ and $\text{w}={w}_{1}\text{i}+{w}_{2}\text{j}+{w}_{3}\text{k}$ is the determinant of the $3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}3$ matrix formed by the components of the vectors:

$\text{u}·\left(\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{w}\right)=|\begin{array}{ccc}{u}_{1}\hfill & {u}_{2}\hfill & {u}_{3}\hfill \\ {v}_{1}\hfill & {v}_{2}\hfill & {v}_{3}\hfill \\ {w}_{1}\hfill & {w}_{2}\hfill & {w}_{3}\hfill \end{array}|.$

## Proof

The calculation is straightforward.

$\begin{array}{cc}\hfill \text{u}·\left(\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{w}\right)& =⟨{u}_{1},{u}_{2},{u}_{3}⟩·⟨{v}_{2}{w}_{3}-{v}_{3}{w}_{2},\text{−}{v}_{1}{w}_{3}+{v}_{3}{w}_{1},{v}_{1}{w}_{2}-{v}_{2}{w}_{1}⟩\hfill \\ & ={u}_{1}\left({v}_{2}{w}_{3}-{v}_{3}{w}_{2}\right)+{u}_{2}\left(\text{−}{v}_{1}{w}_{3}+{v}_{3}{w}_{1}\right)+{u}_{3}\left({v}_{1}{w}_{2}-{v}_{2}{w}_{1}\right)\hfill \\ & ={u}_{1}\left({v}_{2}{w}_{3}-{v}_{3}{w}_{2}\right)-{u}_{2}\left({v}_{1}{w}_{3}-{v}_{3}{w}_{1}\right)+{u}_{3}\left({v}_{1}{w}_{2}-{v}_{2}{w}_{1}\right)\hfill \\ & =|\begin{array}{ccc}{u}_{1}\hfill & {u}_{2}\hfill & {u}_{3}\hfill \\ {v}_{1}\hfill & {v}_{2}\hfill & {v}_{3}\hfill \\ {w}_{1}\hfill & {w}_{2}\hfill & {w}_{3}\hfill \end{array}|\hfill \end{array}$

## Calculating the triple scalar product

Let $\text{u}=⟨1,3,5⟩,\text{v}=⟨2,-1,0⟩\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\text{w}=⟨-3,0,-1⟩.$ Calculate the triple scalar product $\text{u}·\left(\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{w}\right).$

$\begin{array}{cc}\hfill \text{u}·\left(\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{w}\right)& =|\begin{array}{ccc}\hfill 1& \hfill 3& \hfill 5\\ \hfill 2& \hfill -1& \hfill 0\\ \hfill -3& \hfill 0& \hfill -1\end{array}|\hfill \\ & =1|\begin{array}{cc}\hfill -1& \hfill 0\\ \hfill 0& \hfill -1\end{array}|-3|\begin{array}{cc}\hfill 2& \hfill 0\\ \hfill -3& \hfill -1\end{array}|+5|\begin{array}{cc}\hfill 2& \hfill -1\\ \hfill -3& \hfill 0\end{array}|\hfill \\ & =\left(1-0\right)-3\left(-2-0\right)+5\left(0-3\right)\hfill \\ & =1+6-15=-8.\hfill \end{array}$

Calculate the triple scalar product $\text{a}·\left(\text{b}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{c}\right),$ where $\text{a}=⟨2,-4,1⟩,$ $\text{b}=⟨0,3,-1⟩,$ and $\text{c}=⟨5,-3,3⟩.$

$17$

When we create a matrix from three vectors, we must be careful about the order in which we list the vectors. If we list them in a matrix in one order and then rearrange the rows, the absolute value of the determinant remains unchanged. However, each time two rows switch places, the determinant changes sign:

$|\begin{array}{ccc}{a}_{1}\hfill & {a}_{2}\hfill & {a}_{3}\hfill \\ {b}_{1}\hfill & {b}_{2}\hfill & {b}_{3}\hfill \\ {c}_{1}\hfill & {c}_{2}\hfill & {c}_{3}\hfill \end{array}|=d\phantom{\rule{2em}{0ex}}|\begin{array}{ccc}{b}_{1}\hfill & {b}_{2}\hfill & {b}_{3}\hfill \\ {a}_{1}\hfill & {a}_{2}\hfill & {a}_{3}\hfill \\ {c}_{1}\hfill & {c}_{2}\hfill & {c}_{3}\hfill \end{array}|=\text{−}d\phantom{\rule{2em}{0ex}}|\begin{array}{ccc}{b}_{1}\hfill & {b}_{2}\hfill & {b}_{3}\hfill \\ {c}_{1}\hfill & {c}_{2}\hfill & {c}_{3}\hfill \\ {a}_{1}\hfill & {a}_{2}\hfill & {a}_{3}\hfill \end{array}|=d\phantom{\rule{2em}{0ex}}|\begin{array}{ccc}{c}_{1}\hfill & {c}_{2}\hfill & {c}_{3}\hfill \\ {b}_{1}\hfill & {b}_{2}\hfill & {b}_{3}\hfill \\ {a}_{1}\hfill & {a}_{2}\hfill & {a}_{3}\hfill \end{array}|=\text{−}d.$

Verifying this fact is straightforward, but rather messy. Let’s take a look at this with an example:

$\begin{array}{cc}\hfill |\begin{array}{ccc}\hfill 1& 2\hfill & \hfill 1\\ \hfill -2& 0\hfill & \hfill 3\\ \hfill 4& 1\hfill & \hfill -1\end{array}|& =|\begin{array}{cc}0\hfill & \hfill 3\\ 1\hfill & \hfill -1\end{array}|-2|\begin{array}{cc}\hfill -2& \hfill 3\\ \hfill 4& \hfill -1\end{array}|+|\begin{array}{cc}\hfill -2& 0\hfill \\ \hfill 4& 1\hfill \end{array}|\hfill \\ & =\left(0-3\right)-2\left(2-12\right)+\left(-2-0\right)=-3+20-2=15.\hfill \end{array}$

Switching the top two rows we have

$|\begin{array}{ccc}\hfill -2& 0\hfill & \hfill 3\\ \hfill 1& 2\hfill & \hfill 1\\ \hfill 4& 1\hfill & \hfill -1\end{array}|=-2|\begin{array}{cc}2\hfill & \hfill 1\\ 1\hfill & \hfill -1\end{array}|+3|\begin{array}{cc}1\hfill & 2\hfill \\ 4\hfill & 1\hfill \end{array}|=-2\left(-2-1\right)+3\left(1-8\right)=6-21=-15.$

Rearranging vectors in the triple products is equivalent to reordering the rows in the matrix of the determinant. Let $\text{u}={u}_{1}\text{i}+{u}_{2}\text{j}+{u}_{3}\text{k},$ $\text{v}={v}_{1}\text{i}+{v}_{2}\text{j}+{v}_{3}\text{k},$ and $\text{w}={w}_{1}\text{i}+{w}_{2}\text{j}+{w}_{3}\text{k}.$ Applying [link] , we have

$\text{u}·\left(\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{w}\right)=|\begin{array}{ccc}{u}_{1}\hfill & {u}_{2}\hfill & {u}_{3}\hfill \\ {v}_{1}\hfill & {v}_{2}\hfill & {v}_{3}\hfill \\ {w}_{1}\hfill & {w}_{2}\hfill & {w}_{3}\hfill \end{array}|\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\text{u}·\left(\text{w}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}\right)=|\begin{array}{ccc}{u}_{1}\hfill & {u}_{2}\hfill & {u}_{3}\hfill \\ {w}_{1}\hfill & {w}_{2}\hfill & {w}_{3}\hfill \\ {v}_{1}\hfill & {v}_{2}\hfill & {v}_{3}\hfill \end{array}|.$

We can obtain the determinant for calculating $\text{u}·\left(\text{w}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}\right)$ by switching the bottom two rows of $\text{u}·\left(\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{w}\right).$ Therefore, $\text{u}·\left(\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{w}\right)=\text{−}\text{u}·\left(\text{w}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}\right).$

Following this reasoning and exploring the different ways we can interchange variables in the triple scalar product lead to the following identities:

$\begin{array}{ccc}\hfill \text{u}·\left(\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{w}\right)& =\hfill & \text{−}\text{u}·\left(\text{w}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}\right)\hfill \\ \hfill \text{u}·\left(\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{w}\right)& =\hfill & \text{v}·\left(\text{w}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{u}\right)=\text{w}·\left(\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}\right).\hfill \end{array}$

Let $\text{u}$ and $\text{v}$ be two vectors in standard position. If $\text{u}$ and $\text{v}$ are not scalar multiples of each other, then these vectors form adjacent sides of a parallelogram. We saw in [link] that the area of this parallelogram is $‖\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}‖.$ Now suppose we add a third vector $\text{w}$ that does not lie in the same plane as $\text{u}$ and $\text{v}$ but still shares the same initial point. Then these vectors form three edges of a parallelepiped    , a three-dimensional prism with six faces that are each parallelograms, as shown in [link] . The volume of this prism is the product of the figure’s height and the area of its base. The triple scalar product of $\text{u},\text{v},$ and $\text{w}$ provides a simple method for calculating the volume of the parallelepiped defined by these vectors.

## Volume of a parallelepiped

The volume of a parallelepiped with adjacent edges given by the vectors $\text{u},\text{v},\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\text{w}$ is the absolute value of the triple scalar product:

$V=|\text{u}·\left(\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{w}\right)|.$

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