# 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:

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

what is variations in raman spectra for nanomaterials
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
can you provide the details of the parametric equations for the lines that defince doubly-ruled surfeces (huperbolids of one sheet and hyperbolic paraboloid). Can you explain each of the variables in the equations?