# 2.4 The cross product  (Page 4/16)

 Page 4 / 16

## Using expansion along the first row to compute a $3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}3$ Determinant

Evaluate the determinant $|\begin{array}{ccc}\hfill 2& \hfill 5& \hfill -1\\ \hfill -1& \hfill 1& \hfill 3\\ \hfill -2& \hfill 3& \hfill 4\end{array}|.$

We have

$\begin{array}{cc}\hfill |\begin{array}{ccc}\hfill 2& \hfill 5& \hfill -1\\ \hfill -1& \hfill 1& \hfill 3\\ \hfill -2& \hfill 3& \hfill 4\end{array}|& =2|\begin{array}{cc}1\hfill & 3\hfill \\ 3\hfill & 4\hfill \end{array}|-5|\begin{array}{cc}-1\hfill & 3\hfill \\ -2\hfill & 4\hfill \end{array}|-1|\begin{array}{cc}-1\hfill & 1\hfill \\ -2\hfill & 3\hfill \end{array}|\hfill \\ & =2\left(4-9\right)-5\left(-4+6\right)-1\left(-3+2\right)\hfill \\ & =2\left(-5\right)-5\left(2\right)-1\left(-1\right)=-10-10+1\hfill \\ & =-19.\hfill \end{array}$

Evaluate the determinant $|\begin{array}{ccc}\hfill 1& \hfill -2& \hfill -1\\ \hfill 3& \hfill 2& \hfill -3\\ \hfill 1& \hfill 5& \hfill 4\end{array}|.$

$40$

Technically, determinants are defined only in terms of arrays of real numbers. However, the determinant notation provides a useful mnemonic device for the cross product formula.

## Rule: cross product calculated by a determinant

Let $\text{u}=⟨{u}_{1},{u}_{2},{u}_{3}⟩$ and $\text{v}=⟨{v}_{1},{v}_{2},{v}_{3}⟩$ be vectors. Then the cross product $\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}$ is given by

$\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}=|\begin{array}{ccc}\text{i}\hfill & \text{j}\hfill & \text{k}\hfill \\ {u}_{1}\hfill & {u}_{2}\hfill & {u}_{3}\hfill \\ {v}_{1}\hfill & {v}_{2}\hfill & {v}_{3}\hfill \end{array}|=|\begin{array}{cc}{u}_{2}\hfill & {u}_{3}\hfill \\ {v}_{2}\hfill & {v}_{3}\hfill \end{array}|\text{i}-|\begin{array}{cc}{u}_{1}\hfill & {u}_{3}\hfill \\ {v}_{1}\hfill & {v}_{3}\hfill \end{array}|\text{j}+|\begin{array}{cc}{u}_{1}\hfill & {u}_{2}\hfill \\ {v}_{1}\hfill & {v}_{2}\hfill \end{array}|\text{k}.$

## Using determinant notation to find $\text{p}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{q}$

Let $\text{p}=⟨-1,2,5⟩$ and $\text{q}=⟨4,0,-3⟩.$ Find $\text{p}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{q}.$

We set up our determinant by putting the standard unit vectors across the first row, the components of $\text{u}$ in the second row, and the components of $\text{v}$ in the third row. Then, we have

$\begin{array}{cc}\hfill \text{p}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{q}& =|\begin{array}{ccc}\hfill \text{i}& \hfill \text{j}& \hfill \text{k}\\ \hfill -1& \hfill 2& \hfill 5\\ \hfill 4& \hfill 0& \hfill -3\end{array}|=|\begin{array}{cc}\hfill 2& \hfill 5\\ \hfill 0& \hfill -3\end{array}|\text{i}-|\begin{array}{cc}\hfill -1& \hfill 5\\ \hfill 4& \hfill -3\end{array}|\text{j}+|\begin{array}{cc}\hfill -1& \hfill 2\\ \hfill 4& \hfill 0\end{array}|\text{k}\hfill \\ & =\left(-6-0\right)\text{i}-\left(3-20\right)\text{j}+\left(0-8\right)\text{k}\hfill \\ & =-6\text{i}+17\text{j}-8\text{k}.\hfill \end{array}$

Notice that this answer confirms the calculation of the cross product in [link] .

Use determinant notation to find $\text{a}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{b},$ where $\text{a}=⟨8,2,3⟩$ and $\text{b}=⟨-1,0,4⟩.$

$8\text{i}-35\text{j}+2\text{k}$

## Using the cross product

The cross product is very useful for several types of calculations, including finding a vector orthogonal to two given vectors, computing areas of triangles and parallelograms, and even determining the volume of the three-dimensional geometric shape made of parallelograms known as a parallelepiped . The following examples illustrate these calculations.

## Finding a unit vector orthogonal to two given vectors

Let $\text{a}=⟨5,2,-1⟩$ and $\text{b}=⟨0,-1,4⟩.$ Find a unit vector orthogonal to both $\text{a}$ and $\text{b}.$

The cross product $\text{a}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{b}$ is orthogonal to both vectors $\text{a}$ and $\text{b}.$ We can calculate it with a determinant:

$\begin{array}{cc}\hfill \text{a}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{b}& =|\begin{array}{ccc}\hfill \text{i}& \hfill \text{j}& \hfill \text{k}\\ \hfill 5& \hfill 2& \hfill -1\\ \hfill 0& \hfill -1& \hfill 4\end{array}|=|\begin{array}{cc}\hfill 2& \hfill -1\\ \hfill -1& \hfill 4\end{array}|\text{i}-|\begin{array}{cc}\hfill 5& \hfill -1\\ \hfill 0& \hfill 4\end{array}|\text{j}+|\begin{array}{cc}\hfill 5& \hfill 2\\ \hfill 0& \hfill -1\end{array}|\text{k}\hfill \\ & =\left(8-1\right)\text{i}-\left(20-0\right)\text{j}+\left(-5-0\right)\text{k}\hfill \\ & =7\text{i}-20\text{j}-5\text{k}.\hfill \end{array}$

Normalize this vector to find a unit vector in the same direction:

$‖\text{a}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{b}‖=\sqrt{{\left(7\right)}^{2}+{\left(-20\right)}^{2}+{\left(-5\right)}^{2}}=\sqrt{474}.$

Thus, $⟨\frac{7}{\sqrt{474}},\frac{-20}{\sqrt{474}},\frac{-5}{\sqrt{474}}⟩$ is a unit vector orthogonal to $\text{a}$ and $\text{b}.$

Find a unit vector orthogonal to both $\text{a}$ and $\text{b},$ where $\text{a}=⟨4,0,3⟩$ and $\text{b}=⟨1,1,4⟩.$

$⟨\frac{-3}{\sqrt{194}},\frac{-13}{\sqrt{194}},\frac{4}{\sqrt{194}}⟩$

To use the cross product for calculating areas, we state and prove the following theorem.

## Area of a parallelogram

If we locate vectors $\text{u}$ and $\text{v}$ such that they form adjacent sides of a parallelogram, then the area of the parallelogram is given by $‖\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}‖$ ( [link] ).

## Proof

We show that the magnitude of the cross product is equal to the base times height of the parallelogram.

$\begin{array}{cc}\hfill \text{Area of a parallelogram}& =\text{base}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{height}\hfill \\ & =‖\text{u}‖\left(‖\text{v}‖\text{sin}\phantom{\rule{0.2em}{0ex}}\theta \right)\hfill \\ & =‖\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}‖\hfill \end{array}$

## Finding the area of a triangle

Let $P=\left(1,0,0\right),Q=\left(0,1,0\right),\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}R=\left(0,0,1\right)$ be the vertices of a triangle ( [link] ). Find its area.

We have $\stackrel{\to }{PQ}=⟨0-1,1-0,0-0⟩=⟨-1,1,0⟩$ and $\stackrel{\to }{PR}=⟨0-1,0-0,1-0⟩=⟨-1,0,1⟩.$ The area of the parallelogram with adjacent sides $\stackrel{\to }{PQ}$ and $\stackrel{\to }{PR}$ is given by $‖\stackrel{\to }{PQ}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{PR}‖\text{:}$

$\begin{array}{ccc}\hfill \stackrel{\to }{PQ}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{PR}& =\hfill & |\begin{array}{ccc}\hfill \text{i}& \hfill \text{j}& \hfill \text{k}\\ \hfill -1& \hfill 1& \hfill 0\\ \hfill -1& \hfill 0& \hfill 1\end{array}|=\left(1-0\right)\text{i}-\left(-1-0\right)\text{j}+\left(0-\left(-1\right)\right)\text{k}=\text{i}+\text{j}+\text{k}\hfill \\ \hfill ‖\stackrel{\to }{PQ}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{PR}‖& =\hfill & ‖⟨1,1,1⟩‖=\sqrt{{1}^{2}+{1}^{2}+{1}^{2}}=\sqrt{3}.\hfill \end{array}$

The area of $\text{Δ}PQR$ is half the area of the parallelogram, or $\sqrt{3}\text{/}2.$

Find the area of the parallelogram $PQRS$ with vertices $P\left(1,1,0\right),Q\left(7,1,0\right),R\left(9,4,2\right),$ and $S\left(3,4,2\right).$

$6\sqrt{13}$

anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
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 to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials​ and their applications of sensors.
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Got questions? Join the online conversation and get instant answers!