# 1.8 Scalar product (application)

 Page 1 / 1
Solving problems is an essential part of the understanding process.

Questions and their answers are presented here in the module text format as if it were an extension of the treatment of the topic. The idea is to provide a verbose explanation, detailing the application of theory. Solution presented is, therefore, treated as the part of the understanding process – not merely a Q/A session. The emphasis is to enforce ideas and concepts, which can not be completely absorbed unless they are put to real time situation.

## Representative problems and their solutions

We discuss problems, which highlight certain aspects of the scalar vector product. For this reason, questions are categorized in terms of the characterizing features of the subject matter :

• Angle between two vectors
• Condition of perpendicular vectors
• Component as scalar product
• Nature of scalar product
• Scalar product of a vector with itself
• Evaluation of dot product

## Angle between two vectors

Problem : Find the angle between vectors 2 i + j k and i k .

Solution : The cosine of the angle between two vectors is given in terms of dot product as :

$\begin{array}{l}\mathrm{cos}\theta =\frac{\mathbf{a}\mathbf{.}\mathbf{b}}{\mathrm{ab}}\end{array}$

Now,

$\begin{array}{l}\mathbf{a}\mathbf{.}\mathbf{b}=\left(2\mathbf{i}+\mathbf{j}-\mathbf{k}\right)\mathbf{.}\left(2\mathbf{i}-\mathbf{k}\right)\end{array}$

Ignoring dot products of different unit vectors (they evaluate to zero), we have :

$\begin{array}{l}\mathbf{a}\mathbf{.}\mathbf{b}=2\mathbf{i}\mathbf{.}\mathbf{i}+\left(-\mathbf{k}\right)\mathbf{.}\left(-\mathbf{k}\right)=2+1=3\\ a=\surd \left({2}^{2}+{1}^{2}+{1}^{2}\right)=\surd 6\\ b=\surd \left({1}^{2}+{1}^{2}\right)=\surd 2\\ \mathrm{ab}=\surd 6\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}\surd 2=\surd \left(12\right)=2\surd 3\end{array}$

Putting in the expression of cosine, we have :

$\begin{array}{l}\mathrm{cos}\theta =\frac{\mathbf{a}\mathbf{.}\mathbf{b}}{\mathrm{ab}}=\frac{3}{2\surd 3}=\frac{\surd 3}{2}=\mathrm{cos}30°\\ \theta =30°\end{array}$

## Condition of perpendicular vectors

Problem : Sum and difference of two vectors a and b are perpendicular to each other. Find the relation between two vectors.

Solution : The sum a + b and difference a - b are perpendicular to each other. Hence, their dot product should evaluate to zero.

$\begin{array}{l}\left(\mathbf{a}+\mathbf{b}\right)\mathbf{.}\left(\mathbf{a}-\mathbf{b}\right)=0\end{array}$

Using distributive property,

$\begin{array}{l}\mathbf{a}\mathbf{.}\mathbf{a}-\mathbf{a}\mathbf{.}\mathbf{b}+\mathbf{b}\mathbf{.}\mathbf{a}-\mathbf{b}\mathbf{.}\mathbf{b}=0\end{array}$

Using commutative property, a.b = b.a , Hence,

$\begin{array}{l}\mathbf{a}\mathbf{.}\mathbf{a}-\mathbf{b}\mathbf{.}\mathbf{b}=0\\ {a}^{2}-{b}^{2}=0\\ a=b\end{array}$

It means that magnitudes of two vectors are equal. See figure below for enclosed angle between vectors, when vectors are equal :

## Component as scalar product

Problem : Find the components of vector 2 i + 3 j along the direction i + j .

Solution : The component of a vector “ a ” in a direction, represented by unit vector “ n ” is given by dot product :

$\begin{array}{l}{a}_{n}=\mathbf{a}\mathbf{.}\mathbf{n}\end{array}$

Thus, it is clear that we need to find the unit vector in the direction of i + j . Now, the unit vector in the direction of the vector is :

$\begin{array}{l}n=\frac{\mathbf{i}+\mathbf{j}}{|\mathbf{i}+\mathbf{j}|}\end{array}$

Here,

$\begin{array}{l}|\mathbf{i}+\mathbf{j}|=\surd \left({1}^{2}+{1}^{2}\right)=\surd 2\end{array}$

Hence,

$\begin{array}{l}n=\frac{1}{\surd 2}\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}\left(\mathbf{i}+\mathbf{j}\right)\end{array}$

The component of vector 2 i + 3 j in the direction of “ n ” is :

$\begin{array}{l}{a}_{n}=\mathbf{a}\mathbf{.}\mathbf{n}=\left(2\mathbf{i}+3\mathbf{j}\right)\mathbf{.}\frac{1}{\surd 2}\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}\left(\mathbf{i}+\mathbf{j}\right)\end{array}$

$\begin{array}{l}⇒{a}_{n}=\frac{1}{\surd 2}\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}\left(2\mathbf{i}+3\mathbf{j}\right)\mathbf{.}\left(\mathbf{i}+\mathbf{j}\right)\\ ⇒{a}_{n}=\frac{1}{\surd 2}\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}\left(2x1+3x1\right)\\ ⇒{a}_{n}=\frac{5}{\surd 2}\end{array}$

## Nature of scalar product

Problem : Verify vector equality B = C , if A.B = A.C .

Solution : The given equality of dot products is :

$\begin{array}{l}\mathbf{A}\mathbf{.}\mathbf{B}=\mathbf{A}\mathbf{.}\mathbf{C}\end{array}$

The equality will result if B = C . We must, however, understand that dot product is not a simple algebraic product of two numbers (read magnitudes). The angle between two vectors plays a role in determining the magnitude of the dot product. Hence, it is entirely possible that vectors B and C are different yet their dot products with common vector A are equal.

We can attempt this question mathematically as well. Let ${\theta }_{1}$ and ${\theta }_{2}$ be the angles for first and second pairs of dot products. Then,

$\begin{array}{l}\mathbf{A}\mathbf{.}\mathbf{B}=\mathbf{A}\mathbf{.}\mathbf{C}\end{array}$

$\begin{array}{l}\mathrm{AB cos}{\theta }_{1}=\mathrm{AC cos}{\theta }_{2}\end{array}$

If ${\theta }_{1}={\theta }_{2}$ , then $B=C$ . However, if ${\theta }_{1}\ne {\theta }_{2}$ , then $B\ne C$ .

## Scalar product of a vector with itself

Problem : If | a + b | = | a b |, then find the angle between vectors a and b .

Solution : A question that involves modulus or magnitude of vector can be handled in specific manner to find information about the vector (s). The specific identity that is used in this circumstance is :

$\begin{array}{l}\mathbf{A}\mathbf{.}\mathbf{A}={A}^{2}\end{array}$

We use this identity first with the sum of the vectors ( a + b ),

$\begin{array}{l}\left(\mathbf{a}+\mathbf{b}\right)\mathbf{.}\left(\mathbf{a}+\mathbf{b}\right)={|\mathbf{a}+\mathbf{b}|}^{2}\end{array}$

Using distributive property,

$\begin{array}{l}⇒\mathbf{a}\mathbf{.}\mathbf{a}+\mathbf{b}\mathbf{.}\mathbf{a}+\mathbf{a}\mathbf{.}\mathbf{b}+\mathbf{b}\mathbf{.}\mathbf{b}={a}^{2}+{b}^{2}+2ab\mathrm{cos}\theta ={|\mathbf{a}+\mathbf{b}|}^{2}\\ ⇒{|\mathbf{a}+\mathbf{b}|}^{2}={a}^{2}+{b}^{2}+2ab\mathrm{cos}\theta \end{array}$

Similarly, using the identity with difference of the vectors (a-b),

$\begin{array}{l}⇒{|\mathbf{a}-\mathbf{b}|}^{2}={a}^{2}+{b}^{2}-2ab\mathrm{cos}\theta \end{array}$

It is, however, given that :

$\begin{array}{l}⇒|\mathbf{a}+\mathbf{b}|=|\mathbf{a}-\mathbf{b}|\end{array}$

Squaring on either side of the equation,

$\begin{array}{l}⇒{|\mathbf{a}+\mathbf{b}|}^{2}={|\mathbf{a}-\mathbf{b}|}^{2}\end{array}$

Putting the expressions,

$\begin{array}{l}⇒{a}^{2}+{b}^{2}+2ab\mathrm{cos}\theta ={a}^{2}+{b}^{2}-2ab\mathrm{cos}\theta \\ ⇒4ab\mathrm{cos}\theta =0\\ ⇒\mathrm{cos}\theta =0\\ ⇒\theta =90°\end{array}$

Note : We can have a mental picture of the significance of this result. As given, the magnitude of sum of two vectors is equal to the magnitude of difference of two vectors. Now, we know that difference of vectors is similar to vector sum with one exception that one of the operand is rendered negative. Graphically, it means that one of the vectors is reversed.

Reversing one of the vectors changes the included angle between two vectors, but do not change the magnitudes of either vector. It is, therefore, only the included angle between the vectors that might change the magnitude of resultant. In order that magnitude of resultant does not change even after reversing direction of one of the vectors, it is required that the included angle between the vectors is not changed. This is only possible, when included angle between vectors is 90°. See figure.

Problem : If a and b are two non-collinear unit vectors and | a + b | = √3, then find the value of expression :

$\begin{array}{l}\left(\mathbf{a}-\mathbf{b}\right)\mathbf{.}\left(2\mathbf{a}+\mathbf{b}\right)\end{array}$

Solution : The given expression is scalar product of two vector sums. Using distributive property we can expand the expression, which will comprise of scalar product of two vectors a and b .

$\begin{array}{l}\left(\mathbf{a}-\mathbf{b}\right)\mathbf{.}\left(2\mathbf{a}+\mathbf{b}\right)=2\mathbf{a}\mathbf{.}\mathbf{a}+\mathbf{a}\mathbf{.}\mathbf{b}-\mathbf{b}\mathbf{.}2\mathbf{a}+\left(-\mathbf{b}\right)\mathbf{.}\left(-\mathbf{b}\right)=2{a}^{2}-\mathbf{a}\mathbf{.}\mathbf{b}-{b}^{2}\end{array}$

$\begin{array}{l}⇒\left(\mathbf{a}-\mathbf{b}\right)\mathbf{.}\left(2\mathbf{a}+\mathbf{b}\right)=2{a}^{2}-{b}^{2}-ab\mathrm{cos}\theta \end{array}$

We can evaluate this scalar product, if we know the angle between them as magnitudes of unit vectors are each 1. In order to find the angle between the vectors, we use the identity,

$\begin{array}{l}\mathbf{A}\mathbf{.}\mathbf{A}={A}^{2}\end{array}$

Now,

$\begin{array}{l}{|\mathbf{a}+\mathbf{b}|}^{2}=\left(\mathbf{a}+\mathbf{b}\right)\mathbf{.}\left(\mathbf{a}+\mathbf{b}\right)={a}^{2}+{b}^{2}+2ab\mathrm{cos}\theta =1+1+2\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}1\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}1\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}\mathrm{cos}\theta \end{array}$

$\begin{array}{l}⇒{|\mathbf{a}+\mathbf{b}|}^{2}=2+2\mathrm{cos}\theta \end{array}$

It is given that :

$\begin{array}{l}{|\mathbf{a}+\mathbf{b}|}^{2}={\left(\surd 3\right)}^{2}=3\end{array}$

Putting this value,

$\begin{array}{l}⇒2\mathrm{cos}\theta ={|\mathbf{a}+\mathbf{b}|}^{2}-2=3-2=1\end{array}$

$\begin{array}{l}⇒\mathrm{cos}\theta =\frac{1}{2}\\ ⇒\theta =60°\end{array}$

Using this value, we now proceed to find the value of given identity,

$\begin{array}{l}\left(\mathbf{a}-\mathbf{b}\right)\mathbf{.}\left(2\mathbf{a}+\mathbf{b}\right)=2{a}^{2}-{b}^{2}-ab\mathrm{cos}\theta =2\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}{1}^{2}-{1}^{2}-1\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}1\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}\mathrm{cos}60°\end{array}$

$\begin{array}{l}⇒\left(\mathbf{a}-\mathbf{b}\right)\mathbf{.}\left(2\mathbf{a}+\mathbf{b}\right)=\frac{1}{2}\end{array}$

## Evaluation of dot product

Problem : In an experiment of light reflection, if a , b and c are the unit vectors in the direction of incident ray, reflected ray and normal to the reflecting surface, then prove that :

$\begin{array}{l}⇒\mathbf{b}=\mathbf{a}-2\left(\mathbf{a}\mathbf{.}\mathbf{c}\right)\mathbf{c}\end{array}$

Solution : Let us consider vectors in a coordinate system in which “x” and “y” axes of the coordinate system are in the direction of reflecting surface and normal to the reflecting surface respectively as shown in the figure.

We express unit vectors with respect to the incident and reflected as :

$\begin{array}{l}\mathbf{a}=\mathrm{sin}\theta \mathbf{i}-\mathrm{cos}\theta \mathbf{j}\\ \mathbf{b}=\mathrm{sin}\theta \mathbf{i}+\mathrm{cos}\theta \mathbf{j}\end{array}$

Subtracting first equation from the second equation, we have :

$\begin{array}{l}⇒\mathbf{b}-\mathbf{a}=2\mathrm{cos}\theta \mathbf{j}\\ ⇒\mathbf{b}=\mathbf{a}+2\mathrm{cos}\theta \mathbf{j}\end{array}$

Now, we evaluate dot product, involving unit vectors :

$\begin{array}{l}\mathbf{a}\mathbf{.}\mathbf{c}=1\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}1\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}\mathrm{cos}\left(180°-\theta \right)=-\mathrm{cos}\theta \end{array}$

Substituting for cosθ, we have :

$\begin{array}{l}⇒\mathbf{b}=\mathbf{a}-2\left(\mathbf{a}\mathbf{.}\mathbf{c}\right)\mathbf{c}\end{array}$

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
what is nanomaterials​ and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
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 anyone tell who founded equations of motion !?
n=a+b/T² find the linear express
Quiklyyy
Moment of inertia of a bar in terms of perpendicular axis theorem
How should i know when to add/subtract the velocities and when to use the Pythagoras theorem?
Centre of mass of two uniform rods of same length but made of different materials and kept at L-shape meeting point is origin of coordinate
A balloon is released from the ground which rises vertically up with acceleration 1.4m/sec^2.a ball is released from the balloon 20 second after the balloon has left the ground. The maximum height reached by the ball from the ground is
work done by frictional force formula
Torque
Why are we takingspherical surface area in case of solid sphere
In all situatuons, what can I generalize?
the body travels the distance of d=( 14+- 0.2)m in t=( 4.0 +- 0.3) s calculate it's velocity with error limit find Percentage error
Explain it ?Fy=?sN?mg=0?N=mg?s