# 1.7 Scalar (dot) product  (Page 5/5)

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

This is known as cosine law of triangle. Curiously, we may pay attention to first two equations above. As a matter of fact, second equation gives the square of the magnitude of resultant of two vectors a and b .

## Differentiation and dot product

Differentiation of a vector expression yields a vector. Consider a vector expression given as :

$\begin{array}{l}\mathbf{a}=\left({x}^{2}+2x+3\right)\mathbf{i}\end{array}$

The derivative of the vector with respect to x is :

$\begin{array}{l}{\mathbf{a}}^{\text{'}}=\left(2x+2\right)\mathbf{i}\end{array}$

As the derivative is a vector, two vector expressions with dot product is differentiated in a manner so that dot product is retained in the final expression of derivative. For example,

$\begin{array}{l}\frac{d}{dx}\left(\mathbf{a}\mathbf{.}\mathbf{b}\right)={\mathbf{a}}^{\text{'}}\mathbf{b}+\mathbf{a}{\mathbf{b}}^{\text{'}}\end{array}$

## Exercises

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

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 :

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

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.

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

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

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,

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

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

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 :

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

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 :

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

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Source:  OpenStax, Physics for k-12. OpenStax CNX. Sep 07, 2009 Download for free at http://cnx.org/content/col10322/1.175
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