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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.
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 :
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,
$$\begin{array}{l}\mathbf{a}\mathbf{.}\mathbf{b}=(2\mathbf{i}+\mathbf{j}-\mathbf{k})\mathbf{.}(2\mathbf{i}-\mathbf{k})\end{array}$$
Ignoring dot products of different unit vectors (they evaluate to zero), we have :
$$\begin{array}{l}\mathbf{a}\mathbf{.}\mathbf{b}=2\mathbf{i}\mathbf{.}\mathbf{i}+(-\mathbf{k})\mathbf{.}(-\mathbf{k})=2+1=3\\ a=\surd ({2}^{2}+{1}^{2}+{1}^{2})=\surd 6\\ b=\surd ({1}^{2}+{1}^{2})=\surd 2\\ \mathrm{ab}=\surd 6\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}\surd 2=\surd \left(12\right)=2\surd 3\end{array}$$
Putting in the expression of cosine, we have :
$$\begin{array}{l}\mathrm{cos}\theta =\frac{\mathbf{a}\mathbf{.}\mathbf{b}}{\mathrm{ab}}=\frac{3}{2\surd 3}=\frac{\surd 3}{2}=\mathrm{cos}30\xb0\\ \theta =30\xb0\end{array}$$
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.
$$\begin{array}{l}(\mathbf{a}+\mathbf{b})\mathbf{.}(\mathbf{a}-\mathbf{b})=0\end{array}$$
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 :
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,
$$\begin{array}{l}|\mathbf{i}+\mathbf{j}|=\surd ({1}^{2}+{1}^{2})=\surd 2\end{array}$$
Hence,
$$\begin{array}{l}n=\frac{1}{\surd 2}\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}(\mathbf{i}+\mathbf{j})\end{array}$$
The component of vector 2 i + 3 j in the direction of “ n ” is :
$$\begin{array}{l}{a}_{n}=\mathbf{a}\mathbf{.}\mathbf{n}=(2\mathbf{i}+3\mathbf{j})\mathbf{.}\frac{1}{\surd 2}\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}(\mathbf{i}+\mathbf{j})\end{array}$$
$$\begin{array}{l}\Rightarrow {a}_{n}=\frac{1}{\surd 2}\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}(2\mathbf{i}+3\mathbf{j})\mathbf{.}(\mathbf{i}+\mathbf{j})\\ \Rightarrow {a}_{n}=\frac{1}{\surd 2}\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}(2x1+3x1)\\ \Rightarrow {a}_{n}=\frac{5}{\surd 2}\end{array}$$
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 $B\ne C$ .
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 ),
$$\begin{array}{l}(\mathbf{a}+\mathbf{b})\mathbf{.}(\mathbf{a}+\mathbf{b})={|\mathbf{a}+\mathbf{b}|}^{2}\end{array}$$
Using distributive property,
$$\begin{array}{l}\Rightarrow \mathbf{a}\mathbf{.}\mathbf{a}+\mathbf{b}\mathbf{.}\mathbf{a}+\mathbf{a}\mathbf{.}\mathbf{b}+\mathbf{b}\mathbf{.}\mathbf{b}={a}^{2}+{b}^{2}+2ab\mathrm{cos}\theta ={|\mathbf{a}+\mathbf{b}|}^{2}\\ \Rightarrow {|\mathbf{a}+\mathbf{b}|}^{2}={a}^{2}+{b}^{2}+2ab\mathrm{cos}\theta \end{array}$$
Similarly, using the identity with difference of the vectors (a-b),
$$\begin{array}{l}\Rightarrow {|\mathbf{a}-\mathbf{b}|}^{2}={a}^{2}+{b}^{2}-2ab\mathrm{cos}\theta \end{array}$$
It is, however, given that :
$$\begin{array}{l}\Rightarrow |\mathbf{a}+\mathbf{b}|=|\mathbf{a}-\mathbf{b}|\end{array}$$
Squaring on either side of the equation,
$$\begin{array}{l}\Rightarrow {|\mathbf{a}+\mathbf{b}|}^{2}={|\mathbf{a}-\mathbf{b}|}^{2}\end{array}$$
Putting the expressions,
$$\begin{array}{l}\Rightarrow {a}^{2}+{b}^{2}+2ab\mathrm{cos}\theta ={a}^{2}+{b}^{2}-2ab\mathrm{cos}\theta \\ \Rightarrow 4ab\mathrm{cos}\theta =0\\ \Rightarrow \mathrm{cos}\theta =0\\ \Rightarrow \theta =90\xb0\end{array}$$
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 :
$$\begin{array}{l}(\mathbf{a}-\mathbf{b})\mathbf{.}(2\mathbf{a}+\mathbf{b})\end{array}$$
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 .
$$\begin{array}{l}(\mathbf{a}-\mathbf{b})\mathbf{.}(2\mathbf{a}+\mathbf{b})=2\mathbf{a}\mathbf{.}\mathbf{a}+\mathbf{a}\mathbf{.}\mathbf{b}-\mathbf{b}\mathbf{.}2\mathbf{a}+(-\mathbf{b})\mathbf{.}(-\mathbf{b})=2{a}^{2}-\mathbf{a}\mathbf{.}\mathbf{b}-{b}^{2}\end{array}$$
$$\begin{array}{l}\Rightarrow (\mathbf{a}-\mathbf{b})\mathbf{.}(2\mathbf{a}+\mathbf{b})=2{a}^{2}-{b}^{2}-ab\mathrm{cos}\theta \end{array}$$
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,
$$\begin{array}{l}{|\mathbf{a}+\mathbf{b}|}^{2}=(\mathbf{a}+\mathbf{b})\mathbf{.}(\mathbf{a}+\mathbf{b})={a}^{2}+{b}^{2}+2ab\mathrm{cos}\theta =1+1+2\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}1\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}1\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}\mathrm{cos}\theta \end{array}$$
$$\begin{array}{l}\Rightarrow {|\mathbf{a}+\mathbf{b}|}^{2}=2+2\mathrm{cos}\theta \end{array}$$
It is given that :
$$\begin{array}{l}{|\mathbf{a}+\mathbf{b}|}^{2}={(\surd 3)}^{2}=3\end{array}$$
Putting this value,
$$\begin{array}{l}\Rightarrow 2\mathrm{cos}\theta ={|\mathbf{a}+\mathbf{b}|}^{2}-2=3-2=1\end{array}$$
$$\begin{array}{l}\Rightarrow \mathrm{cos}\theta =\frac{1}{2}\\ \Rightarrow \theta =60\xb0\end{array}$$
Using this value, we now proceed to find the value of given identity,
$$\begin{array}{l}(\mathbf{a}-\mathbf{b})\mathbf{.}(2\mathbf{a}+\mathbf{b})=2{a}^{2}-{b}^{2}-ab\mathrm{cos}\theta =2\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}{1}^{2}-{1}^{2}-1\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}1\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}\mathrm{cos}60\xb0\end{array}$$
$$\begin{array}{l}\Rightarrow (\mathbf{a}-\mathbf{b})\mathbf{.}(2\mathbf{a}+\mathbf{b})=\frac{1}{2}\end{array}$$
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 :
$$\begin{array}{l}\Rightarrow \mathbf{b}=\mathbf{a}-2(\mathbf{a}\mathbf{.}\mathbf{c})\mathbf{c}\end{array}$$
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 :
$$\begin{array}{l}\Rightarrow \mathbf{b}-\mathbf{a}=2\mathrm{cos}\theta \mathbf{j}\\ \Rightarrow \mathbf{b}=\mathbf{a}+2\mathrm{cos}\theta \mathbf{j}\end{array}$$
Now, we evaluate dot product, involving unit vectors :
$$\begin{array}{l}\mathbf{a}\mathbf{.}\mathbf{c}=1\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}1\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}\mathrm{cos}(180\xb0-\theta )=-\mathrm{cos}\theta \end{array}$$
Substituting for cosθ, we have :
$$\begin{array}{l}\Rightarrow \mathbf{b}=\mathbf{a}-2(\mathbf{a}\mathbf{.}\mathbf{c})\mathbf{c}\end{array}$$
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