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The two alternate ways of evaluating dot product of two vectors indicate that the product is commutative i.e. independent of the order of two vectors :
A block of mass “m” moves from point A to B along a smooth plane surface under the action of force as shown in the figure. Find the work done if it is defined as :
W = F. Δ x
where F and Δ x are force and displacement vectors.
Expanding the expression of work, we have :
$$\begin{array}{l}W=\mathbf{F}\mathbf{.}\Delta \mathbf{x}=F\Delta x\mathrm{cos}\theta \end{array}$$
Here, F = 10 N, Δx = 10 m and cosθ = cos60° = 0.5.
$$\begin{array}{l}\Rightarrow W=10x10x0.5=\mathrm{50\; J}\end{array}$$
The value of dot product is maximum for the maximum value of cosθ. Now, the maximum value of cosine is $\mathrm{cos}0\xb0=1$ . For this value, dot product simply evaluates to the product of the magnitudes of two vectors.
$$\begin{array}{l}{(\mathbf{a}\mathbf{.}\mathbf{b})}_{\mathrm{max}}=ab\end{array}$$
For $\theta =180\xb0,\mathrm{cos}180\xb0=-1$ and
$$\begin{array}{l}\mathbf{a}\mathbf{.}\mathbf{b}=-ab\end{array}$$
Thus, we see that dot product can evaluate to negative value as well. This is a significant result as many scalar quantities in physics are given negative value. The work done, for example, can be negative, when displacement is in the opposite direction to the component of force along that direction.
The scalar product evaluates to zero for θ = 90° and 270° as cosine of these angles are zero. These results have important implication for unit vectors. The dot products of same unit vector evaluates to 1.
$$\begin{array}{l}\mathbf{i}\mathbf{.}\mathbf{i}=\mathbf{j}\mathbf{.}\mathbf{j}=\mathbf{k}\mathbf{.}\mathbf{k}=1\end{array}$$
The dot products of combination of different unit vectors evaluate to zero.
$$\begin{array}{l}\mathbf{i}\mathbf{.}\mathbf{j}=\mathbf{j}\mathbf{.}\mathbf{k}=\mathbf{k}\mathbf{.}\mathbf{i}=0\end{array}$$
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 :
$$\begin{array}{l}\mathrm{cos}\theta =\frac{\mathbf{a}\mathbf{.}\mathbf{b}}{\mathrm{ab}}=\frac{3}{2\surd 3}=\frac{\surd 3}{2}=\mathrm{cos}30\xb0\\ \theta =30\xb0\end{array}$$
Two vectors in component forms are written as :
$$\begin{array}{l}\mathbf{a}={a}_{x}\mathbf{i}+{a}_{y}\mathbf{j}+{a}_{z}\mathbf{k}\\ \mathbf{b}={b}_{x}\mathbf{i}+{b}_{y}\mathbf{j}+{b}_{z}\mathbf{k}\end{array}$$
In evaluating the product, we make use of the fact that multiplication of the same unit vectors is 1, while multiplication of different unit vectors is zero. The dot product evaluates to scalar terms as :
A closer look at the expansion of dot product of two vectors reveals that the expression is very similar to the expression for a component of a vector. The expression of the dot product is :
$$\begin{array}{l}\mathbf{a}\mathbf{.}\mathbf{b}=ab\mathrm{cos}\theta \end{array}$$
On the other hand, the component of a vector in a given direction is :
$$\begin{array}{l}{a}_{x}=a\mathrm{cos}\theta \end{array}$$
Comparing two equations, we can define component of a vector in a direction given by unit vector " n " as :
This is a very general and useful relation to determine component of a vector in any direction. Only requirement is that we should know the unit vector in the direction in which component is to be determined.
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 :
$$\begin{array}{l}{a}_{n}=\mathbf{a}\mathbf{.}\mathbf{n}=(2\mathbf{i}+3\mathbf{j})\mathbf{.}\frac{1}{\surd 2}\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}(\mathbf{i}+\mathbf{j})\end{array}$$
$$\begin{array}{l}\Rightarrow {a}_{n}=\frac{1}{\surd 2}\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}(2\mathbf{i}+3\mathbf{j})\mathbf{.}(\mathbf{i}+\mathbf{j})\\ \Rightarrow {a}_{n}=\frac{1}{\surd 2}\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}(2x1+3x1)\\ \Rightarrow {a}_{n}=\frac{5}{\surd 2}\end{array}$$
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