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In this section, we summarize the properties of dot product as discussed above. Besides, some additional derived attributes are included for reference.
1: Dot product is commutative
This means that the dot product of vectors is not dependent on the sequence of vectors :
$$\begin{array}{l}\mathbf{a}\mathbf{.}\mathbf{b}=\mathbf{b}\mathbf{.}\mathbf{a}\end{array}$$
We must, however, be careful while writing sequence of dot product. For example, writing a sequence involving three vectors like a.b.c is incorrect. For, dot product of any two vectors is a scalar. As dot product is defined for two vectors (not one vector and one scalar), the resulting dot product of a scalar ( a.b ) and that of third vector c has no meaning.
2: Distributive property of dot product :
$$\begin{array}{l}\mathbf{a}\mathbf{.}(\mathbf{b}+\mathbf{c})=\mathbf{a}\mathbf{.}\mathbf{b}+\mathbf{a}\mathbf{.}\mathbf{c}\end{array}$$
3: The dot product of a vector with itself is equal to the square of the magnitude of the vector.
$$\begin{array}{l}\mathbf{a}\mathbf{.}\mathbf{a}=a\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}a\mathrm{cos\theta}={a}^{2}\mathrm{cos}0\xb0={a}^{2}\end{array}$$
4: The magnitude of dot product of two vectors can be obtained in either of the following manner :
$$\begin{array}{l}\mathbf{a}\mathbf{.}\mathbf{b}=ab\mathrm{cos}\theta \\ \mathbf{a}\mathbf{.}\mathbf{b}=ab\mathrm{cos}\theta =a\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}\left(b\mathrm{cos}\theta \right)=a\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}\text{component of}\phantom{\rule{2pt}{0ex}}\mathbf{b}\phantom{\rule{2pt}{0ex}}\text{along}\phantom{\rule{2pt}{0ex}}\mathbf{a}\phantom{\rule{2pt}{0ex}}\\ \mathbf{a}\mathbf{.}\mathbf{b}=ab\mathrm{cos}\theta =\left(a\mathrm{cos}\theta \right)\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}b=b\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}\text{component of}\phantom{\rule{2pt}{0ex}}\mathbf{a}\phantom{\rule{2pt}{0ex}}\text{along}\phantom{\rule{2pt}{0ex}}\mathbf{b}\phantom{\rule{2pt}{0ex}}\end{array}$$
The dot product of two vectors is equal to the algebraic product of magnitude of one vector and component of second vector in the direction of first vector.
5: The cosine of the angle between two vectors can be obtained in terms of dot product as :
$$\begin{array}{l}\mathbf{a}\mathbf{.}\mathbf{b}=ab\mathrm{cos\theta}\end{array}$$
$$\begin{array}{l}\Rightarrow \mathrm{cos\theta}=\frac{\mathbf{a}\mathbf{.}\mathbf{b}}{ab}\end{array}$$
6: The condition of two perpendicular vectors in terms of dot product is given by :
$$\begin{array}{l}\mathbf{a}\mathbf{.}\mathbf{b}=ab\mathrm{cos}90\xb0=0\end{array}$$
7: Properties of dot product with respect to unit vectors along the axes of rectangular coordinate system are :
$$\begin{array}{l}\mathbf{i}\mathbf{.}\mathbf{i}=\mathbf{j}\mathbf{.}\mathbf{j}=\mathbf{k}\mathbf{.}\mathbf{k}=1\\ \mathbf{i}\mathbf{.}\mathbf{j}=\mathbf{j}\mathbf{.}\mathbf{k}=\mathbf{k}\mathbf{.}\mathbf{i}=0\end{array}$$
8: Dot product in component form is :
$$\begin{array}{l}\mathbf{a}\mathbf{.}\mathbf{b}={a}_{x}{b}_{x}+{a}_{y}{b}_{y}+{a}_{z}{b}_{z}\end{array}$$
9: The dot product does not yield to cancellation. For example, if a.b = a.c , then we can not conclude that b = c . Rearranging, we have :
$$\begin{array}{l}\mathbf{a}\mathbf{.}\mathbf{b}-\mathbf{a}\mathbf{.}\mathbf{c}=0\\ \mathbf{a}\mathbf{.}(\mathbf{b}-\mathbf{c})=0\end{array}$$
This means that a and ( b - c ) are perpendicular to each other. In turn, this implies that ( b - c ) is not equal to zero (null vector). Hence, b is not equal to c as we would get after cancellation.
We can understand this difference with respect to cancellation more explicitly by working through the problem given here :
Problem : Verify vector equality B = C , if A.B = A.C .
Solution : The given equality of dot products is :
$$\begin{array}{l}\mathbf{A}\mathbf{.}\mathbf{B}=\mathbf{A}\mathbf{.}\mathbf{C}\end{array}$$
We should 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. 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 $B\ne C$ .
Law of cosine relates sides of a triangle with one included angle. We can determine this relationship using property of a dot product. Let three vectors are represented by sides of the triangle such that closing side is the sum of other two vectors. Then applying triangle law of addition :
$$\begin{array}{l}\mathbf{c}=(\mathbf{a}+\mathbf{b})\end{array}$$
We know that the dot product of a vector with itself is equal to the square of the magnitude of the vector. Hence,
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