1.7 Scalar (dot) product  (Page 4/5)

Attributes of scalar (dot) product

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.

$$\begin{array}{l}\mathbf{a}\mathbf{.}\mathbf{a}=axa\mathrm{cos\theta }=a^2\cos 0°=a^2\end{array}$$

4: The magnitude of dot product of two vectors can be obtained in either of the following manner :

$$\begin{array}{l}\mathbf{a}\mathbf{.}\mathbf{b}=ab\cos \theta \\ \mathbf{a}\mathbf{.}\mathbf{b}=ab\cos \theta =ax\left(b\cos \theta \right)=ax\text{component of}\mathbf{b}\text{along}\mathbf{a}\\ \mathbf{a}\mathbf{.}\mathbf{b}=ab\cos \theta =\left(a\cos \theta \right)xb=bx\text{component of}\mathbf{a}\text{along}\mathbf{b}\end{array}$$

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 :

$$\begin{array}{l}\mathbf{a}\mathbf{.}\mathbf{b}=ab\cos 90°=0\end{array}$$

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 :

Law of cosine and dot product

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,