### 1 Definition of the scalar product

Consider the two vectors $\underset{̲}{a}$ and $\underset{̲}{b}$ shown in Figure 29.

Figure 29 :

Note that the tails of the two vectors coincide and that the angle between the vectors is labelled $\theta$ . Their scalar product, denoted by $\underset{̲}{a}\cdot \underset{̲}{b}$ , is defined as the product $|\underset{̲}{a}|\phantom{\rule{0.3em}{0ex}}|\underset{̲}{b}|\phantom{\rule{0.3em}{0ex}}cos\theta$ . It is very important to use the dot in the formula. The dot is the specific symbol for the scalar product, and is the reason why the scalar product is also known as the dot product . You should not use a $×$ sign in this context because this sign is reserved for the vector product which is quite different.

The angle $\theta$ is always chosen to lie between 0 and $\pi$ , and the tails of the two vectors must coincide. Figure 30 shows two incorrect ways of measuring $\theta$ .

Figure 30 :

##### Key Point 9

The scalar product of $\underset{̲}{a}$ and $\underset{̲}{b}$ is:

$\phantom{\rule{1em}{0ex}}\underset{̲}{a}\cdot \underset{̲}{b}=|\underset{̲}{a}|\phantom{\rule{0.3em}{0ex}}|\underset{̲}{b}|\phantom{\rule{0.3em}{0ex}}cos\theta$
We can remember this formula as:

“The modulus of the first vector, multiplied by the modulus of the second vector,

multiplied by the cosine of the angle between them.”

Clearly $\underset{̲}{b}\cdot \underset{̲}{a}=|\underset{̲}{b}|\phantom{\rule{0.3em}{0ex}}|\underset{̲}{a}|\phantom{\rule{0.3em}{0ex}}cos\theta$ and so

$\phantom{\rule{2em}{0ex}}\underset{̲}{a}\cdot \underset{̲}{b}=\underset{̲}{b}\cdot \underset{̲}{a}.$

Thus we can evaluate a scalar product in any order: the operation is commutative .

##### Example 8

Vectors $\underset{̲}{a}$ and $\underset{̲}{b}$ are shown in the Figure 31. Vector $\underset{̲}{a}$ has modulus 6 and vector $\underset{̲}{b}$ has modulus 7 and the angle between them is $6{0}^{\circ }$ . Calculate $\underset{̲}{a}.\underset{̲}{b}$ .

Figure 31

##### Solution

The angle between the two vectors is $6{0}^{\circ }$ . Hence

$\phantom{\rule{2em}{0ex}}\underset{̲}{a}\cdot \underset{̲}{b}=|\underset{̲}{a}|\phantom{\rule{0.3em}{0ex}}|\underset{̲}{b}|\phantom{\rule{0.3em}{0ex}}cos\theta =\left(6\right)\left(7\right)cos6{0}^{\circ }=21$

The scalar product of $\underset{̲}{a}$ and $\underset{̲}{b}$ is 21. Note that a scalar product is always a scalar.

##### Example 9

Find $\underset{̲}{i}\cdot \underset{̲}{i}$ where $\underset{̲}{i}$ is the unit vector in the direction of the positive $x$ axis.

##### Solution

Because $\underset{̲}{i}$ is a unit vector its modulus is 1. Also, the angle between $\underset{̲}{i}$ and itself is zero. Therefore

$\phantom{\rule{2em}{0ex}}\underset{̲}{i}.\underset{̲}{i}=\left(1\right)\left(1\right)cos{0}^{\circ }=1$

So the scalar product of $\underset{̲}{i}$ with itself equals 1. It is easy to verify that $\underset{̲}{j}.\underset{̲}{j}=1$ and $\underset{̲}{k}.\underset{̲}{k}=1$ .

##### Example 10

Find $\underset{̲}{i}\cdot \underset{̲}{j}$ where $\underset{̲}{i}$ and $\underset{̲}{j}$ are unit vectors in the directions of the $x$ and $y$ axes.

##### Solution

Because $\underset{̲}{i}$ and $\underset{̲}{j}$ are unit vectors they each have a modulus of 1. The angle between the two vectors is $9{0}^{\circ }$ . Therefore

$\phantom{\rule{2em}{0ex}}\underset{̲}{i}\cdot \underset{̲}{j}=\left(1\right)\left(1\right)cos9{0}^{\circ }=0$

That is $\underset{̲}{i}\cdot \underset{̲}{j}=0$ .

The following results are easily verified:

##### Key Point 10

$\begin{array}{rcll}\underset{̲}{i}\cdot \underset{̲}{i}& =& \underset{̲}{j}\cdot \underset{̲}{j}=\underset{̲}{k}\cdot \underset{̲}{k}=1& \text{}\\ \underset{̲}{i}\cdot \underset{̲}{j}& =& \underset{̲}{j}\cdot \underset{̲}{i}=0& \text{}\\ \underset{̲}{i}\cdot \underset{̲}{k}& =& \underset{̲}{k}\cdot \underset{̲}{i}=0& \text{}\\ \underset{̲}{j}\cdot \underset{̲}{k}& =& \underset{̲}{k}\cdot \underset{̲}{j}=0& \text{}\end{array}$

Generally, whenever any two vectors are perpendicular to each other their scalar product is zero because the angle between the vectors is $9{0}^{\circ }$ and $cos9{0}^{\circ }=0$ .

##### Key Point 11

The scalar product of perpendicular vectors is zero.