### 1 Definition of the scalar product

Consider the two vectors $\underset{\u0332}{a}$ and $\underset{\u0332}{b}$ 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
$\underset{\u0332}{a}\cdot \underset{\u0332}{b}$
, is defined as the product
$\left|\underset{\u0332}{a}\right|\phantom{\rule{0.3em}{0ex}}\left|\underset{\u0332}{b}\right|\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
$\times $
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
$\underset{\u0332}{a}$
and
$\underset{\u0332}{b}$
is:

“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{\u0332}{b}\cdot \underset{\u0332}{a}=\left|\underset{\u0332}{b}\right|\phantom{\rule{0.3em}{0ex}}\left|\underset{\u0332}{a}\right|\phantom{\rule{0.3em}{0ex}}cos\theta $ and so

$\phantom{\rule{2em}{0ex}}\underset{\u0332}{a}\cdot \underset{\u0332}{b}=\underset{\u0332}{b}\cdot \underset{\u0332}{a}.$

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

##### Example 8

Vectors $\underset{\u0332}{a}$ and $\underset{\u0332}{b}$ are shown in the Figure 31. Vector $\underset{\u0332}{a}$ has modulus 6 and vector $\underset{\u0332}{b}$ has modulus 7 and the angle between them is $6{0}^{\circ}$ . Calculate $\underset{\u0332}{a}.\underset{\u0332}{b}$ .

**
Figure 31
**

##### Solution

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

$\phantom{\rule{2em}{0ex}}\underset{\u0332}{a}\cdot \underset{\u0332}{b}=\left|\underset{\u0332}{a}\right|\phantom{\rule{0.3em}{0ex}}\left|\underset{\u0332}{b}\right|\phantom{\rule{0.3em}{0ex}}cos\theta =\left(6\right)\left(7\right)cos6{0}^{\circ}=21$

The scalar product of $\underset{\u0332}{a}$ and $\underset{\u0332}{b}$ is 21. Note that a scalar product is always a scalar.

##### Example 9

Find $\underset{\u0332}{i}\cdot \underset{\u0332}{i}$ where $\underset{\u0332}{i}$ is the unit vector in the direction of the positive $x$ axis.

##### Solution

Because $\underset{\u0332}{i}$ is a unit vector its modulus is 1. Also, the angle between $\underset{\u0332}{i}$ and itself is zero. Therefore

$\phantom{\rule{2em}{0ex}}\underset{\u0332}{i}.\underset{\u0332}{i}=\left(1\right)\left(1\right)cos{0}^{\circ}=1$

So the scalar product of $\underset{\u0332}{i}$ with itself equals 1. It is easy to verify that $\underset{\u0332}{j}.\underset{\u0332}{j}=1$ and $\underset{\u0332}{k}.\underset{\u0332}{k}=1$ .

##### Example 10

Find $\underset{\u0332}{i}\cdot \underset{\u0332}{j}$ where $\underset{\u0332}{i}$ and $\underset{\u0332}{j}$ are unit vectors in the directions of the $x$ and $y$ axes.

##### Solution

Because $\underset{\u0332}{i}$ and $\underset{\u0332}{j}$ are unit vectors they each have a modulus of 1. The angle between the two vectors is $9{0}^{\circ}$ . Therefore

$\phantom{\rule{2em}{0ex}}\underset{\u0332}{i}\cdot \underset{\u0332}{j}=\left(1\right)\left(1\right)cos9{0}^{\circ}=0$

That is $\underset{\u0332}{i}\cdot \underset{\u0332}{j}=0$ .

The following results are easily verified:

##### Key Point 10

$$\begin{array}{rcll}\underset{\u0332}{i}\cdot \underset{\u0332}{i}& =& \underset{\u0332}{j}\cdot \underset{\u0332}{j}=\underset{\u0332}{k}\cdot \underset{\u0332}{k}=1& \text{}\\ \underset{\u0332}{i}\cdot \underset{\u0332}{j}& =& \underset{\u0332}{j}\cdot \underset{\u0332}{i}=0& \text{}\\ \underset{\u0332}{i}\cdot \underset{\u0332}{k}& =& \underset{\u0332}{k}\cdot \underset{\u0332}{i}=0& \text{}\\ \underset{\u0332}{j}\cdot \underset{\u0332}{k}& =& \underset{\u0332}{k}\cdot \underset{\u0332}{j}=0& \text{}\end{array}$$

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.