### 2 A formula for finding the scalar product

We can use the results summarized in Key Point 10 to obtain a formula for finding a scalar product when the vectors are given in Cartesian form. We consider vectors in the $xy$ plane. Suppose $\underset{̲}{a}={a}_{1}\underset{̲}{i}+{a}_{2}\underset{̲}{j}$ and $\underset{̲}{b}={b}_{1}\underset{̲}{i}+{b}_{2}\underset{̲}{j}$ . Then

$\begin{array}{rcll}\underset{̲}{a}\cdot \underset{̲}{b}& =& \left({a}_{1}\underset{̲}{i}+{a}_{2}\underset{̲}{j}\right)\cdot \left({b}_{1}\underset{̲}{i}+{b}_{2}\underset{̲}{j}\right)& \text{}\\ & =& {a}_{1}\underset{̲}{i}\cdot \left({b}_{1}\underset{̲}{i}+{b}_{2}\underset{̲}{j}\right)+{a}_{2}\underset{̲}{j}\cdot \left({b}_{1}\underset{̲}{i}+{b}_{2}\underset{̲}{j}\right)& \text{}\\ & =& {a}_{1}{b}_{1}\underset{̲}{i}\cdot \underset{̲}{i}+{a}_{1}{b}_{2}\underset{̲}{i}\cdot \underset{̲}{j}+{a}_{2}{b}_{1}\underset{̲}{j}\cdot \underset{̲}{i}+{a}_{2}{b}_{2}\underset{̲}{j}\cdot \underset{̲}{j}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$

Using the results in Key Point 10 we can simplify this to give the following formula:

##### Key Point 12
$\underset{̲}{a}\cdot \underset{̲}{b}={a}_{1}{b}_{1}+{a}_{2}{b}_{2}$

Thus to find the scalar product of two vectors their $\underset{̲}{i}$ components are multiplied together, their $\underset{̲}{j}$ components are multiplied together and the results are added.

##### Example 11

If $\underset{̲}{a}=7\underset{̲}{i}+8\underset{̲}{j}$ and $\underset{̲}{b}=5\underset{̲}{i}-2\underset{̲}{j}$ , find the scalar product $\underset{̲}{a}\cdot \underset{̲}{b}$ .

##### Solution

We use Key Point 12:

$\phantom{\rule{2em}{0ex}}\underset{̲}{a}\cdot \underset{̲}{b}=\left(7\underset{̲}{i}+8\underset{̲}{j}\right)\cdot \left(5\underset{̲}{i}-2\underset{̲}{j}\right)=\left(7\right)\left(5\right)+\left(8\right)\left(-2\right)=35-16=19$

The formula readily generalises to vectors in three dimensions as follows:

##### Key Point 13
$\underset{̲}{a}\cdot \underset{̲}{b}={a}_{1}{b}_{1}+{a}_{2}{b}_{2}+{a}_{3}{b}_{3}$
##### Example 12

If $\underset{̲}{a}=5\underset{̲}{i}+3\underset{̲}{j}-2\underset{̲}{k}$ and $\underset{̲}{b}=8\underset{̲}{i}-9\underset{̲}{j}+11\underset{̲}{k}$ , find $\underset{̲}{a}\cdot \underset{̲}{b}$ .

##### Solution

We use the formula in Key Point 13:

$\phantom{\rule{2em}{0ex}}\underset{̲}{a}\cdot \underset{̲}{b}=\left(5\right)\left(8\right)+\left(3\right)\left(-9\right)+\left(-2\right)\left(11\right)=40-27-22=-9$

Note again that the result is a scalar: there are no $\underset{̲}{i}$ ’s, $\underset{̲}{j}$ ’s, or $\underset{̲}{k}$ ’s in the answer.

If $\underset{̲}{p}=4\underset{̲}{i}-3\underset{̲}{j}+7\underset{̲}{k}$ and $\underset{̲}{q}=6\underset{̲}{i}-\underset{̲}{j}+2\underset{̲}{k}$ , find $\underset{̲}{p}\cdot \underset{̲}{q}$ .

Use Key Point 13:

41

If $\underset{̲}{r}=3\underset{̲}{i}+2\underset{̲}{j}+9\underset{̲}{k}$ find $\underset{̲}{r}\cdot \underset{̲}{r}$ . Show that this is the same as ${\left|\underset{̲}{r}\right|}^{2}$ .

$\underset{̲}{r}\cdot \underset{̲}{r}=\left(3\underset{̲}{i}+2\underset{̲}{j}+9\underset{̲}{k}\right)\cdot \left(3\underset{̲}{i}+2\underset{̲}{j}+9\underset{̲}{k}\right)=3\underset{̲}{i}\cdot 3\underset{̲}{i}+3\underset{̲}{i}\cdot 2\underset{̲}{j}+\cdots =9+0+\cdots =94$ .

$\left|\underset{̲}{r}\right|=\sqrt{9+4+81}=\sqrt{94}$ , hence ${\left|\underset{̲}{r}\right|}^{2}=\underset{̲}{r}\cdot \underset{̲}{r}$ .

The above result is generally true:

##### Key Point 14

For any vector $\underset{̲}{r}$ , ${\left|\underset{̲}{r}\right|}^{2}=\underset{̲}{r}\cdot \underset{̲}{r}$