### 4 Using the scalar product to find the angle between vectors

We have two distinct ways of calculating the scalar product of two vectors. From Key Point 9 $\underset{̲}{a}\cdot \underset{̲}{b}=|\underset{̲}{a}|\phantom{\rule{0.3em}{0ex}}|\underset{̲}{b}|\phantom{\rule{0.3em}{0ex}}cos\theta$ whilst from Key Point 13 $\underset{̲}{a}\cdot \underset{̲}{b}={a}_{1}{b}_{1}+{a}_{2}{b}_{2}+{a}_{3}{b}_{3}$ . Both methods of calculating the scalar product are entirely equivalent and will always give the same value for the scalar product. We can exploit this correspondence to find the angle between two vectors. The following example illustrates the procedure to be followed.

##### Example 14

Find the angle between the vectors $\underset{̲}{a}=5\underset{̲}{i}+3\underset{̲}{j}-2\underset{̲}{k}$ and $\underset{̲}{b}=8\underset{̲}{i}-9\underset{̲}{j}+11\underset{̲}{k}$ .

##### Solution

The scalar product of these two vectors has already been found in Example 12 to be $-9$ . The modulus of $\underset{̲}{a}$ is $\sqrt{{5}^{2}+{3}^{2}+{\left(-2\right)}^{2}}=\sqrt{38}$ . The modulus of $\underset{̲}{b}$ is $\sqrt{{8}^{2}+{\left(-9\right)}^{2}+1{1}^{2}}=\sqrt{266}$ . Substituting these values for $\underset{̲}{a}\cdot \underset{̲}{b},\phantom{\rule{1em}{0ex}}\left|\underset{̲}{a}\right|$ and $\underset{̲}{b}$ into the formula for the scalar product we find

$\begin{array}{rcll}\underset{̲}{a}\cdot \underset{̲}{b}& =& |\underset{̲}{a}|\phantom{\rule{0.3em}{0ex}}|\underset{̲}{b}|\phantom{\rule{0.3em}{0ex}}cos\theta & \text{}\\ -9& =& \sqrt{38}\phantom{\rule{0.3em}{0ex}}\sqrt{266}cos\theta \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$

from which

$\phantom{\rule{2em}{0ex}}cos\theta =\frac{-9}{\sqrt{38}\sqrt{266}}=-0.0895$

so that $\phantom{\rule{1em}{0ex}}\theta ={cos}^{-1}\left(-0.0895\right)=95.1{4}^{\circ }$

In general, the angle between two vectors can be found from the following formula:

##### Key Point 16

The angle $\theta$ between vectors $\underset{̲}{a}$ , $\underset{̲}{b}$ is such that:

$cos\theta =\frac{\underset{̲}{a}\cdot \underset{̲}{b}}{|\underset{̲}{a}\left|\phantom{\rule{0.3em}{0ex}}\right|\underset{̲}{b}|}$
##### Exercises
1. If $\underset{̲}{a}=2\underset{̲}{i}-5\underset{̲}{j}$ and $\underset{̲}{b}=3\underset{̲}{i}+2\underset{̲}{j}$ find $\underset{̲}{a}\cdot \underset{̲}{b}$ and verify that $\underset{̲}{a}\cdot \underset{̲}{b}=\underset{̲}{b}\cdot \underset{̲}{a}$ .
2. Find the angle between $\underset{̲}{p}=3\underset{̲}{i}-\underset{̲}{j}$ and $\underset{̲}{q}=-4\underset{̲}{i}+6\underset{̲}{j}$ .
3. Use the definition of the scalar product to show that if two vectors are perpendicular, their scalar product is zero.
4. If $\underset{̲}{a}$ and $\underset{̲}{b}$ are perpendicular, simplify $\left(\underset{̲}{a}-2\underset{̲}{b}\right)\cdot \left(3\underset{̲}{a}+5\underset{̲}{b}\right)$ .
5. If $\underset{̲}{p}=\underset{̲}{i}+8\underset{̲}{j}+7\underset{̲}{k}$ and $\underset{̲}{q}=3\underset{̲}{i}-2\underset{̲}{j}+5\underset{̲}{k}$ , find $\underset{̲}{p}\cdot \underset{̲}{q}$ .
6. Show that the vectors $\frac{1}{2}\underset{̲}{i}+\underset{̲}{j}$ and $2\underset{̲}{i}-\underset{̲}{j}$ are perpendicular.
7. The work done by a force $\underset{̲}{F}$ in moving a body through a displacement $\underset{̲}{r}$ is given by $\underset{̲}{F}\cdot \underset{̲}{r}$ . Find the work done by the force $\underset{̲}{F}=3\underset{̲}{i}+7\underset{̲}{k}$ if it causes a body to move from the point with coordinates $\left(1,1,2\right)$ to the point $\left(7,3,5\right)$ .
8. Find the angle between the vectors $\underset{̲}{i}-\underset{̲}{j}-\underset{̲}{k}$ and $2\underset{̲}{i}+\underset{̲}{j}+2\underset{̲}{k}$ .
1. $-4$ .
2. $142.{1}^{\circ }$ ,
3. This follows from the fact that $cos\theta =0$ since $\theta =9{0}^{\circ }$ .
4. $3{a}^{2}-10{b}^{2}$ .
5. 22.
6. This follows from the scalar product being zero.
7. 39 units.
8. $101.{1}^{\circ }$