Show that the dot product of any two unit vectors is the sum of the product of the components?

Oct 19, 2017

We have:

$\boldsymbol{\underline{A}} = {A}_{1} \boldsymbol{\underline{i}} + {A}_{2} \boldsymbol{\underline{j}} + {A}_{3} \boldsymbol{\underline{k}}$
$\boldsymbol{\underline{B}} = {B}_{1} \boldsymbol{\underline{i}} + {B}_{2} \boldsymbol{\underline{j}} + {B}_{3} \boldsymbol{\underline{k}}$

So then by definition the dot product is

$\boldsymbol{\underline{A}} \cdot \boldsymbol{\underline{B}} = \left({A}_{1} \boldsymbol{\underline{i}} + {A}_{2} \boldsymbol{\underline{j}} + {A}_{3} \boldsymbol{\underline{k}}\right) \cdot \left({B}_{1} \boldsymbol{\underline{i}} + {B}_{2} \boldsymbol{\underline{j}} + {B}_{3} \boldsymbol{\underline{k}}\right)$

Using the distributive property of the dot product:

$\boldsymbol{\underline{A}} \cdot \boldsymbol{\underline{B}} = {A}_{1} \boldsymbol{\underline{i}} \cdot \left({B}_{1} \boldsymbol{\underline{i}} + {B}_{2} \boldsymbol{\underline{j}} + {B}_{3} \boldsymbol{\underline{k}}\right) +$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus {A}_{2} \boldsymbol{\underline{j}} \cdot \left({B}_{1} \boldsymbol{\underline{i}} + {B}_{2} \boldsymbol{\underline{j}} + {B}_{3} \boldsymbol{\underline{k}}\right) +$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus {A}_{3} \boldsymbol{\underline{k}} \cdot \left({B}_{1} \boldsymbol{\underline{i}} + {B}_{2} \boldsymbol{\underline{j}} + {B}_{3} \boldsymbol{\underline{k}}\right)$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = {A}_{1} \boldsymbol{\underline{i}} \cdot {B}_{1} \boldsymbol{\underline{i}} + {A}_{1} \boldsymbol{\underline{i}} \cdot {B}_{2} \boldsymbol{\underline{j}} + {A}_{1} \boldsymbol{\underline{i}} \cdot {B}_{3} \boldsymbol{\underline{k}} +$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus {A}_{2} \boldsymbol{\underline{j}} \cdot {B}_{1} \boldsymbol{\underline{i}} + {A}_{2} \boldsymbol{\underline{j}} \cdot {B}_{2} \boldsymbol{\underline{j}} + {A}_{2} \boldsymbol{\underline{j}} \cdot {B}_{3} \boldsymbol{\underline{k}} +$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus {A}_{3} \boldsymbol{\underline{k}} \cdot {B}_{1} \boldsymbol{\underline{i}} + {A}_{3} \boldsymbol{\underline{k}} \cdot {B}_{2} \boldsymbol{\underline{j}} + {A}_{3} \boldsymbol{\underline{k}} \cdot {B}_{3} \boldsymbol{\underline{k}}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = {A}_{1} {B}_{1} \boldsymbol{\underline{i}} \cdot \boldsymbol{\underline{i}} + {A}_{1} {B}_{2} \boldsymbol{\underline{i}} \cdot \boldsymbol{\underline{j}} + {A}_{1} {B}_{3} \boldsymbol{\underline{i}} \cdot \boldsymbol{\underline{k}} +$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus {A}_{2} {B}_{1} \boldsymbol{\underline{j}} \cdot \boldsymbol{\underline{i}} + {A}_{2} {B}_{2} \boldsymbol{\underline{j}} \cdot \boldsymbol{\underline{j}} + {A}_{2} {B}_{3} \boldsymbol{\underline{j}} \cdot \boldsymbol{\underline{k}} +$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus {A}_{3} {B}_{1} \boldsymbol{\underline{k}} \cdot \boldsymbol{\underline{i}} + {A}_{3} {B}_{2} \boldsymbol{\underline{k}} \cdot \boldsymbol{\underline{j}} + {A}_{3} {B}_{3} \boldsymbol{\underline{k}} \cdot \boldsymbol{\underline{k}}$

We now use the property that the dot product of a unit vector with itself is unity whilst the dot product of the unit vector with any other perpendicular vector is zero:

$\boldsymbol{\underline{A}} \cdot \boldsymbol{\underline{B}} = {A}_{1} {B}_{1} \left(1\right) + {A}_{1} {B}_{2} \left(0\right) + {A}_{1} {B}_{3} \left(0\right) +$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus {A}_{2} {B}_{1} \left(0\right) + {A}_{2} {B}_{2} \left(1\right) + {A}_{2} {B}_{3} \left(0\right) +$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus {A}_{3} {B}_{1} \left(0\right) + {A}_{3} {B}_{2} \left(0\right) + {A}_{3} {B}_{3} \left(1\right)$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = {A}_{1} {B}_{1} + {A}_{2} {B}_{2} + {A}_{3} {B}_{3}$ QED