Editing Dot product (section)

== Properties==
The dot product fulfills the following properties if <math>\mathbf{a}</math>, <math>\mathbf{b}</math>, <math>\mathbf{c}</math> and <math>\mathbf{d}</math> are real [[vector (geometry)|vectors]] and <math>\alpha</math>, <math>\beta</math>, <math>\gamma</math> and <math>\delta</math> are [[scalar (mathematics)|scalars]].<ref name="Lipschutz2009" /><ref name="Spiegel2009" />

; [[Commutative]] : <math display="block"> \mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a} ,</math> which follows from the definition (<math>\theta</math> is the angle between <math>\mathbf{a}</math> and <math>\mathbf{b}</math>):<ref>{{cite web|last=Nykamp|first=Duane|title=The dot product|url=https://mathinsight.org/dot_product|access-date=September 6, 2020|website=Math Insight}}</ref> <math display="block"> \mathbf{a} \cdot \mathbf{b} = \left\| \mathbf{a} \right\| \left\| \mathbf{b} \right\| \cos \theta = \left\| \mathbf{b} \right\| \left\| \mathbf{a} \right\| \cos \theta = \mathbf{b} \cdot \mathbf{a} .</math> The commutative property can also be easily proven with the algebraic definition, and in [[Inner product space|more general spaces]] (where the notion of angle might not be geometrically intuitive but an analogous product can be defined) the angle between two vectors can be defined as
<math display="block"> \theta = \operatorname{arccos}\left( \frac{\mathbf{a}\cdot\mathbf{b}}{\left\|\mathbf{a}\right\| \left\|\mathbf{b}\right\|} \right). </math>
; [[bilinear form|Bilinear]] (additive, distributive and scalar-multiplicative in both arguments) : <math display="block"> 
\begin{align}
(\alpha \mathbf{a} + \beta\mathbf{b})&\cdot (\gamma\mathbf{c}+\delta\mathbf{d}) \\  
&=\alpha\gamma(\mathbf{a}\cdot\mathbf{c}) + \alpha\delta(\mathbf{a}\cdot\mathbf{d}) +\beta\gamma(\mathbf{b}\cdot\mathbf{c}) +\beta\delta(\mathbf{b}\cdot\mathbf{d}) . \end{align}</math>
; Not [[associative]] : Because the dot product is not defined between a scalar <math>\mathbf{a}\cdot\mathbf{b}</math> and a vector <math>\mathbf{c},</math> associativity is meaningless.<ref>Weisstein, Eric W. "Dot Product". From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/DotProduct.html</ref> However, bilinearity implies <math display="block">c (\mathbf{a} \cdot \mathbf{b}) = (c\mathbf{a})\cdot\mathbf{b} = \mathbf{a}\cdot(c\mathbf{b}).</math> This property is sometimes called the "associative law for scalar and dot product",<ref name="BanchoffWermer1983">{{cite book|author1=T. Banchoff|author2=J. Wermer | title=Linear Algebra Through Geometry|year=1983|publisher=Springer Science & Business Media| isbn=978-1-4684-0161-5 | page=12| url=https://archive.org/details/linearalgebrathr00banc_0/page/12/mode/2up}}</ref> and one may say that "the dot product is associative with respect to scalar multiplication".<ref name="BedfordFowler2008">{{cite book | author1=A. Bedford|author2=Wallace L. Fowler|title=Engineering Mechanics: Statics|year=2008|publisher=Prentice Hall | isbn=978-0-13-612915-8 | edition=5th | page=60}}</ref>
; [[Orthogonal]] : Two non-zero vectors <math>\mathbf{a}</math> and <math>\mathbf{b}</math> are ''orthogonal'' if and only if <math>\mathbf{a} \cdot \mathbf{b} = 0</math>.
; No [[cancellation law|cancellation]]
: Unlike multiplication of ordinary numbers, where if <math>ab=ac</math>, then <math>b</math> always equals <math>c</math> unless <math>a</math> is zero, the dot product does not obey the [[cancellation law]]: {{pb}} If <math>\mathbf{a}\cdot\mathbf{b}=\mathbf{a}\cdot\mathbf{c}</math> and <math>\mathbf{a}\neq\mathbf{0}</math>, then we can write: <math>\mathbf{a}\cdot(\mathbf{b}-\mathbf{c}) = 0</math> by the [[distributive law]]; the result above says this just means that <math>\mathbf{a}</math> is perpendicular to <math>(\mathbf{b}-\mathbf{c})</math>, which still allows <math>(\mathbf{b}-\mathbf{c})\neq\mathbf{0}</math>, and therefore allows <math>\mathbf{b}\neq\mathbf{c}</math>.
; [[Product rule]] : If <math>\mathbf{a}</math> and <math>\mathbf{b}</math> are vector-valued [[differentiable function]]s, then the derivative ([[Notation for differentiation#Lagrange's notation|denoted by a prime]] <math>{}'</math>) of <math>\mathbf{a}\cdot\mathbf{b}</math> is given by the rule <math display="block">(\mathbf{a}\cdot\mathbf{b})' = \mathbf{a}'\cdot\mathbf{b} + \mathbf{a}\cdot\mathbf{b}'.</math>

=== Application to the law of cosines ===
[[File:Dot product cosine rule.svg|100px|thumb|Triangle with vector edges '''a''' and '''b''', separated by angle ''θ'']]
{{main|Law of cosines}}

Given two vectors <math>{\color{red}\mathbf{a}}</math> and <math>{\color{blue}\mathbf{b}}</math> separated by angle <math>\theta</math> (see the upper image), they form a triangle with a third side <math>{\color{orange}\mathbf{c}} = {\color{red}\mathbf{a}} - {\color{blue}\mathbf{b}}</math>. Let <math>\color{red}a</math>, <math>\color{blue}b</math> and <math>\color{orange}c</math> denote the lengths of <math>{\color{red}\mathbf{a}}</math>, <math>{\color{blue}\mathbf{b}}</math>, and <math>{\color{orange}\mathbf{c}}</math>, respectively. The dot product of this with itself is:
<math display="block">
\begin{align}
\mathbf{\color{orange}c} \cdot \mathbf{\color{orange}c} & = ( \mathbf{\color{red}a} - \mathbf{\color{blue}b}) \cdot ( \mathbf{\color{red}a} - \mathbf{\color{blue}b} ) \\
 & = \mathbf{\color{red}a} \cdot \mathbf{\color{red}a} - \mathbf{\color{red}a} \cdot \mathbf{\color{blue}b} - \mathbf{\color{blue}b} \cdot \mathbf{\color{red}a} + \mathbf{\color{blue}b} \cdot \mathbf{\color{blue}b} \\
 & = {\color{red}a}^2 - \mathbf{\color{red}a} \cdot \mathbf{\color{blue}b} - \mathbf{\color{red}a} \cdot \mathbf{\color{blue}b} + {\color{blue}b}^2 \\
 & = {\color{red}a}^2 - 2 \mathbf{\color{red}a} \cdot \mathbf{\color{blue}b} + {\color{blue}b}^2 \\
{\color{orange}c}^2 & = {\color{red}a}^2 + {\color{blue}b}^2 - 2 {\color{red}a} {\color{blue}b} \cos \mathbf{\color{purple}\theta} \\
\end{align}
</math>
which is the [[law of cosines]].
{{clear}}