# Dot product of two vectors

How does one show that the dot product of two vectors is $A \cdot B = |A| |B| \cos (\Theta)$?

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The question is unclear. Do you mean "How does one show that the dot product..."? – joriki Mar 6 '11 at 9:44
Yes same meaning, thanks. – user7821 Mar 6 '11 at 9:45
Isn't this the definition of the dot product? Do you have another definition for the dot product in mind, and want to show that that definition is equivalent to this one? Because that's a more sensible question. – Deepak Apr 12 at 14:47

Think about a triangle with sidelengths $|\textbf{a}|,|\textbf{b}|,|\textbf{c}|$. Then we can use the law of cosines.

\begin{align} |\textbf{c}|^2&=|\textbf{a}|^2+|\textbf{b}|^2-2|\textbf{a}||\textbf{b}| \cos \theta \\ \implies 2|\textbf{a}||\textbf{b}| \cos \theta &= |\textbf{a}|^2+|\textbf{b}|^2-|\textbf{c}| = \textbf{a}\cdot \textbf{a} + \textbf{b} \cdot \textbf{b} - \textbf{c}\cdot \textbf{c} \end{align}

By the properties of the dot product and from the fact that $\textbf{c}=\textbf{b}-\textbf{a}$ we find that

\begin{align} \textbf{c}\cdot \textbf{c} &=(\textbf{b}-\textbf{a}) \cdot (\textbf{b}-\textbf{a}) \\ &=(\textbf{b}-\textbf{a})\cdot \textbf{b} - (\textbf{b}-\textbf{a}) \cdot \textbf{a} \\ &= \textbf{b}\cdot \textbf{b} - \textbf{a}\cdot \textbf{b} - \textbf{b}\cdot \textbf{a} + \textbf{a}\cdot \textbf{a}. \end{align}

By substituting $\textbf{c}\cdot \textbf{c}$, we get \begin{align} 2|\textbf{a}||\textbf{b}| \cos \theta &= \textbf{a}\cdot \textbf{a} + \textbf{b} \cdot \textbf{b} - (\textbf{b}\cdot \textbf{b} - \textbf{a}\cdot \textbf{b} - \textbf{b}\cdot \textbf{a} + \textbf{a}\cdot \textbf{a}) \\ &=2 \textbf{a}\cdot \textbf{b}. \end{align}

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One way of showing this requires taking a Geometric look at stuff:

Think of $\vec{a},\vec{b}$ as the vertices of a triangle with one corner in the origin and sides of length $|\vec a|,|\vec b|,|\vec{a-b}|$. Now, use the Law of cosines, and inner product properties (over $\mathbb R$) to calculate $|\vec{a-b}|$:

The law of cosines gives you: $$|\vec{a-b}|^2=|\vec{a}|^2+|\vec{b}|^2-2|\vec{a}||\vec{b}|\cos\theta(a,b)$$

Inner product gives: $$|\vec{a-b}|^2=\langle\vec{a-b},\vec{a-b}\rangle=|\vec{a}|^2+|\vec{b}|^2-2\langle\vec{a},\vec{b}\rangle$$

from there on it's an easy proof (I left you the technical details).

Hope that helps

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Wow, \vec{a-b} looks bad. Yet another reason to prefer either underline or bold to denote vectors (\overrightarrow{a-b} doesn't look quite right either) – kahen Mar 6 '11 at 10:50