<?xml version="1.0" encoding="UTF-8"?>

<record version="1" id="207">
 <title>dot product example</title>
 <name>DotProductExample</name>
 <created>2006-07-22 15:56:56</created>
 <modified>2006-07-22 15:56:56</modified>
 <type>Example</type>
<parent id="205">dot product</parent>
 <creator id="1" name="bloftin"/>
 <modifier id="1" name="bloftin"/>
 <author id="1" name="bloftin"/>
 <classification>
	<category scheme="msc" code="02."/>
 </classification>
 <preamble>% this is the default PlanetMath preamble.  as your knowledge
% of TeX increases, you will probably want to edit this, but
% it should be fine as is for beginners.

% almost certainly you want these
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}

% used for TeXing text within eps files
%\usepackage{psfrag}
% need this for including graphics (\includegraphics)
%\usepackage{graphicx}
% for neatly defining theorems and propositions
%\usepackage{amsthm}
% making logically defined graphics
%\usepackage{xypic}

% there are many more packages, add them here as you need them

% define commands here</preamble>
 <content>{\bf Examples}

(1) Calculate $ \mathbf{A} \cdot \mathbf{B} $ with

$ \mathbf{A} = 5 \mathbf{\hat{i}} + 2 \mathbf{\hat{j}} - \mathbf{\hat{k}}$ \\
$ \mathbf{B} = 3 \mathbf{\hat{i}} - 3 \mathbf{\hat{j}} + 3\mathbf{\hat{k}}$ \\

answer:

$ \mathbf{A} \cdot \mathbf{B} = (5)(3) + (2)(-3) + (-1)(+3) $ \\
$ \mathbf{A} \cdot \mathbf{B} = 15 - 6 - 3  = 6$ \\

(2) Find the angle between the above vectors.

answer:

We know their dot product, so we just need to calculate their magnitudes

$ \left | \mathbf{A} \right | = \sqrt{A_x^2 + A_y^2 + A_z^2} = \sqrt{5^2 + 2^2 +(-1)^2} $ \\
$ \left | \mathbf{A} \right | = \sqrt{25 + 4 + 1} = \sqrt{30} = 5.48 $ \\
\\

$ \left | \mathbf{B} \right | = \sqrt{3^2 + (-3)^2 +(3)^2} $ \\
$ \left | \mathbf{B} \right | = \sqrt{9 + 9 + 9} = \sqrt{27} = 5.2 $ \\
\\

Finally


$ \mathbf{A} \cdot \mathbf{B} = \left | \mathbf{A} \right | \left | \mathbf{B} \right | \cos \theta$

$ \theta = cos^{-1} \left ( \frac{ \mathbf{A} \cdot \mathbf{B} }{ \left | \mathbf{A} \right | \left | \mathbf{B} \right |} \right )$ \\

$ \theta = cos^{-1} \left ( \frac{ 6 }{ (5.48)(5.2)} \right) $ \\
$ \theta = cos^{-1} (0.21) = 77.9^o $</content>
</record>
