Dot Products

My complicated derivation


There’s a simpler derivation using

  1. the geometric defintion
  2. linearity from the definition
  3. orthogonality of the canonical basic vectors $\vec{i}$, $\vec{j}$ and $\vec{k}$.

If I had read that, I wouldn’t have type this document in LaTeX. The following content was transcribed from a PDF file that I made three days ago.


Recall: $\sum_{i=m}^{n} a_{i}=a_{m}+a_{m+1}+a_{m+2}+\cdots +a_{n-1}+a_{n},$ where $m$, $n$ are integers.

Example: $\sum_{i=3}^6 i^2 = 3^2 + 4^2 + 5^2 + 6^2 = 86$

Settings: let ${\color{red}{\vec{a} = \overrightarrow{OP_1} = (a_1, a_2, a_3) = a_1 \vec{i} + a_2 \vec{j} + a_3 \vec{k}}}$ and ${\color{blue}{\vec{b} = \overrightarrow{OP_2} = (b_1, b_2, b_3) = b_1 \vec{i} + b_2 \vec{j} + b_3 \vec{k} }}$ be two 3D-vectors, and $\gamma = \angle {\color{red}{P_1}}O{\color{blue}{P_2}}$.

Two vectors in 3D-space

Two vectors in 3D-space

Two vectors in 3D-space

\begin{align} {\color{red}{\vec{a}}} \cdot {\color{blue}{\vec{b}}} &= \sum_{i=1}^3 {\color{red}{a_i}} {\color{blue}{b_i}} \tag{alg} \\ {\color{red}{\vec{a}}} \cdot {\color{blue}{\vec{b}}} &= |{\color{red}{\vec{a}}}| \, |{\color{blue}{\vec{b}}}| \, \cos {\color{mygreen}{\gamma}} \tag{geom} \end{align}

WLOG assume $\boxed{|{\color{red}{\vec{a}}}| = |{\color{blue}{\vec{b}}}| = 1}$, where $|{\color{red}{\vec{a}}}| = \sqrt{\sum_{i=1}^3 {\color{red}{a_i}}^2}$.

\[ \left| \overrightarrow{{\color{red}{P_1}}{\color{blue}{P_2}}} \right| = \sqrt{\sum_{i=1}^3 ({\color{red}{a_i}} - {\color{blue}{b_i}})^2} = \sqrt{2\left(1 - \boxed{\sum_{i=1}^3 {\color{red}{a_i}} {\color{blue}{b_i}}} \right)} \]
Top view of two vectors in 3D-space

Top view of two vectors in 3D-space

Top view of two vectors in 3D-space

Rotate $\angle {\color{red}{P_1}}O{\color{blue}{P_2}}$ so that the image $\angle {\color{red}{P_1’}}O{\color{blue}{P_2}}’$ lies on the $xy$-plane and ${\color{red}{\overrightarrow{OP_1’}}}$ coincides with the positive $x$-axis.

\begin{align} \left| \overrightarrow{{\color{red}{P_1'}}{\color{blue}{P_2'}}} \right| &= \sqrt{({\color{red}{1}}-{\color{blue}{\cos\gamma}})^2 + ({\color{red}{0}}-{\color{blue}{\sin\gamma}})^2} \\ &= \sqrt{1 - 2\cos\gamma + \cos^2\gamma + \sin^2\gamma} \\ &= \sqrt{2(1 - \boxed{\cos\gamma})} \end{align}

Since rotation doesn’t change angle size and side lengths, $\left| \overrightarrow{{\color{red}{P_1}}{\color{blue}{P_2}}} \right| =\left| \overrightarrow{{\color{red}{P_1’}}{\color{blue}{P_2’}}} \right|$, so the special case $\boxed{|{\color{red}{\vec{a}}}|=|{\color{blue}{\vec{b}}}|=1}$ is done.

Exercise: prove the general case.

LaTeX skills used

I don’t know why \begin{align*} doesn’t work despite the surrounding <div> tags.

The LaTeX analog of CSS float to the right is a wrap figure.


I forgot how to define color without importing an external package.


With used in standalone documents, the dvipsnames option has to be global.


The math3d config is copied from the book TikZ pour l’impatient.


I tried to keep things within one page for printing purposes. I omitted the date and my name in favor of a simple title done quickly.

  {\Huge Dot product}

I took away the paragraph indent \parindent and I reduced the vertical spacing between adjacent lines.


TikZ source code

First figure

{x= {(-0.353cm,-0.353cm)}, z={(0cm,1cm)},y={(1cm,0cm)}}}
  \coordinate[label=left:$O$] (O) at (0,0,0);
  \coordinate[label=right:${\color{red}{P_1}}$] (P1) at (1.2,1,0.5);
  \coordinate[label=above:${\color{blue}{P_2}}$] (P2) at (-0.8,0.9,0.7);
  \draw[-latex] (O)--(1,0,0) node[pos=1.1]{$\vec{i}$};
  \draw[-latex] (O)--(0,1,0) node[pos=1.1]{$\vec{j}$};
  \draw[-latex] (O)--(0,0,1) node[pos=1.1]{$\vec{k}$};
  \draw[-latex,red] (O) -- (P1) node[midway,sloped,below] {$\vec{a}$};
  \draw[-latex,blue] (O) -- (P2) node[midway,sloped,above] {$\vec{b}$};
    \clip (O) -- (P1) -- (P2) -- cycle;
    \draw[mygreen] (O) circle [x radius=0.4, y radius=0.2];
  \draw ($ (O)!.25!($ (P1)!.5!(P2) $) $) node [right,mygreen]{$\gamma$};

Second picture

  \coordinate [label=below left:$O$] (O) at (0,0);
  \coordinate [label=below right:${\color{red}{P_1'=(1,0)}}$] (P1d) at (1,0);
  \coordinate [label=above right:${\color{blue}{P_2'=(\cos\gamma,\sin\gamma)}}$] (P2d) at (67:1);
  \draw[-latex] (-1.5,0) -- (1.5,0) node[pos=1.1] {$x$};
  \draw[-latex] (0,-1.5) -- (0,1.5) node[pos=1.1] {$y$};
  \draw[-latex,red] (O) -- (P1d);
  \draw[-latex,blue] (O) -- (P2d);
  \draw[->,mygreen] (0.3,0) arc (0:67:0.3) node[midway,sloped,above] {$\gamma$};

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