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I'm trying to draw the picture below but have been unsuccessful (I drew part a, can't do part b)

This is what I used to draw part a

\coordinate (A) at (0,0) {};
\coordinate (B) at (2,0) {};
\coordinate (C) at (1,1) {};
\coordinate (D) at (3,1) {};

\draw[thick, black, dotted]  (4,0) -- (5,1) node[sloped,midway,above] {};
\draw[thick, black, dotted]   (1,1) -- (5,1) node[sloped,midway,above] {};
\draw[->,  thick, black,   arrows={-latex}]  (0,0) -- (4,0) node[sloped,right=-0.7cm,above] {$\mathsf{A}$};
\draw[->,  thick, black,  arrows={-latex}]  (0,0) -- (1,1) node[sloped,midway,above=-0.1cm] {$\mathsf{B}$};
\draw[->,  thick, black, arrows={-latex}]  (0,0) -- (5,1) node[sloped,midway,above=-0.1cm] {$\mathsf{A+B}$};

Image I'm trying to reproduce

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Welcome to TeX.SX! You can have a look at our starter guide to familiarize yourself further with our format. – yo' Jul 31 '14 at 22:23

One possibility:



  (5,0) coordinate (X) --
  (0,0) coordinate (O);

  coordinate (A) at (20:4.5cm)  
  coordinate (B) at (55:3cm)
  coordinate (AB) at (75:5cm);

\foreach \Coor in {A,B,AB}
  \draw[->,thick,shorten >= 2pt] (O) -- (\Coor);
  \node[circle,fill,inner sep=1.5pt,label={above right:$\Coor$}] 
    at (\Coor) {};
\node[draw,circle,fill=white,inner sep=1.5pt,label={below left:$O$}] {};

  (O) -- 
    node[sloped,above=5pt] {(length $A$)$\cdot$(length $B$)} 

  pic[draw,line width=0.7pt,angle radius=15mm] {angle={X--O--A}}    
  pic[draw,line width=0.35pt,angle radius=15mm] {angle={A--O--AB}}    
  pic[draw,line width=0.35pt,angle radius=18mm] {angle={X--O--B}}    
  pic[draw,line width=0.7pt,angle radius=18mm] {angle={B--O--AB}};    


enter image description here

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I wonder if determining of coordinates A, B and AB as given in your answer do have some advantages over just saying: \coordinate (A) at (...);\coordinate (B) at (...);\coordinate (AB) at (...);? – Zarko Aug 1 '14 at 14:35

Another possibility could be:

\begin{tikzpicture}[font=\boldmath, >=latex, decoration = {brace,raise=4pt},line width=0.7pt]
\foreach \n/\s/\l/\p/\a in {o/draw/{$0$}/{0,0}/left,A/fill/{$A$}/{20:5}/right,B/fill/{$B$}/{45:3.5}/right,AB/fill/{$AB$}/{65:6}/right}
    \node (\n)[\s,circle,scale=0.3,label={\a:\l}]at (\p){};
\draw (o)  edge[dashed] ++(5,0)
      [->] edge (A)
      [->] edge (B)
      [->] edge (AB);
\draw[decorate,thin] (o) -- node[sloped,above=5pt] {(length $A$)$\cdot$(length $B$)} (AB);
\foreach \t/\a/\b/\r in {/0/20/1.6,thin/20/65/1.6,thin/0/45/2,/45/65/2}
    \draw [->,\t](\a:\r) arc (\a:\b:\r);

With the output:

enter image description here

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