From: Andrew Lorimer <andrew@lorimer.id.au>
Date: Sat, 25 May 2019 12:26:11 +0000 (+1000)
Subject: [spec] finalise complex numbers section
X-Git-Tag: yr12~121
X-Git-Url: https://git.lorimer.id.au/notes.git/diff_plain/f4f68705f28943ec51033727121c6ea70573523d

[spec] finalise complex numbers section
---

diff --git a/spec/spec-collated.pdf b/spec/spec-collated.pdf
index 49943ba..136eeb8 100644
Binary files a/spec/spec-collated.pdf and b/spec/spec-collated.pdf differ
diff --git a/spec/spec-collated.tex b/spec/spec-collated.tex
index 72544ce..d61f60e 100644
--- a/spec/spec-collated.tex
+++ b/spec/spec-collated.tex
@@ -1,13 +1,17 @@
 \documentclass[a4paper]{article}
 \usepackage[a4paper,margin=2cm]{geometry}
 \usepackage{multicol}
+\usepackage{multirow}
 \usepackage{amsmath}
 \usepackage{amssymb}
 \usepackage{harpoon}
 \usepackage{tabularx}
+\usepackage[dvipsnames, table]{xcolor}
 \usepackage{graphicx}
 \usepackage{wrapfig}
 \usepackage{tikz}
+\usetikzlibrary{calc}
+\usepgflibrary{arrows.meta}
 \usepackage{fancyhdr}
 \pagestyle{fancy}
 \fancyhead[LO,LE]{Year 12 Specialist}
@@ -19,6 +23,10 @@
 \setlength\fboxsep{0pt} \setlength\fboxrule{.2pt} % for the \fboxes
 \newcommand*\leftlap[3][\,]{#1\hphantom{#2}\mathllap{#3}}
 \newcommand*\rightlap[2]{\mathrlap{#2}\hphantom{#1}}
+\newcolumntype{L}[1]{>{\hsize=#1\hsize\raggedright\arraybackslash}X}%
+\newcolumntype{R}[1]{>{\hsize=#1\hsize\raggedleft\arraybackslash}X}%
+\definecolor{cas}{HTML}{e6f0fe}
+\linespread{1.5}
 
 \begin{document}
 
@@ -28,18 +36,43 @@
 
     \[\mathbb{C}=\{a+bi:a,b\in\mathbb{R}\}\]
 
+    \begin{align*}
+      \text{Cartesian form: } & a+bi\\
+      \text{Polar form: } & r\operatorname{cis}\theta
+    \end{align*}
+
     \subsection*{Operations}
 
-      \begin{align*}
-        z_1 \pm z_2&=(a \pm c)(b \pm d)i\\
-        k \times z &= ka + kbi\\
-        z_1 \cdot z_2 &= ac-bd+(ad+bc)i\\
-        z_1 \div z_2 &= (z_1 \overline{z_2}) \div |z_2|^2
-      \end{align*}
+      \begin{tabularx}{\columnwidth}{R{0.33}|X|X}
+        & \textbf{Cartesian} & \textbf{Polar} \\
+        \hline
+        \(z_1 \pm z_2\) & \((a \pm c)(b \pm d)i\) & convert to \(a+bi\)\\
+        \hline
+        \(+k \times z\) & \multirow{2}{\hsize}{\(ka \pm kbi\)} & \(kr\operatorname{cis} \theta\)\\
+        \cline{1-1}\cline{3-3}
+        \(-k \times z\) & & \(kr \operatorname{cis}(\theta\pm \pi)\)\\
+        \hline
+        \(z_1 \cdot z_2\) & \(ac-bd+(ad+bc)i\) & \(r_1r_2 \operatorname{cis}(\theta_1 + \theta_2)\)\\
+        \hline
+        \(z_1 \div z_2\) & \((z_1 \overline{z_2}) \div |z_2|^2\) & \(\left(\frac{r_1}{r_2}\right) \operatorname{cis}(\theta_1 - \theta_2)\)
+      \end{tabularx}
+
+      \subsubsection*{Scalar multiplication in polar form}
+      
+        For \(k \in \mathbb{R}^+\):
+        \[k\left(r \operatorname{cis}\theta\right)=kr \operatorname{cis}\theta\]
+
+        \noindent For \(k \in \mathbb{R}^-\):
+        \[k\left(r \operatorname{cis}\theta\right)=kr \operatorname{cis}\left(\begin{cases}\theta - \pi & 0<\operatorname{Arg}(z)\le \pi \\ \theta + \pi & -\pi<\operatorname{Arg}(z)\le 0\end{cases}\right)\]
 
     \subsection*{Conjugate}
 
-      \[\overline{z} = a \pm bi\]
+      \begin{align*}
+        \overline{z} &= a \mp bi\\
+        &= r \operatorname{cis}(-\theta)
+      \end{align*}
+
+      \noindent \colorbox{cas}{On CAS:} \verb|conjg(a+bi)|
 
       \subsubsection*{Properties}
 
@@ -68,8 +101,8 @@
 
       \begin{align*}
         z^{-1}&=\frac{a-bi}{a^2+b^2}\\
-        &=\frac{\overline{z}}{|z|^2}
-        a
+        &=\frac{\overline{z}}{|z|^2}a\\
+        &=r \operatorname{cis}(-\theta)
       \end{align*}
 
     \subsection*{Dividing over \(\mathbb{C}\)}
@@ -81,20 +114,21 @@
         & \qquad \text{(rationalise denominator)}
       \end{align*}
 
-    \subsection*{Argand planes}
-
-      \begin{tikzpicture}\begin{scope}[thick,font=\scriptsize]
-        \draw [->] (-1.5,0) -- (1.5,0) node [above left]  {$\operatorname{Re}(z)$};
-        \draw [->] (0,-1.5) -- (0,1.5) node [below right] {$\operatorname{Im}(z)$};
+    \subsection*{Polar form}
 
-        % If you only want a single label per axis side:
-        \draw (1,-3pt) -- (1,0pt)   node [below] {$1$};
-        \draw (-1,-3pt) -- (-1,0pt) node [below] {$-1$};
-        \draw (-3pt,1) -- (0pt,1)   node [left] {$i$};
-        \draw (-3pt,-1) -- (0pt,-1) node [left] {$-i$};
-      \end{scope}\end{tikzpicture}
+      \begin{align*}
+        z&=r\operatorname{cis}\theta\\
+        &=r(\cos \theta + i \sin \theta)
+      \end{align*}
 
-      Multiplication by \(i \implies\) anticlockwise rotation of \(\frac{\pi}{2}\)
+      \begin{itemize}
+        \item{\(r=|z|=\sqrt{\operatorname{Re}(z)^2 + \operatorname{Im}(z)^2}\)}
+        \item{\(\theta = \operatorname{arg}(z)\) \quad \colorbox{cas}{On CAS:} \verb|arg(a+bi)|}
+        \item{\(\operatorname{Arg}(z) \in (-\pi,\pi)\) \quad \bf{(principal argument)}}
+        \item{\colorbox{cas}{Convert on CAS:}\\ \verb|compToTrig(a+bi)| \(\iff\) \verb|cExpand{r·cisX}|}
+        \item{Multiple representations:\\\(r\operatorname{cis}\theta=r\operatorname{cis}(\theta+2n\pi)\) with \(n \in \mathbb{Z}\) revolutions}
+        \item{\(\operatorname{cis}\pi=-1,\qquad \operatorname{cis}0=1\)}
+      \end{itemize}
 
     \subsection*{de Moivres' theorem}
 
@@ -104,9 +138,7 @@
     
       Include \(\pm\) for all solutions, incl. imaginary
 
-\newcolumntype{R}{>{\raggedleft\arraybackslash}X}
-\newcolumntype{L}{>{\raggedright\arraybackslash}X}
-      \begin{tabularx}{\columnwidth}{rX}
+      \begin{tabularx}{\columnwidth}{ R{0.55} X  }
         \hline
         Sum of squares & \(\begin{aligned} 
         z^2 + a^2 &= z^2-(ai)^2\\
@@ -116,27 +148,341 @@
         \hline
         Division & \(P(z)=D(z)Q(z)+R(z)\) \\
         \hline
-        \parbox[t]{2cm}{Remainder} & Let \(\alpha \in \mathbb{C}\). Remainder of \(P(z) \div (z-\alpha)\) is \(P(\alpha)\)\\
+        Remainder theorem & Let \(\alpha \in \mathbb{C}\). Remainder of \(P(z) \div (z-\alpha)\) is \(P(\alpha)\)\\
+        \hline
+        Factor theorem & \(z-\alpha\) is a factor of \(P(z) \iff P(\alpha)=0\) for \(\alpha \in \mathbb{C}\)\\
         \hline
-\end{tabularx}
+        Conjugate root theorem & \(P(z)=0 \text{ at } z=a\pm bi\) (\(\implies\) both \(z_1\) and \(\overline{z_1}\) are solutions)
+      \end{tabularx}
+
+    \subsection*{Roots}
+
+      \(n\)th roots of \(z=r\operatorname{cis}\theta\) are:
+
+      \[z = r^{\frac{1}{n}} \operatorname{cis}\left(\frac{\theta+2k\pi}{n}\right)\]
+
+      \begin{itemize}
+
+        \item{Same modulus for all solutions}
+        \item{Arguments are separated by \(\frac{2\pi}{n}\)}
+        \item{Solutions of \(z^n=a\) where \(a \in \mathbb{C}\) lie on the circle \(x^2+y^2=\left(|a|^{\frac{1}{n}}\right)^2\) \quad (intervals of \(\frac{2\pi}{n}\))}
+      \end{itemize}
+
+      \noindent For \(0=az^2+bz+c\), use quadratic formula:
+
+      \[z=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
 
-\subsection*{Roots}
+    \subsection*{Fundamental theorem of algebra}
 
-\(n\)th roots of \(z=r\operatorname{cis}\theta\) are:
+      A polynomial of degree \(n\) can be factorised into \(n\) linear factors in \(\mathbb{C}\):
+
+        \[\implies P(z)=a_n(z-\alpha_1)(z-\alpha_2)(z-\alpha_3)\dots(z-\alpha_n)\]
+        \[\text{ where } \alpha_1,\alpha_2,\alpha_3,\dots,\alpha_n \in \mathbb{C}\]
+
+    \subsection*{Argand planes}
+    
+      \begin{center}\begin{tikzpicture}[scale=2]
+        \draw [->] (-0.2,0) -- (1.5,0) node [right]  {$\operatorname{Re}(z)$};
+        \draw [->] (0,-0.2) -- (0,1.5) node [above] {$\operatorname{Im}(z)$};
+        \coordinate (P) at (1,1);
+        \coordinate (a) at (1,0);
+        \coordinate (b) at (0,1);
+        \coordinate (O) at (0,0);
+        \draw (0,0) -- (P) node[pos=0.5, above left]{\(r\)} node[pos=1, right]{\(\begin{aligned}z&=a+bi\\&=r\operatorname{cis}\theta\end{aligned}\)};
+        \draw [gray, dashed] (1,1) -- (1,0) node[black, pos=1, below]{\(a\)};
+        \draw [gray, dashed] (1,1) -- (0,1) node[black, pos=1, left]{\(b\)};
+        \begin{scope}
+          \path[clip] (O) -- (P) -- (a);
+          \fill[red, opacity=0.5, draw=black] (O) circle (2mm);
+          \node at ($(O)+(20:3mm)$) {$\theta$};
+        \end{scope}
+        \filldraw (P) circle (0.5pt);
+      \end{tikzpicture}\end{center}
+
+      \begin{itemize}
+        \item{Multiplication by \(i \implies\) CCW rotation of \(\frac{\pi}{2}\)}
+        \item{Addition: \(z_1 + z_2 \equiv\) \overrightharp{\(Oz_1\)} + \overrightharp{\(Oz_2\)}}
+      \end{itemize}
+
+    \subsection*{Sketching complex graphs}
+      
+      \subsubsection*{Linear}
+
+        \begin{itemize}
+          \item{\(\operatorname{Re}(z)=c\) or \(\operatorname{Im}(z)=c\) (perpendicular bisector)}
+          \item{\(\operatorname{Im}(z)=m\operatorname{Re}(z)\)}
+          \item{\(|z+a|=|z+b| \implies 2(a-b)x=b^2-a^2\)}
+        \end{itemize}
+
+      \subsubsection*{Circles}
+
+        \begin{itemize}
+          \item \(|z-z_1|^2=c^2|z_2+2|^2\)
+          \item \(|z-(a+bi)|=c\)
+        \end{itemize}
+
+      \noindent \textbf{Loci} \qquad \(\operatorname{Arg}(z)<\theta\)
+
+        \begin{center}\begin{tikzpicture}[scale=2,mydot/.style={circle, fill=white, draw, outer sep=0pt, inner sep=1.5pt}]
+          \draw [->] (0,0) -- (1,0) node [right]  {$\operatorname{Re}(z)$};
+          \draw [->] (0,-0.5) -- (0,1) node [above] {$\operatorname{Im}(z)$};
+          \draw [<-, dashed, thick, blue] (-1,0) -- (0,0);
+          \draw [->, thick, blue] (0,0) -- (1,1);
+          \fill [gray, opacity=0.2, domain=-1:1, variable=\x] (-1,-0.5) -- (-1,0) -- (0, 0) -- (1,1) -- (1,-0.5) -- cycle;
+          \begin{scope}
+            \path[clip] (0,0) -- (1,1) -- (1,0);
+            \fill[red, opacity=0.5, draw=black] (0,0) circle (2mm);
+            \node at ($(0,0)+(20:3mm)$) {$\theta$};
+          \end{scope}
+          \node [font=\footnotesize] at (0.5,-0.25) {\(\operatorname{Arg}(z)<\theta\)};
+          \node [blue, mydot] {};
+        \end{tikzpicture}\end{center}
+
+      \noindent \textbf{Rays} \qquad \(\operatorname{Arg}(z-b)=\theta\)
+
+        \begin{center}\begin{tikzpicture}[scale=2,mydot/.style={circle, fill=white, draw, outer sep=0pt, inner sep=1.5pt}]
+          \draw [->] (-0.75,0) -- (1.5,0) node [right]  {$\operatorname{Re}(z)$};
+          \draw [->] (0,-1) -- (0,1) node [above] {$\operatorname{Im}(z)$};
+          \draw [->, thick, brown] (-0.25,0) -- (-0.75,-1);
+          \node [above, font=\footnotesize] at (-0.25,0) {\(\frac{1}{4}\)};
+          \begin{scope}
+            \path[clip] (-0.25,0) -- (-0.75,-1) -- (0,0);
+            \fill[orange, opacity=0.5, draw=black] (-0.25,0) circle (2mm);
+          \end{scope}
+          \node at (-0.08,-0.3) {\(\frac{\pi}{8}\)};
+          \node [font=\footnotesize, left] at (-0.75,-1) {\(\operatorname{Arg}(z+\frac{1}{4})=\frac{\pi}{8}\)};
+          \node [brown, mydot] at (-0.25,0) {};
+          \draw [<->, thick, green] (0,-1) -- (1.5,0.5) node [pos=0.25, black, font=\footnotesize, right] {\(|z-2|=|z-(1+i)|\)};
+          \node [left, font=\footnotesize] at (0,-1) {\(-1\)};
+          \node [below, font=\footnotesize] at (1,0) {\(1\)};
+        \end{tikzpicture}\end{center}
+
+    \section{Vectors}
 
-\[z = r^{\frac{1}{n}} \operatorname{cis}\left(\frac{\theta+2k\pi}{n}\right)\]
 
 \begin{itemize}
+\item
+  \textbf{vector:} a directed line segment\\
+\item
+  arrow indicates direction
+\item
+  length indicates magnitude
+\item
+  notated as \(\vec{a}, \widetilde{A}, \overrightharp{a}\)
+\item
+  column notation: \(\begin{bmatrix}  x \\ y  \end{bmatrix}\)
+\item
+  vectors with equal magnitude and direction are equivalent
+\end{itemize}
+
+%\includegraphics[width=0.2\textwidth,height=\textheight]{graphics/vectors-intro.png}
+
+\subsection{Vector addition}
+
+\(\boldsymbol{u} + \boldsymbol{v}\) can be represented by drawing each
+ vector head to tail then joining the lines.\\
+Addition is commutative (parallelogram)
+
+\subsection{Scalar multiplication}
+
+For \(k \in \mathbb{R}^+\), \(k\boldsymbol{u}\) has the same direction
+as \(\boldsymbol{u}\) but length is multiplied by a factor of \(k\).
+
+When multiplied by \(k < 0\), direction is reversed and length is
+multplied by \(k\).
+
+\subsection{Vector subtraction}
+
+To find \(\boldsymbol{u} - \boldsymbol{v}\), add \(\boldsymbol{-v}\) to
+\(\boldsymbol{u}\)
+
+\subsection{Parallel vectors}
+
+Same or opposite direction
+
+\[\boldsymbol{u} || \boldsymbol{v} \iff \boldsymbol{u} = k \boldsymbol{v} \text{ where } k \in \mathbb{R} \setminus \{0\}\]
+
+\subsection{Position vectors}
+
+Vectors may describe a position relative to \(O\).
+
+For a point \(A\), the position vector is \(\overrightharp{OA}\)
+
+\subsection{Linear combinations of non-parallel
+vectors}
+
+If two non-zero vectors \(\boldsymbol{a}\) and \(\boldsymbol{b}\) are
+not parallel, then:
+
+\[m\boldsymbol{a} + n\boldsymbol{b} = p \boldsymbol{a} + q \boldsymbol{b}\quad \therefore \quad m = p, \> n = q\]
+
+%\includegraphics[width=0.2,height=\textheight]{graphics/parallelogram-vectors.jpg}
+%\includegraphics[width=1]{graphics/vector-subtraction.jpg}
+
+\subsection{Column vector notation}
 
-  \item{Same modulus for all solutions}
-  \item{Arguments are separated by \(\frac{2\pi}{n}\)}
+A vector between points \(A(x_1,y_1), \> B(x_2,y_2)\) can be represented
+as \(\begin{bmatrix}x_2-x_1\\ y_2-y_1 \end{bmatrix}\)
 
-\item{Solutions of \(z^n=a\) where \(a \in \mathbb{C}\) lie on the circle \(x^2+y^2=\left(|a|^{\frac{1}{n}}\right)^2\)}
+\subsection{Component notation}
+
+A vector \(\boldsymbol{u} = \begin{bmatrix}x\\ y \end{bmatrix}\) can be
+written as \(\boldsymbol{u} = x\boldsymbol{i} + y\boldsymbol{j}\).\\
+\(\boldsymbol{u}\) is the sum of two components \(x\boldsymbol{i}\) and
+\(y\boldsymbol{j}\)\\
+Magnitude of vector
+\(\boldsymbol{u} = x\boldsymbol{i} + y\boldsymbol{j}\) is denoted by
+\(|u|=\sqrt{x^2+y^2}\)
+
+Basic algebra applies:\\
+\((x\boldsymbol{i} + y\boldsymbol{j}) + (m\boldsymbol{i} + n\boldsymbol{j}) = (x + m)\boldsymbol{i} + (y+n)\boldsymbol{j}\)\\
+Two vectors equal if and only if their components are equal.
+
+\subsection{Unit vector \(|\hat{\boldsymbol{a}}|=1\)}
+\begin{equation}\begin{split}\hat{\boldsymbol{a}} & = {1 \over {|\boldsymbol{a}|}}\boldsymbol{a} \\ & = \boldsymbol{a} \cdot {|\boldsymbol{a}|}\end{split}\end{equation}
+
+  \subsection*{Scalar/dot product \(\boldsymbol{a} \cdot \boldsymbol{b}\)}
+
+\[\boldsymbol{a} \cdot \boldsymbol{b} = a_1 b_1 + a_2 b_2\]
+
+\textbf{on CAS:} \texttt{dotP({[}a\ b\ c{]},\ {[}d\ e\ f{]})}
+
+\subsection{Scalar product properties}
+
+\begin{enumerate}
+\item
+  \(k(\boldsymbol{a\cdot b})=(k\boldsymbol{a})\cdot \boldsymbol{b}=\boldsymbol{a}\cdot (k\boldsymbol{b})\)
+\item
+  \(\boldsymbol{a \cdot 0}=0\)
+\item
+  \(\boldsymbol{a \cdot (b + c)}=\boldsymbol{a \cdot b + a \cdot c}\)
+\item
+  \(\boldsymbol{i \cdot i} = \boldsymbol{j \cdot j} = \boldsymbol{k \cdot k}= 1\)
+\item
+  If \(\boldsymbol{a} \cdot \boldsymbol{b} = 0\), \(\boldsymbol{a}\) and
+  \(\boldsymbol{b}\) are perpendicular
+\item
+  \(\boldsymbol{a \cdot a} = |\boldsymbol{a}|^2 = a^2\)
+\end{enumerate}
+
+For parallel vectors \(\boldsymbol{a}\) and \(\boldsymbol{b}\):\\
+\[\boldsymbol{a \cdot b}=\begin{cases}
+|\boldsymbol{a}||\boldsymbol{b}| \hspace{2.8em} \text{if same direction}\\
+-|\boldsymbol{a}||\boldsymbol{b}| \hspace{2em} \text{if opposite directions}
+\end{cases}\]
+
+\subsection{Geometric scalar products}
+
+\[\boldsymbol{a} \cdot \boldsymbol{b} = |\boldsymbol{a}| |\boldsymbol{b}| \cos \theta\]
+
+where \(0 \le \theta \le \pi\)
+
+\subsection{Perpendicular vectors}
+
+If \(\boldsymbol{a} \cdot \boldsymbol{b} = 0\), then
+\(\boldsymbol{a} \perp \boldsymbol{b}\) (since \(\cos 90 = 0\))
+
+\subsection{Finding angle between
+vectors}
+
+\textbf{positive direction}
+
+\[\cos \theta = {{\boldsymbol{a} \cdot \boldsymbol{b}} \over {|\boldsymbol{a}| |\boldsymbol{b}|}} = {{a_1 b_1 + a_2 b_2} \over {|\boldsymbol{a}| |\boldsymbol{b}|}}\]
+
+\textbf{on CAS:} \texttt{angle({[}a\ b\ c{]},\ {[}a\ b\ c{]})} (Action
+-\textgreater{} Vector -\textgreater{} Angle)
+
+\subsection{Angle between vector and
+axis}
+
+Direction of a vector can be given by the angles it makes with
+\(\boldsymbol{i}, \boldsymbol{j}, \boldsymbol{k}\) directions.
+
+For
+\(\boldsymbol{a} = a_1 \boldsymbol{i} + a_2 \boldsymbol{j} + a_3 \boldsymbol{k}\)
+which makes angles \(\alpha, \beta, \gamma\) with positive direction of
+\(x, y, z\) axes:
+\[\cos \alpha = {a_1 \over |\boldsymbol{a}|}, \quad \cos \beta = {a_2 \over |\boldsymbol{a}|}, \quad \cos \gamma = {a_3 \over |\boldsymbol{a}|}\]
+
+\textbf{on CAS:} \texttt{angle({[}a\ b\ c{]},\ {[}1\ 0\ 0{]})} for angle
+between \(a\boldsymbol{i} + b\boldsymbol{j} + c\boldsymbol{k}\) and
+\(x\)-axis
+
+\subsection{Vector projections}
+
+Vector resolute of \(\boldsymbol{a}\) in direction of \(\boldsymbol{b}\)
+is magnitude of \(\boldsymbol{a}\) in direction of \(\boldsymbol{b}\):
+
+\[\boldsymbol{u}={{\boldsymbol{a}\cdot\boldsymbol{b}}\over |\boldsymbol{b}|^2}\boldsymbol{b}=\left({\boldsymbol{a}\cdot{\boldsymbol{b} \over |\boldsymbol{b}|}}\right)\left({\boldsymbol{b} \over |\boldsymbol{b}|}\right)=(\boldsymbol{a} \cdot \hat{\boldsymbol{b}})\hat{\boldsymbol{b}}\]
+
+\subsection{Scalar resolute of \(\boldsymbol{a}\) on \(\boldsymbol{b}\)}
+
+\[r_s = |\boldsymbol{u}| = \boldsymbol{a} \cdot \hat{\boldsymbol{b}}\]
+
+\subsection{Vector resolute of \(\boldsymbol{a} \perp \boldsymbol{b}\)}
+
+\[\boldsymbol{w} = \boldsymbol{a} - \boldsymbol{u} \> \text{ where } \boldsymbol{u} \text{ is projection } \boldsymbol{a} \text{ on } \boldsymbol{b}\]
+
+\subsection{Vector proofs}
+
+\subsubsection{Concurrent lines}
+
+\(\ge\) 3 lines intersect at a single point
+
+\subsubsection{Collinear points}
+
+\(\ge\) 3 points lie on the same line\\
+\(\implies \vec{OC} = \lambda \vec{OA} + \mu \vec{OB}\) where
+\(\lambda + \mu = 1\). If \(C\) is between \(\vec{AB}\), then
+\(0 < \mu < 1\)\\
+Points \(A, B, C\) are collinear iff
+\(\vec{AC}=m\vec{AB} \text{ where } m \ne 0\)
+
+\subsubsection{Useful vector properties}
+
+\begin{itemize}
+\item
+  If \(\boldsymbol{a}\) and \(\boldsymbol{b}\) are parallel, then
+  \(\boldsymbol{b}=k\boldsymbol{a}\) for some
+  \(k \in \mathbb{R} \setminus \{0\}\)
+\item
+  If \(\boldsymbol{a}\) and \(\boldsymbol{b}\) are parallel with at
+  least one point in common, then they lie on the same straight line
+\item
+  Two vectors \(\boldsymbol{a}\) and \(\boldsymbol{b}\) are
+  perpendicular if \(\boldsymbol{a} \cdot \boldsymbol{b}=0\)
+\item
+  \(\boldsymbol{a} \cdot \boldsymbol{a} = |\boldsymbol{a}|^2\)
 \end{itemize}
 
-\subsubsection*{Conjugate root theorem}
+\subsection{Linear dependence}
+
+Vectors \(\boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c}\) are linearly
+dependent if they are non-parallel and:
+
+\[k\boldsymbol{a}+l\boldsymbol{b}+m\boldsymbol{c} = 0\]
+\[\therefore \boldsymbol{c} = m\boldsymbol{a} + n\boldsymbol{b} \quad \text{(simultaneous)}\]
+
+\(\boldsymbol{a}, \boldsymbol{b},\) and \(\boldsymbol{c}\) are linearly
+independent if no vector in the set is expressible as a linear
+combination of other vectors in set, or if they are parallel.
+
+Vector \(\boldsymbol{w}\) is a linear combination of vectors
+\(\boldsymbol{v_1}, \boldsymbol{v_2}, \boldsymbol{v_3}\)
+
+\subsection{Three-dimensional vectors}
+
+Right-hand rule for axes: \(z\) is up or out of page.
+
+%\includegraphics{graphics/vectors-3d.png}
+
+\subsection{Parametric vectors}
+
+Parametric equation of line through point \((x_0, y_0, z_0)\) and
+parallel to \(a\boldsymbol{i} + b\boldsymbol{j} + c\boldsymbol{k}\) is:
+
+\begin{equation}\begin{cases}x = x_o + a \cdot t \\ y = y_0 + b \cdot t \\ z = z_0 + c \cdot t\end{cases}\end{equation}
 
-If \(a+bi\) is a solution to \(P(z)=0\), then the conjugate \(\overline{z}=a-bi\) is also a solution.
 
-\end{multicols}
+  \end{multicols}
 \end{document}