\usepackage{graphicx}
\usepackage{wrapfig}
\usepackage{tikz}
+\usepackage{tikz-3dplot}
\usetikzlibrary{calc}
+\usetikzlibrary{angles}
\usepgflibrary{arrows.meta}
\usepackage{fancyhdr}
\pagestyle{fancy}
\newcolumntype{R}[1]{>{\hsize=#1\hsize\raggedleft\arraybackslash}X}%
\definecolor{cas}{HTML}{e6f0fe}
\linespread{1.5}
+\newcommand{\midarrow}{\tikz \draw[-triangle 90] (0,0) -- +(.1,0);}
\begin{document}
\subsection*{Operations}
- \begin{tabularx}{\columnwidth}{R{0.33}|X|X}
+\definecolor{shade1}{HTML}{ffffff}
+\definecolor{shade2}{HTML}{e6f2ff}
+ \definecolor{shade3}{HTML}{cce2ff}
+ \begin{tabularx}{\columnwidth}{r|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}{\(ka \pm kbi\)} & \(kr\operatorname{cis} \theta\)\\
+ \(+k \times z\) & \multirow{2}{*}{\(ka \pm kbi\)} & \(kr\operatorname{cis} \theta\)\\
\cline{1-1}\cline{3-3}
\(-k \times z\) & & \(kr \operatorname{cis}(\theta\pm \pi)\)\\
\hline
\[k\left(r \operatorname{cis}\theta\right)=kr \operatorname{cis}\theta\]
\noindent For \(k \in \mathbb{R}^-\):
- \[k\left(r \operatorname{cis}\theta\right)=kr \operatorname{cis}\left(\begin{cases}\theta - \pi & 0<\operatorname{Arg}(z)\le \pi \\ \theta + \pi & -\pi<\operatorname{Arg}(z)\le 0\end{cases}\right)\]
+ \[k\left(r \operatorname{cis}\theta\right)=kr \operatorname{cis}\left(\begin{cases}\theta - \pi & |0<\operatorname{Arg}(z)\le \pi \\ \theta + \pi & |-\pi<\operatorname{Arg}(z)\le 0\end{cases}\right)\]
\subsection*{Conjugate}
&= r \operatorname{cis}(-\theta)
\end{align*}
- \noindent \colorbox{cas}{On CAS:} \verb|conjg(a+bi)|
+ \noindent \colorbox{cas}{On CAS: \texttt{conjg(a+bi)}}
\subsubsection*{Properties}
\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{\(\theta = \operatorname{arg}(z)\) \quad \colorbox{cas}{On CAS: \texttt{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}
\begin{scope}
\path[clip] (0,0) -- (1,1) -- (1,0);
\fill[red, opacity=0.5, draw=black] (0,0) circle (2mm);
- \node at ($(0,0)+(20:3mm)$) {$\theta$};
+ \node at ($(0,0)+(20:3mm)$) {$\frac{\pi}{4}$};
\end{scope}
- \node [font=\footnotesize] at (0.5,-0.25) {\(\operatorname{Arg}(z)<\theta\)};
+ \node [font=\footnotesize] at (0.5,-0.25) {\(\operatorname{Arg}(z)\le\frac{\pi}{4}\)};
\node [blue, mydot] {};
\end{tikzpicture}\end{center}
\end{tikzpicture}\end{center}
\section{Vectors}
+\begin{center}\begin{tikzpicture}
+ \draw [->] (-0.5,0) -- (3,0) node [right] {\(x\)};
+ \draw [->] (0,-0.5) -- (0,3) node [above] {\(y\)};
+ \draw [orange, ->, thick] (0.5,0.5) -- (2.5,2.5) node [pos=0.5, above] {\(\vec{u}\)};
+ \begin{scope}[very thick, every node/.style={sloped,allow upside down}]
+ \draw [gray, dashed, thick] (0.5,0.5) -- (2.5,0.5) node [pos=0.5] {\midarrow} node[black, pos=0.5, below]{\(x\vec{i}\)};
+ \draw [gray, dashed, thick] (2.5,0.5) -- (2.5,2.5) node [pos=0.5] {\midarrow};
+ \end{scope}
+ \node[black, right] at (2.5,1.5) {\(y\vec{j}\)};
+\end{tikzpicture}\end{center}
-\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}
+\subsection*{Column notation}
-%\includegraphics[width=0.2\textwidth,height=\textheight]{graphics/vectors-intro.png}
+\[\begin{bmatrix}x\\ y \end{bmatrix} \iff x\boldsymbol{i} + y\boldsymbol{j}\]
+\(\begin{bmatrix}x_2-x_1\\ y_2-y_1 \end{bmatrix}\) \quad between \(A(x_1,y_1), \> B(x_2,y_2)\)
-\subsection{Vector addition}
+\subsection*{Scalar multiplication}
-\(\boldsymbol{u} + \boldsymbol{v}\) can be represented by drawing each
- vector head to tail then joining the lines.\\
-Addition is commutative (parallelogram)
+\[k\cdot (x\boldsymbol{i}+y\boldsymbol{j})=kx\boldsymbol{i}+ky\boldsymbol{j}\]
-\subsection{Scalar multiplication}
+\noindent For \(k \in \mathbb{R}^-\), direction is reversed
-For \(k \in \mathbb{R}^+\), \(k\boldsymbol{u}\) has the same direction
-as \(\boldsymbol{u}\) but length is multiplied by a factor of \(k\).
+\subsection*{Vector addition}
+\begin{center}\begin{tikzpicture}[scale=1]
+ \coordinate (A) at (0,0);
+ \coordinate (B) at (2,2);
+ \draw [->, thick, red] (0,0) -- (2,2) node [pos=0.5, below right] {\(\vec{u}=2\vec{i}+2\vec{j}\)};
+ \draw [->, thick, blue] (2,2) -- (1,4) node [pos=0.5, above right] {\(\vec{v}=-\vec{i}+2\vec{j}\)};
+ \draw [->, thick, orange] (0,0) -- (1,4) node [pos=0.5, left] {\(\vec{u}+\vec{v}=\vec{i}+4\vec{j}\)};
+\end{tikzpicture}\end{center}
-When multiplied by \(k < 0\), direction is reversed and length is
-multplied by \(k\).
+\[(x\boldsymbol{i}+y\boldsymbol{j}) \pm (a\boldsymbol{i}+b\boldsymbol{j})=(x \pm a)\boldsymbol{i}+(y \pm b)\boldsymbol{j}\]
-\subsection{Vector subtraction}
+\begin{itemize}
+ \item Draw each vector head to tail then join lines
+ \item Addition is commutative (parallelogram)
+ \item \(\boldsymbol{u}-\boldsymbol{v}=\boldsymbol{u}+(-\boldsymbol{v})\)
+\end{itemize}
-To find \(\boldsymbol{u} - \boldsymbol{v}\), add \(\boldsymbol{-v}\) to
-\(\boldsymbol{u}\)
+\subsection*{Magnitude}
-\subsection{Parallel vectors}
+\[|(x\boldsymbol{i} + y\boldsymbol{j})|=\sqrt{x^2+y^2}\]
-Same or opposite direction
+\subsection*{Parallel vectors}
\[\boldsymbol{u} || \boldsymbol{v} \iff \boldsymbol{u} = k \boldsymbol{v} \text{ where } k \in \mathbb{R} \setminus \{0\}\]
-\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:
-
-\[m\boldsymbol{a} + n\boldsymbol{b} = p \boldsymbol{a} + q \boldsymbol{b}\quad \therefore \quad m = p, \> n = q\]
-
+For parallel vectors \(\boldsymbol{a}\) and \(\boldsymbol{b}\):\\
+\[\boldsymbol{a \cdot b}=\begin{cases}
+|\boldsymbol{a}||\boldsymbol{b}| \hspace{2.8em} \text{if same direction}\\
+-|\boldsymbol{a}||\boldsymbol{b}| \hspace{2em} \text{if opposite directions}
+\end{cases}\]
%\includegraphics[width=0.2,height=\textheight]{graphics/parallelogram-vectors.jpg}
%\includegraphics[width=1]{graphics/vector-subtraction.jpg}
-\subsection{Column vector notation}
-
-A vector between points \(A(x_1,y_1), \> B(x_2,y_2)\) can be represented
-as \(\begin{bmatrix}x_2-x_1\\ y_2-y_1 \end{bmatrix}\)
-
-\subsection{Component notation}
+\subsection*{Perpendicular vectors}
-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
-\(|u|=\sqrt{x^2+y^2}\)
+\[\boldsymbol{a} \perp \boldsymbol{b} \iff \boldsymbol{a} \cdot \boldsymbol{b} = 0\ \quad \text{(since \(\cos 90 = 0\))}\]
-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 \(|\hat{\boldsymbol{a}}|=1\)}
+\[\begin{split}\hat{\boldsymbol{a}} & = {1 \over {|\boldsymbol{a}|}}\boldsymbol{a} \\ & = \boldsymbol{a} \cdot {|\boldsymbol{a}|}\end{split}\]
-\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 product \(\boldsymbol{a} \cdot \boldsymbol{b}\)}
- \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{]})}
+\begin{center}\begin{tikzpicture}[scale=2]
+ \draw [->] (0,0) -- (1,0.5) node [pos=0.5, above left] {\(\boldsymbol{b}\)};
+ \draw [->] (0,0) -- (1,0) node [pos=0.5, below] {\(\boldsymbol{a}\)};
+ \begin{scope}
+ \path[clip] (1,0.5) -- (1,0) -- (0,0);
+ \fill[orange, opacity=0.5, draw=black] (0,0) circle (2mm);
+ \node at ($(0,0)+(15:4mm)$) {\(\theta\)};
+ \end{scope}
+\end{tikzpicture}\end{center}
+\begin{align*}\boldsymbol{a} \cdot \boldsymbol{b} &= a_1 b_1 + a_2 b_2 \\ &= |\boldsymbol{a}| |\boldsymbol{b}| \cos \theta \\ &\quad (\> 0 \le \theta \le \pi) \text{ - from cosine rule}\end{align*}
+\noindent\colorbox{cas}{On CAS: \texttt{dotP({[}a\ b\ c{]},\ {[}d\ e\ f{]})}}
-\subsection{Scalar product properties}
+\subsubsection*{Properties}
\begin{enumerate}
\item
\item
\(\boldsymbol{a \cdot 0}=0\)
\item
- \(\boldsymbol{a \cdot (b + c)}=\boldsymbol{a \cdot b + a \cdot c}\)
+ \(\boldsymbol{a} \cdot (\boldsymbol{b} + \boldsymbol{c})=\boldsymbol{a} \cdot \boldsymbol{b} + \boldsymbol{a} \cdot \boldsymbol{c}\)
\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
+ \(\boldsymbol{a} \cdot \boldsymbol{b} = 0 \quad \implies \quad \boldsymbol{a} \perp \boldsymbol{b}\)
\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}
+\subsection*{Angle between vectors}
\[\cos \theta = {{\boldsymbol{a} \cdot \boldsymbol{b}} \over {|\boldsymbol{a}| |\boldsymbol{b}|}} = {{a_1 b_1 + a_2 b_2} \over {|\boldsymbol{a}| |\boldsymbol{b}|}}\]
-\textbf{on CAS:} \texttt{angle({[}a\ b\ c{]},\ {[}a\ b\ c{]})} (Action
--\textgreater{} Vector -\textgreater{} Angle)
+\noindent \colorbox{cas}{On CAS:} \texttt{angle([a b c], [a b c])}
-\subsection{Angle between vector and
-axis}
+(Action \(\rightarrow\) Vector \(\rightarrow\)Angle)
-Direction of a vector can be given by the angles it makes with
-\(\boldsymbol{i}, \boldsymbol{j}, \boldsymbol{k}\) directions.
+\subsection*{Angle between vector and axis}
-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
+\noindent For\(\boldsymbol{a} = a_1 \boldsymbol{i} + a_2 \boldsymbol{j} + a_3 \boldsymbol{k}\)
+which makes angles \(\alpha, \beta, \gamma\) with positive side 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
+\noindent \colorbox{cas}{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}
+\subsection*{Projections \& resolutes}
-\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{tikzpicture}[scale=3]
+ \draw [->, purple] (0,0) -- (1,0.5) node [pos=0.5, above left] {\(\boldsymbol{a}\)};
+ \draw [->, orange] (0,0) -- (1,0) node [pos=0.5, below] {\(\boldsymbol{u}\)};
+ \draw [->, blue] (1,0) -- (2,0) node [pos=0.5, below] {\(\boldsymbol{b}\)};
+ \begin{scope}
+ \path[clip] (1,0.5) -- (1,0) -- (0,0);
+ \fill[orange, opacity=0.5, draw=black] (0,0) circle (2mm);
+ \node at ($(0,0)+(15:4mm)$) {\(\theta\)};
+ \end{scope}
+ \begin{scope}[very thick, every node/.style={sloped,allow upside down}]
+ \draw [gray, dashed, thick] (1,0) -- (1,0.5) node [pos=0.5] {\midarrow} node[black, pos=0.5, right, rotate=-90]{\(\boldsymbol{w}\)};
+ \end{scope}
+\draw (0,0) coordinate (O)
+ (1,0) coordinate (A)
+ (1,0.5) coordinate (B)
+ pic [draw,red,angle radius=2mm] {right angle = O--A--B};
+\end{tikzpicture}
+
+\subsubsection*{\(\parallel\boldsymbol{b}\) (vector projection/resolute)}
+\begin{align*}
+ \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}}
+\end{align*}
+
+\subsubsection*{\(\perp\boldsymbol{b}\) (perpendicular projection)}
+\[\boldsymbol{w} = \boldsymbol{a} - \boldsymbol{u}\]
+
+\subsubsection*{\(|\boldsymbol{u}|\) (scalar resolute)}
+\begin{align*}
+ r_s &= |\boldsymbol{u}|\\
+ &= \boldsymbol{a} \cdot \hat{\boldsymbol{b}}\\
+ &=\frac{\boldsymbol{a}\cdot\boldsymbol{b}}{|\boldsymbol{b}|}
+\end{align*}
+
+\subsubsection*{Rectangular (\(\parallel,\perp\)) components}
+
+\[\boldsymbol{a}=\frac{\boldsymbol{a}\cdot\boldsymbol{b}}{\boldsymbol{b}\cdot\boldsymbol{b}}\boldsymbol{b}+\left(\boldsymbol{a}-\frac{\boldsymbol{a}\cdot\boldsymbol{b}}{\boldsymbol{b}\cdot\boldsymbol{b}}\boldsymbol{b}\right)\]
+
+
+\subsection*{Vector proofs}
+
+\textbf{Concurrent:} intersection of \(\ge\) 3 lines
+
+\begin{tikzpicture}
+ \draw [blue] (0,0) -- (1,1);
+ \draw [red] (1,0) -- (0,1);
+ \draw [brown] (0.4,0) -- (0.6,1);
+ \filldraw (0.5,0.5) circle (2pt);
+\end{tikzpicture}
+
+\subsubsection*{Collinear points}
+
+\(\ge\) 3 points lie on the same line
+
+\begin{tikzpicture}
+ \draw [purple] (0,0) -- (4,1);
+ \filldraw (2,0.5) circle (2pt) node [above] {\(C\)};
+ \filldraw (1,0.25) circle (2pt) node [above] {\(A\)};
+ \filldraw (3,0.75) circle (2pt) node [above] {\(B\)};
+ \coordinate (O) at (2.8,-0.2);
+ \node at (O) [below] {\(O\)};
+ \begin{scope}[->, orange, thick]
+ \draw (O) -- (2,0.5) node [pos=0.5, above, font=\footnotesize, black] {\(\boldsymbol{c}\)};
+ \draw (O) -- (1,0.25) node [pos=0.5, below, font=\footnotesize, black] {\(\boldsymbol{a}\)};
+ \draw (O) -- (3,0.75) node [pos=0.5, right, font=\footnotesize, black] {\(\boldsymbol{b}\)};
+ \end{scope}
+\end{tikzpicture}
+
+\begin{align*}
+ \text{e.g. Prove that}\\
+ \overrightharp{AC}=m\overrightharp{AB} \iff \boldsymbol{c}&=(1-m)\boldsymbol{a}+m\boldsymbol{b}\\
+ \implies \boldsymbol{c} &= \overrightharp{OA} + \overrightharp{AC}\\
+ &= \overrightharp{OA} + m\overrightharp{AB}\\
+ &=\boldsymbol{a}+m(\boldsymbol{b}-\boldsymbol{a})\\
+ &=\boldsymbol{a}+m\boldsymbol{b}-m\boldsymbol{a}\\
+ &=(1-m)\boldsymbol{a}+m{b}
+\end{align*}
+
+\begin{align*}
+ \text{Also, } \implies \overrightharp{OC} &= \lambda \vec{OA} + \mu \overrightharp{OB} \\
+ \text{where } \lambda + \mu &= 1\\
+ \text{If } C \text{ lies along } \overrightharp{AB}, & \implies 0 < \mu < 1
+\end{align*}
+
+
+ \subsubsection*{Useful vector properties}
\begin{itemize}
\item
\(\boldsymbol{a} \cdot \boldsymbol{a} = |\boldsymbol{a}|^2\)
\end{itemize}
-\subsection{Linear dependence}
+\subsection*{Linear dependence}
Vectors \(\boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c}\) are linearly
dependent if they are non-parallel and:
Vector \(\boldsymbol{w}\) is a linear combination of vectors
\(\boldsymbol{v_1}, \boldsymbol{v_2}, \boldsymbol{v_3}\)
-\subsection{Three-dimensional vectors}
+\subsection*{Three-dimensional vectors}
Right-hand rule for axes: \(z\) is up or out of page.
-%\includegraphics{graphics/vectors-3d.png}
+\tdplotsetmaincoords{60}{120}
+\begin{center}\begin{tikzpicture} [scale=3, tdplot_main_coords, axis/.style={->,thick},
+vector/.style={-stealth,red,very thick},
+vector guide/.style={dashed,gray,thick}]
+
+%standard tikz coordinate definition using x, y, z coords
+\coordinate (O) at (0,0,0);
+
+%tikz-3dplot coordinate definition using x, y, z coords
+
+\pgfmathsetmacro{\ax}{1}
+\pgfmathsetmacro{\ay}{1}
+\pgfmathsetmacro{\az}{1}
+
+\coordinate (P) at (\ax,\ay,\az);
+
+%draw axes
+\draw[axis] (0,0,0) -- (1,0,0) node[anchor=north east]{$x$};
+\draw[axis] (0,0,0) -- (0,1,0) node[anchor=north west]{$y$};
+\draw[axis] (0,0,0) -- (0,0,1) node[anchor=south]{$z$};
+
+%draw a vector from O to P
+\draw[vector] (O) -- (P);
+
+%draw guide lines to components
+\draw[vector guide] (O) -- (\ax,\ay,0);
+\draw[vector guide] (\ax,\ay,0) -- (P);
+\draw[vector guide] (P) -- (0,0,\az);
+\draw[vector guide] (\ax,\ay,0) -- (0,\ay,0);
+\draw[vector guide] (\ax,\ay,0) -- (0,\ay,0);
+\draw[vector guide] (\ax,\ay,0) -- (\ax,0,0);
+\node[tdplot_main_coords,above right]
+at (\ax,\ay,\az){(\ax, \ay, \az)};
+\end{tikzpicture}\end{center}
-\subsection{Parametric vectors}
+\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: