From: Andrew Lorimer 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?ds=inline [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}