From: Andrew Lorimer Date: Sun, 26 May 2019 12:09:17 +0000 (+1000) Subject: [spec] compile vectors notes X-Git-Tag: yr12~120 X-Git-Url: https://git.lorimer.id.au/notes.git/diff_plain/9f1619c0b70d3e3c8ed4aeaf54d93c7749db5eb2?ds=inline [spec] compile vectors notes --- diff --git a/spec/spec-collated.pdf b/spec/spec-collated.pdf index 136eeb8..5cbbb9b 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 d61f60e..616033b 100644 --- a/spec/spec-collated.tex +++ b/spec/spec-collated.tex @@ -10,7 +10,9 @@ \usepackage{graphicx} \usepackage{wrapfig} \usepackage{tikz} +\usepackage{tikz-3dplot} \usetikzlibrary{calc} +\usetikzlibrary{angles} \usepgflibrary{arrows.meta} \usepackage{fancyhdr} \pagestyle{fancy} @@ -27,6 +29,7 @@ \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} @@ -43,12 +46,15 @@ \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 @@ -63,7 +69,7 @@ \[k\left(r \operatorname{cis}\theta\right)=kr \operatorname{cis}\theta\] \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} @@ -72,7 +78,7 @@ &= 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} @@ -123,7 +129,7 @@ \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} @@ -232,9 +238,9 @@ \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} @@ -258,96 +264,85 @@ \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 @@ -355,90 +350,123 @@ Two vectors equal if and only if their components are equal. \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 @@ -455,7 +483,7 @@ Points \(A, B, C\) are collinear iff \(\boldsymbol{a} \cdot \boldsymbol{a} = |\boldsymbol{a}|^2\) \end{itemize} -\subsection{Linear dependence} +\subsection*{Linear dependence} Vectors \(\boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c}\) are linearly dependent if they are non-parallel and: @@ -470,13 +498,46 @@ 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} +\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: