Explanation related to vectors

main.tex

@@ -1,4 +1,4 @@
-\documentclass{article}
+\documentclass{book}
 \usepackage[ipa]{luatexja-preset}
 \usepackage{amssymb,amsfonts}
 \usepackage{graphicx}
@@ -10,5 +10,6 @@
 
 \begin{document}
 \maketitle
+\input{vector.tex}
 \input{trigonometry.tex}
 \end{document}

trigonometry.tex

@@ -1,5 +1,5 @@
-\section{三角関数}
-\begin{figure}\caption{$\sin$と$\cos$の定義（その1）}
+\chapter{三角関数}
+\begin{figure}[ht]\caption{$\sin$と$\cos$の定義（その1）}
 \begin{tikzpicture}
   \coordinate(A) at (0,0);
   \coordinate(B) at (3,0);
@@ -15,7 +15,7 @@ \section{三角関数}
   \draw[decorate, decoration={brace, mirror, raise=4pt}](C)--(A);
 \end{tikzpicture}
 \end{figure}
-\begin{figure}\caption{$\sin$と$\cos$の定義（その2）}
+\begin{figure}[ht]\caption{$\sin$と$\cos$の定義（その2）}
 \begin{tikzpicture}
   \def\a{55}
   \draw(0,0) circle(2);
@@ -29,7 +29,7 @@ \section{三角関数}
   \draw(2,0)node[below right]{$1$};
 \end{tikzpicture}
 \end{figure}
-\begin{figure}\caption{$-\theta$と$\pi-\theta$についての$\sin$と$\cos$}
+\begin{figure}[ht]\caption{$-\theta$と$\pi-\theta$についての$\sin$と$\cos$}
 \begin{tikzpicture}
   \def\a{50}
   \draw[->] (-2.5,0)--(2.5,0);
@@ -51,7 +51,7 @@ \section{三角関数}
   \fill(-\a:2)circle(2pt)node[right,font=\small]{$(\cos\theta,-\sin\theta)$};
 \end{tikzpicture}
 \end{figure}
-\begin{figure}\caption{$\frac{\pi}{2}-\theta$についての$\sin$と$\cos$}
+\begin{figure}[ht]\caption{$\frac{\pi}{2}-\theta$についての$\sin$と$\cos$}
 \begin{tikzpicture}
   \def\a{25}
   \draw[->] (-0.3,0)--(4.5,0);
@@ -69,7 +69,7 @@ \section{三角関数}
   \fill(90-\a:4)circle(2pt)node[above right,font=\small]{$(\sin\theta,\cos\theta)$};
 \end{tikzpicture}
 \end{figure}
-\begin{figure}\caption{正弦定理}
+\begin{figure}[ht]\caption{正弦定理}
 \begin{tikzpicture}
   \def\a{-130}
   \def\b{-10}
@@ -101,7 +101,7 @@ \section{三角関数}
   \draw(0,-2.5)node{$2R\sin A=a$};
 \end{tikzpicture}
 \end{figure}
-\begin{figure}\caption{余弦定理}
+\begin{figure}[ht]\caption{余弦定理}
 \begin{tikzpicture}
   \coordinate(A) at (0,0);
   \coordinate(B) at (5,0);

vector.tex

@@ -0,0 +1,101 @@
+\chapter{ベクトル}
+\begin{figure}[ht]\caption{等しいベクトルの例}
+\begin{tikzpicture}
+  \draw[-Stealth](1,1)--(4,3);
+  \draw[-Stealth](3,0)--(6,2);
+  \filldraw(1,1)circle(1pt)node[below]{A};
+  \filldraw(4,3)circle(1pt)node[right]{B};
+  \filldraw(3,0)circle(1pt)node[below]{C};
+  \filldraw(6,2)circle(1pt)node[right]{D};
+  \draw(2.5,2)node[above left]{$\overrightarrow{\mathrm{AB}}$};
+  \draw(4.5,1)node[above left]{$\overrightarrow{\mathrm{CD}}$};
+\end{tikzpicture}
+\end{figure}
+\begin{figure}[ht]\caption{逆ベクトル}
+\begin{tikzpicture}
+  \draw[-Stealth](1,1)--(4,3)node[midway, above left]{$\bm{a}$};
+  \draw[-Stealth](4.2,2.8)--(1.2,0.8)node[midway, below right]{$-\bm{a}$};
+\end{tikzpicture}
+\end{figure}
+\begin{figure}[ht]\caption{ベクトルの実数倍}
+\begin{tikzpicture}
+  \draw[-Stealth](1,0)--(1+3,2)node[midway, above left]{$\bm{a}$};
+  \draw[-Stealth](3,0)--(3+6,4)node[midway, above left]{$2\bm{a}$};
+  \draw[-Stealth](6,0)--(6+3/2,1)node[midway, above left]{$\displaystyle\frac{1}{2}\bm{a}$};
+  \draw[-Stealth](8,0)--(8-3/2,-1)node[midway, below right]{$\displaystyle-\frac{1}{2}\bm{a}$};
+\end{tikzpicture}
+\end{figure}
+\begin{figure}[ht]\caption{ベクトルの和（その1）}
+\begin{tikzpicture}
+  \draw[-Stealth](0,0)--(4,2)node[midway, above left]{$\bm{a}$};
+  \draw[-Stealth](4,2)--(3,5)node[midway, above right]{$\bm{b}$};
+  \draw[-Stealth](0,0)--(3,5)node[midway, left]{$\bm{a}+\bm{b}$};
+\end{tikzpicture}
+\end{figure}
+\begin{figure}[ht]\caption{ベクトルの和（その2）}
+\begin{tikzpicture}
+  \draw[-Stealth](0,0)--(4,2)node[midway, above left]{$\bm{a}$};
+  \draw[-Stealth](0,0)--(-1,3)node[midway, above right]{$\bm{b}$};
+  \draw[-Stealth](0,0)--(3,5)node[midway, left]{$\bm{a}+\bm{b}$};
+  \draw[dashed](4,2)--(3,5);
+  \draw[dashed](-1,3)--(3,5);
+\end{tikzpicture}
+\end{figure}
+\begin{figure}[ht]\caption{$\bm{a}+\bm{b}=\bm{b}+\bm{a}$}
+\begin{tikzpicture}
+  \draw[-Stealth](1,0)--(4,3)node[midway, below right]{$\bm{a}$};
+  \draw[-Stealth](4,3)--(2,4)node[midway, above right]{$\bm{b}$};
+  \draw[-Stealth](1,0)--(2,4)node[midway, right]{$\bm{a}+\bm{b}$}node[midway, left]{$\bm{b}+\bm{a}$};
+  \draw[-Stealth](1,0)--(-1,1)node[midway, right]{$\bm{b}$};
+  \draw[-Stealth](-1,1)--(2,4)node[midway, left]{$\bm{a}$};
+\end{tikzpicture}
+\end{figure}
+\begin{figure}[ht]\caption{$(\bm{a}+\bm{b})+\bm{c}=\bm{a}+(\bm{b}+\bm{c})$}
+\begin{tikzpicture}
+  \draw[-Stealth](0,0)--(3,1)node[midway, right]{$\bm{a}$};
+  \draw[-Stealth](3,1)--(5,3)node[midway, right]{$\bm{b}$};
+  \draw[-Stealth](5,3)--(4,6)node[midway, right]{$\bm{c}$};  
+  \draw[-Stealth](0,0)--(5,3)node[midway, left]{$\bm{a}+\bm{b}$};
+  \draw[-Stealth](3,1)--(4,6)node[midway, right]{$\bm{b}+\bm{c}$};
+  \draw[-Stealth](0,0)--(4,6)node[midway, left]{
+    \begin{tabular}{l}
+     $(\bm{a}+\bm{b})+\bm{c}$\\
+      $=\bm{a}+(\bm{b}+\bm{c})$
+    \end{tabular}};  
+\end{tikzpicture}
+\end{figure}
+\begin{figure}[ht]\caption{ベクトルの大きさ（2次元）}
+\begin{tikzpicture}
+  \draw(3,0)--(0,0)node[midway, below=5pt]{$a_x$};
+  \draw[decorate, decoration={brace, raise=4pt}](3,0)--(0,0);
+  \draw(3,2)--(3,0)node[midway, right=5pt]{$a_y$};
+  \draw[decorate, decoration={brace, raise=4pt}](3,2)--(3,0);
+  \draw[decorate, decoration={brace, raise=4pt}](0,0)--(3,2)
+  node[midway, sloped, above=5pt]{$\sqrt{a_x^2+a_y^2}$};
+  \draw[-Stealth](0,0)--(3,2)node[midway, sloped, below]{$\bm{a}$};
+\end{tikzpicture}
+\end{figure}
+\begin{figure}[ht]\caption{ベクトルの大きさ（3次元）}
+\begin{tikzpicture}
+  \draw(0,0)--(4,0)node[midway, below=5pt]{$a_x$};
+  \draw[decorate, decoration={brace, raise=4pt}](4,0)--(0,0);
+  \draw[dashed](0,0)--(1,2);
+  \draw[dashed](1,2)--(1,5);
+  \draw(0,0)--(0,3);
+  \draw(0,3)--(1,5);
+  \draw(0,3)--(4,3);
+  \draw(1,5)--(5,5);
+  \draw(5,2)--(5,5)node[midway, sloped, below=5pt]{$a_z$};
+  \draw[decorate, decoration={brace, raise=4pt}](5,5)--(5,2);
+  \draw[dashed](1,2)--(5,2);
+  \draw(4,0)--(4,3);
+  \draw(4,0)--(5,2)node[midway, sloped, below=5pt]{$a_y$};
+  \draw[decorate, decoration={brace, raise=4pt}](5,2)--(4,0);
+  \draw(4,3)--(5,5);
+  \draw[-Stealth, style=thick](0,0)--(5,5)node[midway, sloped, above=5pt]{$\sqrt{a_x^2+a_y^2+a_z^2}$}node[below=2pt,pos=0.8]{$\bm{a}$};
+  \draw[decorate, decoration={brace, raise=4pt}](0,0)--(5,5);
+  \draw[dashed](0,0)--(5,2)node[midway, sloped, above=5pt]{\tiny$\sqrt{a_x^2+a_y^2}$};
+  \draw[decorate, decoration={brace, raise=4pt}](0,0)--(5,2);
+\end{tikzpicture}
+\end{figure}
+