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		\vskip 1em%
		{\large \@date}%
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	\par
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%opening
\newcommand{\snovdn}{Skalarni produkt vektorjev }
\newcommand{\predmdn}{mat}
\newcommand{\predmkaj}{domace\_naloge}
\newcommand{\stevilkadn}{26}
\newcommand{\cm}[1]{\SI{#1}{\centi\meter}}
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\makeatletter
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\title{%
	\snovdn --- \stevilkadn. domača naloga
	\\
	\large Matematika, Gimnazija Bežigrad}
\author{\begin{tabular}{rl}
		\textbf{Profesor:} & prof. Vilko Domajnko \\
		\textbf{Avtor:} & Anton Luka Šijanec, 2. a
%		\textbf{Avtor:} & Anton Luka Šijanec \\ & Member 2 \\ & Member 3
\end{tabular}}
\newcommand\hcancel[2][black]{\setbox0=\hbox{$0#2$}%
\rlap{\raisebox{.45\ht0}{\textcolor{#1}{\rule{\wd0}{1pt}}}}#2} 
% \everymath{\displaystyle} % https://tex.stackexchange.com/a/32847/212260
\begin{document}
\maketitle
\begin{abstract}
Ta dokument vsebuje navodila in rešitve domačih nalog snovi \textit{\snovdn}pri matematiki, ki sem jih spisal sam.
\end{abstract}
\paragraph{Navodilo naloge} \textbf{vaje2:} 83 / 567, 568, 569
%\tableofcontents
\begin{enumerate}[label=\textbf{\arabic*.}]
	\setcounter{enumi}{566}
	\item Izračunaj skalarni produkt označenih vektorjev na dva načina (po definiciji in z uporabo pravokotne projekcije enega vektorja na drugi vektor):
		\begin{tasks}[label=\textbf{\xslalph*)}](2)
			\task v enakostraničnem trikotniku s stranico, dolgo \cm{3},
				\begin{tikzpicture}[vect/.style={->,shorten >=0pt,>=latex'}]
					\tkzDefPoint(0,0){A} \tkzDefPoint(3,0){B} \tkzDefPoint(60:3){C}
					\tkzDrawPolygon[fill=red!30, opacity=.3](A,B,C)
					\tkzDrawSegments[vect](A,B A,C)
				\end{tikzpicture}
				$$\Vec{c}\cdot\Vec{b}=|\Vec{c}|\cdot|\Vec{b}|=9\cdot\cos{\ang{60}}=4,5$$
			\task v enakostraničnem trikotniku s stranico, dolgo \cm{3},
				\begin{tikzpicture}[vect/.style={->,shorten >=0pt,>=latex'}]
					\tkzDefPoint(0,0){A} \tkzDefPoint(3,0){B} \tkzDefPoint(60:3){C}
					\tkzDrawPolygon[fill=red!30, opacity=.3](A,B,C)
					\tkzDrawSegments[vect](C,A B,C)
				\end{tikzpicture}
				$$\Vec{a}\cdot\Vec{b}=|\Vec{a}|\cdot|\Vec{b}|\cdot\cos{\left(\ang{180}-\ang{60}\right)}=-4,5$$
			\task v pravokotniku s stranicama, dolgima \cm{6} in \cm{2},
				\begin{tikzpicture}[vect/.style={->,shorten >=0pt,>=latex'}]
					\tkzDefPoint(0,0){A} \tkzDefPoint(6,0){B} \tkzDefPoint(6,2){C} \tkzDefPoint(0,2){D}
					\tkzDrawPolygon[fill=red!30, opacity=.3](A,B,C,D)
					\tkzDrawSegments[vect](A,B B,C)
				\end{tikzpicture}
				$$\Vec{a}\cdot\Vec{b}=|\Vec{a}|\cdot|\Vec{b}|\cdot\cos{\ang{90}}=0$$
			\task v pravokotniku s stranicama, dolgina \cm{6} in \cm{2},
				\begin{tikzpicture}[vect/.style={->,shorten >=0pt,>=latex'}]
					\tkzDefPoint(0,0){A} \tkzDefPoint(6,0){B} \tkzDefPoint(6,2){C} \tkzDefPoint(0,2){D}
					\tkzDrawPolygon[fill=red!30, opacity=.3](A,B,C,D)
					\tkzDrawSegments[vect](A,B A,C)
				\end{tikzpicture}
				$$\vektor{b}\cdot\vektor{e}=|\vektor{b}|\cdot|\vektor{e}|\cdot\cos{\left(\sqrt{\cm{6}^2+\cm{2}^2}\right)}\Dot=11,926966$$
			\task v pravilnem 6-kotniku s stranico, dolgo \cm{2},
			
				\begin{tikzpicture}[vect/.style={->,shorten >=0pt,>=latex'}]
					\tkzDefPoint(0,0){A} \tkzDefPoint(2,0){B}
					\tkzDefRegPolygon[name=P,sides=6,side](A,B)
					\tkzDrawPolygon[fill=red!30, opacity=.3](P1,P2,P3,P4,P5,P6)
					\tkzDrawSegments[vect](P2,P1 P2,P3)
				\end{tikzpicture}
				$$\vektor{P_1P_2}\cdot\vektor{P_2P_3}=|\vektor{P_1P_2}|\cdot|\vektor{P_2P_3}|\cdot\cos{\ang{120}}=-2$$
			\task v pravilnem 6-kotniku s stranico, dolgo \cm{2}.
			
				\begin{tikzpicture}[vect/.style={->,shorten >=0pt,>=latex'}]
					\tkzDefPoint(0,0){A} \tkzDefPoint(2,0){B}
					\tkzDefRegPolygon[name=P,sides=6,side](A,B)
					\tkzDrawPolygon[fill=red!30, opacity=.3](P1,P2,P3,P4,P5,P6)
					\tkzDrawSegments[vect](P1,P6 P3,P2)
				\end{tikzpicture}
				$$\Vec{P_6P_1}\cdot\Vec{P_3P_2}=|\Vec{P_6P_1}|\cdot|\Vec{P_3P_2}|\cdot\cos{\ang{120}}=-2$$
		\end{tasks}
	\item Kocka $ABCDA'B'C'D'$ ima \cm{10} dolg rob. Izračunaj $\vektor{AB}\cdot\vektor{AC}$, $\vektor{AB}\cdot\vektor{AC'}$,
		$\vektor{AD'}\cdot\vektor{AC'}$, $\vektor{BD'}\cdot\vektor{DB'}$.
		$$\vektor{AB}\cdot\vektor{AC}=\cm{10}\cdot\cm{10}\cdot\cos{\ang{45}}=50$$
		$$\vektor{AB}\cdot\vektor{AC'}=\cm{10}\cdot\sqrt{\cm{10}^2+\cm{10}^2}\cdot\cos{\ang{45}}=70,710678$$
		$$\vektor{AD'}\cdot\vektor{AC'}=\cm{10}\cdot\sqrt{\cm{10}^2+\cm{10}^2}\cdot\cos{\ang{45}}=100$$
		$$\vektor{BD'}\cdot\vektor{DB'}=\sqrt{\cm{10}^2+\cm{10}^2}\cdot\sqrt{\cm{10}^2+\cm{10}^2}\cos\ang{45}=100$$
	\item Tetraeder $ABCD$ ima \cm{10} dolg rob. Izračunaj $\vektor{AB}\cdot\vektor{AC}$, $\vektor{AB}\cdot\vektor{AD}$,
		$\vektor{AD}\cdot\vektor{AD}$.
		\def\kot{70,528779}
		$$\vektor{AB}\cdot\vektor{AC}=\cm{10}\cdot\cm{10}\cdot\cos\arccos\frac{1}{3}=33,\overline{3}$$
		$$\vektor{AB}\cdot\vektor{AD}=\cm{10}\cdot\cm{10}\cdot\cos\arccos\frac{1}{3}=33,\overline{3}$$
		$$\vektor{AD}\cdot\vektor{AD}=100$$
\end{enumerate}
\section{Zaključek}
Ta dokument je informativne narave in se lahko še spreminja. Najnovejša različica, torej PDFji in
\hologo{LaTeX}\footnote{Za izdelavo dokumenta potrebujete \texttt{TeXLive 2020}.}
izvorna koda, zgodovina sprememb in prejšnje različice, je na voljo v mojem šolskem Git repozitoriju na
\url{https://git.sijanec.eu/sijanec/sola-gimb-2} v mapi 
\href{https://git.sijanec/sola-gimb-2/src/branch/master/\predmdn/\predmkaj/\stevilkadn/}{/\predmdn/\predmkaj/\stevilkadn/}. Povezava za ogled zadnje različice tega dokumenta v PDF obliki je  \url{http://razor.arnes.si/~asija3/files/sola/gimb/2/\predmdn/\predmkaj/\stevilkadn/dokument.pdf} in/ali \url{https://git.sijanec.eu/sijanec/sola-gimb-2/raw/branch/master/\predmdn/\predmkaj/\stevilkadn/dokument.pdf}.
\if\razhroscevanje1
\vfill
\section*{Razhroščevalne informacije}
Te informacije so generirane, ker je omogočeno razhroščevanje. Pred objavo dokumenta izklopite razhroščevanje. To naredite tako, da nastavite ukaz \texttt{razhroscevanje} na 0 v začetku dokumenta.

Grafi imajo natančnost \functionSamples\space točk na graf.

Konec generiranja dokumenta: \today\ ob \currenttime\footnote{To ne nakazuje dejanskega časa, ko je bil dokument napisan, temveč čas, ko je bi dokument generiran v PDF/DVI obliko. Isto velja za datum v glavi dokumenta. Če berete direktno iz LaTeX datoteke, bo to vedno današnji datum.}%\input|"date -Ins"

Dokument se je generiral R0qK1KR2 \SI{}{\second}.
\fi
%	\item $$$$ aaasecgeninsaaa R0qK1KR2
\end{document}
