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%opening
\newcommand{\snovdn}{Računanje s komponentami }
\newcommand{\predmdn}{mat}
\newcommand{\predmkaj}{domace\_naloge}
\newcommand{\stevilkadn}{22}
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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}}
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% \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}
%\tableofcontents
\begin{enumerate}[label=\textbf{\arabic*.}]
\setcounter{enumi}{12}
\item Določi koordinate točke $M$, tako da bo točka $Z(-7, 8, -10)$ razpolovišče daljice $LM$, če je $L(-4, 2, -8)$.
\begin{center}
skica:
\begin{tikzpicture}
\tkzDefPoint(0,0){L} \tkzDefPoint(2,0){Z} \tkzDefPoint(4,0){M}
\tkzDrawLine(L,M)
\tkzDrawPoints(L,Z,M)
\tkzLabelPoints(L,Z,M)
\end{tikzpicture}
\end{center}
$$
Z\left(-7, 8, -10\right) = Z\left(\frac{-4+M_1}{2}, \frac{2+M_2}{2}, \frac{-8+M_3}{2}\right)
$$
$$
\frac{-4+M_1}{2} = -7 \wedge
\frac{2+M_2}{2} = 8 \wedge
\frac{-8+M_3}{2} = -10
\rightarrow M\left(M_1, M_2, M_3\right) = M\left(-10, 14, -12\right)
$$
\paragraph{Rešitev} $M\left(-10, 14, -12\right)$
\item Dani sta točki $A\left(4, -7, -1\right)$ in $B\left(-4, -3, -8\right)$. Določi točko $C$, tako da bo točka $T\left(1, -1, 3\right)$ težišče trikotnika $ABC$.
\begin{center}
skica:
\begin{tikzpicture}
\tkzDefPoint(0,0){A} \tkzDefPoint(4,0){B} \tkzDefPoint(3,3){C}
\tkzDefMidPoint(A,B) \tkzGetPoint{M} \tkzDefMidPoint(B,C) \tkzGetPoint{N} \tkzDefMidPoint(C,A) \tkzGetPoint{O}
\tkzDrawPolygon[fill=red!30, opacity=.3](A,B,C)
\tkzDrawLine(M,C) \tkzDrawLine(N,A) \tkzDrawLine(O,B)
\tkzInterLL(N,A)(O,B) \tkzGetPoint{T}
\tkzDrawPoints(A,B,C,T,M) \tkzLabelPoints(A,B,C,T,M)
\end{tikzpicture}
\end{center}
$$
M = M\left(\frac{4+\left(-4\right)}{2}, \frac{-7+\left(-3\right)}{2}, \frac{-1+\left(-8\right)}{2}\right) = M\left(0, -5, 4,5\right)
$$
Vemo, da težišče razpolovi težiščnico na tri enake dele. Naredimo vektor $\vektor{MT}$ in ga pomnožimo s skalarjem tri, da dobimo vektor $\vektor{MC}$.
$$
\vektor{MT} = \Vec{T} - \Vec{M} = \left(1-0, -1-\left(-5\right), 3-4,5\right) = \left(1, 4, -1,5\right) \rightarrow \vektor{MC} = 3\vektor{MT} = \left(3, 12, -4,5\right)
$$
$$
\rightarrow r_C = r_M + \vektor{DC} = \left(0+3, -5+12, 4,5+\left(-4,5\right)\right) = C\left(3, 7, 0\right)
$$
\parahraph{Rešitev} $C\left(3, 7, 0\right)$
\item Točka $P$ leži na daljici $AB$ tako, da je $|AP|:|PB|=1:1$. Zapiši koordinate točke $P$, če je $A\left(-6, -13, 2\right)$ in $B\left(-13, 5, 4\right)$. Koordinate zaokroži na dve decimalni mesti.
\begin{center}
skica:
\begin{tikzpicture}
\tkzDefPoint(0,0){A} \tkzDefPoint(2,0){P} \tkzDefPoint(4,0){B}
\tkzDrawLine(A,B)
\tkzDrawPoints(A,P,B)
\tkzLabelPoints(A,P,B)
\end{tikzpicture}
\end{center}
$$
P = \left(\frac{-6+\left(-13\right)}{2}, \frac{-13+5}{2}, \frac{2+4}{2}\right) = \left(-9,5, -4, 3\right)
$$
\paragraph{Rešitev} Točka $P$ ima koordinate $P \left(-9,5, -4, 3\right)$
\item Dane so točke $A\left(6, 5, 2\right)$, $B\left(6, 2, -5\right)$, $C\left(-7, -2, -6\right)$. Določi koordinate točke $D$, tako da bo štirikotnik $ABCD$ paralelogram.
\begin{center}
skica:
\begin{tikzpicture}
\tkzDefPoint(0,0){A} \tkzDefPoint(2,0){B} \tkzDefPoint(3,2){C} \tkzDefPoint(1,2){D}
\tkzDrawPolygon[fill=red!30, opacity=0.3](A,B,C,D)
\tkzDrawPoints(A,B,C,D) \tkzLabelPoints(A,B,C,D)
\end{tikzpicture}
$$
\vektor{AD} = \vektor{BC} = \Vec{C}-\Vec{B} = \left(-7-6, -2-2, -6-\left(-5\right)\right) = \left(-13, -4, -1\right)
$$
$$
\rightarrow r_D = r_A + \vektor{AD} = (-13+6, -4+5, -1+2) = (-7, 1, 1)
$$
\end{center}
\paragraph{Rešitev} $D\left(-7, 1, 1\right)$
\item Zapiši prvo komponento vektorja $\Vec{b}=\left(x, -4\right)$, da bo vzporeden z vektorjem $\Vec{a}=\left(6, -2\right)$.
$$
\Vec{b} = k\cdot\Vec{a} = (6k, -2k) \wedge -2 \cdot 2 = -4 \rightarrow k = -4 \rightarrow \Vec{b} = (12, -4)
$$
\paragraph{Rešitev} $x = 12$
\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}
