path: root/mat/domace_naloge/28/dokument.tex

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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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\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}: 93 / 612, 622, 632, 642, 652 \textit{(+ 617, 627)}
%\tableofcontents
\begin{enumerate}[label=\textbf{\arabic*.}]
	\setcounter{enumi}{611}
	\item Izračunaj.
		\begin{tasks}[label=\textbf{\xslalph*)}](2)
			\task $(2,3)+(4,5)=(6,8)$ \\
						$(2,3)-(4,5)=(-2,-2)$ \\
						$(2,3)\cdot(4,5)=23$ \\
						$2\cdot(4,5)=(8,10)$
			\task $(1,2,3)+(9,8,7)=(10,10,10)$ \\
						$(1,2,3)-(9,8,7)=(-8,-6,-4)$ \\
						$(1,2,3)\cdot(9,8,7)=9+16+21$ \\
						$5\cdot(9,8,7)=(45,40,35)$
		\end{tasks}
	\item Izračunaj.
		\begin{tasks}[label=\textbf{\xslalph*)}](2)
			\task $(3,7)\cdot(2,9)=69$
			\task $(-6,10)\cdot(3,2)=2$
			\task $(1,2,3)\cdot(4,5,6)=32$
			\task $(-2,2,-4)\cdot(1,0,3)=-14$
			\task $\left(5,0,1\frac{1}{2}\right)\cdot\left(1\frac{1}{2},\frac{2}{7},3\right)=12$
			\task $\left(\sqrt{2}+\sqrt{3},2\sqrt{3}-1,-\sqrt{2}\right)\cdot\left(\sqrt{3}-\sqrt{2},\sqrt{2},\sqrt{3}\right)
				=1-\sqrt{2}+\sqrt{6}$
			\task $\left(2\Vec{i}+3\Vec{j}+\Vec{k}\right)\cdot\left(5\Vec{i}+\Vec{j}+3\Vec{k}\right)=16$
			\task $\left(\Vec{i}-\Vec{j}+2\Vec{k}\right)\cdot\left(\Vec{i}+3\Vec{j}-3\Vec{k}\right)=-8$
			\task $\left(5\Vec{i}+2\Vec{j}\right)\cdot\left(4\Vec{i}-3\Vec{j}\right)=14$
			\task $\left(\Vec{i}-5\Vec{j}\right)\cdot\left(-\Vec{i}+3\Vec{j}\right)=-16$
			\task $\left(2\Vec{i}-3\Vec{k}+\Vec{j}\right)\cdot\left(2\Vec{j}-3\Vec{k}\right)=11$
			\task $\left(3\Vec{j}-2\Vec{i}-\Vec{k}\right)\cdot\left(\Vec{i}+\frac{1}{2}\Vec{j}+\frac{3}{2}\Vec{k}\right)=-2$
		\end{tasks}
	\setcounter{enumi}{616}
	\item Dana sta vektorja $\Vec{m}=(3,5)$ in $\Vec{n}=(-3,2)$. Vektorja nariši v koordinatni sistem in na sekundo natančno izračunaj kot med njima.
		\begin{multicols}{2}
			$$\Vec{m}\cdot\Vec{n}=|\Vec{m}|\cdot|\Vec{n}|\cdot\cos\alpha$$
			$$\Vec{m}\cdot\Vec{n}=1$$
			$$1=\sqrt{3^2+5^2}\cdot\sqrt{\left(-3\right)^2+2^2}\cdot\cos\alpha$$
			$$\alpha=\arccos\frac{1}{\sqrt{3^2+5^2}\cdot\sqrt{\left(-3\right)^2+2^2}}$$
			$$\alpha=\ang{87,273689}$$
			\vfill\null\columnbreak
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				\tkzInit[xmax=5,ymax=5,ymin=-5,xmin=-5]
				\tkzGrid
				\tkzDrawXY[>=latex]
				\tkzDefPoint(3,5){M}
				\tkzDefPoint(-3,2){N}
				\tkzDefPoint(0,0){O}
				\tkzDrawPoints(O,M,N)
				\tkzShowPointCoord[xlabel=$3$, ylabel=$5$](M)
				\tkzShowPointCoord[xlabel=$-3$, ylabel=$2$](N)
				\tkzLabelPoint(M){$M (3,5)$}
				\tkzLabelPoint[above left](N){$N (-3,2)$}
				\tkzLabelPoints(O)
				\tkzDrawSegments[->](O,M O,N)
				\tkzLabelLine[above left](O,M){$\Vec{m}$}
				\tkzLabelLine[below left](O,N){$\Vec{n}$}
				\tkzMarkAngle[mark=none,->](M,O,N)
				\tkzFillAngle[fill=red!30,opacity=.3](M,O,N)
				\tkzLabelAngle[pos=1.25](M,O,N){$\alpha$}
			\end{tikzpicture}
		\end{multicols}
	\setcounter{enumi}{621}
	\item Točke $A(-2,4,1)$, $B(0,3,1)$ in $C(3,-1,0)$ so oglišča trikotnika v prostoru. Izračunaj kot $BCA$.
		\paragraph{Kot $CAB$:}
		$$
			\overline{AC}^2=\overline{CB}^2+\overline{BA}^2-2\cdot\overline{CB}\cdot\overline{BA}\cdot\cos\beta
		$$
		$$
			\sqrt{(-5)^2+(5)^2+(1)^2}^2=\sqrt{(3)^2+(-4)^2+(-1)^2}^2+\sqrt{(2)^2+(-1)^2+(0)^2}^2
		$$
		$$
			-2\cdot\sqrt{(3)^2+(-4)^2+(-1)^2}\cdot\sqrt{(2)^2+(-1)^2+(0)^2}\cdot\cos\beta
		$$
		$$
			51=26+5-2\sqrt{26}\sqrt{5}\cos\beta
		$$
		$$
			20=-2\sqrt{26}\sqrt{5}\cos\beta
		$$
		$$
			-10=\sqrt{26}\sqrt{5}\cos\beta
		$$
		$$
			-0,877058=\cos\beta
		$$
		$$
			\beta=\arccos-0,877058=\ang{151,289}
		$$
		\paragraph{Kot $BCA$:}
		$$
			\overline{AB}^2=\overline{AC}^2+\overline{CB}^2-2\cdot\overline{AC}\cdot\overline{CB}\cdot\cos\gamma
		$$
		$$
			5=51+26-2\sqrt{51}\sqrt{26}\cos\gamma
		$$
		$$
			36=\sqrt{51}\sqrt{26}\cos\gamma
		$$
		$$
			\arccos0,988623=\gamma=\ang{8,650931}
		$$
		\paragraph{Kot $BAC$:}
		$$
			\alpha=\ang{180}-(\ang{151,289}+\ang{8,650931})=\ang{20,059583}
		$$
	\setcounter{enumi}{626}
	\item Kolikšen ostri kot oklepa vektor z abscisno osjo, kolikšen z ordinatno in kolikšen z aplikatno osjo?
		\begin{tasks}[label=\textbf{\xslalph*)}](2)
			\task $\Vec{a}=(-3,2,5)$
				\paragraph{Z abscisno:}
					$$(-1,0,0)\cdot(-3,2,5)=-1\cdot\sqrt{(-3)^2+2^2+5^2}\cdot\cos\alpha$$
					$$3=6,164414\cos\alpha$$
					$$\arccos0,486664=\alpha=\ang{60,878}$$
				\paragraph{Z ordinatno:}
					$$(0,1,0)\cdot(-3,2,5)=6,164414\cos\beta$$
					$$\arccos\frac{2}{6,164414}=\beta=\arccos0,324=\ang{71,068}$$
				\paragraph{Z aplikatno:}
					$$\arccos\frac{5}{6,164414}=\gamma=\arccos0,811107=\ang{35,795759915}$$
			\task $\Vec{b}=2\Vec{i}+3\Vec{k}-\Vec{j}$
				\paragraph{Z abscisno:}
					$$(1,0,0)\cdot(2,-1,3)=\sqrt{2^2+(-1)^2+3^2}^2\cdot\cos\delta$$
					$$\arccos\frac{2}{3,731657}=\delta=\arccos0,534552=\ang{57,688}$$
				\paragraph{Z ordinatno:}
					$$\arccos\frac{-1}{-3,731657}=\epsilon=\arccos0,267261=\ang{74,498640}$$
				\paragraph{Z aplikatno:}
					$$\arccos\frac{3}{3,731657}=\zeta=\arccos0,801783726=\ang{36,699225}$$
					% yeet
		\end{tasks}
	\setcounter{enumi}{631}
	\item Dani sta točki $A(3,-1,0)$ in $B(-2,2,2)$. Zapiši enotski vektor $\Vec{e}$ v smeri vektorja $\vektor{AB}$.
		$$\Vec{e}=\vektor{AB}\cdot\frac{1}{\sqrt{5^2+(-3)^2+(-2)^2}}=0,162221\cdot\vektor{AB}$$
	\setcounter{enumi}{641}
	\newpage
	\item Naj bo $\Vec{a}=(1,k,1)$ in $\Vec{b}=(2-k,1,-k^2)$. Določi:
		\begin{enumerate}[label=\textbf{\xslalph*)}]
			\item $k\in\mathbb{R}$ tako, da bosta vektorja $\Vec{a}$ in $\Vec{b}$ pravokotna,
				$$(1,k,1)\cdot(2-k,1,-k^2)=0=(2-k^2) \rightarrow k=\sqrt{2}$$
			\item $k\in\mathbb{R}$ tako, da bo vektor $\Vec{a}$ oklepal s ordinatno osjo kot $\ang{60}$,
				$$
					(0,1,0)\cdot(1,k,1)=\sqrt{2+k^2}\frac{1}{2}
				$$
				$$
					2k=\sqrt{2+k^2} \rightarrow 4k^2=2+k^2 \rightarrow 3k^2=2 \rightarrow k^2=\frac{2}{3}
				$$
				$$
					k=\sqrt{\frac{2}{3}}
				$$
			\item $k\in\mathbb{N}$ tako, da bo dolžina vektorja $\Vec{a}$ enaka $3\sqrt{3}$,
				$$
					3\sqrt{3}=\sqrt{2+k^2} \rightarrow 25=k^2 \rightarrow k=5
				$$
			\item $k\in\mathbb{Z}$ tako, da bo skalarni produkt $\Vec{a}\cdot\Vec{b}$ naravno število.
				$$(1,k,1)\cdot(2-k,1,-k^2)=2-k+k-k^2=2-k^2 \rightarrow k=1$$
		\end{enumerate}
	\setcounter{enumi}{651}
	\item Točke $A(2,3,4)$, $B(-2,3,2)$, $C(2,0,2)$ so oglišča trikotnika. Določi točko:
		\begin{enumerate}[label=\textbf{\xslalph*)}]
			\item $D$ tako, da bo $ABCD$ paralelogram,
				$$
					D=C-(B-A)=C-B+A=(2,0,2)-(-2,3,2)+(2,3,4)=(6,0,4)
				$$
			\item $E$ na daljici $AB$ tako, da bo $|AE|:|AB|=3:4$,
				$$
					E=A+\frac{3}{4}(B-A)=(2,3,4)+\frac{3}{4}(-4,0,-2)=\left(-1,3,2\frac{1}{2}\right)
				$$
			\item $F$, ki je težišče trikotnika $ABC$,
				$$
					M=A+\frac{1}{2}(B-A)=(2,3,4)+\frac{1}{2}(-4,0,-2)=\left(0,3,3\right)
				$$
				$$
					F=M+\frac{1}{3}(C-M)=(0,3,3)+\frac{1}{3}(2,-3,-1)=\left(\frac{2}{3},2,2\frac{2}{3}\right)
				$$
			\item $G$ tako, da bo točka $C$ težišče trikotnika $ABG$.
				$$
					G=M+3\cdot(C-M)=(0,3,3)+(6,-9,-3)=(6,-6,0)
				$$
		\end{enumerate}
\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}.
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\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
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