1 files changed, 124 insertions, 26 deletions

diff --git a/mat/domace_naloge/20/dokument.tex b/mat/domace_naloge/20/dokument.tex
index bf0f096..ee74c49 100644
--- a/mat/domace_naloge/20/dokument.tex
+++ b/mat/domace_naloge/20/dokument.tex
@@ -134,39 +134,137 @@ Ta dokument vsebuje domačo nalogo, ki zajema snov \textit{\snovdn}pri matematik
 \tableofcontents
 \section{Vaje \textit{Matematika 2}: stran 50, dve sodi in dve lihi nalogi 317-336}
 \begin{enumerate}[label=\textbf{\arabic*.}] % 318, 329, 320, 325
-	\setcounter{enumi}{317}
-	\item Nariši vektor $\vektor{a}$, dolg \cm{3}. Nariši še vektorje:
-		\footnote{Za sezname ne uporabljam več \texttt{enumerate}, ampak \texttt{tasks} (Deutsche Kvalität)}
-		\begin{primerTasks}(8)
-			\task $\displaystyle\frac{1}{3}\vektor{a}$ % a
-			\task $\displaystyle\frac{4}{5}\vektor{a}$ % b
-			\task $\displaystyle-\frac{3}{7}\vektor{a}$ % c
-			\task $\displaystyle-\frac{3}{4}\vektor{a}$ % č
-			\task $\displaystyle\frac{5}{4}\vektor{a}$ % d
-			\task $\displaystyle\frac{7}{5}\vektor{a}$ % e
-			\task $\displaystyle-\frac{3}{2}\vektor{a}$ % f
-			\task $\displaystyle-\frac{8}{5}\vektor{a}$ % g
-			\task $\displaystyle\sqrt{2}\vektor{a}$ % h
-			\task $\displaystyle\sqrt{5}\vektor{a}$ % i
-			\task $\displaystyle-\sqrt{7}\vektor{a}$ % j
-			\task $\displaystyle-\sqrt{10}\vektor{a}$ % k
-			\task $\displaystyle\frac{\sqrt{3}}{4}\vektor{a}$ % l
-			\task $\displaystyle\frac{\sqrt{6}}{7}\vektor{a}$ % m
-			\task $\displaystyle-\frac{\sqrt{12}}{3}\vektor{a}$ % n
-			\task $\displaystyle-\frac{\sqrt{15}}{8}\vektor{a}$ % o
-		\end{primerTasks}
+	\if\false1
+		\setcounter{enumi}{317}
+		\item Nariši vektor $\vektor{a}$, dolg \cm{3}. Nariši še vektorje:
+			\footnote{Za sezname ne uporabljam več \texttt{enumerate}, ampak \texttt{tasks} (Deutsche Kvalität)}
+			\begin{primerTasks}(8)
+				\task $\displaystyle\frac{1}{3}\Vec{a}$ % a
+				\task $\displaystyle\frac{4}{5}\Vec{a}$ % b
+				\task $\displaystyle-\frac{3}{7}\Vec{a}$ % c
+				\task $\displaystyle-\frac{3}{4}\Vec{a}$ % č
+				\task $\displaystyle\frac{5}{4}\Vec{a}$ % d
+				\task $\displaystyle\frac{7}{5}\Vec{a}$ % e
+				\task $\displaystyle-\frac{3}{2}\Vec{a}$ % f
+				\task $\displaystyle-\frac{8}{5}\Vec{a}$ % g
+				\task $\displaystyle\sqrt{2}\Vec{a}$ % h
+				\task $\displaystyle\sqrt{5}\Vec{a}$ % i
+				\task $\displaystyle-\sqrt{7}\Vec{a}$ % j
+				\task $\displaystyle-\sqrt{10}\Vec{a}$ % k
+				\task $\displaystyle\frac{\sqrt{3}}{4}\Vec{a}$ % l
+				\task $\displaystyle\frac{\sqrt{6}}{7}\Vec{a}$ % m
+				\task $\displaystyle-\frac{\sqrt{12}}{3}\Vec{a}$ % n
+				\task $\displaystyle-\frac{\sqrt{15}}{8}\Vec{a}$ % o
+			\end{primerTasks}
+			\begin{center}
+				\begin{tikzpicture}[vect/.style={->,shorten >=3pt,>=latex'}]
+					% 0	
+					\tkzDefPoint(0,0){A} \tkzDefPoint(3,0){B}
+					\tkzDrawPoints[red](A,B) \tkzDrawSegments[vect](A,B)
+					\tkzLabelLine[above](A,B){$\Vec{a}$}
+					% a
+					\tkzDefPoint(30:3){aA}
+					\tkzDrawLine(A,aA) \tkzDrawLine(aA,B)
+					\tkzDefPoint(1,0){a1} \tkzDefPoint(30:1){a2} \tkzDefPoint(30:2){a3}
+					\tkzDefPoint(2,0){a4}
+					\tkzCompass(A,a2) \tkzCompass(a2,a3) \tkzDrawLine(a3,a4)
+					\tkzDrawLine(a1,a2) \tkzDrawPoints[red](a1) \tkzDrawSegment[vect](A,a1)
+					\tkzLabelLine[below](A,a1){$\frac{1}{3}\Vec{a}$}
+					% b
+					\tkzDefPoint(30:(3/5)){b1} \tkzDefPoint(30:(3/5)*2){b2}
+					\tkzDefPoint(30:(3/5)*3){b3} \tkzDefPoint(30:(3/5)*4){b4}
+					\tkzCompass(A,b1) \tkzCompass(b1,b2) \tkzCompass(b2,b3)
+					\tkzCompass(b3,b4) \tkzCompass(b4,aA)
+					\tkzDefPoint((3/5)*4,0){b5} \tkzDrawLine(b4,b5) \tkzDrawPoints[red](b5)
+					\tkzDrawSegments[vect](A,b5)
+					\tkzLabelLine[below](A,b5){$\frac{4}{5}\Vec{a}$}
+				\end{tikzpicture}
+			\end{center}
+		\fi
+	\setcounter{enumi}{318}
+	\item Nariši trikotnik $ABC$ s podatki $a=\cm{4}$, $b=\cm{6}$, $c=\cm{7}$. Nariši še $\displaystyle\frac{1}{3}\vektor{AC}+\sqrt{2}\vektor{AB}$. % 319
 		\begin{center}
-			\begin{tikzpicture}
-				\tkzDefPoint(0,0){A} \tkzDefPoint(0,3){B}
-				\tkzDrawPoints(A,B)
+			\begin{tikzpicture}[vect/.style={->,shorten >=3pt,>=latex'}]
+				\tkzDefPoint(0,0){A} \tkzDefPoint(7,0){B}
+				\tkzDefPoint(6,0){bh} \tkzDefPoint(7-4,0){ah}
+				\tkzDefPoint(0,6){3h} \tkzDefPoint(0,4){2h} \tkzDefPoint(0,2){1h}
+				\tkzCompass(A,B) \tkzDrawLine(A,B) % \tkzLabelLine[below](A,B){$c$}
+				\tkzInterCC(A,bh)(B,ah) \tkzGetPoints{C}{CX}
+				\tkzDrawPolygon[fill=red!30, opacity=.3](A,B,C)
+				\tkzLabelLine[below right](A,C){$\vektor{AC}$}
+				\tkzDrawLine(A,3h) \tkzCompass(A,1h) \tkzCompass(1h,2h)
+				\tkzCompass(2h,3h) \tkzDrawLine(3h,C)
+				\tkzDefPointWith[colinear=at 1h](3h,C) \tkzGetPoint{Z}
+				\tkzInterLL(A,C)(1h,Z) \tkzGetPoint{T}
+				\tkzLabelLine[above](A,T){$\frac{1}{3}\vektor{AC}$}
+				% sedaj pa sqrt(2)AB
+				\tkzDefPoint(0,7){P} \tkzCompass(A,P)
+				\tkzDefPoint(7,7){K} \tkzDrawLine(B,K) \tkzDrawLine(P,K)
+				\tkzInterLC(A,B)(A,K) \tkzGetPoints{K2X}{K2} \tkzCompass(A,K2)
+				\tkzLabelLine[below](A,K2){$\sqrt{2}\vektor{AB}$}
+				% sedaj pa seštevek
+				\tkzDefPointWith[colinear=at B](A,C) \tkzGetPoint{S}
+				\tkzLabelLine[above left](A,S){$\frac{1}{3}\vektor{AC}+\sqrt{AB}$}
+				\tkzDrawSegments[vect](A,C A,T, A,K A,K2)
+				\tkzDrawSegments[red, vect](A,S)
+				\tkzMarkRightAngles(A,P,K P,K,B K,B,A B,A,P)
+				\tkzDrawPoints[red](A,B,C,T,K,K2,S) \tkzLabelPoints(A,B,C)
 			\end{tikzpicture}
 		\end{center}
-	\setcounter{enumi}{319}
+	\newpage
 	\item Nariši pravokotnik $ABCD$ s podatki: $a=\cm{6}$, $b=\cm{2}$. Nariši še $\displaystyle-\frac{3}{2}\vektor{AC}+\frac{\sqrt{2}}{2}\vektor{BC}$.
+		\begin{center}
+			\begin{tikzpicture}[vect/.style={->,shorten >=3pt,>=latex'}]
+				\tkzDefPoint(0,0){A} \tkzDefPoint(6,2){C} \tkzDefPoint(6,0){B}
+				\tkzDefPoint(0,2){D}
+				\tkzDrawPolygon[fill=red!30, opacity=.3](A,B,C,D)
+				\tkzDefPoint(0,(2/3)){a1} \tkzDefPoint(0,(2/3)*2){a2}
+				\tkzCompass(A,a1) \tkzCompass(a1,a2) \tkzCompass(a2,D)
+				\tkzDefPointWith[colinear=at a1](D,C) \tkzGetPoint{Z}
+				\tkzInterLL(A,C)(a1,Z) \tkzGetPoint{V}
+				\tkzDrawLine(a1,V) \tkzLabelLine[above left](A,C){$\vektor{AC}$}
+				\tkzLabelLine[below](V,C){$-\frac{2}{3}\vektor{AC}$}
+				% zdaj pa še koren iz dve polovic vektorja BC
+				\tkzDefPoint(6-2,0){K} \tkzDefPoint(6-2,2){P}
+				\tkzCompass(B,K) \tkzCompass(C,P)
+				\tkzInterLC(C,B)(C,K) \tkzGetPoints{K2X}{K2}
+				\tkzDefMidPoint(C,K2) \tkzGetPoint{K22}
+				\tkzLabelLine[left](C,K22){$\frac{\sqrt{2}}{2}\vektor{CB}$}
+				\tkzDefPointWith[colinear=at K22](C,V) \tkzGetPoint{S}
+				\tkzDrawPoints[red](A,B,C,D,V,K2,K22,S) \tkzLabelPoints[](A,B,C,D)
+				\tkzDrawSegments[vect](A,C C,V C,K2 C,K22)
+				\tkzDrawSegments[vect, red](C,S)
+				\tkzMarkRightAngles(C,P,K P,K,B K,B,C B,C,P)
+			\end{tikzpicture}
+		\end{center}
 	\setcounter{enumi}{324}
-	\item Naj bo $\vektor{m}=2\vektor{a}+3\vektor{b}$ in $\vektor{n}=\vektor{a}-2\vektor{b}$. Izrazi vektor $2\vektor{m}-\frac{1}{2}\vektor{n}$ z vektorjema $\vektor{a}$ in $\vektor{b}$.
+	\item Naj bo $\Vec{m}=2\Vec{a}+3\Vec{b}$ in $\Vec{n}=\Vec{a}-2\Vec{b}$. Izrazi vektor $2\Vec{m}-\frac{1}{2}\Vec{n}$ z vektorjema $\Vec{a}$ in $\Vec{b}$.
+		$$
+			2\Vec{a}+3\Vec{b}-(\frac{1}{2}(\Vec{a}-2\Vec{b})) = 
+			2\Vec{a} + 3\Vec{b} - \frac{1}{2}\Vec{a} + \frac{1}{2} \cdot 2\Vec{b} =
+			1\frac{1}{2}\Vec{a}+4\Vec{b}
+		$$
 	\setcounter{enumi}{328}
 	\item Nariši daljico $AB$, dolgo \cm{6}. Na njen nariši tako točno $M$, da velja $|AM|:|MB|=1:2$, in tako točno $N$, da velja $|AN|:|AB|=1:2$. Vektorja $\vektor{AM}$ in $\vektor{AN}$ izrazi z vektorjem $\vektor{AB}$.
+		\begin{center}
+			\begin{tikzpicture}[vect/.style={->,shorten >=3pt,>=latex'}]
+				\tkzDefPoint(0,0){A} \tkzDefPoint(6,0){B}
+				\tkzCompass(A,B) \tkzDrawLine(A,B) % \tkzLabelLine[below](A,B){$AB$}
+				\tkzDefPoint(30:6){3h} \tkzDefPoint(30:4){2h} \tkzDefPoint(30:2){1h}
+				\tkzDrawLine(A,3h) \tkzDrawLine(3h,B)
+				\tkzCompass(A,1h) \tkzCompass(1h,2h) \tkzCompass(2h,3h)
+				\tkzDefPoint((6/3)*1,0){M} \tkzDrawLine(h1,M)
+				\tkzDefMidPoint(A,B) \tkzGetPoint{N}
+				\tkzDrawPoints[red](A,B,M,N) \tkzLabelPoints[below left](A,B,M,N)
+				\tkzDrawSegments[vect](A,M A,N)
+			\end{tikzpicture}
+		\end{center}
+		$$
+			\vektor{AM} = \frac{1}{3}\vektor{AB}
+		$$
+		$$
+			\vektor{AN} = \frac{1}{2}\vektor{AB}
+		$$
 \end{enumerate}